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+[
+    {
+        "Problem": "A permutation of the set of positive integers $[n] = \\{1, 2, \\ldots, n\\}$ is a sequence $(a_1, a_2, \\ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$. Let $P(n)$ be the number of permutations of $[n]$ for which $ka_k$ is a perfect square for all $1\\leq k\\leq n$. Find with proof the smallest $n$ such that $P(n)$ is a multiple of $2010$.",
+        "Solutions": "We claim that the smallest $n$ is $67^2 = \\boxed{4489}$.",
+        "Solution_1": "Let $S = \\{1, 4, 9, \\ldots\\}$ be the set of positive perfect squares. We claim that the relation $R = \\{(j, k)\\in [n]\\times[n]\\mid jk\\in S\\}$ is an equivalence relation on $[n]$.\nWe are restricted to permutations for which $ka_k \\in S$, in other words to permutations that send each element of $[n]$ into its equivalence class. Suppose there are $N$ equivalence classes: $C_1, \\ldots, C_N$. Let $n_i$ be the number of elements of $C_i$, then $P(n) = \\prod_{i=1}^{N} n_i!$.\nNow $2010 = 2\\cdot 3\\cdot 5\\cdot 67$. In order that $2010\\mid P(n)$, we must have $67\\mid n_m!$ for the class $C_m$ with the most elements. This means $n_m\\geq 67$, since no smaller factorial will have $67$ as a factor. This condition is sufficient, since $n_m!$ will be divisible by $30$ for $n_m\\geq 5$, and even more so $n_m\\geq 67$.\nThe smallest element $g_m$ of the equivalence class $C_m$ is square-free, since if it were divisible by the square of a prime, the quotient would be a smaller element of $C_m$. Also, each prime $p$ that divides $g_m$ divides all the other elements $k$ of $C_m$, since $p^2\\mid kg_m$ and thus $p\\mid k$. Therefore $g_m\\mid k$ for all $k\\in C_m$. The primes that are not in $g_m$ occur an even number of times in each $k\\in C_m$.\nThus the equivalence class $C_m = \\{g_mk^2\\leq n\\}$. With $g_m = 1$, we get the largest possible $n_m$. This is just the set of squares in $[n]$, of which we need at least $67$, so $n\\geq 67^2$. This condition is necessary and sufficient.",
+        "Solution_2": "This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as \"equivalence relation\":\nIt is possible to write all positive integers $n$ in the form $p\\cdot m^2$, where $m^2$ is the largest perfect square dividing $n$, so $p$ is not divisible by the square of any prime. Obviously, one working permutation of $[n]$ is simply $(1, 2, \\ldots, n)$; this is acceptable, as $ka_k$ is always $k^2$ in this sequence.\nLemma 1. We can permute any numbers that, when each divided by the largest perfect square that divides it, yield equal quantities $p$.\nProof. Let $p_k$ and $m_k$ be the values of $p$ and $m$, respectively, for a given $k$ as defined above, such that $p$ is not divisible by the square of any prime. We can obviously permute two numbers which have the same $p$, since if $p_j = p_w$ where $j$ and $w$ are 2 values of $k$, then $j\\cdot w = p_j^2\\cdot m_j^2\\cdot m_w^2$, which is a perfect square. This proves that we can permute any numbers with the same value of $p$.\nEnd Lemma\nLemma 2. We will prove the converse of Lemma 1: Let one number have a $p$ value of $\\phi$ and another, $\\gamma$. $\\phi\\cdot f$ and $\\gamma\\cdot g$ are both perfect squares.\nProof. $\\phi\\cdot f$ and $\\gamma\\cdot g$ are both perfect squares, so for $\\phi\\cdot \\gamma$ to be a perfect square, if $g$ is greater than or equal to $f$, $g/f$ must be a perfect square, too. Thus $g$ is $f$ times a square, but $g$ cannot divide any squares besides $1$, so $g = 1f$; $g = f$. Similarly, if $f\\geq g$, then $f = g$ for our rules to keep working.\nEnd Lemma\nWe can permute $l$ numbers with the same $p$ in $l!$ ways. We must have at least 67 numbers with a certain $p$ so our product will be divisible by 67. Obviously, then it will also be divisible by 2, 3, and 5, and thus 2010, as well. Toms as $h$, in general, we need numbers all the way up to $h\\cdot 67^2$, so obviously, $67^2$ is the smallest such number such that we can get a $67!$ term; here 67 $p$ terms are 1. Thus we need the integers $1, 2, \\ldots, 67^2$, so $67^2$, or $\\boxed{4489}$, is the answer.",
+        "Solution_3_.28Not_a_formal_proof_but_understandable.29": "We claim the smallest $n$ is $67^2 = \\boxed{4489}$\nLooking at small cases we see that $P(n)$ changes every time n increases to a perfect square. We find that we can not permute the non squares around because $n$ (not a perfect square) $*a_w$ where $w \\neq n$ will not give a perfect square. But we can permute the perfect squares around to other perfect squares since a square times a square makes another square we can do this (# of squares below $n$)! times. So now we need to find $n = a^2$ such that $a!$ is divisible by $2010$ which largest prime factor is $67$ so, $n$ is $n$ is $67^2 = \\boxed{4489}$\n~bjump",
+        "Solution_Number_Sense": "We have to somehow calculate the number of permutations for a given $n$. How in the world do we do this? Because we want squares, why not call a number $k=m*s^2$, where $s$ is the largest square that allows $m$ to be non-square?  $m=1$ is the only square $m$ can be, which only happens if $k$ is a perfect square.\nFor example, $126 = 14 * 3^2$, therefore in this case $k=126, m = 14, s = 3$.\nI will call a permutation of the numbers $P$, while the original $1, 2, 3, 4, ...$ I will call $S$.\nNote that essentially we are looking at \"pairing up\" elements between $P$ and $S$ such that the product of $P_k$ and $S_k$ is a perfect square. How do we do this? Using the representation above.\nEach square has to have an even exponent of every prime represented in its prime factorization. Therefore, we can just take all exponents of the primes $mod 2$ and if there are any odd numbers, those are the ones we have to match- in effect, they are the $m$ numbers mentioned at the beginning.\nBy listing the $m$ values, in my search for \"dumb\" or \"obvious\" ideas I am pretty confident that only values with identical $m$s can be matched together. With such a solid idea let me prove it.\nIf we were to \"pair up\" numbers with different $m$s, take for example $S_{18}$ with an $m$ of $2$ and $P_{18}=26$ with an $m$ of $26$, note that their product gives a supposed $m$ of $13$ because the $2$ values cancel out. But then, what happens to the extra $13$ left? It doesn't make a square, contradiction. To finish up this easy proof, note that if a \"pair\" has different $m$ values, and the smaller one is $m_1$, in order for the product to leave a square, the larger $m$ value has to have not just $m_1$ but another square inside it, which is absurd because we stipulated at the beginning that $m$ was square-free except for the trivial multiplication identity, 1.\nNow, how many ways are there to do this? If there are $c_1$ numbers with $m=1$, there are clearly $(c_1)!$ ways of sorting them. The same goes for $m=2, 3, etc.$ by this logic. Note that the $P(n)$ as stated by the problem requires a $67$ thrown in there because $2010=2*3*5*67$, so there has to be a $S_n$ with 67 elements with the same $m$. It is evident that the smallest $n$ will occur when $m=1$, because if $m$ is bigger we would have to expand $n$ to get the same number of $m$ values. Finally, realize that the only numbers with $m=1$ are square numbers! So our smallest $n=67^2=4489$, and we are done.\nI relied on looking for patterns a lot in this problem. When faced with combo/number theory, it is always good to draw a sketch. Never be scared to try a problem on the USAJMO. It takes about 45 minutes. Well, it's 2010 and a number 1. Cheers!\n-expiLnCalc",
+        "Solution_4": "It's well known that there exists $f(n)$ and $g(n)$ such that $n = f(n) \\cdot g(n)$, no square divides $f(n)$ other than 1, and $g(n)$ is a perfect square.\nWe prove first: If $f(k) = f(a_k)$, $k \\cdot a_k$ is a perfect square.\n$k \\cdot a_k = f(k) \\cdot g(k) \\cdot f(a_k) \\cdot g(a_k) = f(k)^2 \\cdot g(k) \\cdot g(a_k)$, which is a perfect square.\nWe will now prove: If $k \\cdot a_k$ is a perfect square, $f(k) = f(a_k)$.\nWe do proof by contrapositive: If $f(k) \\neq f(a_k)$, $k \\cdot a_k$ is not a perfect square.\n$v_p(k)$ is the p-adic valuation of k. (Basically how many factors of p you can take out of k)\nNote that if $f(k) \\neq f(a_k)$, By the Fundamental Theorem of Arithmetic, $f(k)$ and $f(a_k)$'s prime factorization are different, and thus there exists a prime p, such that $v_p(f(k)) \\neq v_p(f(a_k))$. Also, since $f(k)$ and $f(a_k)$ is squarefree, $v_p(k), v_p(a_k) \\leq 1$. Thus, $v_p(k \\cdot a_k) = 1$, making $k \\cdot a_k$ not a square.\nThus, we can only match k with $a_k$ if they have the same f value. Thus, to find P(k), we can do it by f value, permuting the $a_k$ with f value 1, then 2, ... Thus, our answer is:\n$P(n) = \\prod_{1 \\leq i \\leq n, g(i) = 1} \\left\\lfloor \\sqrt{\\frac{n}{i}} \\right \\rfloor !$\nFor all $n < 67^2$, $P(n)$ doesn't have a factor of 67. However, if $n = 67^2$, the first term will be a multiple of 2010, and thus the answer is $67^2 = \\boxed{4489}$\n-Alexlikemath\nAlternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.",
+        "Solution_5.28stupid_but_simple.29": "The answer is 4489.(67 squared)\nProof\nWe start with trying to fulfill ways to do $P(n)$. When n is 1 there is one way to do it and fulfill $ka_k$ as a perfect square. Same with the numbers 2 and 3. The order for both of these number are all in increasing order   Now when it comes to 4 we could do  and  This means when n is a perfect square it increases where $k$ and $a_k$ are perfect squares. The prime factorization of 2010 is:  Because we only care about 67, the biggest divisor we do 67 squared because $n$ has to be a square.\n-Multpi12"
+    },
+    {
+        "Problem": "Let $n > 1$ be an integer. Find, with proof, all sequences\n$x_1, x_2, \\ldots, x_{n-1}$ of positive integers with the following\nthree properties:",
+        "Solution": "The sequence is $2, 4, 6, \\ldots, 2n-2$.",
+        "Solution_2": "The claim is that in this sequence, if there are $2$ elements $a,b$ where $a,b<n$, such that $\\gcd(a,b)=1$, then the sequence contains every number less than $2n$.\nProof: Let $a$ and $b$ be the numbers less than $n$ such that $\\gcd(a,b)=1$.\nWe take this sequence modulo $n$. This means that if $x_i$ is an element in this sequence then $-x_i$ is as well.\n$a,b,2n-a,2n-b$ are all elements in the sequence. Clearly, one of $2n-a+b$ and $2n-b+a$ is less than $2n$, which means that $\\pm (a-b)$ are in this sequence modulo $n$.\nNow we want to show every number is achievable. We have already established that $a$ and $b$ are relatively prime, so by euclidean algorithm, if we take the positive difference of $a'$ and $b'$ every time, we will get that $1$ is in our sequence. Then, we can simply add or subtract $1$ as many times from $a$ as desired to get every single number.\nWe have proved that there are no two numbers that can be relatively prime in our sequence, implying that no two consecutive numbers can be in this sequence. Because our sequence has $n-1$ terms, our sequence must be one of $2,4,6,..,2(n-1)$ or $1,3,5,...$, the latter obviously fails, so $2,4,6...2(n-1)$ is our only possible sequence.",
+        "Solution_3": "We can add $x_1$ to every expression in property (a) to get  Therefore, we have $n-2$ distinct (because all the $x_i$ are distinct) expressions of the form $x_1+x_i$ for $i = 1, 2, \\dots n-1$ that are all less than $2n$, which means these $n-2$ expressions are also equal to some $x_k$.\nNow, we have that the $x_i$ are all positive, so $x_1>0$. Adding $x_1$ to both sides, $2x_1>x_1$. Therefore, since the $n-2$ expressions are all at least $2x_1$, they have to be equal to $x_2, x_3, \\dots, x_{n-1}$.\nSince there are $n-2$ distinct expressions $x_1+x_i$ equal to $n-2$ distinct expressions $x_j$, we have that each $x_1+x_i$ is equal to one $x_j$. Using the orders of the $x_1+x_i$ and the $x_j$, we find that\n\nThis gives us\n\nNow, we can use property (b) with $i=1$, so $x_1 + x_{n-1} = 2n$, which means $x_1 = 2$. This gives the sequence \n-sixeoneeight"
+    },
+    {
+        "Problem": "Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter\n$AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto\nlines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle\nformed by lines $PQ$ and $RS$ is half the size of $\\angle XOZ$, where\n$O$ is the midpoint of segment $AB$.",
+        "Solution_1": "Let $\\alpha = \\angle BAZ$, $\\beta = \\angle ABX$.\nSince $XY$ is a chord of the circle with diameter $AB$,\n$\\angle XAY = \\angle XBY = \\gamma$. From the chord $YZ$,\nwe conclude $\\angle YAZ = \\angle YBZ = \\delta$.\nTriangles $BQY$ and $APY$ are both right-triangles, and share the\nangle $\\gamma$, therefore they are similar, and so the ratio $PY : YQ = AY : YB$. Now by Thales' theorem the angles $\\angle AXB = \\angle AYB = \\angle AZB$ are all right-angles. Also, $\\angle PYQ$,\nbeing the fourth angle in a quadrilateral with 3 right-angles is\nagain a right-angle.  Therefore $\\triangle PYQ \\sim \\triangle AYB$ and\n$\\angle YQP = \\angle YBA = \\gamma + \\beta$.\nSimilarly, $RY : YS = AY : YB$, and so $\\angle YRS = \\angle YAB = \\alpha + \\delta$.\nNow $RY$ is perpendicular to $AZ$ so the direction $RY$ is $\\alpha$ counterclockwise from the vertical, and since $\\angle YRS = \\alpha + \\delta$ we see that $SR$ is $\\delta$ clockwise from the vertical. (Draw an actual vertical line segment if necessary.)\nSimilarly, $QY$ is perpendicular to $BX$ so the direction $QY$ is $\\beta$ clockwise from the vertical, and since $\\angle YQP$ is $\\gamma + \\beta$ we see that $QY$ is $\\gamma$ counterclockwise from the vertical.\nTherefore the lines $PQ$ and $RS$ intersect at an angle $\\chi = \\gamma + \\delta$. Now by the central angle theorem $2\\gamma = \\angle XOY$\nand $2\\delta = \\angle YOZ$, and so $2(\\gamma + \\delta) = \\angle XOZ$,\nand we are done.\nNote that $RTQY$ is a quadrilateral whose angles sum to 360°; can you find a faster approach using this fact?",
+        "Solution_2": "\nLet $T$ be the foot of the perpendicular from $Y$ to $\\overline{AB}$, let $O$ be the center of the semi-circle.\nSince we have a semi-circle, if we were to reflect it over $\\overline {AB}$, we would have a full circle, with $\\triangle{AXB}$ and $\\triangle{AZB}$ inscribed in it. Now, notice that $Y$ is a point on that full circle, so we can say that $T$ lies on the Simson Line $\\overline{PQ}$ from $Y$ to $\\triangle AXB$ and that it also lies on the Simson line $\\overline {RS}$ from $Y$ to $\\triangle AZB$. Thus, $T$ lies on two distinct lines in a plane, which means that $T=\\overline{PQ}\\cap\\overline{RS}$. Therefore, it suffices to show that $\\angle PTS=\\tfrac{1}{2}\\angle XOZ$.\nSince $m\\angle YTA + m\\angle YPA = 90^\\circ + 90^\\circ = 180^\\circ$ and $m \\angle YTB + m \\angle YSB = 90^\\circ + 90^\\circ = 180^\\circ$, we know that $TAPY$ and $TBSY$ are cyclic quadrilaterals.\nWe use this fact to get  \\\\\nNow note that $\\angle XAY$ is the inscribed angle of minor arc $\\overset{\\huge\\frown}{PY}$, and $\\angle XOY$ is the central angle of minor arc $\\overset{\\huge\\frown}{AB}$, so $\\angle XAY = \\frac{\\overset{\\huge\\frown}{PY}}{2} = \\frac{\\angle XOY}{2}$. Similarly, $\\angle YBZ = \\frac{\\overset{\\huge\\frown}{YZ}}{2}=\\frac{\\angle YOZ}{2}$. Thus we can say\nCombining statements $(1)$ and $(2)$, we can say that $\\angle PTS = \\frac{\\angle XOZ}{2}$, as desired. $\\square$\n~thinker123"
+    },
+    {
+        "Problem": "A triangle is called a parabolic triangle if its vertices lie on a\nparabola $y = x^2$. Prove that for every nonnegative integer $n$, there\nis an odd number $m$ and a parabolic triangle with vertices at three\ndistinct points with integer coordinates with area $(2^nm)^2$.",
+        "Solution": "Let the vertices of the triangle be $(a, a^2), (b, b^2), (c, c^2)$.\nThe area of the triangle is the absolute value of $A$ in the equation:\nIf we choose $a < b < c$, $A > 0$ and gives the actual area. Furthermore,\nwe clearly see that the area does not change when we subtract the same\nconstant value from each of $a$, $b$ and $c$. Thus, all possible areas\ncan be obtained with $a = 0$, in which case $A = \\frac{1}{2}bc(c-b)$.\nIf a particular choice of $b$ and $c$ gives an area $A = (2^nm)^2$,\nwith $n$ a positive integer and $m$ a positive odd integer, then setting\n$b' = 4b$, $c' = 4c$ gives an area $A' = 4^3 A = 8^2 A = (2^{n+3}m)^2$.\nTherefore, if we can find solutions for $n = 0$, $n = 1$ and $n = 2$,\nall other solutions can be generated by repeated multiplication\nof $b$ and $c$ by a factor of $4$.\nSetting $b=1$ and $c=2$, we get $A=1 = (2^0\\cdot1)^2$, which yields\nthe $n=0$ case.\nSetting $b=1$ and $c=9$, we get $A = 9\\cdot4 = (2^1\\cdot3)^2$, which yields\nthe $n=1$ case.\nSetting $b=1$ and $c=5$, we get $A=1\\cdot2\\cdot5 = 10$. Multiplying these\nvalues of $b$ and $c$ by $10$, we get $b'=10$, $c'=50$,\n$A'=10\\cdot20\\cdot50 = 100^2 = (2^2\\cdot5^2)^2$,\nwhich yields the $n=2$ case. This completes the construction.",
+        "Solution_2": "We proceed via induction on n. Notice that we prove instead a stronger result: there exists a parabolic triangle with area $(2^nm)^2$ with two of the vertices sharing the same y-coordinate.\nBASE CASE:\nIf $n = 0$, consider the parabolic triangle $ABC$ with $A(0, 0), B(1, 1), C(-1, 1)$ that has area $1/2 \\cdot 1 \\cdot 2 = 1$, so that $n = 0$ and $m = 1$.\nIf $n = 1$, let $ABC = A(5, 25), B(4, 16), C(-4, 16)$. Because $ABC$ has area $1/2 \\cdot 8 \\cdot 9 = 36$, we set $n = 1$ and $m = 3$.\nIf $n = 2$, consider the triangle formed by $A(21, 441), B(3, 9), C(-3, 9)$. It is parabolic and has area $1/2 \\cdot 6 \\cdot 432 = 1296 = 36^2$, so $n = 2$ and $m = 9$.\nINDUCTIVE STEP:\nIf $n = k$ produces parabolic triangle $ABC$ with $A(a, a^2), B(b, b^2),$ and $C(-b, b^2)$, consider $A$'$B$'$C$' with vertices $A(4a, 16a^2)$, $B(4b, 16b^2)$, and $C(-4b, 16b^2)$. If $ABC$ has area $(2^km)^2$, then $A$'$B$'$C$' has area $(2^{k+3}m)^2$, which is easily verified using the $1/2 \\cdot\\text{base} \\cdot \\text{height}$ formula for triangle area. This completes the inductive step for $k \\implies k+3$.\nHence, for every nonnegative integer $n$, there exists an odd $m$ and a parabolic triangle with area $(2^nm)^2$ with two vertices sharing the same ordinate. The problem statement is a direct result of this result.         -MathGenius_",
+        "Solution_3_.28without_induction.29": "First, consider triangle with vertices $(0,0)$, $(1,1)$, $(-1,1)$. This has area $1$ so $n=0$ case is satisfied.\nThen, consider triangle with vertices $(a,a^2), (-a,a^2), (b,b^2)$, and set $a=2^{2n}$ and $b=2^{4n-2}+1$.\nThe area of this triangle is $\\frac{1}{2} \\cdot base \\cdot height=a(b^2-a^2)$.\nWe have that $b^2-a^2=2^{8n-4}+2^{4n-1}+1-2^{4n}=2^{8n-4}-2^{4n-1}+1=(2^{4n-2}-1)^2$\nWe desire $A=a(b^2-a^2)=2^{2n}m^2$, or $2^{4n-2}-1=m$, and $m$ is clearly always odd for positive $n$, completing the proof.",
+        "Solution_4": "We simply need to provide an example for all $n$ that satisfies the condition, and we do so.\nLet $a = 2^{2n+1}+1$. Then consider the triangle with coordinates $(0,0), (a,a^2), (a^2,a^4) = (x_1, y_1), (x_2, y_2), (x_3, y_3)$.\nBy the shoelace formula, this triangle has area which clearly can be written in the form $(2^n \\times m)^2 = 2^{2n} \\times m^2$, where $m^2 = (2^{2n+1}+1)^4$ or $m=(2^{2n+1}+1)^2$. Now, we just have to prove that $(2^{2n+1}+1)^2$ is always odd. This is true because $2^{2n+1}$ is even (because it's a power of $2$), so $2^{2n+1}+1$ is odd, and since an odd number squared is odd, the whole thing is odd as well. And we are done, since we have proved that for all $n$, we can show that there exists such a triangle by merely providing an example. $\\square$\n~thinker123"
+    },
+    {
+        "Problem": "Two permutations $a_1, a_2, \\ldots, a_{2010}$ and\n$b_1, b_2, \\ldots, b_{2010}$ of the numbers $1, 2, \\ldots, 2010$\nare said to intersect if $a_k = b_k$ for some value of $k$ in the\nrange $1 \\le k\\le 2010$.  Show that there exist $1006$ permutations\nof the numbers $1, 2, \\ldots, 2010$ such that any other such\npermutation is guaranteed to intersect at least one of these $1006$\npermutations.",
+        "Solution_1": "Let $n$ be a positive integer. Let $m$ be the smallest positive integer with\n$m > n - m$.  Since $n > n - n = 0$, $m \\le n$. Let $N = \\{1, \\ldots, n\\}$\nbe the set of positive integers from $1$ to $n$.  Let $M \\subset N$,\nbe $M = \\{1, \\ldots, m\\}$.\nLet $P_n$ be the set of of permutations of $N$.\nLet $C_m$ be the set of cyclic permutations of $M$, there are $m$\ncyclic permutations in all, and $C_m$ acts transitively on $M$, i.e.\nfor every pair of elements $a,b \\in M$, there is an element of $C_m$\nthat maps $a$ to $b$.\nLet $C'_m \\subset P_n$ be the permutations in $P_n$ that leave $N\\setminus M$\nfixed, and restricted to $M$ yield one of the permutations in $C_m$.\nThere is a natural one-to-one correspondence between $C'_m$ and $C_m$.\nWe claim that the $m$ permutations $C'_m$ intersect every permutation in\n$P_n$.\nSuppose, to the contrary, that there exists a permutation $p \\in P_n$\nthat does not intersect any permutation in $C'_m$. Since $C'_m$ acts\ntransitively on $M \\subset N$ the permutation $p$ cannot send any element of\n$M$ to any other element of $M$, therefore it must send all the\nelements of $M$ to $N\\setminus M$, but since $N\\setminus M$ has $n - m$ elements and $m > n - m$, this is impossible\nby the pigeonhole principle. Therefore such a permutation cannot\nexist, and the permutations in $C'_m$ intersect every permutation in\n$P_n$.\nFor $n = 2010$ we get $m = 1006$, which is the required special\ncase of the general result above.",
+        "Solution_2": "I construct the following permutations by continuously rotating the first 1006 numbers):\n(1, 2, 3 ... 1005, 1006, 1007 ... 2009, 2010)\n(2, 3, 4 ... 1006, 1, 1007 ... 2009, 2010)\n...\n...\n...\n(1006, 1, 2 ... 1005, 1007, ... 2009, 2010)\n\nI claim that these permutations above satisfy the property that any other permutation will intersect with at least one of them.\nProof:\nAssume for the sake of contradiction that there exists a permutation that won't intersect with these, say $(a_1, a_2, a_3 ... a_{2010})$.\nLemma:\nIf there exists a k such that $1 \\leq k \\leq 1006$ $a_k \\leq 1006$, we will get an intersection with the permutations.\nNote that for any 2 distinct permutations in our list, the numbers in kth index must be different, since each permutation is a rotation of an previous permutation. Also, the numbers can't exceed 1006, so, each number must occur exactly once in the kth index.\nEnd Lemma\nUsing the lemma, in order to avoid intersections, we need $1007 \\leq a_k \\leq 2010$, for all $1 \\leq k \\leq 1006$.\nBut, there are 1004 numbers such that $1007 \\leq n \\leq 2010$, but we need to have 1004 numbers to fill in 1006 spots, and thus, by pigeonhole principle(the numbers are the holes, the spots are the pigeons), there must be 2 spots that have the same number! However, in a permutation, all numbers must be distinct, so we have a contradiction.\nSo, we have shown such a set of permutations exists satisfying that any permutation is guaranteed to intersect one of them, and the proof is complete. $\\blacksquare$\n-Alexlikemath"
+    },
+    {
+        "Problem": "Let $ABC$ be a triangle with $\\angle A = 90^{\\circ}$. Points $D$\nand $E$ lie on sides $AC$ and $AB$, respectively, such that $\\angle ABD = \\angle DBC$ and $\\angle ACE = \\angle ECB$. Segments $BD$ and\n$CE$ meet at $I$.  Determine whether or not it is possible for\nsegments $AB, AC, BI, ID, CI, IE$ to all have integer lengths.",
+        "Solution": "We know that angle $BIC = 135^{\\circ}$, as the other two angles in triangle $BIC$ add to $45^{\\circ}$. Assume that only $AB, AC, BI$, and $CI$ are integers. Using the Law of Cosines on triangle BIC,\n$BC^2 = BI^2 + CI^2 - 2BI\\cdot CI \\cdot \\cos 135^{\\circ}$. Observing that $BC^2 = AB^2 + AC^2$ is an integer and that $\\cos 135^{\\circ} = -\\frac{\\sqrt{2}}{2},$ we have\nand therefore,\nThe LHS ($\\sqrt{2}$) is irrational, while the RHS is the quotient of the division of two integers and thus is rational. Clearly, there is a contradiction. Therefore, it is impossible for $AB, AC, BI$, and $CI$ to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.",
+        "Solution_2": "The answer is no.\nSuppose otherwise. It is easy to see (through simple angle chasing) that $\\angle DIC=45^{\\circ}$. Also, since $I$ is the incenter, we have $\\angle IAC = 45^{\\circ}$. Using the Law of Cosines, we have  so that $CD$ is irrational. But $\\triangle IAC \\sim \\triangle DIC$, thus $IC^2=CD\\cdot AC$, implying that $CD$ is rational, contradiction. $\\blacksquare$",
+        "Solution_3": "The result can be also proved without direct appeal to trigonometry,\nvia just the angle bisector theorem and the structure of Pythagorean\ntriples. (This is a lot more work).\nA triangle in which all the required lengths are integers exists if and\nonly if there exists a triangle in which $AB$ and $AC$ are\nrelatively-prime integers and the lengths of the segments $BI, ID, CI, IE$ are all rational\n(we divide all the lengths by the $\\gcd(AB, AC)$ or\nconversely multiply all the lengths by the least common multiple\nof the denominators of the rational lengths).\nSuppose there exists a triangle in which the lengths $AB$ and $AC$ are\nrelatively-prime integers and the lengths $IB, ID, CI, IE$ are all rational.\nSince $CE$ is the bisector of $\\angle ACB$, by the angle bisector\ntheorem, the ratio $IB : ID = CB : CD$, and since $BD$ is the\nbisector of $\\angle ABC$, $CB : CD = (AB + BC) : AC$. Therefore,\n$IB : ID = (AB + BC) : AC$. Now $IB : ID$ is by assumption rational,\nso $(AB + BC) : AC$ is rational, but $AB$ and $AC$ are assumed integers\nso $BC : AC$ must also be rational. Since $BC$ is the hypotenuse of\na right-triangle, its length is the square root of an integer,\nand thus either an integer or irrational, so $BC$ must be an integer.\nWith $AB$ and $AC$ relatively-prime, we conclude that the side\nlengths of $\\triangle ABC$ must be a Pythagorean triple: $(2pq, p^2 - q^2, p^2 + q^2)$, with $p > q$ relatively-prime positive integers\nand $p+q$ odd.\nWithout loss of generality, $AC = 2pq, AB = p^2 - q^2, BC = p^2+q^2$.\nBy the angle bisector theorem,\nSince $\\triangle CAE$ is a right-triangle, we have:\nand so $CE$ is rational if and only if $2p^2 + 2q^2$ is a perfect square.\nAlso by the angle bisector theorem,\nand therefore, since $\\triangle DAB$ is a right-triangle, we have:\nand so $BD$ is rational if and only if $p^2 + q^2$ is a perfect square.\nCombining the conditions on $CE$ and $BD$, we see that\n$2p^2+2q^2$ and $p^2+q^2$ must both be perfect squares. If it were so,\ntheir ratio, which is $2$, would be the square of a rational number,\nbut $\\sqrt{2}$ is irrational, and so the assumed triangle cannot exist.",
+        "Solution_4": "We proceed by contradiction.\nFTSOC, let $AB, BI, ID$ have integer lengths. Then $BD = BI + ID \\in \\mathbb{Z}$ as well. By trigonometry, \nRearranging we find $\\cos{(\\frac{\\angle ABC}{2})} = \\frac{AB}{BD} \\in \\mathbb{Q}$. As $0 < \\angle ABC < \\frac{\\pi}{2}$ so $0 < \\frac{\\angle ABC}{2} < \\frac{\\pi}{4}$ but there are no possible angles in the interval that result in a rational cosine by Niven's theorem. So we have contradiction.\n~Aaryabhatta1."
+    },
+    {
+        "Problem": "Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square.",
+        "Solution": "The answer is $n=1$, which is easily verified to be a valid integer $n$.\nNotice that  Then for $n\\geq 2$, we have $2^n+7^n\\equiv 3,5 \\pmod{12}$ depending on the parity of $n$. But perfect squares can only be $0,1,4,9\\pmod{12}$, contradiction. Therefore, we are done. $\\blacksquare$",
+        "Solution_1": "Let $2^n + 12^n + 2011^n = x^2$. \nThen $(-1)^n + 1 \\equiv x^2 \\pmod {3}$. \nSince all perfect squares are congruent to 0 or 1 modulo 3, this means that n must be odd. \nProof by Contradiction:\nWe wish to show that the only value of $n$ that satisfies is $n = 1$. \nAssume that $n \\ge 2$. \nThen consider the equation\n$2^n + 12^n = x^2 - 2011^n$. \nFrom modulo 2, we easily know x is odd. Let $x = 2a + 1$, where a is an integer. \n$2^n + 12^n = 4a^2 + 4a + 1 - 2011^n$. \nDividing by 4,\n$2^{n-2} + 3^n \\cdot 4^{n-1} = a^2 + a + \\dfrac {1}{4} (1 - 2011^n)$. \nSince $n \\ge 2$, $n-2 \\ge 0$, so $2^{n-2}$ similarly, the entire LHS is an integer, and so are $a^2$ and $a$. \nThus, $\\dfrac {1}{4} (1 - 2011^n)$ must be an integer. \nLet $\\dfrac {1}{4} (1 - 2011^n) = k$. \nThen we have $1- 2011^n = 4k$. \n$1- 2011^n \\equiv 0 \\pmod {4}$. \n$(-1)^n \\equiv 1 \\pmod {4}$. \nThus, n is even. \nHowever, it has already been shown that $n$ must be odd.  This is a contradiction.  Therefore, $n$ is not greater than or equal to 2, and must hence be less than 2.  The only positive integer less than 2 is 1.",
+        "Solution_2": "If $n = 1$, then $2^n + 12^n + 2011^n = 2025 = 45^2$, a perfect square.\nIf $n > 1$ is odd, then $2^n + 12^n + 2011^n \\equiv 0 + 0 + (-1)^n \\equiv 3 \\pmod{4}$.\nSince all perfect squares are congruent to $0,1 \\pmod{4}$, we have that $2^n+12^n+2011^n$ is not a perfect square for odd $n > 1$.\nIf $n = 2k$ is even, then $(2011^k)^2 < 2^{2k}+12^{2k}+2011^{2k}$ $= 4^k + 144^k + 2011^{2k} <$ $2011^k + 2011^k + 2011^{2k} < (2011^k+1)^2$.\nSince $(2011^k)^2 < 2^n+12^n+2011^n < (2011^k+1)^2$, we have that $2^n+12^n+2011^n$ is not a perfect square for even $n$.\nThus, $n = 1$ is the only positive integer for which $2^n + 12^n + 2011^n$ is a perfect square.\n",
+        "Solution_3": "Looking at residues mod 3, we see that $n$ must be odd, since even values of $n$ leads to $2^n + 12^n + 2011^n = 2 \\pmod{3}$.  Also as shown in solution 2, for $n>1$, $n$ must be even.  Hence, for $n>1$, $n$ can neither be odd nor even.  The only possible solution is then $n=1$, which indeed works.",
+        "Solution_4": "Take the whole expression mod 12. Note that the perfect squares can only be of the form 0, 1, 4 or 9 (mod 12). Note that since the problem is asking for positive integers, $12^n$ is always divisible by 12, so this will be disregarded in this process. If $n$ is even, then $2^n \\equiv 4 \\pmod{12}$ and $2011^n \\equiv 7^n \\equiv 1 \\pmod {12}$. Therefore, the sum in the problem is congruent to $5 \\pmod {12}$, which cannot be a perfect square. Now we check the case for which $n$ is an odd number greater than 1. Then $2^n \\equiv 8 \\pmod{12}$ and $2011^n \\equiv 7^n \\equiv 7 \\pmod {12}$. Therefore, this sum would be congruent to $3 \\pmod {12}$, which cannot be a perfect square. The only case we have not checked is $n=1$. If $n=1$, then the sum in the problem is equal to $2+12+2011=2025=45^2$. Therefore the only possible value of $n$ such that $2^n+12^n+2011^n$ is a perfect square is $n=1$.",
+        "Solution_5": "We will first take the expression modulo $3$. We get $2^n+12^n+2011^n \\equiv -1^n+1^n \\pmod 3$.\nLemma 1: All perfect squares are equal to $0$ or $1$ modulo $3$.\nWe can prove this by testing the residues modulo $3$. We have $0^2 \\equiv 0 \\pmod 3$, $1^2 \\equiv 1 \\pmod 3$, and $2^2 \\equiv 1 \\pmod 3$, so the lemma is true.\nWe know that if $n$ is odd, $-1^n+1^n \\equiv 0 \\pmod 3$, which satisfies the lemma's conditions. However, if $n$ is even, we get $2 \\pmod 3$, which does not satisfy the lemma's conditions. So, we can conclude that $n$ is odd.\nNow, we take the original expression modulo $4$. For right now, we will assume that $n>1$, and test $n=1$ later. For $n>1$, $2^n \\equiv 0 \\pmod 4$, so $2^n+12^n+2011^n=-1^n \\pmod 4$.\nLemma 2: All perfect squares are equal to $0$ or $1$ modulo $4$.\nWe can prove this by testing the residues modulo $4$. We have $0^2 \\equiv 0 \\pmod 4$, $1^2 \\equiv 1 \\pmod 4$, $2^2 \\equiv 0 \\pmod 4$, and $3^2 \\equiv 1 \\pmod 4$, so the lemma is true.\nWe know that if $n$ is even, $-1^n \\equiv 0 \\pmod 4$, which satisfies the lemma's conditions. However, if $n$ is odd, $-1^n \\equiv -1 \\equiv 3 \\pmod 4$, which does not satisfy the lemma's conditions. Therefore, $n$ must be even.\nHowever, a number cannot be even and odd at the same time, so this is impossible. Now, we only have to test $n=1$. We know that $2^1+12^1+2011^1=45^2$, so the only integer is $\\boxed{n=1}$.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Let $a$, $b$, $c$ be positive real numbers such that $a^2 + b^2 + c^2 + (a + b + c)^2 \\le 4$.  Prove that",
+        "Solution_1": "Since\n\nit is natural to consider a change of variables:\n\nwith the inverse mapping given by:\n\nWith this change of variables, the constraint becomes\n\nwhile the left side of the inequality we need to prove is now\nTherefore it remains to prove that\nWe note that the product of the three (positive) terms is 1/8, therefore by AM-GM their mean is at least 1/2, and thus their sum is at least 3/2 and we are done.",
+        "Solution_2": "Rearranging the condition yields that\nNow note that\nSumming this for all pairs of $\\{ a,b,c \\}$ gives that\nBy AM-GM. Dividing by $2$ gives the desired inequality.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "For a point $P = (a, a^2)$ in the coordinate plane, let $\\ell(P)$ denote the line passing through $P$ with slope $2a$. Consider the set of triangles with vertices of the form $P_1 = (a_1,  a_1^2)$, $P_2 = (a_2, a_2^2)$, $P_3 = (a_3, a_3^2)$, such that the intersections of the lines $\\ell(P_1)$, $\\ell(P_2)$, $\\ell(P_3)$ form an equilateral triangle $\\triangle$. Find the locus of the center of $\\triangle$ as $P_1P_2P_3$ ranges over all such triangles.",
+        "Solution_1": "Note that the lines $l(P_1), l(P_2), l(P_3)$ are  respectively. It is easy to deduce that the three points of intersection are  The slopes of each side of this equilateral triangle are  and we want to find the locus of  Define the three complex numbers $w_n = 1+2a_ni$ for $n=1,2,3$. Then note that the slope - that is, the imaginary part divided by the real part - of all $w_n^3$ is constant, say it is $k$. Then for $n=1,2,3$,\n\nRearranging, we get thatorNote that this is a cubic, and the roots are $a_1,a_2$ and $a_3$ which are all distinct, and so there are no other roots. Using Vieta's, we get that\nand\nObviously all values of $k$ are possible, and so our answer is the line  $\\blacksquare$\n~ cocohearts",
+        "Solution_2": "Note that all the points $P=(a,a^2)$ belong to the parabola $y=x^2$ which we will denote $p$. This parabola has a focus $F=\\left(0,\\frac{1}{4}\\right)$ and directrix $y=-\\frac{1}{4}$ which we will denote $d$. We will prove that the desired locus is $d$.\nFirst note that for any point $P$ on $p$, the line $\\ell(P)$ is the tangent line to $p$ at $P$. This is because $\\ell(P)$ contains $P$ and because $[\\frac{d}{dx}] x^2=2x$. If you don't like calculus, you can also verify that $\\ell(P)$ has equation $y=2a(x-a)+a^2$ and does not intersect $y=x^2$ at any point besides $P$. Now for any point $P$ on $p$ let $P'$ be the foot of the perpendicular from $P$ onto $d$. Then by the definition of parabolas, $PP'=PF$. Let $q$ be the perpendicular bisector of $\\overline{P'F}$. Since $PP'=PF$, $q$ passes through $P$. Suppose $K$ is any other point on $q$ and let $K'$ be the foot of the perpendicular from $K$ to $d$. Then in right $\\Delta KK'P'$, $KK'$ is a leg and so $KK'<KP'=KF$. Therefore $K$ cannot be on $p$. This implies that $q$ is exactly the tangent line to $p$ at $P$, that is $q=\\ell(P)$. So we have proved Lemma 1: If $P$ is a point on $p$ then $\\ell(P)$ is the perpendicular bisector of $\\overline{P'F}$.\nWe need another lemma before we proceed. Lemma 2: If $F$ is on the circumcircle of $\\Delta XYZ$ with orthocenter $H$, then the reflections of $F$ across $\\overleftrightarrow{XY}$, $\\overleftrightarrow{XZ}$, and $\\overleftrightarrow{YZ}$ are collinear with $H$.\nProof of Lemma 2: Say the reflections of $F$ and $H$ across $\\overleftrightarrow{YZ}$ are $C'$ and $J$, and the reflections of $F$ and $H$ across $\\overleftrightarrow{XY}$ are $A'$ and $I$. Then we angle chase $\\angle JYZ=\\angle HYZ=\\angle HXZ=\\angle JXZ=m(JZ)/2$ where $m(JZ)$ is the measure of minor arc $JZ$ on the circumcircle of $\\Delta XYZ$. This implies that $J$ is on the circumcircle of $\\Delta XYZ$, and similarly $I$ is on the circumcircle of $\\Delta XYZ$. Therefore $\\angle C'HJ=\\angle FJH=m(XF)/2$, and $\\angle A'HX=\\angle FIX=m(FX)/2$. So $\\angle C'HJ = \\angle A'HX$. Since $J$, $H$, and $X$ are collinear it follows that $C'$, $H$ and $A'$ are collinear. Similarly, the reflection of $F$ over $\\overleftrightarrow{XZ}$ also lies on this line, and so the claim is proved.\nNow suppose $A$, $B$, and $C$ are three points of $p$ and let $\\ell(A)\\cap\\ell(B)=X$, $\\ell(A)\\cap\\ell(C)=Y$, and $\\ell(B)\\cap\\ell(C)=Z$. Also let $A''$, $B''$, and $C''$ be the midpoints of $\\overline{A'F}$, $\\overline{B'F}$, and $\\overline{C'F}$ respectively. Then since $\\overleftrightarrow{A''B''}\\parallel \\overline{A'B'}=d$ and $\\overleftrightarrow{B''C''}\\parallel \\overline{B'C'}=d$, it follows that $A''$, $B''$, and $C''$ are collinear. By Lemma 1, we know that $A''$, $B''$, and $C''$ are the feet of the altitudes from $F$ to $\\overline{XY}$, $\\overline{XZ}$, and $\\overline{YZ}$. Therefore by the Simson Line Theorem, $F$ is on the circumcircle of $\\Delta XYZ$. If $H$ is the orthocenter of $\\Delta XYZ$, then by Lemma 2, it follows that $H$ is on $\\overleftrightarrow{A'C'}=d$. It follows that the locus described in the problem is a subset of $d$.\nSince we claim that the locus described in the problem is $d$, we still need to show that for any choice of $H$ on $d$ there exists an equilateral triangle with center $H$ such that the lines containing the sides of the triangle are tangent to $p$. So suppose $H$ is any point on $d$ and let the circle centered at $H$ through $F$ be $O$. Then suppose $A$ is one of the intersections of $d$ with $O$. Let $\\angle HFA=3\\theta$, and construct the ray through $F$ on the same halfplane of $\\overleftrightarrow{HF}$ as $A$ that makes an angle of $2\\theta$ with $\\overleftrightarrow{HF}$. Say this ray intersects $O$ in a point $B$ besides $F$, and let $q$ be the perpendicular bisector of $\\overline{HB}$. Since $\\angle HFB=2\\theta$ and $\\angle HFA=3\\theta$, we have $\\angle BFA=\\theta$. By the inscribed angles theorem, it follows that $\\angle AHB=2\\theta$. Also since $HF$ and $HB$ are both radii, $\\Delta HFB$ is isosceles and $\\angle HBF=\\angle HFB=2\\theta$. Let $P_1'$ be the reflection of $F$ across $q$. Then $2\\theta=\\angle FBH=\\angle C'HB$, and so $\\angle C'HB=\\angle AHB$. It follows that $P_1'$ is on $\\overleftrightarrow{AH}=d$, which means $q$ is the perpendicular bisector of $\\overline{FP_1'}$.\nLet $q$ intersect $O$ in points $Y$ and $Z$ and let $X$ be the point diametrically opposite to $B$ on $O$. Also let $\\overline{HB}$ intersect $q$ at $M$. Then $HM=HB/2=HZ/2$. Therefore $\\Delta HMZ$ is a $30-60-90$ right triangle and so $\\angle ZHB=60^{\\circ}$. So $\\angle ZHY=120^{\\circ}$ and by the inscribed angles theorem, $\\angle ZXY=60^{\\circ}$. Since $ZX=ZY$ it follows that $\\Delta ZXY$ is and equilateral triangle with center $H$.\nBy Lemma 2, it follows that the reflections of $F$ across $\\overleftrightarrow{XY}$ and $\\overleftrightarrow{XZ}$, call them $P_2'$ and $P_3'$, lie on $d$. Let the intersection of $\\overleftrightarrow{YZ}$ and the perpendicular to $d$ through $P_1'$ be $P_1$, the intersection of $\\overleftrightarrow{XY}$ and the perpendicular to $d$ through $P_2'$ be $P_2$, and the intersection of $\\overleftrightarrow{XZ}$ and the perpendicular to $d$ through $P_3'$ be $P_3$. Then by the definitions of $P_1'$, $P_2'$, and $P_3'$ it follows that $FP_i=P_iP_i'$ for $i=1,2,3$ and so $P_1$, $P_2$, and $P_3$ are on $p$. By lemma 1, $\\ell(P_1)=\\overleftrightarrow{YZ}$, $\\ell(P_2)=\\overleftrightarrow{XY}$, and $\\ell(P_3)=\\overleftrightarrow{XZ}$. Therefore the intersections of $\\ell(P_1)$, $\\ell(P_2)$, and $\\ell(P_3)$ form an equilateral triangle with center $H$, which finishes the proof.\n--Killbilledtoucan",
+        "Solution_3": "Note that the lines $l(P_1), l(P_2), l(P_3)$ are  respectively. It is easy to deduce that the three points of intersection are  The slopes of each side of this equilateral triangle are  and we want to find the locus of  We know that  for some $\\theta.$ Therefore, we can use the tangent addition formula to deduce  and  Now we show that $\\frac{a_1+a_2+a_3}{3}$ can be any real number. Let's say  for some real number $k.$ Multiplying both sides by $2-\\tan^2\\theta$ and rearranging yields a cubic in $\\tan\\theta.$ Clearly this cubic has at least one real solution. As $\\tan \\theta$ can take on any real number, all values of $k$ are possible, and our answer is the line  Of course, as the denominator could equal 0, we must check $\\tan \\theta=\\pm \\frac{1}{\\sqrt{3}}.$  The left side is nonzero, while the right side is zero, so these values of $\\theta$ do not contribute to any values of $k.$ So, our answer remains the same. $\\blacksquare$ ~ Benq"
+    },
+    {
+        "Problem": "A word is defined as any finite string of letters.  A word is a palindrome if it reads the same backwards as forwards.  Let a sequence of words $W_0$, $W_1$, $W_2$, $\\dots$ be defined as follows: $W_0 = a$, $W_1 = b$, and for $n \\ge 2$, $W_n$ is the word formed by writing $W_{n - 2}$ followed by $W_{n - 1}$.  Prove that for any $n \\ge 1$, the word formed by writing $W_1$, $W_2$, $\\dots$, $W_n$ in succession is a palindrome.",
+        "Solution": "Let $r$ be the reflection function on the set of words, namely $r(a_1\\dots a_n) = a_n \\dots a_1$ for all words $a_1 \\dots a_n$, $n\\ge 1$. Then the following property is evident (e.g. by mathematical induction):\n$r(w_1 \\dots w_k) = r(w_k) \\dots r(w_1)$, for any words $w_1, \\dots, w_k$, $k \\ge 1$.\nWe use mathematical induction to prove the statement of the problem. First, $W_1 = b$, $W_1W_2 = bab$, $W_1W_2W_3 = babbab$  are palindromes. Second, suppose $n\\ge 3$, and that the words $W_1 W_2 \\dots W_k$  ($k = 1$, $2$, $\\dots$, $n$) are all palindromes, i.e. $r(W_1W_2\\dots W_k) = W_1W_2\\dots W_k$. Now, consider the word $W_1 W_2 \\dots W_{n+1}$:\n\n\n\n\n\nBy the principle of mathematical induction, the statement of the problem is proved. Lightest 21:54, 1 April 2012 (EDT)\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Points $A$, $B$, $C$, $D$, $E$ lie on a circle $\\omega$ and point $P$ lies outside the circle.  The given points are such that (i) lines $PB$ and $PD$ are tangent to $\\omega$, (ii) $P$, $A$, $C$ are collinear, and (iii) $\\overline{DE} \\parallel \\overline{AC}$.  Prove that $\\overline{BE}$ bisects $\\overline{AC}$.",
+        "Solution": "Since $\\overline{DE} \\parallel \\overline{AC}$, we have that $AD=CE\\rightarrow \\angle ABD=\\angle CBE$. But it is well known that $ABCD$ is a harmonic quadrilateral, thus $\\overline{BD}$ is a symmedian of triangle $ABC$, from which it follows that $\\overline{BE}$ is a median of $\\triangle ABC$. $\\blacksquare$",
+        "Solution_1": "Connect segment PO, and name the interaction of PO and the circle as point M.\nSince PB and PD are tangent to the circle, it's easy to see that M is the midpoint of arc BD.\n∠ BOA = 1/2 arc AB + 1/2 arc CE\nSince AC // DE, arc AD = arc CE,\nthus, ∠ BOA = 1/2 arc AB + 1/2 arc AD = 1/2 arc BD = arc BM = ∠ BOM\nTherefore, PBOM is cyclic, ∠ PFO = ∠ OBP = 90°, AF = AC (F is the interaction of BE and AC)\nBE bisects AC, proof completed!\n~ MVP Harry",
+        "Solution_2": "Let $O$ be the center of the circle, and let $X$ be the intersection of $AC$ and $BE$. Let $\\angle OPA$ be $x$ and $\\angle OPD$ be $y$.\n$\\angle OPB = \\angle OPD = y$, $\\angle BED = \\frac{\\angle DOB}{2} = 90-y$, \n$\\angle ODE = \\angle PDE - 90 = 90-x-y$\n$\\angle OBE = \\angle PBE - 90 = x = \\angle OPA$\nThus $PBXO$ is a cyclic quadrilateral and $\\angle OXP = \\angle OBP = 90$ and so $X$ is the midpoint of chord $AC$.\n~pandadude",
+        "Solution_3": "This is the solution from EGMO Problem 1.43 page 242\nLet $O$ be the center of the circle, and let $M$ be the midpoint of $AC$.  Let $\\theta$ denote the circle with diameter $OP$.  Since $\\angle OBP = \\angle OMP = \\angle ODP = 90^\\circ$, $B$, $D$, and $M$ all lie on $\\theta$.\n\nSince quadrilateral $BOMP$ is cyclic, $\\angle BMP = \\angle BOP$.  Triangles $BOP$ and $DOP$ are congruent, so $\\angle BOP = \\angle BOD/2 = \\angle BED$, so $\\angle BMP = \\angle BED$. Because $AC$ and $DE$ are parallel, $M$ lies on $BE$ (using Euclid's Parallel Postulate).\n-Evan Chen (vEnhance)",
+        "Solution_4": "Note that by Lemma 9.9 of EGMO, $(A,C;B,D)$ is a harmonic bundle. We project through $E$ onto $\\overline{AC}$,\n\nWhere $P_{\\infty}$ is the point at infinity for parallel lines $\\overline{DE}$ and $\\overline{AC}$. Thus, we get $\\frac{MA}{MC}=-1$, and $M$ is the midpoint of $AC$. ~novus677\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Consider the assertion that for each positive integer $n \\ge 2$, the remainder upon dividing $2^{2^n}$ by $2^n-1$ is a power of 4. Either prove the assertion or find (with proof) a counter-example.",
+        "Solution": "We will show that $n = 25$ is a counter-example.\nSince $\\textstyle 2^n \\equiv 1 \\pmod{2^n - 1}$, we see that for any integer $k$, $\\textstyle 2^{2^n} \\equiv 2^{(2^n - kn)} \\pmod{2^n-1}$. Let $0 \\le m < n$ be the residue of $2^n \\pmod n$. Note that since $\\textstyle m < n$ and $\\textstyle n \\ge 2$, necessarily $\\textstyle 2^m < 2^n -1$, and thus the remainder in question is $\\textstyle 2^m$. We want to show that $\\textstyle 2^m \\pmod {2^n-1}$ is an odd power of 2 for some $\\textstyle n$, and thus not a power of 4.\nLet $\\textstyle n=p^2$ for some odd prime $\\textstyle p$. Then $\\textstyle \\varphi(p^2) = p^2 - p$. Since 2 is co-prime to $\\textstyle p^2$, we have\n\nand thus\nTherefore, for a counter-example, it suffices that $\\textstyle 2^p \\pmod{p^2}$ be odd. Choosing $\\textstyle p=5$, we have $\\textstyle 2^5 = 32 \\equiv 7 \\pmod{25}$. Therefore, $\\textstyle 2^{25} \\equiv 7 \\pmod{25}$ and thus\n\nSince $\\textstyle 2^7$ is not a power of 4, we are done.",
+        "Solution_2": "Lemma (useful for all situations): If $x$ and $y$ are positive integers such that $2^x - 1$ divides $2^y - 1$, then $x$ divides $y$.\nProof: $2^y \\equiv 1 \\pmod{2^x - 1}$. Replacing the $1$ with a $2^x$ and dividing out the powers of two should create an easy induction proof which will be left to the reader as an Exercise.\nConsider $n = 25$. We will prove that this case is a counterexample via contradiction.\nBecause $4 = 2^2$, we will assume there exists a positive integer $k$ such that $2^{2^n} - 2^{2k}$ divides $2^n - 1$ and $2^{2k} < 2^n - 1$. Dividing the powers of $2$ from LHS gives $2^{2^n - 2k} - 1$ divides $2^n - 1$. Hence, $2^n - 2k$ divides $n$. Because $n = 25$ is odd, $2^{24} - k$ divides $25$. Euler's theorem gives $2^{24} \\equiv 2^4 \\equiv 16 \\pmod{25}$ and so $k \\ge 16$. However, $2^{2k} \\geq 2^{32} > 2^{25} - 1$, a contradiction. Thus, $n = 25$ is a valid counterexample.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Given a triangle $ABC$, let $P$ and $Q$ be points on segments $\\overline{AB}$ and $\\overline{AC}$, respectively, such that $AP = AQ$.  Let $S$ and $R$ be distinct points on segment $\\overline{BC}$ such that $S$ lies between $B$ and $R$, $\\angle{BPS} = \\angle{PRS}$, and $\\angle{CQR} = \\angle{QSR}$.  Prove that $P$, $Q$, $R$, $S$ are concyclic (in other words, these four points lie on a circle).",
+        "Solution": "Since $\\angle BPS = \\angle PRS$, the circumcircle of triangle $PRS$ is tangent to $AB$ at $P$.  Similarly, since $\\angle CQR = \\angle QSR$, the circumcircle of triangle $QRS$ is tangent to $AC$ at $Q$.\n\nFor the sake of contradiction, suppose that the circumcircles of triangles $PRS$ and $QRS$ are not the same circle.  Since $AP = AQ$, $AP^2 = AQ^2$, so the power of A with respect to both circumcircles is the same. Thus, $A$ lies on the radical axis of both circles.  However, both circles pass through $R$ and $S$, so the radical axis of both circles is $RS$.  Hence, $A$ lies on $RS$, which is a contradiction.\nTherefore, the two circumcircles are the same circle.  In other words, $P$, $Q$, $R$, and $S$ all lie on the same circle.",
+        "Solution_2": "Note that (as in the first solution) the circumcircle of triangle $PRS$ is tangent to $AB$ at $P$.  Similarly, since $\\angle CQR = \\angle QSR$, the circumcircle of triangle $QRS$ is tangent to $AC$ at $Q$.\nNow, suppose these circumcircles are not the same circle. They already intersect at $R$ and $S$, so they cannot intersect anymore. Thus, AS must touch the two circumcircles at points $M$ and $N$, with $M$ on the circumcircle of triangle $PRS$. By Power of a Point, $AQ^2 = AM \\cdot AS$ and $AP^2 = AN \\cdot AS$. Hence, because $AP = AQ$, $AM = AN$, a contradiction because then, as they lie on the same line segment, M and N must be the same point! (Note line segment, not line.) Hence, the two circumcircles are the same circle."
+    },
+    {
+        "Problem": "Find all integers $n \\ge 3$ such that among any $n$ positive real numbers $a_1$, $a_2$, $\\dots$, $a_n$ with\n\nthere exist three that are the side lengths of an acute triangle.",
+        "Solution": "Without loss of generality, assume that the set $\\{a\\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \\le n \\cdot a_1.$ Now set $b_i \\equiv \\frac{a_i}{a_1},$ and since a triangle with sidelengths from $\\{a\\}$ will be similar to the corresponding triangle from $\\{b\\},$ we simply have to show the existence of acute triangles in $\\{b\\}.$ Note that $b_1 = 1$ and for all $i$, $b_i \\le n.$\nNow three arbitrary sidelengths $x$, $y$, and $z$, with $x \\le y \\le z,$ will form a valid triangle if and only if $x+y>z.$ Furthermore, this triangle will be acute if and only if $x^2 + y^2 > z^2.$ However, the first inequality can actually be inferred from the second, since $x+y>z \\longrightarrow x^2 + y^2 +2xy > z^2$ and $2xy$ is trivially greater than $0.$ So we just need to find all $n$ such that there is necessarily a triplet of $b$'s for which $b_i^2 + b_j^2 > b_k^2$ (where $b_i < b_j < b_k$).\nWe now make another substitution: $c_i \\equiv b_i ^2.$ So $c_1 = 1$ and for all $i$, $c_i \\le n^2.$ Now we examine the smallest possible sets $\\{c\\}$ for small $n$ for which the conditions of the problem are not met. Note that by smallest, we mean the set whose greatest element is as small as possible. If $n=3$, then the smallest possible set, call it $\\{s_3\\},$ is trivially $\\{1,1,2\\}$, since $c_1$ and $c_2$ are obviously minimized and $c_3$ follows as minimal. Using this as the base case, we see inductively that in general $\\{s_n\\}$ is the set of the first $n$ Fibonacci numbers. To show this note that if $\\{s_n\\} = \\{F_0, F_1, ... F_n\\}$, then $\\{s_{n+1}\\} = \\{F_0, F_1, ... F_n, c_{n+1}\\}.$ The smallest possible value for $c_{n+1}$ is the sum of the two greatest values of $\\{s_n\\}$ which are $F_{n-1}$ and $F_n$. But these sum to $F_{n+1}$ so $\\{s_{n+1}\\} = \\{F_0, F_1, ... F_{n+1}\\}$ and our induction is complete.\nNow since we know that the Fibonacci set is the smallest possible set which does not satisfy the conditions of the problem, then any set $\\{c\\}$ whose greatest term is less than $F_{n-1}$ must satisfy the conditions. And since $\\{c\\}$ is bounded between $1$ and $n^2$, then the conditions of the problem are met if and only if $F_{n-1} > n^2$. The first $n$ for which this restriction is satisfied is $n=13$ and the exponential behavior of the Fibonacci numbers ensure that every $n$ greater than $13$ will also satisfy this restriction. So the final solution set is $\\boxed{\\{n \\ge 13\\}}$.",
+        "Solution_2": "Outline:\n1. Define the Fibonacci numbers to be $F_1 = F_2 = 1$ and $F_k = F_{k-1} + F_{k-2}$ for $k \\ge 3$.\n2. If the chosen $n$ is such that $F_n \\le n^2$, then choose the sequence $a_n$ such that $a_k = \\sqrt{F_k}$ for $1 \\le k \\le n$. It is easy to verify that such a sequence satisfies the condition that the largest term is less than or equal to $n$ times the smallest term. Also, because for any three terms $x = \\sqrt{F_a}, y = \\sqrt{F_b}, z = \\sqrt{F_c}$ with $a<b<c$, $x^2 + y^2 = F_a + F_b \\le F_{b-1} + F_b = F_{b+1} \\le F_c = z^2$, x, y, z do not form an acute triangle. Thus, all $n$ such that $F_n \\le n^2$ do not work.\n3. It is easy to observe via a contradiction argument that all $n$ such that $F_n > n^2$ produce an acute triangle. (If, without loss of generality, $a_n$ is an increasing sequence, such that no three (in particular, consecutive) terms form an acute triangle, then $a_2^2 \\ge F_1a_1^2, a_3^2 \\ge a_2^2 + a_1^2 \\ge F_2a^2$, and by induction $a_n^2 > F_na_1^2$, a contradiction to the condition's inequality.)\n4. Note that $F_{12} = 144 = 12^2$ and $F_{13} = 233 > 169 = 13^2$. It is easily to verify through strong induction that all $n$ greater than 12 make $F_n > n^2$. Thus, $\\boxed{n \\ge 13}$ is the desired solution set."
+    },
+    {
+        "Problem": "Let $a$, $b$, $c$ be positive real numbers.  Prove that",
+        "Solution": "By the Cauchy-Schwarz inequality,\n\nso\n\nSince $a^2 + b^2 + c^2 \\ge ab + ac + bc$,\n\nHence,\nAgain by the Cauchy-Schwarz inequality,\n\nso\n\nSince $a^2 + b^2 + c^2 \\ge ab + ac + bc$,\n\nHence,\nTherefore,",
+        "Solution_2": "Titu's Lemma: The sum of multiple fractions in the form $\\frac{a_n^2}{b_n}$ where $a_n$ and $b_n$ are sequences of real numbers is greater than of equal to the square of the sum of all $a_i$ divided by the sum of all $b_i$, where i is a whole number less than n+1. Titu's Lemma can be proved using the Cauchy-Schwarz Inequality after multiplying out the denominator of the RHS.\nConsider the LHS of the proposed inequality. Split up the numerators and multiply both sides of the fraction by either a or 3a to make the LHS\n\n(Cyclic notation is a special notation in which all permutations of (a,b,c) is the summation command.)\nThen use Titu's Lemma on all terms:  owing to the fact that $a^2+b^2+c^2 \\ge ab+bc+ca$, which is actually equivalent to $(a-b)^2 + (b-c)^2 + (c-a)^2 \\ge 0$!",
+        "Solution_3": "We proceed to prove that\n(then the inequality in question is just the cyclic sum of both sides, since \n)\nIndeed, by AP-GP, we have\n\nand\n\nSumming up, we have\n\nwhich is equivalent to:\n\nDividing $36(5a+b)$ from both sides, the desired inequality is proved.\n--Lightest 15:31, 7 May 2012 (EDT)",
+        "Solution_4": "By Cauchy-Schwarz,\n\n\n(by AM-GM)  as desired.",
+        "Solution_5": "We note that if we can prove that there exists a real number $x$ such that  for all positive reals $a$ and $b$, then we are done since both sides are a cyclic sum of $a,b$, and $c$. Note that if we multiply both $a$ and $b$ by a constant $r$, the left and right sides will both multiply by $r^2$ (since the numerator on the left will multiply by $r^3$ and the denominator will multiply by $r$. This means that if the inequality is satisfied for an ordered pair $(a,b)$, then it is also satisfied for $(ar,br)$ for any positive real $r$. Thus, without loss of generality we can let $b=1$,and our inequality becomes  Expanding this and rearranging we get $a^3(1-5x)+a^2(-x)+a(5x-\\frac{10}{3})+x+\\frac{7}{3}\\geq0$. Since $1$ is a root of the left side, we can factor this as  Note that for this to always be nonnegative, we must have $a^2(1-5x)+a(1-6x)-x-\\frac{7}{3}$ be positive when $a>1$ and negative when $a<1$. Thus, it must have a root at $a=1$, so we plug in $a=1$ is a root and find that we must have $x=-\\frac{1}{36}$. When $x=-\\frac{1}{36}$, the expression factors as $(a-1)^2(\\frac{41}{36}a+\\frac{83}{36})\\geq0$ which obviously is nonnegative for all positive reals $a$, so we are done.\n~john0512, bronzetruck2016\n",
+        "Solution_6": "$LHS=\\sum_{cyc}\\frac{a^3+3b^3}{5a+b}=\\sum_{cyc}\\frac{a^3}{5a+b}+\\sum_{cyc}\\frac{3b^3}{5a+b}=\\sum_{cyc}\\frac{a^4}{5a^2+ab}+\\sum_{cyc}\\frac{9b^4}{15ab+3b^2}$ which tells that $LHS\\geq \\frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2)+ab+bc+ac}+\\frac{3(a^2+b^2+c^2)}{(a^2+b^2+c^2)+5(ab+bc+ac)}\\geq (\\frac{1}{2}+\\frac{1}{6})(a^2+b^2+c^2)=RHS$, our proof is done\n~bluesoul"
+    },
+    {
+        "Problem": "Let $\\alpha$ be an irrational number with $0 < \\alpha < 1$, and draw a circle in the plane whose circumference has length 1.  Given any integer $n \\ge 3$, define a sequence of points $P_1$, $P_2$, $\\dots$, $P_n$ as follows.  First select any point $P_1$ on the circle, and for $2 \\le k \\le n$ define $P_k$ as the point on the circle for which the length of arc $P_{k - 1} P_k$ is $\\alpha$, when travelling counterclockwise around the circle from $P_{k - 1}$ to $P_k$.  Supose that $P_a$ and $P_b$ are the nearest adjacent points on either side of $P_n$.  Prove that $a + b \\le n$.",
+        "Solution": "Use mathematical induction. For $n=3$ it is true because one point can't be closest to $P_3$ in both ways, and that $1+2\\le 3$. Suppose that for some $n$, the nearest adjacent points $P_a$ and $P_b$ on either side of $P_n$ satisfy  $a+b \\le n$. Then consider the nearest adjacent points $P_c$ and $P_d$ on either side of $P_{n+1}$. It is by the assumption of the nearness we can see that either $(c,d)=(a+1,b+1)$ still holds, or $P_1$ jumps into the interior of the arc $P_{a+1}P_{n}P_{b+1}$, so that $c$ or $d$ equals to $1$. Let's consider the following two cases.\n(i) Suppose $a+b=n$.\nSince the length of the arc $P_nP_a$ is $\\{(a-n)\\alpha\\}$ (where $\\{x\\}$ equals to $x$ subtracted by the greatest integer not exceeding $x$) and length of the arc $P_bP_n$ is $\\{(n-b)\\alpha\\} = \\{a\\alpha\\}$, we now consider a point $P_0$ which is defined by $P_1$ traveling clockwise on the circle such that the length of arc $P_0P_1$ is $\\alpha$. We claim that $P_0$ is in the interior of the arc $P_bP_nP_a$. Algebraically, it is equivalent to either $\\{0-n\\alpha\\} < \\{a\\alpha-n\\alpha\\}$ or $\\{n\\alpha -0 \\} < \\{n\\alpha - b\\alpha\\} = \\{a\\alpha\\}$.\nSuppose the latter fails, i.e. $\\{n\\alpha\\} \\ge \\{a\\alpha\\}$. Then suppose $n\\alpha = m_1 + r_1$ and $a\\alpha = m_2 + r_2$, where $m_1$, $m_2$ are integers and $0< r_2\\le r_1 <1$ ($r_2$ is not zero because $a\\alpha$ is irrational). We now have \n and\nTherefore $P_0$ is either closer to $P_n$ than $P_a$ on the $P_a$ side, or closer to $P_n$ than $P_b$ on the $P_b$ side. In other words, $P_1$ is the closest adjacent point of $P_{n+1}$ on the $P_{a+1}$ side, or the closest adjacent point of $P_{n+1}$ on the $P_{b+1}$ side. Hence $P_c$ or $P_d$ is $P_1$, therefore $c+d \\le n+1$.\n(ii) Suppose $a+b\\le n-1$\nThen either $c+d = (a+1)+(b+1) \\le n+1$ when $c=a+1$ and $d=b+1$, or $c+d \\le 1+n$ when one of $P_c$ or $P_d$ is $P_1$.\nIn either case, $c+d\\le n+1$ is true.\n--Lightest 20:27, 6 May 2012 (EDT)"
+    },
+    {
+        "Problem": "For distinct positive integers $a$, $b < 2012$, define $f(a,b)$ to be the number of integers $k$ with $1 \\le k < 2012$ such that the remainder when $ak$ divided by 2012 is greater than that of $bk$ divided by 2012.  Let $S$ be the minimum value of $f(a,b)$, where $a$ and $b$ range over all pairs of distinct positive integers less than 2012.  Determine $S$.",
+        "Solution_1": "First we'll show that $S \\geq 502$, then we'll find an example $(a, b)$ that have $f(a, b)=502$.\nLet $x_k$ be the remainder when $ak$ is divided by 2012, and let $y_k$ be defined similarly for $bk$. First, we know that, if $x_k > y_k >0$, then $x_{2012-k} \\equiv a(2012-k) \\equiv 2012-ak \\equiv 2012-x_k \\pmod {2012}$ and $y_{2012-k} \\equiv 2012-y_k \\pmod {2012}$. This implies that, since $2012 - x_k \\neq 0$ and $2012 -y_k \\neq 0$, $x_{2012-k} < y_{2012-k}$. Similarly, if $0< x_k < y_k$ then $x_{2012-k} > y_{2012-k}$, establishing a one-to-one correspondence between the number of $k$ such that $x_k < y_k$. Thus, if $n$ is the number of $k$ such that $x_k \\neq y_k$ and $y_k \\neq 0 \\neq x_k$, then $S \\geq \\frac{1}{2}n$. Now I'll show that $n \\geq 1004$.\nIf $gcd(k, 2012)=1$, then I'll show you that $x_k \\neq y_k$. This is actually pretty clear; assume that's not true and set up a congruence relation:\n\nSince $k$ is relatively prime to 2012, it is invertible mod 2012, so we must have $a \\equiv b \\pmod {2012}$. Since $0 < a, b <2012$, this means $a=b$, which the problem doesn't allow, thus contradiction, and $x_k \\neq y_k$. Additionally, if $gcd(k, 2012)=1$, then $x_k \\neq 0 \\neq y_k$, then based on what we know about $n$ from the previous paragraph, $n$ is at least as large as the number of k relatively prime to 2012. Thus, $n \\geq \\phi(2012) = \\phi(503*4) = 1004$. Thus, $S \\geq 502$.\nTo show 502 works, consider $(a, b)=(1006, 2)$. For all even $k$ we have $x_k=0$, so it doesn't count towards $f(1006, 2)$. Additionally, if $k = 503, 503*3$ then $x_k = y_k = 1006$, so the only number that count towards $f(1006, 2)$ are the odd numbers not divisible by 503. There are 1004 such numbers. However, for all such odd k not divisible by 503 (so numbers relatively prime to 2012), we have $x_k \\neq 0 \\neq y_k$ and $2012-k$ is also relatively prime to 2012. Since under those conditions exactly one of $x_k > y_k$ and $x_{2012-k} > y_{2012-k}$ is true, we have at most 1/2 of the 1004 possible k actually count to $f(1006, 2)$, so $\\frac{1004}{2} = 502 \\geq f(1006, 2) \\geq S \\geq 502$, so $S=502$.",
+        "Solution_2": "Let $ak \\equiv r_{a} \\pmod{2012}$ and $bk \\equiv r_{b} \\pmod{2012}$. Notice that this means $a(2012 - k) \\equiv 2012 - r_{a} \\pmod{2012}$ and $b(2012 - k) \\equiv 2012 - r_{b} \\pmod{2012}$. Thus, for every value of $k$ where $r_{a} > r_{b}$, there is a value of $k$ where $r_{b} > r_{a}$. Therefore, we merely have to calculate $\\frac{1}{2}$ times the number of values of $k$ for which $r_{a} \\neq r_{b}$ and $r_{a} \\neq 0$.\nHowever, the answer is NOT $\\frac{1}{2}(2012) = 1006$! This is because we must count the cases where the value of $k$ makes $r_{a} = r_{b}$ or where $r_{a} = 0$.\nSo, let's start counting.\nIf $k$ is even, we have either $a \\equiv 0 \\pmod{1006}$ or $a - b \\equiv 0 \\pmod{1006}$. So, $a = 1006$ or $a = b + 1006$. We have $1005$ even values of $k$ (which is all the possible even values of $k$, since the two above requirements don't put any bounds on $k$ at all).\nIf $k$ is odd, if $k = 503$ or $k = 503 \\cdot 3$, then $a \\equiv 0 \\pmod{4}$ or $a \\equiv b \\pmod{4}$. Otherwise, $ak \\equiv 0 \\pmod{2012}$ or $ak \\equiv bk \\pmod{2012}$, which is impossible to satisfy, given the domain $a, b < 2012$. So, we have $2$ values of $k$.\nIn total, we have $2 + 1005 = 1007$ values of $k$ which makes $r_{a} = r_{b}$ or $r_{a} = 0$, so there are $2011 - 1007 = 1004$ values of $k$ for which $r_{a} \\neq r_{b}$ and $r_{a} \\neq 0$. Thus, by our reasoning above, our solution is $\\frac{1}{2} \\cdot 1004 = \\boxed{502}$.\nSolution by $\\textbf{\\underline{Invoker}}$",
+        "Solution_3": "The key insight in this problem is noticing that when $ak$ is higher than $bk$, $a(2012-k)$ is lower than $b(2012-k)$, except at $2 \\pmod{4}$ residues*. Also, they must be equal many times. $2012=2^2*503$. We should have multiples of $503$. After trying all three pairs and getting $503$ as our answer, we win. But look at the $2\\pmod{4}$ idea. What if we just took $2$ and plugged it in with $1006$?\nWe get $502$.",
+        "Solution_4": "Say that the problem is a race track with $2012$ spots. To intersect the most, we should get next to each other a lot so the negation is high. As $2012=2^2*503$, we intersect at a lot of multiples of $503$."
+    },
+    {
+        "Problem": "Let $P$ be a point in the plane of triangle $ABC$, and $\\gamma$ a line passing through $P$.  Let $A'$, $B'$, $C'$ be the points where the reflections of lines $PA$, $PB$, $PC$ with respect to $\\gamma$ intersect lines $BC$, $AC$, $AB$, respectively.  Prove that $A'$, $B'$, $C'$ are collinear.",
+        "Solution": "By the sine law on triangle $AB'P$,\n\nso\n\nSimilarly,\n\nHence,\nSince angles $\\angle AB'P$ and $\\angle CB'P$ are supplementary or equal, depending on the position of $B'$ on $AC$,\n\nSimilarly,\nBy the reflective property, $\\angle APB'$ and $\\angle BPA'$ are supplementary or equal, so\n\nSimilarly,\n\nTherefore,\n\nso by Menelaus's theorem, $A'$, $B'$, and $C'$ are collinear.",
+        "Solution_2.2C_Barycentric_.28Modified_by_Evan_Chen.29": "We will perform barycentric coordinates on the triangle $PCC'$, with $P=(1,0,0)$, $C'=(0,1,0)$, and $C=(0,0,1)$. Set $a = CC'$, $b = CP$, $c = C'P$ as usual. Since $A$, $B$, $C'$ are collinear, we will define $A = (p : k : q)$ and $B = (p : \\ell : q)$.\nClaim: Line $\\gamma$ is the angle bisector of $\\angle APA'$, $\\angle BPB'$, and $\\angle CPC'$. \nThis is proved by observing that since $A'P$ is the reflection of $AP$ across $\\gamma$, etc.\nThus $B'$ is the intersection of the isogonal of $B$ with respect to $\\angle P$\nwith the line $CA$; that is,\n\nAnalogously, $A'$ is the intersection of the isogonal of $A$ with respect to $\\angle P$\nwith the line $CB$; that is,\n\nThe ratio of the first to third coordinate in these two points\nis both $b^2pq : c^2k\\ell$, so it follows $A'$, $B'$, and $C'$ are collinear.\n~peppapig_"
+    },
+    {
+        "Problem": "Are there integers $a$ and $b$ such that $a^5b+3$ and $ab^5+3$ are both perfect cubes of integers?",
+        "Solution": "No, such integers do not exist.  This shall be proven by contradiction, by showing that if $a^5b+3$ is a perfect cube then $ab^5+3$ cannot be.\nRemark that perfect cubes are always congruent to $0$, $1$, or $-1$ modulo $9$.  Therefore, if $a^5b+3\\equiv 0,1,\\text{ or} -1\\pmod{9}$, then $a^5b\\equiv 5,6,\\text{ or }7\\pmod{9}$.\nIf $a^5b\\equiv 6\\pmod 9$, then note that $3|b$.  (This is because if $3|a$ then $a^5b\\equiv 0\\pmod 9$.)  Therefore $ab^5\\equiv 0\\pmod 9$ and $ab^5+3\\equiv 3\\pmod 9$, contradiction.\nOtherwise, either $a^5b\\equiv 5\\pmod 9$ or $a^5b\\equiv 7\\pmod 9$.  Note that since $a^6b^6$ is a perfect sixth power, and since neither $a$ nor $b$ contains a factor of $3$, $a^6b^6\\equiv 1\\pmod 9$.  If $a^5b\\equiv 5\\pmod 9$, then   Similarly, if $a^5b\\equiv 7\\pmod 9$, then   Therefore $ab^5+3\\equiv 5,7\\pmod 9$, contradiction.\nTherefore no such integers exist.\nAmkan2022",
+        "Solution_2": "We shall prove that such integers do not exist via contradiction.\nSuppose that $a^5b + 3 = x^3$ and $ab^5 + 3 = y^3$ for integers x and y. Rearranging terms gives $a^5b = x^3 - 3$ and $ab^5 = y^3 - 3$. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = $(x^3 - 3)^\\frac{5}{24} (y^3 - 3)^\\frac{-1}{24}$ and b = $(y^3 - 3)^\\frac{5}{24} (x^3 - 3)^\\frac{-1}{24}$. Consider a prime p in the prime factorization of $x^3 - 3$ and $y^3 - 3$. If it has power $r_1$ in $x^3 - 3$ and power $r_2$ in $y^3 - 3$, then $5r_1$ - $r_2$ is a multiple of 24 and $5r_2$ - $r_1$ also is a multiple of 24.\nAdding and subtracting the divisions gives that $r_1$ - $r_2$ divides 12. (actually, $r_1 - r_2$ is a multiple of 4, as you can verify if $\\{^{5r_1 - r_2 = 24}_{5r_2 - r_1 = 48}$. So the rest of the proof is invalid.) Because $5r_1$ - $r_2$ also divides 12, $4r_1$ divides 12 and thus $r_1$ divides 3. Repeating this trick for all primes in $x^3 - 3$, we see that $x^3 - 3$ is a perfect cube, say $q^3$. Then $x^3 - q^3 = 3,$ and $(x-q)(x^2 + xq + q^2) = 3$, so that $x - q = 1$ and $x^2 + xq + q^2 = 3$. Clearly, this system of equations has no integer solutions for $x$ or $q$, a contradiction, hence completing the proof.\nTherefore no such integers exist.",
+        "Solution_3": "Let $a^5b+3=x^3$ and $ab^5+3=y^3$. Then, $a^5b=x^3-3$, $ab^5=y^3-3$, and \nNow take $\\text{mod }9$ (recall that perfect cubes $\\equiv -1,0,1\\pmod{9}$ and perfect sixth powers $\\equiv 0,1\\pmod{9}$) on both sides. There are $3\\times 3=9$ cases to consider on what values $\\text{mod }9$ that $x^3$ and $y^3$ take. Checking these $9$ cases, we see that only $x^3\\equiv y^3\\equiv 0\\pmod{9}$ or $x\\equiv y\\equiv 0\\pmod{3}$ yield a valid residue $\\text{mod }9$ (specifically, $(x^3-3)(y^3-3)\\equiv 0\\pmod{9}$). But this means that $3\\mid ab$, so $729\\mid (ab)^6$ so  contradiction.",
+        "Solution_4": "If $a^5b+3$ is a perfect cube, then $a^5b$ can be one of $5,6,7 \\pmod 9$, so $(a^5b)^5 = a^{25} b^5$ can be one of $5^5 \\equiv 2$, $6^5 \\equiv 0$, or $7^5 \\equiv 4 \\pmod 9$. If $a$ were divisible by $3$, we'd have $a^5 b \\equiv 0 \\pmod 9$, which we've ruled out. So $\\gcd(a,9) = 1$, which means $a^6 \\equiv 1 \\pmod 9$, and therefore $a^{25} b^5 \\equiv ab^5 \\pmod 9$.\nWe've shown that $a b^5$ can be one of $0, 2, 4 \\pmod 9$, so $ab^5 + 3$ can be one of $3, 5, 7 \\pmod 9$. None of these are possibilities for a perfect cube, so if $a^5b+3$ is a perfect cube, $ab^5+3$ cannot be.",
+        "Solution_5": "As in previous solutions, notice $ab^5,a^5b \\equiv 5,6,7 \\pmod 9$. Now multiplying gives $a^6b^6$, which is only $0,1 \\pmod 9$, so after testing all cases we find that $ab^5\\equiv a^5b \\equiv 6 \\mod 9$. Then since $\\phi (9) = 6$, $ab^5\\equiv \\frac{a}{b}\\pmod 9$ and $a^5b \\equiv \\frac{b}{a}\\pmod 9$ (Note that $a,b$ cannot be $0\\pmod 9$). Thus we find that the inverse of $6$ is itself under modulo $9$, a contradiction.",
+        "Solution_6": "I claim there are no such a or b such that both expressions are cubes.\nAssume to the contrary $a^5b +3$ and $ab^5 + 3$ are cubes.\nLemma 1: If $a^5b +3$ and $ab^5 + 3$ are cubes, then $ab^5, a^5b \\equiv 5,7 \\pmod 9$\nProof Since cubes are congruent to any of $0, 1, -1 \\pmod 9$, $ab^5,a^5b \\equiv 5,6,7 \\pmod 9$. But if $ab^5 \\equiv 6 \\pmod 9$, $3|a$, so $a^5b \\equiv 0 \\pmod 9$, contradiction. A similar argument can be made for $ab^5 \\neq 6 \\pmod 9$.\nLemma 2: If k is a perfect 6th power, then $k \\equiv 0,1 \\pmod 9$\nProof: Since cubes are congruent to $0, 1,  -1 \\pmod 9$, we can square, and get 6th powers are congruent to $0, 1 \\pmod 9$.\nSince $ab^5 \\cdot a^5b = a^6 b^6 = (ab)^6$, which is a perfect 6th power, by lemma 2, $ab^5 \\cdot a^5b \\equiv 0,1 \\pmod 9$.\nBut, by lemma 1, $ab^5 \\cdot a^5b \\equiv 5 \\cdot 5, 5 \\cdot 7, 7 \\cdot 7 \\equiv 7, 8, 4 \\pmod 9$.\nSo, $ab^5 \\cdot a^5b$, which is an integer, can't go into any of the possible residue classes modulo 9, without breaking one of these lemmas. This is a contradiction, and the proof is complete. $\\blacksquare$\n-AlexLikeMath"
+    },
+    {
+        "Problem": "Each cell of an $m\\times n$ board is filled with some nonnegative integer. Two numbers in the filling are said to be adjacent if their cells share a common side. (Note that two numbers in cells that share only a corner are not adjacent). The filling is called a garden if it satisfies the following two conditions:\n(i) The difference between any two adjacent numbers is either $0$ or $1$.\n(ii) If a number is less than or equal to all of its adjacent numbers, then it is equal to  $0$ .\nDetermine the number of distinct gardens in terms of $m$ and $n$ .",
+        "Solution": "We claim that any configuration of $0$'s produces a distinct garden.  To verify this claim, we show that, for any cell that is nonzero, the value of that cell is its distance away from the nearest zero, where distance means the shortest chain of adjacent cells connecting two cells.  Now, since we know that any cell with a nonzero value must have a cell adjacent to it that is less than its value, there is a path that goes from this cell to the $0$ that is decreasing, which means that the value of the cell must be its distance from the $0 \\rightarrow$ as the path must end.  From this, we realize that, for any configuration of $0$'s, the value of each of the cells is simply its distance from the nearest $0$, and therefore one garden is produced for every configuration of $0$'s.\nHowever, we also note that there must be at least one $0$ in the garden, as otherwise the smallest number in the garden, which is less than or equal to all of its neighbors, is $>0$, which violates condition $(ii)$.  There are $2^{mn}$ possible configurations of $0$ and not $0$ in the garden, one of which has no $0$'s, so our total amount of configurations is $\\boxed{2^{mn} -1}$"
+    },
+    {
+        "Problem": "In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively.  Let $\\omega_A$, $\\omega_B$, $\\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively.  Given the fact that segment $AP$ intersects $\\omega_A$, $\\omega_B$, $\\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$.",
+        "Solution_1": "\nIn this solution, all lengths and angles are directed.\nFirstly, it is easy to see by that $\\omega_A, \\omega_B, \\omega_C$ concur at a point $M$ (the Miquel point). Let $XM$ meet $\\omega_B, \\omega_C$ again at $D$ and $E$, respectively. Then by Power of a Point, we have  Thusly  But we claim that $\\triangle XDP \\sim \\triangle PBM$. Indeed,  and \nTherefore, $\\frac{XD}{XP} = \\frac{PB}{PM}$. Analogously we find that $\\frac{XE}{XP} = \\frac{PC}{PM}$ and we are done.\ncourtesy v_enhance, minor clarification by integralarefun",
+        "Solution_2": "Diagram\nRefer to the Diagram link.\nBy Miquel's Theorem, there exists a point at which $\\omega_A, \\omega_B, \\omega_C$ intersect. We denote this point by $M.$ Now, we angle chase:\n\nIn addition, we have\n\nNow, by the Ratio Lemma, we have\n (by the Law of Sines in $\\triangle MZY$) (by the Law of Sines in $\\triangle MBC$) by the Ratio Lemma.\nThe proof is complete.",
+        "Solution_3": "Use directed angles modulo $\\pi$.\nLemma. $\\angle{XRY} \\equiv \\angle{XQZ}.$\nProof.\nNow, it follows that (now not using directed angles)\n\nusing the facts that $ARY$ and $APB$, $AQZ$ and $APC$ are similar triangles, and that $\\frac{RA}{\\sin \\angle{RXA}} = \\frac{QA}{\\sin \\angle{QXA}}$ equals twice the circumradius of the circumcircle of $AQR$.",
+        "Solution_4": "We will use some construction arguments to solve the problem. Let $\\angle BAC=\\alpha,$ $\\angle ABC=\\beta,$ $\\angle ACB=\\gamma,$ and let $\\angle APB=\\theta.$ We construct lines through the points $Q,$ and $R$ that intersect with $\\triangle ABC$ at the points $Q$ and $R,$ respectively, and that intersect each other at $T.$ We will construct these lines such that $\\angle CQV=\\angle ARV=\\theta.$\nNow we let the intersections of $AP$ with $RV$ and $QU$ be $Y'$ and $Z',$ respectively. This construction is as follows.\nWe know that $\\angle BRY'=180^\\circ-\\angle ARY'=180^\\circ-\\theta.$ Hence, we have,\nSince the opposite angles of quadrilateral $RY'PB$ add up to $180^\\circ,$ it must be cyclic. Similarly, we can also show that quadrilaterals $CQZ'P,$ and $AQTR$ are also cyclic.\nSince points $Y'$ and $Z'$ lie on $AP,$ we know that,\n\nand that\nHence, the points $Y'$ and $Z'$ coincide with the given points $Y$ and $Z,$ respectively.\nSince quadrilateral $AQTR$ is also cyclic, we have,\nSimilarly, since quadrilaterals $CQZ'P,$ and $AQTR$ are also cyclic, we have,\n\nand,\nSince these three angles are of $\\triangle TY'Z',$ and they are equal to corresponding angles of $\\triangle ABC,$ by AA similarity, we know that $\\triangle TY'Z'\\sim \\triangle ABC.$\nWe now consider the point $X=\\omega_c\\cap AC.$ We know that the points $A,$ $Q,$ $T,$ and $R$ are concyclic. Hence, the points $A,$ $T,$ $X,$ and $R$ must also be concyclic.\nHence, quadrilateral $AQTX$ is cyclic.\n\nSince the angles $\\angle ART$ and $\\angle AXT$ are inscribed in the same arc $\\overarc{AT},$ we have,\nConsider by this result, we can deduce that the homothety that maps $ABC$ to $TY'Z'$ will map $P$ to $X.$ Hence, we have that,\nSince $Y'=Y$ and $Z'=Z$ hence,\nas required.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Let $f(n)$ be the number of ways to write $n$ as a sum of powers of $2$, where we keep track of the order of the summation.  For example, $f(4)=6$ because $4$ can be written as $4$, $2+2$, $2+1+1$, $1+2+1$, $1+1+2$, and $1+1+1+1$.  Find the smallest $n$ greater than $2013$ for which $f(n)$ is odd.",
+        "Solution": "First of all, note that $f(n)$ = $\\sum_{i=0}^{k} f(n-2^{i})$ where $k$ is the largest integer such that $2^k \\le n$.  We let $f(0) = 1$ for convenience.\nFrom here, we proceed by induction, with our claim being that the only $n$ such that $f(n)$ is odd are $n$ representable of the form $2^{a} - 1, a \\in \\mathbb{Z}$\nWe induct on $a$.  It is trivially true for $a = 0$ and $a = 1$.  From here, we show that, if the only numbers $n \\le 2^{a-1} - 1$ where $f(n)$ is odd are of the form described above, then the only numbers $n \\le 2^{a} -1$ that are odd are of that form.  We first consider all numbers $b$, such that $2^{a-1} \\le b \\le 2^{a} - 2$, going from the lower bound to the upper bound (a mini induction, you might say).  We know that $f(b) = \\sum_{i=0}^{a-1} f(b-2^{i})$.  For a number in this summation to be odd, $b - 2^i = 2^m -1 \\rightarrow b = 2^i + 2^m - 1$.  However, we know that $b > 2^{a-1}$, so $m$ must be equal to $a-1$, or else $b$ cannot be in that interval.  Now, from this, we know that $i < a-1$, as $b<2^{a} - 1$.  Therefore, $i$ and $m$ are distinct, and thus $f(b - 2^i)$ and $f(b- 2^{a-1})$ are odd; since there are just two odd numbers, the ending sum for any $b$ is even.  Finally, considering $2^{a} - 1$, the only odd number is $f(2^{a} - 1 - 2^{a-1})$, so the ending sum is odd.  $\\Box$\nThe smallest $n$ greater than $2013$ expressible as $2^d - 1, d \\in \\mathbb{N}$ is $2^{11} -1 = \\boxed{2047}$",
+        "Solution_with_Explanations": "Of course, as with any number theory problem, use actual numbers to start, not variables! By plotting out the first few sums (do it!) and looking for patterns, we observe that $f(n)=\\sum_{\\textrm{power}=0}^{\\textrm{pow}_{\\textrm{larg}}} f(n-2^{\\textrm{power}})$, where $\\textrm{pow}_{\\textrm{larg}}$ represents the largest power of $2$ that is smaller than $n$. I will call this sum the Divine Sign, or DS.\nBut wait a minute... we are trying to determine odd/even of $f(n)$. Why not call all the evens 0 and odds 1, basically using mod 2? Sounds so simple. Draw a small table for the values: as $n$ goes up from $0$, you get: $1,1,0,1,0,0,0,1,0...$. We have to set $f(0)=1$ for this to work. Already it looks like $f(n)$ is only odd if $n=2^{\\textrm{power}}-1$.\nThe only tool here is induction. The base case is clearly established. Then let's assume we successfully made our claim up to $2^n-1$. We need to visit numbers from $2^n$ to $2^{n+1}-1$. Realize that $2^n$ has $0$ for $f$ because there will be two numbers in DS that give a $f$ of one: $2^{n-1}$ and $1$.\nBut to look at whether a value of $f(\\textrm{number})$ is 1 or 0, we need to revisit our first equation. We can answer this rather natural question: When will a number to be inducted upon, say $2^n+k$, ever have a 1 as $f(\\textrm{number})$ in the DS equation? Well- because by our assumption of the claim up to $2^n-1$, we know that the only way for that to happen is if $2^n+k-2^{\\textrm{power}}$ in the DS is equal to $2^{\\textrm{Some power}} - 1$. Clearly $1 \\leq k \\leq 2^n - 1$.\nFinally, we can simplify. Using our last equation, $2^n+k-2^{\\textrm{power}}=2^{\\textrm{Some power}}-1$, regrouping gives $2^n+k=2^{\\textrm{power}}+2^{\\textrm{Some power}}-1$.\nMost importantly, realize that $\\textrm{power}$ can be from $0$ to $n$, because of the restraints on $k$ mentioned earlier. Same with $\\textrm{Some power}$. Immediately at least one of $\\textrm{power}$ and $\\textrm{Some power}$ has to be $n$. If both were smaller, LHS is greater, contradiction. If both were greater, RHS is greater, contradiction.\nTherefore, by setting one of $\\textrm{power}$ or $\\textrm{Some power}$ to $n$, we realize $k=2^{\\textrm{A certain power}}-1$.\nThe conclusion is clear, right? Each $k$ from $1$ to $2^{n-1}-1$ yields two distinct cases: one of $\\textrm{power}$ and $\\textrm{Some power}$ is equal to $n$, while the other is LESS THAN $n$. But for $k=2^n-1$, there is ONE CASE: BOTH values have to equal $n$. Therefore, the only $k$ that has $f(2^n+k)$ as odd must only be $2^n-1$, because the other ones yield a $f$ of 1+1=0 in our mod. That proves our induction for a new power of 2, namely $n+1$, meaning that $f(\\textrm{number})$ is only odd if $\\textrm{number} = 2^{\\textrm{Power of two}} - 1$, and we are almost done...\nThus, the answer is $2^{11}-1=\\boxed{2047}$.\nThis was pretty intuitive, and realizing quite $elementary$ steps about how powers of 2 work gave us the solution! Estimated time: ~50 minutes.\n~expiLnCalc.\nminor edits ~mathboy100\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Quadrilateral $XABY$ is inscribed in the semicircle $\\omega$ with diameter $XY$.  Segments $AY$ and $BX$ meet at $P$.  Point $Z$ is the foot of the perpendicular from $P$ to line $XY$.  Point $C$ lies on $\\omega$ such that line $XC$ is perpendicular to line $AZ$.  Let $Q$ be the intersection of segments $AY$ and $XC$.  Prove that",
+        "Solution": "Using the Law of Sines and simplifying, we have\nIt is easy to see that $APZX$ is cyclic. Also, we are given $XQ\\perp AZ$. Then we have \n and we are done.",
+        "Solution_1": "Let us use coordinates. Let O, the center of the circle, be (0,0). WLOG the radius of the circle is 1, so set Y (1,0) and X (-1,0). Also, for arbitrary constants $a$ and $b$ set A $(\\cos a, \\sin a)$ and B $(\\cos b, \\sin b)$. Now, let's use our coordinate tools. It is easily derived that the equation of $BX$ is $y = \\frac{\\sin b}{1 + \\cos b}(x + 1) = v(x+1)$ and the equation of $AY$ is $y = \\frac{\\sin a}{1 - \\cos a}(x - 1) = u(x-1)$, where $u$ and $v$ are defined appropriately. Thus, by equating the y's in the equation we find the intersection of these lines, $P$, is $\\left(\\frac{u-v}{u+v}\\right), \\frac{2uv}{u+v})$. Also, $Z\\left(\\frac{u-v}{u+v}\\right), 0)$. It shall be left to the reader to find the slope of $AZ$, the coordinates of Q and C, and use the distance formula to verify that $\\frac{BY}{XP} + \\frac{CY}{XQ} = \\frac{AY}{AX}$.",
+        "Solution_2": "First of all\nsince the quadrilateral $APZX$ is cyclic, and triangle $AXQ$ is rectangle, and $CX$ is orthogonal to $AZ$. Now\nbecause $XABY$ is cyclic and we have proved that\nso $BC$ is parallel to $AY$, and  Now by Ptolomey's theorem on $APZX$ we have  we see that triangles $PXZ$ and $QXA$ are similar since  and  is already proven, so  Substituting yields  dividing by $(PX)(XZ)$ We get  Now triangles $AYZ$, and $XYP$ are similar so  but also triangles $XPY$ and $XZB$ are similar and we get  Comparing we have,  Substituting,  Dividing the new relation by $AX$ and multiplying by $XB$ we get  but  since triangles $AXB$ and $QXY$ are similar, because  and  since $CY=AB$ Substituting again we get  Now since triangles $ACQ$ and $XYQ$ are similar we have  and by the similarity of $APB$ and $XPY$, we get  so substituting, and separating terms we get  In the beginning we prove that $AC=BY$ and $AB=CY$ so \n$\\blacksquare$",
+        "Solution_3": "It is obvious that \n\nfor some value $\\alpha$. Also, note that $\\angle BYA=\\alpha$. Set \n \nWe have\n\nand \n\nThis gives\n\nSimilarly, we can deduce that \n\nAdding gives\n",
+        "Solution_4": "First, since $XY$ is the diameter and $A$, $B$, and $C$ lie on the circle, . Next, because $AZ$ and $CY$ are both perpendicular to $CX$, we have $AZ$ to be parallel to $CY$.\nNow looking at quadrilateral $APZX$, we see that this is cyclic because  Set $\\alpha = \\angle{BXA} = \\angle{BYA}$, and $\\beta = \\angle{BXC} = \\angle{BYC}$. \nNow,  since $AZ$ and $CY$ are parallel. \nAlso,  \nThat means  so  This means $\\angle{QXZ} = \\angle{YBC} = \\alpha$, so $BC$ and $AY$ are parallel. \nFinally, we can look at the equation. \nWe know  so $XP = \\frac{AX}{\\cos{\\alpha}}.$\nWe also know  so $XQ = \\frac{AX}{\\cos(\\alpha+\\beta)}.$\nPlugging this into the LHS of the equation, we get \nNow, let $H$ be the point on $AY$ such that $BH$ is perpendicular to $AY$. Also, since $\\angle{AYB} = \\angle{CXY}$, their arcs have equal length, and $AB=CY$. \nNow, the LHS is simplified even more to  which is equal to  which is equal to  This completes the proof.\n~jeteagle\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem_6": "Find all real numbers $x,y,z\\geq 1$ satisfying",
+        "Solution_with_Thought_Process": "Without loss of generality, let $1 \\le x \\le y \\le z$. Then $\\sqrt{x + xyz} = \\sqrt{x - 1} + \\sqrt{y - 1} + \\sqrt{z - 1}$.\nSuppose x = y = z. Then $\\sqrt{x + x^3} = 3\\sqrt{x-1}$, so $x + x^3 = 9x - 9$. It is easily verified that $x^3 - 8x + 9 = 0$ has no solution in positive numbers greater than 1. Thus, $\\sqrt{x + xyz} \\ge \\sqrt{x - 1} + \\sqrt{y - 1} + \\sqrt{z - 1}$ for x = y = z. We suspect if the inequality always holds.\nLet x = 1. Then we have $\\sqrt{1 + yz} \\ge \\sqrt{y-1} + \\sqrt{z-1}$, which simplifies to  and hence  Let us try a few examples: if y = z = 2, we have $3 > 2$; if y = z, we have $y^2 - 2y + 3 \\ge 2(y-1)$, which reduces to $y^2 - 4y + 5 \\ge 0$. The discriminant (16 - 20) is negative, so in fact the inequality is strict. Now notice that yz - y - z + 3 = (y-1)(z-1) + 2. Now we see we can let $u = \\sqrt{(y-1)(z-1)}$! Thus,  and the claim holds for x = 1.\nIf x > 1, we see the $\\sqrt{x - 1}$ will provide a huge obstacle when squaring. But, using the identity $(x+y+z)^2 = x^2 + y^2 + z^2 + xy + yz + xz$:\n\nwhich leads to\n\nAgain, we experiment. If x = 2, y = 3, and z = 3, then $18 > 7 + 4\\sqrt{6}$.\nNow, we see the finish: setting $u = \\sqrt{x-1}$ gives $x = u^2 + 1$. We can solve a quadratic in u! Because this problem is a #6, the crown jewel of USAJMO problems, we do not hesitate in computing the messy computations:\nBecause the coefficient of $u^2$ is positive, all we need to do is to verify that the discriminant is nonpositive:\nLet us try a few examples. If y = z, then the discriminant D = $8(y-1) - 8(y-1) - yz(12 + 4y^2 - 8y - 8(y-1)) = -yz(4y^2 - 16y + 20) = -4yz(y^2 - 4y + 5) < 0$.\nWe are almost done, but we need to find the correct argument. (How frustrating!)\nSuccess! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are no real solutions to the original equation in the hypothesis.\n--Thinking Process by suli"
+    },
+    {
+        "Problem": "Let $a$, $b$, $c$ be real numbers greater than or equal to $1$. Prove that",
+        "Solution": "Since $(a-1)^5\\ge 0$,\n\nor\n\nSince $a^2-5a+10=\\left( a-\\dfrac{5}{2}\\right)^2 +\\dfrac{15}{4}>0$,\n\nAlso note that $10a^2-5a+1=10\\left( a-\\dfrac{1}{4}\\right)^2+\\dfrac{3}{8}> 0$,\nWe conclude\n\nSimilarly, \n\n\nSo \nor\n\nTherefore,"
+    },
+    {
+        "Problem": "Let $\\triangle{ABC}$ be a non-equilateral, acute triangle with $\\angle A=60^{\\circ}$, and let $O$ and $H$ denote the circumcenter and orthocenter of $\\triangle{ABC}$, respectively.\n(a) Prove that line $OH$ intersects both segments $AB$ and $AC$.\n(b) Line $OH$ intersects segments $AB$ and $AC$ at $P$ and $Q$, respectively. Denote by $s$ and $t$ the respective areas of triangle $APQ$ and quadrilateral $BPQC$. Determine the range of possible values for $s/t$.",
+        "Solution": "\nLemma: $H$ is the reflection of $O$ over the angle bisector of $\\angle BAC$ (henceforth 'the' reflection)\nProof: Let $H'$ be the reflection of $O$, and let $B'$ be the reflection of $B$.\nThen reflection takes $\\angle ABH'$ to $\\angle AB'O$.\n$\\Delta ABB'$ is equilateral, and $O$ lies on the perpendicular bisector of $\\overline{AB}$\nIt's well known that $O$ lies strictly inside $\\Delta ABC$ (since it's acute), meaning that $\\angle ABH' = \\angle AB'O = 30^{\\circ},$ from which it follows that $\\overline{BH'} \\perp \\overline{AC}$ . Similarly,  $\\overline{CH'} \\perp \\overline{AB}$. Since $H'$ lies on two altitudes, $H'$ is the orthocenter, as desired.\nEnd Lemma\nSo $\\overline{OH}$ is perpendicular to the angle bisector of $\\angle OAH$, which is the same line as the angle bisector of $\\angle BAC$, meaning that $\\Delta APQ$ is equilateral.\nLet its side length be $s$, and let $PH=t$, where $0 < t < s, t \\neq s/2$ because $O$ lies strictly within $\\angle BAC$, as must $H$, the reflection of $O$. Also, it's easy to show that if $O=H$ in a general triangle, it's equilateral, and we know $\\Delta ABC$ is not equilateral. Hence H is not on the bisector of $\\angle BAC \\implies t \\neq s/2$. Let $\\overrightarrow{BH}$ intersect $\\overline{AC}$ at $P_B$.\nSince $\\Delta HP_BQ$ and $BP_BA$ are 30-60-90 triangles, $AB=2AP_B=2(s-QP_B)=2(s-HQ/2)=2s-HQ=2s-(s-t)=s+t$\nSimilarly, $AC=2s-t$\nThe ratio $\\frac{[APQ]}{[ABC]-[APQ]}$ is $\\frac{AP \\cdot AQ}{AB \\cdot AC - AP \\cdot AQ} = \\frac{s^2}{(s+t)(2s-t)-s^2}$\nThe denominator equals $(1.5s)^2-(.5s-t)^2-s^2$ where $.5s-t$ can equal any value in $(-.5s, .5s)$ except $0$. Therefore, the denominator can equal any value in $(s^2, 5s^2/4)$, and the ratio is any value in $\\boxed{\\left(\\frac{4}{5},1\\right)}.$\nNote: It's easy to show that for any point $H$ on $\\overline{PQ}$ except the midpoint, Points B and C can be validly defined to make an acute, non-equilateral triangle.",
+        "Solution_2": "Let $J$ be the farthest point on the circumcircle of $\\triangle{ABC}$ from line $BC$.\nLemma: Line $AJ$||Line $OH$\nProof: Set $b=-\\dfrac{1}{2}+i\\dfrac{\\sqrt{3}}{2}$ and $c=-\\dfrac{1}{2}-i\\dfrac{\\sqrt{3}}{2}$, and $a$ on the unit circle. It is well known that $o=0$ and $h=a+b+c$, so we have $h-o=a+b+c=a-1=a-j$, so $\\dfrac{a-j}{h-o}$ is real and thus the 2 lines are parallel.\nWLOG let $A$ be in the first quadrant. Clearly by the above lemma $OH$ must intersect line $BC$ closer to $B$ than to $C$. Intersect $AJ$ and $BC$ at $D$ and $OH$ and $BC$ at $E$. We clearly have $0 < \\dfrac{\\overarc{JC}-\\overarc{AB}}{2} = \\dfrac{120-\\overarc{AC}}{2} = \\angle{ADB} = \\angle{OEC} < 30 = \\angle{OBC}$, $OH$ must intersect $AB$. We also have, letting the intersection of line $OH$ and line $AC$ be $Q$, and letting intersection of $OH$ and $AB$ be $P$, $\\angle{OQA} = \\angle{JAB} = \\dfrac{\\overarc{JB}}{2} = 60$. Since $\\angle{OCA}=\\angle{BCA}-\\angle{BCO} < 90-30 =\\angle{OQA}$, and $\\angle{OQC} = 120>\\angle{OAC}$, $OH$ also intersects $AC$. We have $\\angle{OPA}=\\angle{PAJ}=60$, so $\\triangle{APQ}$ is equilateral. Letting $AJ=2x$, and letting the foot of the perpendicular from $O$ to $AJ$ be $L$, we have $OL=\\sqrt{1-x^{2}}$, and since $OL$ is an altitude of $\\triangle{APQ}$, we have $[APQ]=\\dfrac{OL^{2}\\sqrt{3}}{3} = \\dfrac{\\sqrt{3}(1-x^{2})}{3}$. Letting the foot of the perpendicular from $A$ to $OJ$ be $K$, we have $\\triangle{JKA}\\sim \\triangle{JLO}$ by AA with ratio $\\dfrac{JA}{JO} = 2x$. Therefore, $JK = 2x(JL) = 2x^{2}$. Letting $D$ be the foot of the altitude from $J$ to $BC$, we have $KD=JD-JK=\\dfrac{3}{2}-2x^{2}$, since $re(B)=re(C) \\implies JD=re(j)-(-\\dfrac{1}{2})=\\dfrac{3}{2}$. Thus, since $BC=\\sqrt{3}$ we have $[ABC]=\\dfrac{(\\dfrac{3}{2}-2x^{2})(\\sqrt{3})}{2} = \\dfrac{(3-4x^{2})(\\sqrt{3})}{2}$, so $[PQCB]=[ABC]-[APQ]=\\dfrac{(5-8x^{2})(\\sqrt{3})}{12}$, so $\\dfrac{[APQ]}{[PQCB]} = \\dfrac{4-4x^{2}}{5-8x^{2}} = \\dfrac{1}{2}+\\dfrac{3}{10-16x^{2}}$. We have $x=\\sin(\\dfrac{JOA}{2})$, with $0<120-2\\angle{ACB}=\\angle{JOA}<60$, so $x$ can be anything in the interval $(0, \\dfrac{1}{2})$. Therefore, the desired range is $(\\dfrac{4}{5}, 1)$.\nSolution by Shaddoll"
+    },
+    {
+        "Problem": "Let $\\mathbb{Z}$ be the set of integers. Find all functions $f : \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that  for all $x, y \\in \\mathbb{Z}$ with $x \\neq 0$.",
+        "Solution": "Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions.\nLemma 1: $f(0) = 0$.\nProof: Assume the opposite for a contradiction. Plug in $x = 2f(0)$ (because we assumed that $f(0) \\neq 0$), $y = 0$. What you get eventually reduces to:\n\nwhich is a contradiction since the LHS is divisible by 2 but not 4.\nThen plug in $y = 0$ into the original equation and simplify by Lemma 1. We get:\n\nThen:\n\nTherefore, $f(x)$ must be 0 or $x^2$.\nNow either $f(x)$ is $x^2$ for all $x$ or there exists $a \\neq 0$ such that $f(a)=0$. The first case gives a valid solution. In the second case, we let $y = a$ in the original equation and simplify to get:\n\nBut we know that $xf(-x) = \\frac{f(x)^2}{x}$, so:\n\nSince $a$ is not 0, $f(2x)$ is 0 for all $x$ (including 0). Now either $f(x)$ is 0 for all $x$, or there exists some $m \\neq 0$ such that $f(m) = m^2$. Then $m$ must be odd. We can let $x = 2k$ in the original equation, and since $f(2x)$ is 0 for all $x$, stuff cancels and we get:\n\nfor $\\mathbf{k \\neq 0}$.\nNow, let $y = m$ and we get:\n\nNow, either both sides are 0 or both are equal to $m^6$. If both are $m^6$ then:\n\nwhich simplifies to:\n\nSince $k \\neq 0$ and $m$ is odd, both cases are impossible, so we must have:\n\nThen we can let $k$ be anything except 0, and get $f(x)$ is 0 for all $x \\equiv 3 \\pmod{4}$ except $-m^2$.  Also since $x^2f(-x) = f(x)^2$, we have $f(x) = 0 \\Rightarrow f(-x) = 0$, so $f(x)$ is 0 for all $x \\equiv 1 \\pmod{4}$ except $m^2$. So $f(x)$ is 0 for all $x$ except $\\pm m^2$. Since $f(m) \\neq 0$, $m = \\pm m^2$. Squaring, $m^2 = m^4$ and dividing by $m$, $m = m^3$. Since $f(m^3) = 0$, $f(m) = 0$, which is a contradiction for $m \\neq 1$. However, if we plug in $x = 1$ with $f(1) = 1$ and $y$ as an arbitrary large number with $f(y) = 0$ into the original equation, we get $0 = 1$ which is a clear contradiction, so our only solutions are $f(x) = 0$ and $f(x) = x^2$.",
+        "Solution_2": "Given that the range of f consists entirely of integers, it is clear that the LHS must be an integer and $f(yf(y))$ must also be an integer, therefore $\\frac{f(x)^2}{x}$ is an integer. If $x$ divides $f(x)^2$ for all integers $x \\ne 0$, then $x$ must be a factor of $f(x)$, therefore $f(0)=0$. Now, by setting $y=0$ in the original equation, this simplifies to $xf(-x)=\\frac{f(x)^2}{x}$. Assuming $x \\ne 0$, we have $x^2f(-x)=f(x)^2$. Substituting in $-x$ for $x$ gives us $x^2f(x)=f(-x)^2$. Substituting in $\\frac{f(x)^2}{x^2}$ in for $f(-x)$ in the second equation gives us $x^2f(x)=\\frac{f(x)^4}{x^4}$, so $x^6f(x)=f(x)^4$. In particular, if $f(x) \\ne 0$, then we have $f(x)^3=x^6$, therefore $f(x)$ is equivalent to $0$ or $x^2$ for every integer $x$. Now, we shall prove that if for some integer $t \\ne 0$, if $f(t)=0$, then $f(x)=0$ for all integers $x$. If we assume $f(y)=0$ and $y \\ne 0$ in the original equation, this simplifies to $xf(-x)+y^2f(2x)=\\frac{f(x)^2}{x}$. However, since $x^2f(-x)=f(x)^2$, we can rewrite this equation as $\\frac{f(x)^2}{x}+y^2f(2x)=\\frac{f(x)^2}{x}$, $y^2f(2x)$ must therefore be equivalent to $0$. Since, by our initial assumption, $y \\ne 0$, this means that $f(2x)=0$, so, if for some integer $y \\ne 0$, $f(y)=0$, then $f(x)=0$ for all integers $x$. The contrapositive must also be true, i.e. If $f(x) \\ne 0$ for all integers $x$, then there is no integral value of $y \\ne 0$ such that $f(y)=0$, therefore $f(x)$ must be equivalent for $x^2$ for every integer $x$, including $0$, since $f(0)=0$. Thus, $f(x)=0, x^2$ are the only possible solutions.",
+        "Solution_3": "Let's assume $f(0)\\neq 0.$ Substitute $(x,y)=(2f(0),0)$ to get\nThis means that $2(2f(0)-1)$ is a perfect square. However, this is impossible, as it is equivalent to $2\\pmod{4}.$ Therefore, $f(0)=0.$ Now substitute $x\\neq 0, y=0$ to get \nSimilarly, \nFrom these two equations, we can find either $f(x)=f(-x)=0,$ or $f(x)=f(-x)=x^2.$ Both of these are valid solutions on their own, so let's see if there are any solutions combining the two.\nLet's say we can find $f(x)=x^2, f(y)=0,$ and $x,y\\neq 0.$ Then \n (NEEDS FIXING: $f(x)^2/x= x^4/x = x^3$, so the RHS is $0$ instead of $x-x^3$.)\nIf $f(2x)=4x^2,$ then $y^2=\\frac{x-x^3}{4x^2}=\\frac{1-x^2}{4x},$ which is only possible when $y=0.$ This contradicts our assumption. Therefore, $f(2x)=0.$ This forces $x=\\pm 1$ due to the right side of the equation. Let's consider the possibility $f(2)=0, f(1)=1.$ Substituting $(x,y)=(2,1)$ into the original equation yields   which is impossible. So $f(2)=f(-2)=4$ and there are no solutions \"combining\" $f(x)=x^2$ and $f(x)=0.$\nTherefore our only solutions are $\\boxed{f(x)=0}$ and $\\boxed{f(x)=x^2.}$",
+        "Solution_4": "Let the given assertion be $P(x, y)$. We try $P(x, 0)$ and get $xf(2c-x)=f(x)^2/x+c$, where $f(0)=c$. We plug in $x=c$ and get $cf(c)=f(c)^2/c+c$. Rearranging and solving for $c^2$ gives us $c^2=\\frac{f(c)^2}{f(c)-1}$. Obviously, the only $c$ that works such that the RHS is an integer is $c=0$, and thus $f(0)=0$.\nWe use this information on assertion $P(x,0)$ and obtain $xf(-x)=f(x^2)/x$, or $f(-x)=\\frac{f(x)^2}{x^2}$. Thus, $f(x)$ is an even function. It follows that $f(x)=0, x^2$ for each $x$. We now prove that $f(x)=x^2$, f(x)=0$ are the only solutions. [in progress] ~SigmaPiE"
+    },
+    {
+        "Problem": "Let $b\\geq 2$ be an integer, and let $s_b(n)$ denote the sum of the digits of $n$ when it is written in base $b$.  Show that there are infinitely many positive integers that cannot be represented in the form $n+s_b(n)$, where $n$ is a positive integer.",
+        "Solution_1": "Define $S(n) = n + s_b(n)$, and call a number unrepresentable if it cannot equal $S(n)$ for a positive integer $n$.\nWe claim that in the interval $(b^p, b^{p+1}]$ there exists an unrepresentable number, for every positive integer $p$.\nIf $b^{p+1}$ is unrepresentable, we're done. Otherwise, time for our lemma:\nLemma: Define the function $f(p)$ to equal the number of integers x less than $b^p$ such that $S(x) \\ge b^p$. If $b^{p+1} = S(y)$ for some y, then $f(p+1) > f(p)$.\nProof: Let $F(p)$ be the set of integers x less than $b^p$ such that $S(x) \\ge b^p$. Then for every integer in $F(p)$, append the digit $(b-1)$ to the front of it to create a valid integer in $F(p+1)$. Also, notice that $(b-1) \\cdot b^p \\le y < b^{p+1}$. Removing the digit $(b-1)$ from the front of y creates a number that is not in $F(p)$. Hence, $F(p) \\rightarrow F(p+1)$, but there exists an element of $F(p+1)$ not corresponding with $F(p)$, so $f(p+1) > f(p)$.\nNote that our lemma combined with the Pigeonhole Principle essentially proves the claim. Therefore, because there are infinitely many disjoint intervals containing an unrepresentable number, there are infinitely many unrepresentable numbers.",
+        "Solution_2_.28Simple.29": "Let $f(n) = n + s_b(n)$. It is easy to see that $f(b^2) = b^2 + 1$, and $f(b^2 - b + 1) = b^2 - b + 1 + b - 1 + 1 = b^2 + 1$, because the digits of $b^2 - b + 1$ are $b-1$ and $1$. Additionally, if we take any number, let's say $x$, and multiply it by $b^3$, and then add it to $b^2$ and $b^2 - b + 1$, we still obtain $f(b^3x + b^2) = f(b^3x + b^2 - b + 1)$, because we are simply adding $b^3x + x$ to each value.\nThus, we conclude that there are infinitely many numbers that are counted twice in $f(n)$, and that these numbers come $x^3$ apart. Finally, it is clear that for a large $N$, the number of numbers that are counted at least twice is at least $\\left\\lfloor\\frac{N}{x^3}\\right\\rfloor$. Because $f(n) > n$ for all $n$, this means that there are at least $\\left\\lfloor\\frac{N}{x^3}\\right\\rfloor$ numbers uncounted, and because this is unbounded, the proof is complete.\n~mathboy100"
+    },
+    {
+        "Problem": "Let $k$ be a positive integer. Two players $A$ and $B$ play a game on an infinite grid of regular hexagons. Initially all the grid cells are empty. Then the players alternately take turns with $A$ moving first. In his move, $A$ may choose two adjacent hexagons in the grid which are empty and place a counter in both of them. In his move, $B$ may choose any counter on the board and remove it. If at any time there are $k$ consecutive grid cells in a line all of which contain a counter, $A$ wins. Find the minimum value of $k$ for which $A$ cannot win in a finite number of moves, or prove that no such minimum value exists.",
+        "Solution": "The answer is $k=6$. We prove that $A$ can win for $k=5$ (which hence proves it for $k<5$ as well) and show that $B$ can thwart $A$ for $k\\geq 6$.\nArrange the board so that a pair of opposite sides are horizontal. Create a coordinate system on the board by setting the center of some hexagon as the origin and setting the hexagons directly above and above-and-right as $(0, 1)$ and $(1, 0)$, respectively. Then, for example, the below-and-right hexagon touching the origin is $(1, -1)$. So two hexagons touch if their coordinate difference is one of these.\nNow for $k=5$, person $A$ places his counters only in $\\{0, 1, 2, 3, 4\\}\\times\\{0, 1\\}$. Note that if, at $A$'s turn, there are 4 counters in either column, then $A$ can win immediately, so let us assume that in both columns there are at least 2 missing, meaning that at most $6$ counters are on the board. We would like to find when $A$ cannot play under these circumstances. If we look at the disjoint sets\n$\\{(0, 0), (0, 1), (1, 0)\\}, \\{(1, 1), (2, 0), (2, 1)\\}, \\{(3, 0), (3, 1), (4, 0)\\}, \\{(4, 1)\\}$\nwe see that either some set has at least $2$ hexagons without counters, in which case $A$ can move, or all four sets have exactly $1$ missing counter. Similarly for the sets\n$\\{(0, 0)\\}, \\{(0, 1), (1, 0), (1, 1)\\}, \\{(2, 0), (2, 1), (3, 0)\\}, \\{(3, 1), (4, 0), (4, 1)\\}$\nSo both $(0, 0), (4, 1)$ have no token on them. This means that $(0, 1), (1, 0), (3, 1), (4, 0)$ do. Thus $(3, 0)$ and $(1, 1)$ do not. So this is the only situation in which $A$ can neither win immediately nor play in only these 10 hexagons.\nSo $A$ plays only in these 10 hexagons until either he has a win or he can't anymore. If he wins, then we're done. Otherwise, $A$ plays in hexagons $(5, 0)$ and $(5, 1)$. Then $B$ either removes $(5, 1)$ so that $A$ can win at $(2, 0)$, or $B$ removes $(5, 0)$ and $A$ plays at $(5, 0)$ and $(4, 1)$ and then at either $(3, 0)$ or $(1, 1)$ the next turn, or $B$ removes one of $\\{0, 1, 2, 3, 4\\}\\times\\{0, 1\\}$ in which case $A$ can either win immediately at $(3, 0)$ or can play in both columns, and then win the next turn.\nNow if $k\\geq 6$, then if $A$ plays on anything in the lattice generated by $(2, -1)$ and $(1, 1)$, that is $(2a+b, b-a)$ for $a, b$ integers, then $B$ removes it. Otherwise, $B$ removes any of $A$'s counters. This works because in order for $A$ to win, there must be at least 2 counters in this lattice, but $A$ can only put a counter on $1$ at any time, so there's at most 1 on the lattice at any time.\nSo $A$ wins if $k\\leq 5$, and $B$ wins if $k\\geq 6$."
+    },
+    {
+        "Problem": "Let $ABC$ be a triangle with incenter $I$, incircle $\\gamma$ and circumcircle $\\Gamma$.  Let $M,N,P$ be the midpoints of sides $\\overline{BC}$, $\\overline{CA}$, $\\overline{AB}$ and let $E,F$ be the tangency points of $\\gamma$ with $\\overline{CA}$ and $\\overline{AB}$, respectively.  Let $U,V$ be the intersections of line $EF$ with line $MN$ and line $MP$, respectively, and let $X$ be the midpoint of arc $BAC$ of $\\Gamma$.\n(a) Prove that $I$ lies on ray $CV$.\n(b) Prove that line $XI$ bisects $\\overline{UV}$.",
+        "Solution": "(a)\nSolution 1: We will prove this via contradiction: assume that line $IC$ intersects line $MP$ at $Q$ and line $EF$ and $R$, with $R$ and $Q$ not equal to $V$. Let $x = \\angle A/2 = \\angle IAE$ and $y = \\angle C/2 = \\angle ICA$. We know that $\\overline{MP} \\parallel \\overline{AC}$ because $MP$ is a midsegment of triangle $ABC$; thus, by alternate interior angles (A.I.A) $\\angle MVE = \\angle FEA = (180^\\circ - 2x) / 2 = 90^\\circ - x$, because triangle $AFE$ is isosceles. Also by A.I.A, $\\angle MQC = \\angle QCA = y$. Furthermore, because $AI$ is an angle bisector of triangle $AFE$, it is also an altitude of the triangle; combining this with $\\angle QIA = x + y$ from the Exterior Angle Theorem gives $\\angle FRC = 90 - x - y$. Also, $\\angle VRQ = \\angle FRC = 90 - x - y$ because they are vertical angles. This completes part (a).\nSolution 2: First we show that the intersection $V'$of $MP$ with the internal angle bisector of $C$ is the same as the intersection $V''$ of $EF$ with the internal angle bisector of $C.$ Let $D$ denote the intersection of $AB$ with the internal angle bisector of $C,$ and let $a,b,c$ denote the side lengths of $BC, AC, AB.$ By Menelaus on $V'', F, E,$ with respect to $\\triangle ADC,$ \n Similarly,   Since $V'$ and $V''$ divide $CD$ in the same ratio, they must be the same point. Now, since $\\frac{b-a}{b+a}>-1,$ $I$ lies on ray $CV.$ $\\blacksquare$\nSolution 3: By the Iran Lemma, we know $CI, EF, MP$ concur, so Part $\\text{A}$ follows easily.\n(b)\nSolution 1: Using a similar argument to part (a), point U lies on line $BI$. Because $\\angle MVC = \\angle VCA = \\angle MCV$, triangle $VMC$ is isosceles. Similarly, triangle $BMU$ is isosceles, from which we derive that $VM = MC = MB = MU$. Hence, triangle $VUM$ is isosceles.\nNote that $X$ lies on both the circumcircle and the perpendicular bisector of segment $BC$. Let $D$ be the midpoint of $UV$; our goal is to prove that points $X$, $D$, and $I$ are collinear, which equates to proving $X$ lies on ray $ID$.\nBecause $MD$ is also an altitude of triangle $MVU$, and $MD$ and $IA$ are both perpendicular to $EF$, $\\overline{MD} \\parallel \\overline{IA}$. Furthermore, we have $\\angle VMD = \\angle UMD = x$ because $APMN$ is a parallelogram. (incomplete)\nSolution 2: Let $I_A$, $I_B$, and $I_C$ be the excenters of $ABC$. Note that the circumcircle of $ABC$ is the nine-point circle of $I_AI_BI_C$. Since $AX$ is the external angle bisector of $\\angle BAC$, $X$ is the midpoint of $I_BI_C$. Now $UV$ and $I_BI_C$ are parallel since both are perpendicular to the internal angle bisector of $\\angle BAC$. Since $IX$ bisects $I_BI_C$, it bisects $UV$ as well.\nSolution 3: Let $X'$ be the antipode of $X$ with respect to the circumcircle of triangle $ABC$. Then, by the Incenter-Excenter lemma, $X'$ is the center of a circle containing $B$, $I$, and $C$. Because $XX'$ is a diameter, $XB$ and $XC$ are tangent to the aforementioned circle; thus by a well-known symmedian lemma, $XI$ coincides with the $I$-symmedian of triangle $IBC$. From part (a); we know that $BVUC$ is cyclic (we can derive a similar argument for point $U$); thus $XI$ coincides with the median of triangle $VIU$, and we are done.\nSolution 4: Let $M_b, M_c$ be the midpoints of arcs $CA, AB$ respectively, and let $T_a$ be the tangency point between the $A$-mixtilinear incircle of $ABC$ and $\\Gamma$. It's well-known that $T_a \\in XI$, $T_aX$ bisects $M_bM_c$, $AI \\perp EF$, and $AI \\perp M_bM_c$.\nNow, it's easy to see $EF \\parallel M_bM_c$, so $IUV$ and $IM_bM_c$ are homothetic at $I$. But $T_aX \\equiv IX$ bisects $M_bM_c$, so Part $\\text{B}$ follows directly from the homothety. \n~ ike.chen"
+    },
+    {
+        "Problem": "Given a sequence of real numbers, a move consists of choosing two terms and replacing each with their arithmetic mean. Show that there exists a sequence of $2015$ distinct real numbers such that after one initial move is applied to the sequence -- no matter what move -- there is always a way to continue with a finite sequence of moves so as to obtain in the end a constant sequence.",
+        "Solution": "Let the set be ${-1007, -1006, ...0,1,2...1006, 1007}$, namely all the consecutive integers from $-1007$ to $1007$.  Notice that the operation we are applying in this problem does not change the sum or the mean of the set, which is $0$.\nThere are $1007$ pairs of opposite integers $\\{a,-a\\}$. After the first two elements are chosen, there are at least $1005$ such pairs. For each such pair we perform the operation of average, hence reducing these $2010$ elements to $0$. Then use the other $5$ elements together with three $0$'s produced to form the group of eight: ${a_1,a_2,a_3,a_4,a_5,a_6=0,a_7=0,a_8=0}$, and perform the operation in the following order:  where $m_i=\\frac{a_i+a_{i+1}}{2}$. Then, $(m_1,m_2)\\to(m_{11}, m_{11})$ for two groups, $(m_3,m_4)\\to(m_{12}, m_{12})$ for the other two groups, and finally $(m_{11},m_{12})\\to(m_{111}, m_{111})$ for all the eight elements. Since the sum of the eight-group is $0$, $m_{111}$ must also be $0$. Therefore, all the elements are reduced to $0$.\nThe key to the algorithm is to form a $2^k$ subset, which is guaranteed to be reducible to all the members of the same value, namely the mean. Then before that, if we could always choose $M\\ge N-2^k$ members to form pairs, each yielding the average of the total group, then all the members are reduced to the average. Under the condition that two arbitrary elements are chosen first, we need only $N\\ge4$ to guarantee this result. But for $N=2$ the first operation leads to equal elements, so $N=3$ is the only case when all the members may not be reduced to average.\nSidenote: Actually, for $N=3$, the members are all reduced to the average, as the sum of the terms is constant and does not change.",
+        "Solution_2": "Consider any arithmetic sequence. WLOG, let it be $s = (1, 2, 3, \\dots, 2015)$, i.e. $s_i = i\\ \\forall\\ 1\\le i\\le 2015$. Define the move $(x, y)$ as replacing the numbers located at positions $x$ and $y$ with their mean, assuming $x$ and $y$ are distinct. If they are the same integer, define it as not making a move. Now, suppose the initial move $m_0 = (a, b)$. If $a+b=2016$, then $s_a = s_b = \\frac{a+b}{2} = 1008$. Then, applying the moves\n\nwe get $s_1 = s_2 = \\cdots = s_{2015} = 1008$.\nOtherwise, suppose $a+b\\ne 2016$. Then consider the following $3$ moves:\n\nWe have $m_{-1}$ makes $s_{2016-a} = s_{2016-b} = 2016 - \\frac{a+b}{2}$. So, $m_{-2}$ makes\n\nSimilarly for $m_{-3}$ with $s_b$ and $s_{2016-b}$. Then, finishing up with the moves\n\nwe get $s_1 = s_2 = \\cdots = s_{2015} = 1008$.\n\n(95% Sure) Doesn't this solution fail for Case $2$ when $a$ or $b$ is $1008$? (Because $2016-1008 = 1008$ so $m_{-1}$ is basically non-existent.) \n- ike.chen",
+        "Solution_3_.28INCORRECT.29": "Let the set be ${a_1,a_2,\\dots ,a_{2015}}$, where all the terms are nonnegative. Note that the sum of all the terms in this sequence will always be the same after any amount of moves. To prove this, let $i,j$ be integers with $1\\le i<j\\le 2015$, and we have $a_i+a_j = \\frac{a_i+a_j}{2}+\\frac{a_i+a_j}{2}$.\nAlso, $a_ia_j \\le (\\frac{a_i+a_j}{2})(\\frac{a_i+a_j}{2})$ by AM-GM, so the product of all the terms will not decrease after any number of moves. However, the product will only stay the same when $a_i=a_j$, so the product will always increase if $a_i\\ne a_j$.\nFinally, note that $a_1a_2\\dots a_{2015}\\le (\\frac{a_1+a_2+\\dots +a_{2015}}{2015})^{2015}$ by AM-GM, so because $a_1+a_2+\\dots +a_{2015}$ is fixed, there is a maximum product that is reached after a finite number of moves as the product increases. This product is reached when $a_1=a_2=\\dots =a_{2015}$, so we are done.\nThis solution is incorrect; the product may take an infinite number of moves to reach the maximum (for example, consider the sequence $1, 3, 4, 5... 2016$)",
+        "Solution_4": "We claim that the sequence $-1007, -1006, ... ,0, ... , 1006, 1007$ satisifes the conditions\nLet the initial pair that we choose to operate on be denoted as $(a, b)$\nCase 1: $(a, b)$ $\\neq$ $0$\nSub case 1a: if $a + b$ sum to $0$ in this sequence, the pair is really just $(a, -a)$ or $(-a, a)$. If we replace $a$, and $-a$ with their AM then they both become $0$. Notice that for every number in the original sequence, there is either a negative or positive value to that (depending on the sign). So once we convert the initial pair to $0$, we can do so for every other pair, for example let $a_i$ be a number in the sequence, we would operate on $(a_i, -a_i)$ for every number until every number in the sequence becomes $0$. (Note that the amount of operations this would take is finite).\nSub case 1b: If $a + b \\neq 0$ then we can use the same logic, but with one extra step. After we do $(a, b)$ as our initial operation, we do $(-a, -b)$. This is equivalent to $(-a, a)$ and $(-b, b)$ which turns all those values into $0$ (this is also the same as comparing $(a+b)/2$ and $(-a-b)/2$ which cancels out). Then since the only values that remain besides $0$ both have a negative/positive counter part, do the same thing in Sub case 1a; $(a_i, -a_i)$ turning the whole sequence to $0$ in a finite number of operations.\nCase 2: The inital pair is $(0, b)$\nObviously, there is a $-b$ in the sequence, (this will come into play later). Since $b \\neq 0$ our only $0$ vanishes due to the AM, our goal is to make another $0$. Remember that for every number in the original sequence (besides $0$) there is a negative or positive value depending on the numbers sign. So even if $b$ to $-b$ or any other number in the sequence doesn't make $0$, we have another option.\nFind any $(a, -a)$ such that $(a + -a) = 0$ (again this is guaranteed to be in the sequence).\nNow once you have your $0$s, take the $-b$, and take the AM of that and $0$. Then there will be 2 $-b/2$s. This will make $0$s with our two numbers from $(0, b)$ since that produces 2 $b/2$s. Since those cancel and make $0$, everything else is the same positive negative condition, which finishes. ~dukcrantel\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Solve in integers the equation",
+        "Solution": "We first notice that both sides must be integers, so $\\frac{x+y}{3}$ must be an integer.\nWe can therefore perform the substitution $x+y = 3t$ where $t$ is an integer.\nThen:\n\n$4t+1$ is therefore the square of an odd integer and can be replaced with $(2n+1)^2 = 4n^2 + 4n +1$.\nBy substituting using $t = n^2 + n$ we get:\n\nUsing substitution we get the solutions: $(n^3 + 3n^2 - 1, - n^3 + 3n + 1) \\cup ( - n^3 + 3n + 1, n^3 + 3n^2 - 1)$"
+    },
+    {
+        "Problem": "Quadrilateral $APBQ$ is inscribed in circle $\\omega$ with $\\angle P = \\angle Q = 90^{\\circ}$ and $AP = AQ < BP$. Let $X$ be a variable point on segment $\\overline{PQ}$. Line $AX$ meets $\\omega$ again at $S$ (other than $A$). Point $T$ lies on arc $AQB$ of $\\omega$ such that $\\overline{XT}$ is perpendicular to $\\overline{AX}$. Let $M$ denote the midpoint of chord $\\overline{ST}$. As $X$ varies on segment $\\overline{PQ}$, show that $M$ moves along a circle.",
+        "Solution_1": "\nWe will use coordinate geometry.\nWithout loss of generality,\nlet the circle be the unit circle centered at the origin, \n,\nwhere $(1-a)^2+b^2=1$.\nLet angle $\\angle XAB=A$, which is an acute angle, $\\tan{A}=t$, then $X=(1-a,at)$.\nAngle $\\angle BOS=2A$, $S=(-\\cos(2A),\\sin(2A))$.\nLet $M=(u,v)$, then $T=(2u+\\cos(2A), 2v-\\sin(2A))$.\nThe condition $TX \\perp AX$ yields: $(2v-\\sin(2A)-at)/(2u+\\cos(2A)+a-1)=\\cot A.$     (E1)\nUse identities $(\\cos A)^2=1/(1+t^2)$,  $\\cos(2A)=2(\\cos A)^2-1= 2/(1+t^2) -1$, $\\sin(2A)=2\\sin A\\cos A=2t^2/(1+t^2)$, we obtain $2vt-at^2=2u+a$.   (E1')\nThe condition that $T$ is on the circle yields $(2u+\\cos(2A))^2+ (2v-\\sin(2A))^2=1$, namely $v\\sin(2A)-u\\cos(2A)=u^2+v^2$.   (E2)\n$M$ is the mid-point on the hypotenuse of triangle $STX$, hence $MS=MX$, yielding $(u+\\cos(2A))^2+(v-\\sin(2A))^2=(u+a-1)^2+(v-at)^2$.   (E3)\nExpand (E3), using (E2) to replace $2(v\\sin(2A)-u\\cos(2A))$ with $2(u^2+v^2)$, and using (E1') to replace $a(-2vt+at^2)$ with $-a(2u+a)$, and we obtain\n$u^2-u-a+v^2=0$, namely $(u-\\frac{1}{2})^2+v^2=a+\\frac{1}{4}$, which is a circle centered at $(\\frac{1}{2},0)$ with radius $r=\\sqrt{a+\\frac{1}{4}}$.",
+        "Solution_2": "Let the midpoint of $AO$ be $K$. We claim that $M$ moves along a circle with radius $KP$.\nWe will show that $KM^2 = KP^2$, which implies that $KM = KP$, and as $KP$ is fixed, this implies the claim.\n$KM^2 = \\frac{AM^2+OM^2}{2}-\\frac{AO^2}{4}$ by the median formula on $\\triangle AMO$.\n$KP^2 = \\frac{AP^2+OP^2}{2}-\\frac{AO^2}{4}$ by the median formula on $\\triangle APO$.\n$KM^2-KP^2 = \\frac{1}{2}(AM^2+OM^2-AP^2-OP^2)$.\nAs $OP = OT$, $OP^2-OM^2 = MT^2$ from right triangle $OMT$. $(1)$\nBy $(1)$, $KM^2-KP^2 = \\frac{1}{2}(AM^2-MT^2-AP^2)$.\nSince $M$ is the circumcenter of $\\triangle XTS$, and $MT$ is the circumradius, the expression $AM^2-MT^2$ is the power of point $A$ with respect to $(XTS)$. However, as $AX*AS$ is also the power of point $A$ with respect to $(XTS)$, this implies that $AM^2-MT^2=AX*AS$. $(2)$\nBy $(2)$, $KM^2-KP^2 = \\frac{1}{2}(AX*AS-AP^2)$\nFinally, $\\triangle APX \\sim \\triangle ASP$ by AA similarity ($\\angle XAP = \\angle SAP$ and $\\angle APX = \\angle AQP = \\angle ASP$), so $AX*AS = AP^2$. $(3)$\nBy $(3)$, $KM^2-KP^2=0$, so $KM^2=KP^2$, as desired. $QED$"
+    },
+    {
+        "Problem": "Find all functions $f:\\mathbb{Q}\\rightarrow\\mathbb{Q}$ such thatfor all rational numbers $x<y<z<t$ that form an arithmetic progression. ($\\mathbb{Q}$ is the set of all rational numbers.)",
+        "Solution_1": "According to the given, $f(x-a)+f(x+0.5a)=f(x-0.5a)+f(x)$, where x and a are rational. Likewise $f(x-0.5a)+f(x+a)=f(x+0.5a)+f(x)$. Hence $f(x+a)-f(x)= f(x)-f(x-a)$, namely $2f(x)=f(x-a)+f(x+a)$. Let $f(0)=C$, then consider $F(x)=f(x)-C$, where $F(0)=0,$ $2F(x)=F(x-a)+F(x+a)$.\n$F(2x)=F(x)+[F(x)-F(0)]=2F(x)$, \n$F(3x)=F(2x)+[F(2x)-F(x)]=3F(x)$.\nEasily, by induction, $F(nx)=nF(x)$ for all integers $n$.\nTherefore, for nonzero integer m, $(1/m)F(mx)=F(x)$ , namely $F(x/m)=(1/m)F(x)$\nHence $F(n/m)=(n/m)F(1)$. Let $F(1)=k$, we obtain $F(x)=kx$, where $k$ is the slope of the linear functions, and $f(x)=kx+C$.",
+        "Solution_2": "We have  and  Subtracting these two and rearranging gives  and since $f(x+2d)=f(x+d)+f(x)-f(x-d)$ we get  from which we get  Then we have $f(x)+f(y)=f(0)+f(x+y)=2f\\left(\\frac{x+y}{2}\\right)$. Setting $f(0)=c$, we let $f(x)=g(x)+c$ to get $g(x)+g(y)=g(x+y)$. This is Cauchy's functional equation, so it has solutions at $g(x)=kx$, so the answer is $\\boxed{f(x)=kx+c}$."
+    },
+    {
+        "Problem": "Let $ABCD$ be a cyclic quadrilateral. Prove that there exists a point $X$ on segment $\\overline{BD}$ such that $\\angle BAC=\\angle XAD$ and $\\angle BCA=\\angle XCD$ if and only if there exists a point $Y$ on segment $\\overline{AC}$ such that $\\angle CBD=\\angle YBA$ and $\\angle CDB=\\angle YDA$.",
+        "Solution_1": "Note that lines $AC, AX$ are isogonal in $\\triangle ABD$, so an inversion centered at $A$ with power $r^2=AB\\cdot AD$ composed with a reflection about the angle bisector of $\\angle DAB$ swaps the pairs $(D,B)$ and $(C,X)$. Thus, so that $ACBD$ is a harmonic quadrilateral. By symmetry, if $Y$ exists, then $(B,D;A,C)=-1$. We have shown the two conditions are equivalent, whence both directions follow$.\\:\\blacksquare\\:$",
+        "Solution_2": "All angles are directed. Note that lines $AC, AX$ are isogonal in $\\triangle ABD$ and $CD, CE$ are isogonal in $\\triangle CDB$. From the law of sines it follows that\n\nTherefore, the ratio equals $\\frac{AD\\cdot DC}{DB\\cdot BC}.$\nNow let $Y$ be a point of $AC$ such that $\\angle{ABE}=\\angle{CBY}$. We apply the above identities for $Y$ to get that $\\frac{CY}{YA}\\cdot \\frac{CE}{EA}=\\left(\\frac{CD}{DA}\\right)^2$. So $\\angle{CDY}=\\angle{EDA}$, the converse follows since all our steps are reversible.\nBeware that directed angles, or angles $\\bmod$ $180$, are not standard olympiad material. If you use them, provide a definition.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Steve is piling $m\\geq 1$ indistinguishable stones on the squares of an $n\\times n$ grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions $(i, k), (i, l), (j, k), (j, l)$ for some $1\\leq i, j, k, l\\leq n$, such that $i<j$ and $k<l$. A stone move consists of either removing one stone from each of $(i, k)$ and $(j, l)$ and moving them to $(i, l)$ and $(j, k)$ respectively, or removing one stone from each of $(i, l)$ and $(j, k)$ and moving them to $(i, k)$ and $(j, l)$ respectively.\nTwo ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves.\nHow many different non-equivalent ways can Steve pile the stones on the grid?",
+        "Solution": "Let the number of stones in row $i$ be $r_i$ and let the number of stones in column $i$ be $c_i$. Since there are $m$ stones, we must have $\\sum_{i=1}^n r_i=\\sum_{i=1}^n c_i=m$\nLemma 1: If any $2$ pilings are equivalent, then $r_i$ and $c_i$ are the same in both pilings $\\forall i$.\nProof: We suppose the contrary. Note that $r_i$ and $c_i$ remain invariant after each move, therefore, if any of the $r_i$ or $c_i$ are different, they will remain different.\nLemma 2: Any $2$ pilings with the same $r_i$ and $c_i$ $\\forall i$ are equivalent.\nProof: Suppose piling 1 and piling 2 not the same piling. Call a stone in piling 1 wrong if the stone occupies a position such that there are more stones in that position in piling 1 than piling 2. Similarly define a wrong stone in piling 2. Let a wrong stone be at $(a, b)$ in piling 1. Since $c_b$ is the same for both pilings, we must have a wrong stone in piling 2 at column b, say at $(c, b)$, such that $c\\not = a$. Similarly, we must have a wrong stone in piling 1 at row c, say at $(c, d)$ where $d \\not = b$. Clearly, making the move $(a,b);(c,d) \\implies (c,b);(a,d)$ in piling 1 decreases the number of wrong stones in piling 1. Therefore, the number of wrong stones in piling 1 must eventually be $0$ after a sequence of moves, so piling 1 and piling 2 are equivalent.\nLemma 3: Given the sequences $g_i$ and $h_i$ such that $\\sum_{i=1}^n g_i=\\sum_{i=1}^n h_i=m$ and $g_i, h_i\\geq 0 \\forall i$, there is always a piling that satisfies $r_i=g_i$ and $c_i=h_i$ $\\forall i$.\nProof: We take the lowest $i$, $j$, such that $g_i, h_j >0$ and place a stone at $(i, j)$, then we subtract $g_i$ and $h_j$ by $1$ each, until $g_i$ and $h_i$ become $0$ $\\forall i$, which will happen when $m$ stones are placed, because $\\sum_{i=1}^n g_i$ and $\\sum_{i=1}^n h_i$ are both initially $m$ and decrease by $1$ after each stone is placed. Note that in this process $r_i+g_i$ and $c_i+h_i$ remains invariant, thus, the final piling satisfies the conditions above.\nBy the above lemmas, the number of ways to pile is simply the number of ways to choose the sequences $r_i$ and $c_i$ such that $\\sum_{i=1}^n r_i=\\sum_{i=1}^n c_i=m$ and $r_i, c_i \\geq 0 \\forall i$. By stars and bars, the number of ways is $\\binom{n+m-1}{m}^{2}$.\nSolution by Shaddoll"
+    },
+    {
+        "Problem": "The isosceles triangle $\\triangle ABC$, with $AB=AC$, is inscribed in the circle $\\omega$. Let $P$ be a variable point on the arc $\\stackrel{\\frown}{BC}$ that does not contain $A$, and let $I_B$ and $I_C$ denote the incenters of triangles $\\triangle ABP$ and $\\triangle ACP$, respectively.\nProve that as $P$ varies, the circumcircle of triangle $\\triangle PI_BI_C$ passes through a fixed point.",
+        "Solution_1": "\nWe claim that $M$ (midpoint of arc $BC$) is the fixed point.\nWe would like to show that $M$, $P$, $I_B$, $I_C$ are cyclic.\nWe extend $PI_B$ to intersect $\\omega$ again at R.\nWe extend $PI_C$ to intersect $\\omega$ again at S.\nWe invert around a circle centered at $P$ with radius $1$ (for convenience).\n(I will denote X' as the reflection of X for all the points)\nThe problem then becomes: Prove $I_B'$, $I_C'$, and $M'$ are collinear.\nNow we look at triangle $\\triangle PR'S'$. We apply Menelaus (the version where all three points lie outside the triangle).\nIt suffices to show that\nBy inversion, we know $PX' = \\dfrac{1}{PX}$ for any point $X$ and $X'Y' = \\dfrac{XY}{PX \\cdot PY}$ for any points $X$ and $Y$.\nPlugging this into our Menelaus equation we obtain that it suffices to show\n\nWe cancel out the like terms and rewrite. It suffices to show\n\nWe know that $AM$ is the diameter of $\\omega$ because $\\triangle ABC$ is isosceles and $AM$ is the angle bisector. We also know $\\angle RMA = \\dfrac{\\angle ACB}{2} = \\dfrac{\\angle ABC}{2} = \\angle SMA$ so $R$ and $S$ are symmetric with respect to $AM$ so $RM = SM$.\nThus, it suffices to show $\\dfrac{SI_C}{RI_B} = 1$.\nThis is obvious because $RI_B = RA = SA = SI_C$.\nTherefore we are done. $\\blacksquare$",
+        "Solution_2": "We will use complex numbers as mentioned here. Set the circumcircle of $\\triangle ABC$ to be the unit circle. Let  such that  We claim that the circumcircle of $\\triangle PI_BI_C$ passes through $M=-1.$ This is true if  is real. Now observe that  so $k$ is real and we are done. $\\blacksquare$\n",
+        "Solution_3": "\nLet $M$ be the midpoint of arc $BC$. Let $D$ be the midpoint of arc $AB$. Let $E$ be the midpoint of arc $AC$.\nThen, $P, I_B$, and $D$ are collinear and $P, I_C$, and $E$ are collinear.\nWe'll prove $MPI_B I_C$ is cyclic. (Intuition: we'll show that $M$ is the Miquel's point of quadrilateral $DE I_C I_B$.\n$D$ is the center of the circle $A I_B B$ (the $P-$ excenter of $PAB$ is also on the same circle). Therefore $D I_B = DB$. Similarly $E I_C = EC$. Since $AB=AC$, $DB=EC$. Therefore $D I_B = E I_C$. Obviously $ME = MD$ and $\\angle MEI_C = \\angle MEP = \\angle MDP = \\angle MDI_B$. Thus by SAS, $\\triangle MEI_C \\cong \\triangle MDI_B$.\nHence $\\angle I_B M I_C = \\angle DME = \\angle DPE = \\angle I_B P I_C$, so $MPI_B I_C$ is cyclic and we are done.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Prove that there exists a positive integer $n < 10^6$ such that $5^n$ has six consecutive zeros in its decimal representation.",
+        "Solution": "Let digit $1$ of a number be the units digit, digit $2$ be the tens digit, and so on. Let the 6 consecutive zeroes be at digits $k-5$ through digit $k$. The criterion is then obviously equivalent to\n\nWe will prove that $n = 20+2^{19}, k = 20$ satisfies this, thus proving the problem statement (since $n = 20+2^{19} < 10^6$).\nWe want\n\nWe can split this into its prime factors. Obviously, it is a multiple of $5^{20}$, so we only need to consider it $\\mod2^{20}$.\n\n\n($\\varphi$ is the Euler Totient Function.) By Euler's Theorem, since $\\bigl(5, 2^{20}\\bigr)=1$,\n\nso\n\nSince $2^{20} > 10^6, 5^{20} < 10^{14} = 10^{20-6}$,\n\nfor $n = 20+2^{19}$ and $k = 20$, and thus the problem statement has been proven. $\\blacksquare$"
+    },
+    {
+        "Problem": "Let $X_1, X_2, \\ldots, X_{100}$ be a sequence of mutually distinct nonempty subsets of a set $S$. Any two sets $X_i$ and $X_{i+1}$ are disjoint and their union is not the whole set $S$, that is, $X_i\\cap X_{i+1}=\\emptyset$ and $X_i\\cup X_{i+1}\\neq S$, for all $i\\in\\{1, \\ldots, 99\\}$. Find the smallest possible number of elements in $S$.",
+        "Solution_1": "The answer is that $|S| \\ge 8$.\nFirst, we provide a inductive construction for $S = \\left\\{ 1, \\dots, 8 \\right\\}$. Actually, for $n \\ge 4$ we will provide a construction for $S = \\left\\{ 1, \\dots, n \\right\\}$ which has $2^{n-1} + 1$ elements in a line. (This is sufficient, since we then get $129$ for $n = 8$.) The idea is to start with the following construction for $|S| = 4$: Then inductively, we do the following procedure to move from $n$ to $n+1$: take the chain for $n$ elements, delete an element, and make two copies of the chain (which now has even length). Glue the two copies together, joined by $\\varnothing$ in between. Then place the element $n+1$ in alternating positions starting with the first (in particular, this hits $n+1$). For example, the first iteration of this construction gives: \nNow let's check $|S| \\ge 8$ is sufficient. Consider a chain on a set of size $|S| = 7$. (We need $|S| \\ge 7$ else $2^{|S|} < 100$.) Observe that there are sets of size $\\ge 4$ can only be neighbored by sets of size $\\le 2$, of which there are $\\binom 71 + \\binom 72 = 28$. So there are $\\le 30$ sets of size $\\ge 4$. Also, there are $\\binom 73 = 35$ sets of size $3$. So the total number of sets in a chain can be at most $30 + 28 + 35 = 93 < 100$.",
+        "Solution_2": "My proof that $|S|\\ge 8$ is basically the same as the one above. Here is another construction for $|S| = 8$ that I like because it works with remainders and it's pretty intuitive. The basic idea is to assign different subsets to different remainders when divided by particular numbers, and then to use the Chinese Remainder Theorem to show that all of the subsets are distinct. The motivation for this comes from the fact that we want $X_i$ and $X_{i+1}$ to always be disjoint, so remainders are a great way to systematically make that happen, since $i$ and $i+1$ do not have the same remainder modulo any positive integer greater than $1.$ Anyway, here is the construction:\nLet $S = \\left\\{ {1, 2, ..., 8} \\right\\}.$ For $i = 1, 2, ..., 98,$ we will choose which elements of the set $\\left\\{ {1, 2, 3, 4} \\right\\}$ belong to $X_i$ based on the remainder of $i$ modulo $9,$ and we will choose which elements of the set $\\left\\{ {5, 6, 7, 8} \\right\\}$ belong to $X_i$ based on the remainder of $i$ modulo $11.$ We do this as follows:\n\n\nFinally, we specially define $X_{99} = \\left\\{ {1, 2, 3} \\right\\}$ and $X_{100} = \\left\\{ {5, 6, 7} \\right\\}.$\nIt is relatively easy to see that this configuration satisfies all of the desired conditions. We see that $X_{98} = \\left\\{ {4, 7} \\right\\},$ so $X_{98}$ and $X_{99}$ are disjoint, as are $X_{99}$ and $X_{100}.$ The remainder configuration above takes care of the rest, so any two consecutive sets are disjoint. Then, by the Chinese Remainder Theorem, no two integers from $1$ to $98$ have the same combination of residues modulo $9$ and modulo $11,$ so all of the sets $X_i$ are distinct for $i = 1, 2, ..., 98.$ It is also easy to verify that none of these match $X_{99}$ or $X_{100},$ since they all have at most two elements of $\\left\\{ {1, 2, 3, 4} \\right\\}$ and at most two elements of $\\left\\{ {5, 6, 7, 8} \\right\\},$ whereas $X_{99}$ and $X_{100}$ do not satisfy this; hence all of the sets are distinct. Finally, notice that, for any pair of consecutive sets, at least one of them has at most $3$ elements, while the other has at most $4.$ Thus, their union always has at most $7$ elements, so $X_i\\cup X_{i+1}\\neq S$ for all $i = 1, 2, ..., 99.$\nAll of the conditions are satisfied, so this configuration works. We thus conclude that $\\text{min}\\left(\\left|S\\right|\\right) = 8.$\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Find, with proof, the least integer $N$ such that if any $2016$ elements are removed from the set $\\{1, 2,...,N\\}$, one can still find $2016$ distinct numbers among the remaining elements with sum $N$.\n",
+        "Solution": "Since any $2016$ elements are removed, suppose we remove the integers from $1$ to $2016$. Then the smallest possible sum of $2016$ of the remaining elements is \nso clearly $N\\ge 6097392$. We will show that $N=6097392$ works.\n$\\vspace{0.2 in}$\n$\\{1,2\\cdots 6097392\\}$ contain the integers from $1$ to $6048$, so pair these numbers as follows:\n\n\n\n\n\nWhen we remove any $2016$ integers from the set $\\{1,2,\\cdots N\\}$, clearly we can remove numbers from at most $2016$ of the $3024$ pairs above, leaving at least $1008$ complete pairs. To get a sum of $N$, simply take these $1008$ pairs, all of which sum to $6049$. The sum of these $2016$ elements is $1008 \\cdot 6049 = 6097392$, as desired.\n$\\vspace{0.2 in}$\nWe have shown that $N$ must be at least $6097392$, and that this value is attainable. Therefore our answer is $\\boxed{6097392}$.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Let $\\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. Point $H$ is the foot of the perpendicular from $A$ to line $\\overleftrightarrow{BC}$, and points $P$ and $Q$ are the feet of the perpendiculars from $H$ to the lines $\\overleftrightarrow{AB}$ and $\\overleftrightarrow{AC}$, respectively.\nGiven that prove that the points $O,P,$ and $Q$ are collinear.",
+        "Solution_1": "\nIt is well-known that $AH\\cdot 2AO=AB\\cdot AC$ (just use similar triangles or standard area formulas). Then by Power of a Point,\n Consider the transformation $X\\mapsto \\Psi(X)$ which dilates $X$ from $A$ by a factor of $\\dfrac{AB}{AQ}=\\dfrac{AC}{AP}$ and reflects about the $A$-angle bisector. Then $\\Psi(O)$ clearly lies on $AH$, and its distance from $A$ is  so $\\Psi(O)=H$, hence we conclude that $O,P,Q$ are collinear, as desired.",
+        "Solution_2": "We will use barycentric coordinates with respect to $\\triangle ABC.$ The given condition is equivalent to $(\\sin B\\sin C)^2=\\frac{1}{2}.$ Note that  Therefore, we must show that  Expanding, we must prove\nLet $x=e^{iA}, y=e^{iB}, z=e^{iC},$ such that $xyz=-1.$ The left side is equal to  The right side is equal to \n which is equivalent to the left hand side. Therefore, the determinant is $0,$ and $O,P,Q$ are collinear. $\\blacksquare$\n",
+        "Solution_3": "For convenience, let $a, b, c$ denote the lengths of segments $BC, CA, AB,$ respectively, and let $\\alpha, \\beta, \\gamma$ denote the measures of $\\angle CAB, \\angle ABC, \\angle BCA,$ respectively. Let $R$ denote the circumradius of $\\triangle ABC.$\nSince the central angle $\\angle AOB$ subtends the same arc as the inscribed angle $\\angle ACB$ on the circumcircle of $\\triangle ABC,$ we have $\\angle AOB = 2\\gamma.$ Note that $OA = OB,$ so $\\angle OAB = \\angle OBA.$ Thus, $\\angle OAB = \\frac{\\pi}{2} - \\gamma.$ Similarly, one can show that $\\angle OAC = \\frac{\\pi}{2} - \\beta.$ (One could probably cite this as well-known, but I have proved it here just in case.)\nClearly, $AO = R.$ Since $AH^2 = 2\\cdot AO^2,$ we have $AH = \\sqrt{2}R.$ Thus, $AH\\cdot AO = \\sqrt{2}R^2.$\nNote that $AH = b\\sin\\gamma = c\\sin\\beta.$ The Extended Law of Sines states that:\n\nTherefore, $AH = \\frac{bc}{2R} = \\sqrt{2}R.$ Thus, $bc = \\sqrt{2}R^2.$\nSince $\\angle PHA = \\beta$ and $\\angle QHA = \\gamma,$ we have:\n\n\nIt follows that:\n\nWe see that $AP\\cdot AQ = AH\\cdot AO.$\nRearranging $AP\\cdot AQ = AH\\cdot AO,$ we get $\\frac{AP}{AH} = \\frac{AO}{AQ}.$ We also have $\\angle PAH = \\angle OAQ = \\frac{\\pi}{2} - \\beta,$ so $\\triangle PAH\\sim\\triangle OAQ$ by SAS similarity. Thus, $\\angle AOQ = \\angle APH,$ so $\\angle AOQ$ is a right angle.\nRearranging $AP\\cdot AQ = AH\\cdot AO,$ we get $\\frac{AP}{AO} = \\frac{AO}{AH}.$ We also have $\\angle PAO = \\angle HAQ = \\frac{\\pi}{2} - \\gamma,$ so $\\triangle PAO\\sim\\triangle HAQ$ by SAS similarity. Thus, $\\angle AOP = \\angle AQH,$ so $\\angle AOP$ is a right angle.\nSince $\\angle AOP$ and $\\angle AOQ$ are both right angles, we get $\\angle POQ = \\pi,$ so we conclude that $P, O, Q$ are collinear, and we are done. (We also obtain the extra interesting fact that $AO\\perp PQ.$)",
+        "Solution_4": "Draw the altitude from $O$ to $AB$, and let the foot of this altitude be $D$.\nThen, by the Right Triangle Altitude Theorem on triangle $AHB$, we have: $AB\\cdot AP=AH^{2}$.\nSince $OD$ is the perpendicular bisector of $AB$, $2\\cdot AD = AB$.\nSubstituting this into our previous equation gives $2\\cdot AD \\cdot AP = AH^{2}$, which equals $2\\cdot AO^{2}$ by the problem condition.\nThus, $2\\cdot AD\\cdot AP = 2\\cdot AO^{2} \\implies AD\\cdot AP = AO^{2}$.\nAgain, by the Right Triangle Altitude Theorem, angle $AOP$ is right.\nBy dropping an altitude from $O$ to $AC$ and using the same method, we can find that angle $AOQ$ is right. Since $\\angle AOP=\\angle AOQ=90$, $P$, $O$, $Q$ are collinear and we are done.\n~champion999",
+        "Solution_5": "We use complex numbers. Let lower case letters represent their respective upper case points, with $|a| = |b| = |c| = 1$. Spamming the foot from point to segment formula, we obtain   and  We now simplify the given length condition:  We would like to show that $P$, $O$, $Q$ are collinear, or  After some factoring (or expanding) that takes about 15 minutes, this eventually reduces to  which is true. $\\square$\n-MP8148",
+        "Solution_6": "Claim: $\\triangle AOQ \\sim \\triangle AHB$\nProof: We compute the area of $\\triangle ABC$ using two methods. Let $R=AO$ be the circumradius of $\\triangle ABC$.\nFirst, by extended law of sines, $BC=2R \\sin \\angle BAC$. We are also given that $AH= R \\sqrt{2}$.\n$AH \\perp BC$, so\nSecond, we compute the area using $[ABC]=\\frac{1}{2} \\cdot AB \\cdot AC \\sin \\angle BAC$.\nEquating these two expressions for the area of $\\triangle ABC$ and reducing, we get \nBut $2R^2 =2 AO^2 = AH^2$, so $AB \\cdot AC = AH^2 \\sqrt{2}$ and $\\bf{AB =\\frac{AH^2 \\sqrt{2}}{AC}}$.\nSince $\\angle AQH=\\angle AHC =90^\\circ$, and both triangles share the angle $\\angle HAC$, $\\triangle AHC \\sim \\triangle AQH$. This tells us $\\frac{AQ}{AH}=\\frac{AH}{AC}$ or $AQ =\\frac{AH^2}{AC}$.\nSubstituting into the bolded equation, we get $AB=AQ \\sqrt{2}$.\nThe original length condition can be written as $AH= AO \\sqrt{2}$.\nWe also have\nTherefore, by SAS similarity, $\\triangle AOQ \\sim \\triangle AHB$.\nWe can prove analogously that $\\triangle AOP \\sim \\triangle AHC$.\nWe now have $\\angle AOQ=\\angle AHB =90^\\circ$ and $\\angle AOP=\\angle AHC =90^\\circ$.\nThis implies that $\\angle POQ=\\angle POA +\\angle QOA=180^\\circ$ which tells us $P,O,$ and $Q$ are collinear, as desired.$\\square$\n-vvluo\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Find all functions $f:\\mathbb{R}\\rightarrow \\mathbb{R}$ such that for all real numbers $x$ and $y$,",
+        "Solution_1": "Step 1: Set $x = y = 0$ to obtain $f(0) = 0.$\nStep 2: Set $x = 0$ to obtain $f(y)f(-y) = f(y)^2.$\n$\\indent$ In particular, if $f(y) \\ne 0$ then $f(y) = f(-y).$\n$\\indent$ In addition, replacing $y \\to -t$, it follows that $f(t) = 0 \\implies f(-t) = 0$ for all $t \\in \\mathbb{R}.$\nStep 3: Set $x = 3y$ to obtain $\\left[f(y) + 3y^2\\right]f(8y) = f(4y)^2.$\n$\\indent$ In particular, replacing $y \\to t/8$, it follows that $f(t) = 0 \\implies f(t/2) = 0$ for all $t \\in \\mathbb{R}.$\nStep 4: Set $y = -x$ to obtain $f(4x)\\left[f(x) + f(-x) - 2x^2\\right] = 0.$\n$\\indent$ In particular, if $f(x) \\ne 0$, then $f(4x) \\ne 0$ by the observation from Step 3, because $f(4x) = 0 \\implies f(2x) = 0 \\implies f(x) = 0.$ Hence, the above equation implies that $2x^2 = f(x) + f(-x) = 2f(x)$, where the last step follows from the first observation from Step 2.\n$\\indent$ Therefore, either $f(x) = 0$ or $f(x) = x^2$ for each $x.$\n$\\indent$ Looking back on the equation from Step 3, it follows that $f(y) + 3y^2 \\ne 0$ for any nonzero $y.$ Therefore, replacing $y \\to t/4$ in this equation, it follows that $f(t) = 0 \\implies f(2t) = 0.$\nStep 5: If $f(a) = f(b) = 0$, then $f(b - a) = 0.$\n$\\indent$ This follows by choosing $x, y$ such that $x - 3y = a$ and $3x - y = b.$ Then $x + y = \\tfrac{b - a}{2}$, so plugging $x, y$ into the given equation, we deduce that $f\\left(\\tfrac{b - a}{2}\\right) = 0.$ Therefore, by the third observation from Step 4, we obtain $f(b - a) = 0$, as desired.\nStep 6: If $f \\not\\equiv 0$, then $f(t) = 0 \\implies t = 0.$\n$\\indent$ Suppose by way of contradiction that there exists an nonzero $t$ with $f(t) = 0.$ Choose $x, y$ such that $f(x) \\ne 0$ and $x + y = t.$ The following three facts are crucial:\n$\\indent$ 1. $f(y) \\ne 0.$ This is because $(x + y) - y = x$, so by Step 5, $f(y) = 0 \\implies f(x) = 0$, impossible.\n$\\indent$ 2. $f(x - 3y) \\ne 0.$ This is because $(x + y) - (x - 3y) = 4y$, so by Step 5 and the observation from Step 3, $f(x - 3y) = 0 \\implies f(4y) = 0 \\implies f(2y) = 0 \\implies f(y) = 0$, impossible.\n$\\indent$ 3. $f(3x - y) \\ne 0.$ This is because by the second observation from Step 2, $f(3x - y) = 0 \\implies f(y - 3x) = 0.$ Then because $(x + y) - (y - 3x) = 4x$, Step 5 together with the observation from Step 3 yield $f(3x - y) = 0 \\implies f(4x) = 0 \\implies f(2x) = 0 \\implies f(x) = 0$, impossible.\n$\\indent$ By the second observation from Step 4, these three facts imply that $f(y) = y^2$ and $f(x - 3y) = \\left(x - 3y\\right)^2$ and $f(3x - y) = \\left(3x - y\\right)^2.$ By plugging into the given equation, it follows that\n But the above expression miraculously factors into $\\left(x + y\\right)^4$! This is clearly a contradiction, since $t = x + y \\ne 0$ by assumption. This completes Step 6.\nStep 7: By Step 6 and the second observation from Step 4, the only possible solutions are $f \\equiv 0$ and $f(x) = x^2$ for all $x \\in \\mathbb{R}.$ It's easy to check that both of these work, so we're done.",
+        "Solution_2": "Step 1: $x=y=0 \\implies f(0)=0$\nStep 2: $x=0 \\implies f(y)f(-y)=f(y)^{2}$. Now, assume $y \\not = 0$. Then, if $f(y)=0$, we substitute in $-y$ to get $f(y)f(-y)=f(-y)^{2}$, or $f(y)=f(-y)=0$. Otherwise, we divide both sides by $f(y)$ to get $f(y)=f(-y)$. If $y=0$, we obviously have $f(0)=f(0)$. Thus, the function is even.\n.\nStep 3: $y=-x \\implies 2f(4x)(f(x)-x^{2})=0$. Thus, $\\forall x$, we have $f(4x)=0$ or $f(x)=x^{2}$.\nStep 4: We now assume $f(x) \\not = 0$, $x\\not = 0$. We have $f(\\frac{x}{4})=\\frac{x^{2}}{16}$. Now, setting $x=y=\\frac{x}{4}$, we have $f(\\frac{x}{2})=\\frac{x^{2}}{4}$ or $f(\\frac{x}{2})=0$. The former implies that $f(x)=0$ or $x^{2}$. The latter implies that $f(x)=0$ or $f(x)=\\frac{x^{2}}{2}$. Assume the latter. $y=-2x \\implies -\\frac{3y^{2}}{2}f(7y)-{2y^{2}}f(5y)=\\frac{x^{2}}{2}$. Clearly, this implies that $f(x)$ is negative for some $m$. Now, we have $f(\\frac{m}{4})=\\frac{m^{2}}{16} \\implies f(\\frac{m}{2})=0,\\frac{m^{2}}{4} \\implies f(m) \\geq 0$, which is a contradiction. Thus, $\\forall x$$f(x)=0$ or $f(x)=x^{2}$.\nStep 5: We now assume $f(x)=0$, $f(y)=y^{2}$ for some $x,y \\not = 0$. Let $m$ be sufficiently large integer, let $z=|4^{m}x|$ and take the absolute value of $y$(since the function is even). Choose $c$ such that $3z-c=y$. Note that we have $\\frac{c}{z}$~$3$ and $\\frac{y}{z}$~$0$. Note that $f(z)=0$. Now, $x=z, y=c \\implies$ LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to $(z+c)^{4}$~$256z^{4}$. Now if $f(z-3c)=0$, the second term of the LHS/RHS clearly ~0 as $m \\to \\infty$. if $f(z-3c)=0$, then we have LHS/RHS ~ $0$, otherwise, we have LHS/RHS~$\\frac{8^{2}\\cdot 3z^{4}}{256z^{4}}$~$\\frac{3}{4}$, a contradiction, as we're clearly not dividing by $0$, and we should have LHS/RHS=1.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Prove that there are infinitely many distinct pairs $(a,b)$ of relatively prime positive integers $a>1$ and $b>1$ such that $a^b+b^a$ is divisible by $a+b$.",
+        "Solution_1": "Let $a = 2n-1$ and $b = 2n+1$. We see that $(2n \\pm 1)^2 = 4n^2\\pm4n+1 \\equiv 1 \\pmod{4n}$. Therefore, we have $(2n+1)^{2n-1} + (2n-1)^{2n+1} \\equiv 2n + 1 + 2n - 1  = 4n \\equiv 0 \\pmod{4n}$, as desired.\n(Credits to mathmaster2012)",
+        "Solution_2": "Let $x$ be odd where $x>1$. We have $x^2-1=(x-1)(x+1),$ so $x^2-1 \\equiv 0 \\pmod{2x+2}.$ This means that  $x^{x+2}-x^x \\equiv 0 \\pmod{2x+2},$ and since x is odd, $x^{x+2}+(-x)^x \\equiv 0 \\pmod{2x+2},$ or $x^{x+2}+(x+2)^x \\equiv 0 \\pmod{2x+2},$ as desired.",
+        "Solution_3": "Because problems such as this usually are related to expressions along the lines of $x\\pm1$, it's tempting to try these. After a few cases, we see that $\\left(a,b\\right)=\\left(2x-1,2x+1\\right)$ is convenient due to the repeated occurrence of $4x$ when squared and added. We rewrite the given expressions as:  After repeatedly factoring the initial equation,we can get:  Expanding each of the squares, we can compute each product independently then sum them:   Now we place the values back into the expression:  Plugging any positive integer value for $x$ into $\\left(a,b\\right)=\\left(2x-1,2x+1\\right)$ yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs $\\left(a,b\\right)$. $\\square$\n-fatant",
+        "Solution_4": "Let $a = 2x + 1$ and $b = 2x-1$, where $x$ leaves a remainder of $1$ when divided by $4$.We seek to show that $(2x+1)^{2x-1} + (2x-1)^{2x+1} \\equiv 0 \\mod 4x$ because that will show that there are infinitely many distinct pairs $(a,b)$ of relatively prime integers $a>1$ and $b>1$ such that $a^b+b^a$ is divisible by $a+b$.\nClaim 1: $(2x+1)^{2x-1} + (2x-1)^{2x+1} \\equiv 0 \\mod 4$.\nWe have that the remainder when $2x+1$ is divided by $4$ is $3$ and the remainder when $2x-1$ is divided by $4$ is always $1$. Therefore, the remainder when $(2x+1)^{2x-1} + (2x-1)^{2x+1}$ is divided by $4$ is always going to be $(-1)^{2x-1} + 1^{2x+1} = 0$.\nClaim 2: $(2x+1)^{2x-1} + (2x-1)^{2x+1} \\equiv 0 \\mod x$\nWe know that $(2x+1) \\mod x \\equiv 1$ and $(2x-1) \\mod x \\equiv 3$, so the remainder when $(2x+1)^{2x-1} + (2x-1)^{2x+1}$ is divided by $4$ is always going to be $(-1)^{2x-1} + 1^{2x+1} = 0$.\nClaim 3: $(2x+1)^{2x-1} + (2x-1)^{2x+1} \\equiv 0 \\mod 4x$\nTrivial given claim $1,2$. $\\boxed{}$\n~AopsUser101",
+        "Solution_5": "I claim $(a,b) = (2n-1,2n+1)$,  $n (\\in \\mathbb{N}) \\geq 2$ always satisfies above conditions.\nNote: We could have also substituted 2n with 2^n or 4n, 8n, ... any sequence of numbers such that they are all even. The proof will work the same.\nSince there are infinitely many integers larger than or equal to 2, there are infinitely many distinct pairs $(a,b)$.\nWe only need to prove:\n$a^b+b^a \\equiv 0 \\pmod{a+b}$\nWe can expand $a^b + b^a = (2n-1)^{2n+1} + (2n+1)^{2n-1}$ using binomial theorem. However, since $a + b = 2n-1 + 2n+1 = 4n$, all the $2n$ terms (with more than 2 powers of) when evaluated modulo $4n$ equal to 0, and thus can be omitted. We are left with the terms: $(2n+1)(2n)^1-1+(2n-1)(2n)^1+1 = 4n \\cdot 2n$, which is divisible by $4n$.\n$(2n-1)^{2n+1} + (2n+1)^{2n-1} \\equiv (2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \\cdot 2n \\equiv 0 \\pmod{4n}$\n-AlexLikeMath",
+        "Solution_6_.28Motivation_for_Solution.29": "Note that  To get rid of the $a^a+b^a$ part $\\pmod{a+b},$ we can use the sum of powers factorization. However, $a$ must be odd for us to do this. If we assume that $a$ is odd,\n\nBecause $a$ and $b$ are relatively prime, $a+b$ cannot divide $a^a.$ Thus, we have to show that there exists an integer $b$ such that for odd $a,$\n\nSuppose that $a=2n-1.$ To keep the powers small, we try $b=2n+k$ for small values of $k.$ We can find that $b=2n$ does not work. $b=2n+1$ works though, as $a+b=4n$ and\n\nBecause $a$ is odd, $b=a+2$ is relatively prime to $a.$ Thus,  is a solution for positive $n\\ge 2.$ There are infinitely many possible values for $n,$ so the proof is complete. $\\blacksquare$\n~BS2012\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Consider the equation\n(a) Prove that there are infinitely many pairs $(x,y)$ of positive integers satisfying the equation.\n(b) Describe all pairs $(x,y)$ of positive integers satisfying the equation.",
+        "Solution_1": "We have $(3x^3+xy^2)(yx^2+3y^3)=(x-y)^7$, which can be expressed as $xy(3x^2+y^2)(x^2+3y^2)=(x-y)^7$. At this point, we think of substitution. A substitution of form $a=x+y, b=x-y$ is slightly derailed by the leftover x and y terms, so instead, seeing the $xy$ in front, we substitute $x=a+b, y=a-b$. This leaves us with:\n\nRearranging, we have $b^6(8b+1)=a^6$. To satisfy this equation in integers, $8b+1$ must obviously be a $6$th power, and further inspection shows that it must also be odd. Also, since it is a square and all odd squares are 1 mod 8, every odd sixth power gives a solution. Since the problem asks for positive integers, the pair $(a,b)=(0,0)$ does not work. We go to the next highest odd $6$th power, $3^6$ or $729$. In this case, $b=91$, so the LHS is $91^6\\cdot3^6=273^6$, so $a=273$. Using the original substitution yields $(x,y)=(364,182)$ as the first solution. We have shown part a by showing that there are an infinite number of positive integer solutions for $(a,b)$, which can then be manipulated into solutions for $(x,y)$.\nTo solve part (b), we look back at the original method of generating solutions. Define $a_n$ and $b_n$ to be the pair representing the $n$th solution. Since the $n$th odd number is $2n+1$, $b_n=\\frac{(2n+1)^6-1}{8}$. It follows that $a_n=(2n+1)b_n=\\frac{(2n+1)^7-(2n+1)}{8}$. From our original substitution, $(x,y)=\\left(\\frac{(2n+1)^7+(2n+1)^6-2n-2}{8},\\frac{(2n+1)^7-(2n+1)^6-2n}{8}\\right)$.",
+        "Solution_2_.28and_motivation.29": "First, we shall prove a lemma:\nLEMMA:\nPROOF: Expanding and simplifying the right side, we find that \nwhich proves our lemma.\nNow, we have that \nRearranging and getting rid of the denominator, we have that \nFactoring, we have Dividing both sides, we have \nNow, since the LHS is the 6th power of a rational number, and the RHS is an integer, the RHS must be a perfect 6th power. Define $a=\\frac{x+y}{x-y}$. By inspection, $a$ must be a positive odd integer satistisfying $a \\geq 3$. We also have  Now, we can solve for $x$ and $y$ in terms of $a$:\n$x-y=\\frac{a^6-1}{4}$ and $x+y=a(x-y)=\\frac{a(a^6-1)}{4}$.\nNow we have: \nand it is trivial to check that this parameterization works for all such $a$ (to keep $x$ and $y$ integral), which implies part (a).\nMOTIVATION FOR LEMMA:\nI expanded the LHS, noticed the coefficients were $(3,10,3)$, and immediately thought of binomial coefficients. Looking at Pascal's triangle, it was then easy to find and prove the lemma.\n-sunfishho",
+        "Solution_where": "So named because it is a mix of solutions 1 and 2 but differs in other aspects.\nAfter fruitless searching, let $x+y=a$, $x-y=b$. Clearly $a,b>0$. Then,\n\nChange the LHS of the original equation to\n\nand change the RHS to $b^7$. Therefore $\\biggl(\\dfrac{a}{b}\\biggr)^6=4b+1$. Let $n=\\frac{a}{b}$, and note that $n$ is an integer. Therefore $b=\\frac{n^6-1}{4},a=bn=n\\biggl(\\dfrac{n^6-1}{4}\\biggr)$. Because $n^6\\equiv1\\pmod{4}$, $n$ is odd, and thus is greater than $1$, since $b>0$. Therefore, substituting for $x$ and $y$, we get:",
+        "Solution_4": "Part a:\nLet $a = 1 + \\frac{1}{n}$, where $n$ is a positive integer. We will show that there is precisely one solution $(x, y)$ to the equation such that $x = ay$.\nIf $x = ay$, we have\n\nThe numerator is a multiple of $\\frac{1}{n^5}$, so $y$ is an integer multiple of $n^2$. Thus, $x = \\frac{n+1}{n}\\cdot y$ is also an integer, and we conclude that this pair $(x, y)$ satisfies the system of equations. Because this works for any positive integer $n$, we conclude that there are infinitely many solutions to the equation.\nPart b:\nWe will now prove that these are the only possible solutions. Suppose the contrary, that there are solutions with a different ratio between $x$ and $y$. As before, set $a = 1 + \\frac{1}{n}$, but this time $n = \\frac{p}{q}$ in simplest terms. Then,\n\nFor this to be rational, $q$ must divide the expression in brackets and thus must divide $16p^5$. However, $p$ and $q$ are relatively prime, so $q$ must divide $16$, and $p$ is therefore odd.\nNext, set $q = 2^k$, where $k$ is a positive integer between $1$ and $4$, inclusive. We have\n\nThe sum in the brackets must be divisible by $2^{7k}$, which is a power of $2$ greater than or equal to $2^7$. Let $z$ be this sum. Then,\n\nwhere $v$ represents $15\\cdot 2^{2+2k}p^3$ and $w$ represents $15\\cdot 2^{4k}p$. Therefore, we must have\n\nFor $k > 1$, this is equal to $0 \\pmod{32}$, so we conclude that $k = 1$ (and thus $q = 2$). Then,\n\nFor this to be integral, the expression in brackets must be a multiple of $8$. However, there are three terms that are odd in this expression, $p^5$, $15p^3$, and $15p$. Thus, we have a contradiction, and we conclude that the only solutions $(x, y)$ are of the form stated above. $\\square$\n~mathboy100\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "($*$) Let $ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$; let lines $PB$ and $CA$ intersect at $E$; and let lines $PC$ and $AB$ intersect at $F$. Prove that the area of triangle $DEF$ is twice that of triangle $ABC$.\n",
+        "Solution_.28No_Bash.29": "Extend $DP$ to hit $EF$ at $K$. Then note that $[DEF]\\cdot\\frac{AK}{DK}\\cdot\\frac{AB}{AF}\\cdot\\frac{AC}{AE}=[ABC].$ Letting $BF=x$ and $PF=y$, we have that $\\frac{x+AB}y=\\frac{y+PC}x=\\frac{AC}{BP}.$ Solving and simplifying using LoC on $\\triangle BPC$ gives $\\frac{AB}{AF}=\\frac{PC}{PB+PC}.$ Similarly, $\\frac{AC}{AE}=\\frac{PB}{PB+PC}.$\n\nNow we find $\\frac{AK}{DK}.$ Note that $\\frac{AD}{DP}=\\frac{AD}{BD}\\cdot\\frac{BD}{DP}=\\frac{AC}{PB}\\cdot\\frac{AB}{PC}=\\frac{AB^2}{PB\\cdot PC}.$ Now let $E'=DE\\cap AF$ and $F'=DF\\cap AE$. Then by an area/concurrence theorem, we have that $\\frac{DK}{AK}+\\frac{DE'}{EE'}+\\frac{DF'}{FF'}=1,$ or $\\frac{DK}{AK}+(1-\\frac{DP}{AP}-\\frac{DC}{BC})+(1-\\frac{DP}{AP}-\\frac{BD}{BC})=1.$ Thus we have that $\\frac{DK}{AK}=2\\cdot\\frac{DP}{AP}.$\n\nManipulating these gives $\\frac{AK}{DK}=\\frac{(PB+PC)^2}{2\\cdot PB\\cdot PC}.$ Thus $\\frac{AK}{DK}\\cdot\\frac{AB}{AF}\\cdot\\frac{AC}{AE}=\\frac12,$ and we are done.\n~cocohearts",
+        "Solution_1": "WLOG, let $AB = 1$. Let $[ABD] = X, [ACD] = Y$, and  $\\angle BAD = \\theta$. After some angle chasing, we find that $\\angle BCF \\cong \\angle BEC \\cong \\theta$ and $\\angle FBC \\cong \\angle BCE \\cong 120^{\\circ}$. Therefore, $\\triangle FBC$ ~ $\\triangle BCE$.\nLemma 1: If $BF = k$, then $CE = \\frac 1k$.\nThis lemma results directly from the fact that $\\triangle FBC$ ~ $\\triangle BCE$;  $\\frac{BF}{BC} = \\frac{BF}{1} = \\frac{BC}{CE} = \\frac{1}{CE}$, or $CE = \\frac{1}{BF}$.\nLemma 2: $[AEF] = (k+\\frac 1k + 2)(X+Y)$.\nWe see that $[AEF] = (X+Y) \\frac{[AEF]}{[ABC]} = (k+1)(1+\\frac 1k)(X+Y) = (k + \\frac 1k + 2)(X+Y)$, as desired.\nLemma 3: $\\frac{X}{Y} = k$.\nWe see that \n\nHowever, after some angle chasing and by the Law of Sines in $\\triangle BCF$, we have $\\frac{k}{\\sin(\\theta)} = \\frac{1}{\\sin(60^{\\circ} - \\theta)}$, or $k = \\frac{\\sin(\\theta)}{\\sin (60^{\\circ} - \\theta)}$, which implies the result.\nBy the area lemma, we have $[BDF] = kX$ and $[CDF] = \\frac{Y}{k}$.\nWe see that $[DEF] = [AEF] - [ABC] - [BDF] - [DCE] = Xk + Yk + \\frac Xk + \\frac Yk + 2X + 2Y - X - Y - Xk - \\frac Yk = X + Y + \\frac Xk + yk$. Thus, it suffices to show that $X + Y + \\frac Xk + Yk = 2X + 2Y$, or $\\frac Xk + Yk = X + Y$. Rearranging, we find this to be equivalent to $\\frac XY = k$, which is Lemma 3, so the result has been proven.",
+        "Solution_2": "We will use barycentric coordinates and vectors. Let $\\vec{X}$ be the position vector of a point $X.$ The point $(x, y, z)$ in barycentric coordinates denotes the point $x\\vec{A} + y\\vec{B} + z\\vec{C}.$ For all points in the plane of $\\triangle{ABC},$ we have $x + y + z = 1.$ It is clear that $A = (1, 0, 0)$; $B = (0, 1, 0)$; and $C = (0, 0, 1).$\nDefine the point $P$ as $P = \\left(x_P, y_P, z_P\\right).$ The fact that $P$ lies on the circumcircle of $\\triangle{ABC}$ gives us $x^2_P + y^2_P + z^2_P = 1.$ This, along with the condition $x_P + y_P + z_P = 1$ inherent to barycentric coordinates, gives us $x_Py_P + y_Pz_P + z_Px_P = 0.$\nWe can write the equations of the following lines:\nWe can then solve for the points $D, E, F$:\nThe area of an arbitrary triangle $XYZ$ is:\nTo calculate $[DEF],$ we wish to compute $(\\vec{D}\\times\\vec{E}) + (\\vec{E}\\times\\vec{F}) + (\\vec{F}\\times\\vec{D}).$ After a lot of computation, we obtain the following:\nEvaluating the denominator,\nSince $x_P + y_P + z_P = 1$ and $x_Py_P + y_Pz_P + z_Px_P = 0,$ it follows that:\nWe thus conclude that:\nFrom this, it follows that $[DEF] = 2[ABC],$ and we are done.\n",
+        "Solution_3": "We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim  \\triangle{CFB}$ and $\\triangle{BCP} \\sim  \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$.\nThe slope of $FE$ is \n\nIt is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace,\n\n\n\n\n\nSo $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D\nBy Mathdummy.",
+        "Solution_4_Without_the_nasty_computations": "Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths.\nLet $BP = x$ and $CP = y$. Then,\nFrom Law of Cosine, $BC^2 = x^2 + y^2 + xy$.\nFrom Ptolemy's theorem, $AP BC = x AC + y AB$, so $AP = x + y$.\nLemma 1: In Triangle ABC with side lengths $a,b,c$ and $\\angle A =120^o$, the length of the angle bisector of $A$ is \n\nThis can be easily proved with Stewart's and Law of Cosine.\nUsing Lemma 1, we have \n\n\n\nPlug in $AP=x+y$, we get: \n\n\n\nThen\nBy Mathdummy."
+    },
+    {
+        "Problem": "Are there any triples $(a,b,c)$ of positive integers such that $(a-2)(b-2)(c-2) + 12$ is prime that properly divides the positive number $a^2 + b^2 + c^2 + abc - 2017$?",
+        "Solution_1": "The answer is no. Substitute $x=a-2,y=b-2,z=c-2$. This means that $x,y,z\\geq -1$. Then  It is given in the problem that this is positive. Now, suppose for the sake of contradiction that $xyz+12$ is a prime. Clearly $x,y,z\\neq 0$. Then we have  is an integer greater than or equal to $1$. This also implies that $x+y+z > 41$. Since $xyz+12$ is prime, we must have  Additionally, $x, y, z$ must be odd, so that $xyz+12$ is odd while $x+y+z-41,x+y+z+49$ are even. So, if  we must have  Now suppose WLOG that $x=-1$ and $y,z>0$. Then we must have $yz\\leq 10$, impossible since $x+y+z>41$. Again, suppose that $x,y=-1$ and $z>0$. Then we must have  and since in this case we must have $z>43$, this is also impossible.\nThen the final case is when $x,y,z$ are positive odd numbers. Note that if $xyz>x+y+z$ for positive integers $x,y,z$, then $abc>a+b+c$ for positive integers $a,b,c$ where $a>x,b>y,c>z$. Then we only need to prove the case where $x+y+z=43$, since $x+y+z$ is odd. Then one of  is true, implying that $xyz\\leq -11$ or $xyz\\leq 34$. But if $x+y+z=43$, then $xyz$ is minimized when $x=1,y=1,z=41$, so that $xyz\\geq 41$. This is a contradiction, so we are done."
+    },
+    {
+        "Problem": "Let $O$ and $H$ be the circumcenter and the orthocenter of an acute triangle $ABC$. Points $M$ and $D$ lie on side $BC$ such that $BM = CM$ and $\\angle BAD = \\angle CAD$. Ray $MO$ intersects the circumcircle of triangle $BHC$ in point $N$. Prove that $\\angle ADO = \\angle HAN$.",
+        "Solution_1": "It's well known that the reflection of $H$ across $\\overline{BC}$, $H'$, lies on $(ABC)$. Then $(BHC)$ is just the reflection of $(BH'C)$ across $\\overline{BC}$, which is equivalent to the reflection of $(ABC)$ across $\\overline{BC}$. Reflect points $A$ and $N$ across $\\overline{BC}$ to points $A'$ and $N'$, respectively. Then $N'$ is the midpoint of minor arc $\\overarc{BC}$, so $A, D, N'$ are collinear in that order. It suffices to show that $\\angle AA'N'=\\angle ADO$.\nClaim: $\\triangle AA'N' \\sim \\triangle ADO$. The proof easily follows.\nProof: Note that $\\angle BAA'=\\angle CAO=90^{\\circ}-\\angle ABC$. Then we have $\\angle A'AN'=\\angle BAD-\\angle BAA'=\\angle CAD-\\angle CAO=\\angle DAO$. So, it suffices to show that  Notice that $\\triangle ABA' \\sim \\triangle AOC$, so that  Therefore, it suffices to show that  But it is easy to show that $\\triangle BAN'\\sim \\triangle DAC$, implying the result. $\\blacksquare$",
+        "Solution_2": "\nSuppose ray $OM$ intersects the circumcircle of $BHC$ at $N'$, and let the foot of the A-altitude of $ABC$ be $E$. Note that $\\angle BHE=90-\\angle HBE=90-90+\\angle C=\\angle C$. Likewise, $\\angle CHE=\\angle B$. So, $\\angle BHC=\\angle BHE+\\angle CHE=\\angle B+\\angle C$.\n$BHCN'$ is cyclic, so $\\angle BN'C=180-\\angle BHC=180-\\angle B-\\angle C=\\angle A$. Also, $\\angle BAC=\\angle A$. These two angles are on different circles and have the same measure, but they point to the same line $BC$! Hence, the two circles must be congruent. (This is also a well-known result)\nWe know, since $M$ is the midpoint of $BC$, that $OM$ is perpendicular to $BC$. $AH$ is also perpendicular to $BC$, so the two lines are parallel. $AN$ is a transversal, so $\\angle HAN=\\angle ANO$. We wish to prove that $\\angle ANO=\\angle ADO$, which is equivalent to $AOND$ being cyclic.\nNow, assume that ray $OM$ intersects the circumcircle of $ABC$ at a point $P$. Point $P$ must be the midpoint of $\\stackrel{\\frown}{BC}$. Also, since $AD$ is an angle bisector, it must also hit the circle at the point $P$. The two circles are congruent, which implies $MN=MP\\implies ND=DP\\implies$ NDP is isosceles. Angle ADN is an exterior angle, so $\\angle ADN=\\angle DNP+\\angle DPO=2\\angle DPO$. \nAssume WLOG that $\\angle B>\\angle C$. So, $\\angle DPO=\\angle APO=\\frac{\\angle B+\\angle C}{2}-\\angle C=\\frac{\\angle B-\\angle C}{2}$.\nIn addition, $\\angle AON=\\angle AOP=\\angle AOB+\\angle BOP=2\\angle C+\\angle A$. Combining these two equations, $\\angle AON+\\angle ADN=\\angle B-\\angle C+2\\angle C+\\angle A=\\angle A+\\angle B+\\angle C=180$.\nOpposite angles sum to $180$, so quadrilateral $AOND$ is cyclic, and the condition is proved.\n-william122"
+    },
+    {
+        "Problem": "Let $P_1, \\ldots, P_{2n}$ be $2n$ distinct points on the unit circle $x^2 + y^2 = 1$ other than $(1,0)$. Each point is colored either red or blue, with exactly $n$ of them red and exactly $n$ of them blue. Let $R_1, \\ldots, R_n$ be any ordering of the red points. Let $B_1$ be the nearest blue point to $R_1$ traveling counterclockwise around the circle starting from $R_1$. Then let $B_2$ be the nearest of the remaining blue points to $R_2$ traveling counterclockwise around the circle from $R_2$, and so on, until we have labeled all the blue points $B_1, \\ldots, B_n$. Show that the number of counterclockwise arcs of the form $R_i \\rightarrow B_i$ that contain the point $(1,0)$ is independent of the way we chose the ordering $R_1, \\ldots, R_n$ of the red points.",
+        "Solution": "I define a sequence to be, starting at $(1,0)$ and tracing the circle counterclockwise, and writing the color of the points in that order - either R or B. For example, possible sequences include $RB$, $RBBR$, $BBRRRB$, $BRBRRBBR$, etc.\nNote that choosing an $R_1$ is equivalent to choosing an $R$ in a sequence, and $B_1$ is defined as the $B$ closest to $R_1$ when moving rightwards. If no $B$s exist to the right of $R_1$, start from the far left. For example, if I have the above example $RBBR$, and I define the 2nd $R$ to be $R_1$, then the first $B$ will be $B_1$. Because no $R$ or $B$ can be named twice, I can simply remove $R_1$ and $B_1$ from my sequence when I choose them. I define this to be a move. Hence, a possible move sequence of $BBRRRB$ is:\n$BBR_1RRB_1\\implies B_2BRR_2\\implies B_3R_3$\nNote that, if, in a move, $B_n$ appears to the left of $R_n$, then $\\stackrel{\\frown}{R_nB_n}$ intersects $(1,0)$\nNow, I define a commencing $B$ to be a $B$ which appears to the left of all $R$s, and a terminating $R$ to be a $R$ which appears to the right of all $B$s. Let the amount of commencing $B$s be $j$, and the amount of terminating $R$s be $k$, I claim that the number of arcs which cross $(1,0)$ is constant, and it is equal to $\\text{max}(j,k)$. I will show this with induction.\nBase case is when $n=1$. In this case, there are only two possible sequences - $RB$ and $BR$. In the first case, $\\stackrel{\\frown}{R_1B_1}$ does not cross $(1, 0)$, but both $j$ and $k$ are $0$, so $\\text{max}(j,k)=0$. In the second example, $j=1$, $k=1$, so $\\text{max}(j,k)=1$. $\\stackrel{\\frown}{R_1B_1}$ crosses $(1,0)$ since $B_1$ appears to the left of $R_1$, so there is one arc which intersects. Hence, the base case is proved.\nFor the inductive step, suppose that for a positive number $n$, the number of arcs which cross $(1,0)$ is constant, and given by $\\text{max}(j, k)$ for any configuration. Now, I will show it for $n+1$.\nSuppose I first choose $R_1$ such that $B_1$ is to the right of $R_1$ in the sequence. This implies that $\\stackrel{\\frown}{R_1B_1}$ does not cross $(1,0)$. But, neither $R_1$ nor $B_1$ is a commencing $B$ or terminating $R$. These numbers remain constant, and now after this move we have a sequence of length $2n$. Hence, by assumption, the total amount of arcs is $0+\\text{max}(j,k)=\\text{max}(j,k)$.\nThus, this solution probably doesn't work.\nNow suppose that $R_1$ appears to the right of $B_1$, but $B_1$ is not a commencing $B$. This implies that there are no commencing $B$s in the series, because there are no $B$s to the left of $B_1$, so $j=0$. Note that this arc does intersect $(1,0)$, and $R_1$ must be a terminating $R$. $R_1$ must be a terminating $R$ because there are no $B$s to the right of $R_1$, or else that $B$ would be $B_1$. The $2n$ length sequence that remains has $0$ commencing $B$s and $k-1$ terminating $R$s. Hence, by assumption, the total amount of arcs is $1+\\text{max}(0,k-1)=1+k-1=k=\\text{max}(j,k)$.\nFinally, suppose that $R_1$ appears to the right of $B_1$, and $B_1$ is a commencing $B$. We know that this arc will cross $(1,0)$. Analogous to the previous case, $R_1$ is a terminating $R$, so the $2n$ length sequence which remains has $j-1$ commencing $B$s and $k-1$ terminating $R$s. Hence, by assumption, the total amount of arcs is $1+\\text{max}(j-1,k-1)=1+\\text{max}(j,k)-1=\\text{max}(j,k)$.\nThere are no more possible cases, hence the induction is complete, and the number of arcs which intersect $(1,0)$ is indeed a constant which is given by $\\text{max}(j,k)$.\n-william122",
+        "Solution_2": "Lemma: If we switch the ordering of two consecutive $R_k$, $R_{k+1}$, the number of arcs crossing $(1,0)$ stays invariant.\nProof: There are two situations. If the two arcs don't cross this is simple because the actual arcs stay the same, and only the number order of the arcs change. Otherwise, if $B_k$ is to the left of $R_{k+1}$, the two arcs will have some overlap. The order of the points counterclockwise must be RRBB or some rotation of it. Notice how in both the old and the new ordering, the arc between the second R and the first B is counted twice and the arcs between the same colors are only counted once. Thus no matter where $(1,0)$ is the number of arcs containing it will stay invariant.\nLemma 2: We can swap the place of any two points $R_i$ and $R_j$ while keeping the number of crossings invariant.\nProof: By Lemma 1 we can use consecutive arc moves to swap $R_i$ with $R_{i+1}$, then the new $R_{i+1}$ with $R_{i+2}$ and so on until it swaps with $R_j$. Now, we can do the reverse swaps to move the new $R_{j-1}$ to $R_{j-2}$, and so on into the original position of $R_i$. Since the other points were moved once in one direction and once in the opposite, all the other points are kept in place.\nNow, let $I$ be the sequence where $R_1$ through $R_n$ is numbered counterclockwise. We show that any ordering has the same number of crossings as $I$. We can swap the current place of $R_1$ with the desired position in $I$. Now we can ignore $R_1$ and swap the currrent place of $R_2$ with the place of $R_2$ in $I$, and so on until all the points are in the desired position. Thus, all orderings of $R_1$ to $R_n$ have the same number of crossings as $I$.\n-tigershark22"
+    },
+    {
+        "Problem": "For each positive integer $n$, find the number of $n$-digit positive integers that satisfy both of the following conditions:\n$\\bullet$ no two consecutive digits are equal, and\n$\\bullet$ the last digit is a prime.",
+        "Solution_1": "The answer is $\\boxed{\\frac{2}{5}\\left(9^n+(-1)^{n+1}\\right)}$.\nSuppose $a_n$ denotes the number of $n$-digit numbers that satisfy the condition. We claim $a_n=4\\cdot 9^{n-1}-a_{n-1}$, with $a_1=4$.\n$\\textit{Proof.}$ It is trivial to show that $a_1=4$. Now, we can do casework on whether or not the tens digit of the $n$-digit integer is prime. If the tens digit is prime, we can choose the digits before the units digit in $a_{n-1}$ ways and choose the units digit in $3$ ways, since it must be prime and not equal to the tens digit. Therefore, there are $3a_{n-1}$ ways in this case.\nIf the tens digit is not prime, we can use complementary counting. First, we consider the number of $(n-1)$-digit integers that do not have consecutive digits. There are $9$ ways to choose the first digit and $9$ ways to choose the remaining digits. Thus, there are $9^{n-1}$ integers that satisfy this. Therefore, the number of those $(n-1)$-digit integers whose units digit is not prime is $9^{n-1}-a_{n-1}$. It is easy to see that there are $4$ ways to choose the units digit, so there are $4\\left(9^{n-1}-a_{n-1}\\right)$ numbers in this case. It follows that \nand our claim has been proven.\nThen, we can use induction to show that $a_n=\\frac{2}{5}\\left(9^n+(-1)^{n+1}\\right)$. It is easy to see that our base case is true, as $a_1=4$. Then, \nwhich is equal to \nas desired. $\\square$\nSolution by TheUltimate123.",
+        "Solution_2": "The answer is $\\boxed{\\frac{2}{5}\\left(9^n-(-1)^{n}\\right)}$.\nAs in the first solution, let $a_n$ denote the number of $n$-digit numbers that satisfy the condition.  Clearly $a_1 = 4$ and $a_2 = 32$.  We claim that for $n \\geq 3$ we have the recurrence $a_n = 8a_{n-1} + 9a_{n-2}$.\nTo prove this, we split the $n$-digit numbers satisfying the conditions into cases depending on whether or not the second digit is $0$.   If the second digit is nonzero, our number is formed from one of the $a_{n-1}$ numbers with one fewer digit satisfying the conditions, times $8$ possible choices of adding a digit to the left.  If the second digit is zero, our number is formed from one of the $a_{n-2}$ numbers with two fewer digits satisfying the conditions, times $9$ possible choices of adding $0$ and then any nonzero digit to the left.  This proves our claim.\nThis gives us a linear three-term recurrence.  It is well-known that its solution is of the form $c_1r_1^n + c_2r_2^n$, where the $c_i$ are constants to be determined from the initial conditions $a_1$ and $a_2$, and $r_1$ and $r_2$ are the roots of the corresponding quadratic equation $x^2 = 8x + 9$.  We solve and get $(x-9)(x+1) = 0$, so our roots are $r_1 = 9$ and $r_2 = -1$.  Now, we use our conditions $a_1 = 4$ and $a_2 = 32$ to derive the system of linear equations\n\nSolving this system yields $c_1 = \\frac{2}{5}$ and $c_2 = -\\frac{2}{5}$, and we are done.\nSolution by putnam-lowell.",
+        "Solution_3": "The answer is $\\boxed{\\frac{2}{5}\\left(9^n-(-1)^{n}\\right)}$.\nAs in the first solution, let $a_n$ denote the number of $n$-digit numbers that satisfy the condition. Clearly $a_1 = 4$ and $a_2 = 32$.\nFrom here, we proceed by complementary counting. We first count the total amount of numbers that satisfy both bullets but that may have zero as its first digit. The units digit can be one of four primes, and each digit to the left will have 9 choices (any digit but the one that was just used). Then the total for this group of numbers is just $4*9^{n-1}$.\nNow we must subtract all numbers in the above group that have 0 as its first digit. This is just $a_{n-1}$ because for each number in the first group beginning with a zero, we could take away the zero, leaving us with a number that works for the case $n-1$ (this is true because the next number would not be zero, or the consecutive digit requirement would be violated). Then we have the recursive formula $a_n = 4*9^{n-1}-a_{n-1}$.\nTo simplify this, we take a look at the first few terms. We see that\n\nWe see a pattern where $a_n = 4(9^{n-1}-9^{n-2}+9^{n+3}...+(-1)^{n-1}9^0)$ and we can prove that it holds for all $a_n$ because subtracting $a_{n-1}$ from $4(9^n)$ is the same thing as reversing all previous signs of the preceding powers of 9. This constitutes an alternating pattern, which we can calculate as a geometric series. The first term is $4(9^{n-1})$ and the common ratio is $-\\frac{1}{9}$, so\n\nWe are done.\nSolution by aopsal",
+        "Solution_4": "Let $f(n)$ denote the number of $n$-digit positive integers satisfying the conditions listed in the problem.\nClaim 1: $f(n) = 8f(n-1) + 9f(n-2).$\nTo prove this, let $k$ be the leftmost digit of the $n$-digit positive integer. When $k$ ranges from $1$ to $9,$ the allowable second-to-leftmost digits is the set $\\{0, 1, 2, …, 9\\},$ with $k$ excluded. Note that since $1, 2, …, 9$ are all repeated $8$ times and using our definition of $f(n)$ (since $n$-digit positive integers must begin with a positive digit), we have proved that $8f(n-1)$ part. Since $0$ is repeated $9$ times, and since digits that are allowed to be adjacent to the $0$ (the third-to-leftmost digit of our positive integer) are integers ranging from $1$ to $9,$ we have proved the $9f(n-2)$ part similarly. We can see that $\\frac{f(1)}{4} = 1$ and $\\frac{f(2)}{4} = 8,$ by simple computation. These will be our base cases.\nNow, let $s(n) = \\frac{f(n)}{4}.$\nClaim 2: $s(n+1) = 9s(n) + (-1)^n$\nWe can prove this using induction. Let $s(a) = k$ and $s(a+1) = 9k - 1.$ By plugging $k$ and $9k - 1$ into our formula $s(a+2) = 8s(a+1) + 9s(a),$ we obtain Now, by plugging $9k - 1$ and $81k - 8$ into the same formula, we obtain $s(a+3) = 8(81k - 8) + 9(9k - 1) = 729k - 73 = 9s(a+2) - 1.$ Since with $a+2$ and $a+3$ we’re back where we started (this is since $s(a+3) = 9s(a+2) - 1$ just like $s(a+1) = 9s(a) - 1$), and using our base cases of $s(1) = 1$ and $s(2) = 8,$ our induction is complete.\nClaim 3: $s(n) = \\frac{9^n + (-1)^{n+1}}{10}$\nWe can prove that by simply using our base case of $s(1) = 1$ and repeatedly using the distributive property to obtain further values of $s(n).$ For the case where $n$ is even, we can manipulate the summation into $\\sum_{i=0}^{\\frac{n}{2} - 1} 8 \\cdot 9^{2i}$ which equals $\\frac{8 \\cdot 9^{n} - 1}{9^2 - 1} = \\frac{9^n - 1}{10},$ by geometric series. For the case where $n$ is odd, we can manipulate the summation into $1 + \\sum_{i=0}^{\\frac{n-1}{2} - 1} 8 \\cdot 9^{2i + 1}$ which equals Thus, we have proved Claim 3 for both odd and even $n.$\nSince $f(n) = 4s(n),$ our answer is $4 \\cdot \\frac{9^n + (-1)^{n+1}}{10} = \\boxed{ \\frac{2}{5} (9^n + (-1)^{n+1})}.$\n-fidgetboss_4000\n",
+        "Solution_5": "Let $a_n$ be the number of $n$-digit numbers that satisfy the condition that most-left digit different than $0$. Let $b_n$ be the number of $n$-digit numbers that satisfy the condition that most-left digit is $0$. We can write that\n\nand so we find that\n\n$x+1$ is the characteristic polynomial of the homogeneous recurrence relation $a_n + a_{n-1}  = 0$. Then $x=-1$ is a root. Also, by the non-homogeneous part of $4\\cdot 9^{n-1}$, we understand that $x=9$ is another root of the recurrence relation $(1)$ with the complementary solution idea. Therefore, $(a_n)$ sequence in the form \nWe can calculate that $C_1 = \\dfrac{2}{5}, C_2 = - \\dfrac{2}{5}$ with using $a_1=4, a_2=32$. Thus we get:\nLokman GÖKÇE\n"
+    },
+    {
+        "Problem": "Let $a,b,c$ be positive real numbers such that $a+b+c=4\\sqrt[3]{abc}$. Prove that",
+        "Solution_1": "WLOG let $a \\leq b \\leq c$. Add $2(ab+bc+ca)$ to both sides of the inequality and factor to get: \nBy substituting $a+b+c=4\\sqrt[3]{abc}$, we get:\n\nThe last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.",
+        "Solution_2": "WLOG let $a \\geq b \\geq c$. Note that the equations are homogeneous, so WLOG let $c=1$.\nThus, the inequality now becomes $2ab + 2a + 2b + 4 \\geq a^2 + b^2 + 1$, which simplifies to $2(a+b) + 3 \\geq (a-b)^2$.\nNow we will use the condition. Letting $x=a+b$ and $y=a-b$, we have\n$x+1=\\sqrt[3]{16(x^2-y^2)} \\implies 16y^2=-x^3+13x^2-3x-1$.\nPlugging this into the inequality, we have $2x+3 \\geq \\frac{1}{16}(-x^3+13x^2-3x-1) \\implies x^3-13x^2+35x+49 = (x-7)^2(x+1) \\geq 0$, which is true since $x \\geq 0$.\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.",
+        "Solution_3": "https://wiki-images.artofproblemsolving.com//6/69/IMG_8946.jpg"
+    },
+    {
+        "Problem": "($*$) Let $ABCD$ be a quadrilateral inscribed in circle $\\omega$ with $\\overline{AC} \\perp \\overline{BD}$. Let $E$ and $F$ be the reflections of $D$ over lines $BA$ and $BC$, respectively, and let $P$ be the intersection of lines $BD$ and $EF$. Suppose that the circumcircle of $\\triangle EPD$ meets $\\omega$ at $D$ and $Q$, and the circumcircle of $\\triangle FPD$ meets $\\omega$ at $D$ and $R$. Show that $EQ = FR$.\n",
+        "Solution_1": "First we have that $BE=BD=BF$ by the definition of a reflection. Let $\\angle DEB = \\alpha$ and $\\angle DFB = \\beta.$ Since $\\triangle DBE$ is isosceles we have $\\angle BDE = \\alpha.$ Also, we see that $\\angle BDE = \\angle CAB = \\angle CDB = \\alpha,$ using similar triangles and the property of cyclic quadrilaterals. Similarly,  Now, from $BE=BD=BF$ we know that $B$ is the circumcenter of $\\triangle DEF.$ Using the properties of the circumcenter and some elementary angle chasing, we find that\nNow, we claim that $Q$ is the intersection of ray $\\overrightarrow{EB}$ and the circumcircle of $ABCD.$ To prove this, we just need to show that $DEPQ$ is cyclic by this definition of $Q.$ We have that  We also have from before that  so $\\angle DQE=\\angle DPE$ and this proves the claim.\nWe can use a similar proof to show that $F, B, R$ are collinear.\nNow, $DP$ is the radical axis of the circumcircles of $\\triangle EDP$ and $\\triangle FDP.$ Since $B$ lies on $DP,$ and $E, Q$ lie on the circumcircle of $\\triangle EPD$ and $F, R$ lie on the circumcircle of $\\triangle FPD,$ we have that  However, $BF=BE,$ so $BR=BQ.$ Since $E, B, Q$ are collinear and so are $F, B, R$ we can add these $2$ equations to get  which completes the proof.\n~nukelauncher\n(Monday G. Fern)",
+        "Solution_2": "We begin with the following claims.\nClaim. $B$ is the circumcenter of $\\triangle DEF$.\nProof. By reflection $BD=BE=BF$.\nClaim. $A$ is the circumcenter of $\\triangle EPD$.\nProof. First, we have\nThen\nThen $2\\angle ADE = 180^{\\circ}-2(180^{\\circ}-\\angle DPE) \\implies \\angle DAE = 2(180^{\\circ}-\\angle DPE)$. This is enough to imply what we desire. \\newline\nClaim. $C$ is the circumcenter of $\\triangle FPD$.\nProof. Similar to above.\nClaim. $E, B, Q$ are collinear.\nProof. We have\nClaim.$F, B, R$ are collinear.\nProof. Similar to above.\nSince $DEPQ$ is cyclic, $\\triangle EBP \\sim \\triangle DBQ$. However, $BE=BD$, so $BP=BQ$. Similarly, $BP=BR$. Finishing, we have , as desired. $\\blacksquare$ ~MSC\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem_4": "Triangle $ABC$ is inscribed in a circle of radius $2$ with $\\angle ABC \\geq 90^\\circ$, and $x$ is a real number satisfying the equation $x^4 + ax^3 + bx^2 + cx + 1 = 0$, where $a=BC,b=CA,c=AB$. Find all possible values of $x$.\n",
+        "Solution": "Notice that\n\nThus, if $b > \\frac{a^2}{4} + \\frac{c^2}{4},$ then the expression above is strictly greater than $0$ for all $x,$ meaning that $x$ cannot satisfy the equation $x^4 + ax^3 + bx^2 + cx + 1 = 0.$ It follows that $b\\le\\frac{a^2}{4} + \\frac{c^2}{4}.$\nSince $\\angle ABC\\ge 90^{\\circ},$ we have $b^2\\ge a^2 + c^2.$ From this and the above we have $4b\\le a^2 + c^2\\le b^2,$ so $4b\\le b^2.$ This is true for positive values of $b$ if and only if $b\\ge 4.$ However, since $\\triangle ABC$ is inscribed in a circle of radius $2,$ all of its side lengths must be at most the diameter of the circle, so $b\\le 4.$ It follows that $b=4.$\nWe know that $4b\\le a^2 + c^2\\le b^2.$ Since $4b = b^2 = 16,$ we have $4b = a^2 + c^2 = b^2 = 16.$\nThe equation $x^4 + ax^3 + bx^2 + cx + 1 = 0$ can be rewritten as $\\left(x^2 + \\frac{a}{2}x\\right)^2 + \\left(\\frac{c}{2}x + 1\\right)^2 = 0,$ since $b = \\frac{a^2}{4} + \\frac{c^2}{4}.$ This has a real solution if and only if the two separate terms have zeroes in common. The zeroes of $\\left(x^2 + \\frac{a}{2}x\\right)^2$ are $0$ and $-\\frac{a}{2},$ and the zero of $\\left(\\frac{c}{2}x + 1\\right)^2 = 0$ is $-\\frac{2}{c}.$ Clearly we cannot have $0=-\\frac{2}{c},$ so the only other possibility is $-\\frac{a}{2} = -\\frac{2}{c},$ which means that $ac = 4.$\nWe have a system of equations: $ac = 4$ and $a^2 + c^2 = 16.$ Solving this system gives $(a, c) = \\left(\\sqrt{6}+\\sqrt{2}, \\sqrt{6}-\\sqrt{2}\\right), \\left(\\sqrt{6}-\\sqrt{2}, \\sqrt{6}+\\sqrt{2}\\right).$ Each of these gives solutions for $x$ as $-\\frac{\\sqrt{6}+\\sqrt{2}}{2}$ and $-\\frac{\\sqrt{6}-\\sqrt{2}}{2},$ respectively. Now that we know that any valid value of $x$ must be one of these two, we will verify that both of these values of $x$ are valid.\nFirst, consider a right triangle $ABC,$ inscribed in a circle of radius $2,$ with side lengths $a = \\sqrt{6}+\\sqrt{2}, b = 4, c = \\sqrt{6}-\\sqrt{2}.$ This generates the polynomial equation\n\nThis is satisfied by $x=-\\frac{\\sqrt{6}+\\sqrt{2}}{2}.$\nSecond, consider a right triangle $ABC,$ inscribed in a circle of radius $2,$ with side lengths $a = \\sqrt{6}-\\sqrt{2}, b = 4, c = \\sqrt{6}+\\sqrt{2}.$ This generates the polynomial equation\n\nThis is satisfied by $x=-\\frac{\\sqrt{6}-\\sqrt{2}}{2}.$\nIt follows that the possible values of $x$ are $-\\frac{\\sqrt{6}+\\sqrt{2}}{2}$ and $-\\frac{\\sqrt{6}-\\sqrt{2}}{2}.$\nFun fact: these solutions correspond to a $15$-$75$-$90$ triangle.\n(sujaykazi)\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem_5": "Let $p$ be a prime, and let $a_1, \\dots, a_p$ be integers. Show that there exists an integer $k$ such that the numbers produce at least $\\tfrac{1}{2} p$ distinct remainders upon division by $p$.\n",
+        "Solution": "$\\textbf{Lemma: }$ For fixed $i\\neq j,$ where $i, j\\in\\{1, 2, ..., p\\},$ the statement $a_i + ik\\equiv a_j + jk\\text{ (mod } p\\text{)}$ holds for exactly one $k\\in {1, 2, ..., p}.$\n$\\textbf{Proof: }$ Notice that the left side minus the right side is congruent to $(a_i - a_j) + (i - j)k$ modulo $p.$ For this difference to equal $0,$ there is a unique solution for $k$ modulo $p$ given by $k\\equiv (a_j - a_i)(i - j)^{-1}\\text{ (mod } p\\text{)},$ where we have used the fact that every nonzero residue modulo $p$ has a unique multiplicative inverse. Therefore, there is exactly one $k\\in {1, 2, ..., p}$ that satisfies $a_i + ik\\equiv a_j + jk\\text{ (mod } p\\text{)}$ for any fixed $i\\neq j. \\textbf{ End Lemma}$\nSuppose that you have $p$ graphs $G_1, G_2, ..., G_p,$ and graph $G_k$ consists of the vertices $(i, k)$ for all $1\\le i\\le p.$ Within any graph $G_k,$ vertices $(i_1, k)$ and $(i_2, k)$ are connected by an edge if and only if $a_{i_1} + i_1k\\equiv a_{i_2} + i_2k\\text{ (mod } p\\text{)}.$ Notice that the number of disconnected components of any graph $G_k$ equals the number of distinct remainders when divided by $p$ given by the numbers $a_1 + k, a_2 + 2k, ..., a_p + pk.$\nThese $p$ graphs together have exactly one edge for every unordered pair of elements of $\\{1, 2, ..., p\\},$ so they have a total of exactly $\\frac{p(p-1)}{2}$ edges. Therefore, there exists at least one graph $G_k$ that has strictly fewer than $\\frac{p}{2}$ edges, meaning that it has more than $\\frac{p}{2}$ disconnected components. Therefore, the collection of numbers $\\{a_i + ik: 1\\le i\\le p\\}$ for this particular value of $k$ has at least $\\frac{p}{2}$ distinct remainders modulo $p.$ This completes the proof.\n(sujaykazi)\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem_6": "Karl starts with $n$ cards labeled $1,2,3,\\dots,n$ lined up in a random order on his desk. He calls a pair $(a,b)$ of these cards swapped if $a>b$ and the card labeled $a$ is to the left of the card labeled $b$. For instance, in the sequence of cards $3,1,4,2$, there are three swapped pairs of cards, $(3,1)$, $(3,2)$, and $(4,2)$.\nHe picks up the card labeled 1 and inserts it back into the sequence in the opposite position: if the card labeled 1 had $i$ card to its left, then it now has $i$ cards to its right. He then picks up the card labeled $2$ and reinserts it in the same manner, and so on until he has picked up and put back each of the cards $1,2,\\dots,n$ exactly once in that order. (For example, the process starting at $3,1,4,2$ would be $3,1,4,2\\to 3,4,1,2\\to 2,3,4,1\\to 2,4,3,1\\to 2,3,4,1$.)\nShow that no matter what lineup of cards Karl started with, his final lineup has the same number of swapped pairs as the starting lineup.",
+        "Solution": "Note that in this solution, the term \"inversions\" is used synonymously with \"swapped pairs.\"\nWe define a new process $P'$ where, when re-inserting card $i$, we additionally change its label from $i$ to $n+i$. For example, an example of $P'$ also starting with $3142$ is:\n\nNote that now, each step of $P'$ preserves the number of inversions. Moreover, the final configuration of $P'$ is the same as the final configuration of $P$ with all cards incremented by $n$, and thus, of course, has the same number of inversions.\n~ v_enhance (clarified by integralarefun)\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "There are $a+b$ bowls arranged in a row, numbered $1$ through $a+b$, where $a$ and $b$ are given positive integers. Initially, each of the first $a$ bowls contains an apple, and each of the last $b$ bowls contains a pear.\nA legal move consists of moving an apple from bowl $i$ to bowl $i+1$ and a pear from bowl $j$ to bowl $j-1$, provided that the difference $i-j$ is even. We permit multiple fruits in the same bowl at the same time. The goal is to end up with the first $b$ bowls each containing a pear and the last $a$ bowls each containing an apple. Show that this is possible if and only if the product $ab$ is even.",
+        "Solution": "Claim: If $a$ and $b$ are both odd, then the end goal of achieving $b$ pears followed by $a$ apples is impossible.\nProof: Let $A_1$ and $P_1$ denote the number of apples and the number of pears in odd-numbered positions, respectively. We show that $A_1 - P_1$ is invariant. Notice that if $i$ and $j$ are odd, then $A_1$ and $P_1$ both decrease by 1, as one apple and one pear are both moved from odd-numbered positions to even-numbered positions. If $i$ and $j$ are even, then $A_1$ and $P_1$ both increase by 1.\nBecause the starting configuration $aa\\ldots ab\\ldots b$ has $\\left\\lfloor \\frac{a}{2} \\right\\rfloor + 1$ odd-numbered apples and $\\left\\lfloor \\frac{b}{2} \\right\\rfloor$ odd-numbered pears, the initial value of $A_1 - P_1$ is $\\left\\lfloor \\frac{a}{2} \\right\\rfloor - \\left\\lfloor \\frac{b}{2} \\right\\rfloor + 1$. But the desired ending configuration has $\\left\\lfloor \\frac{b}{2}\\right\\rfloor + 1$ odd-numbered pears and $\\left\\lfloor \\frac{a}{2} \\right\\rfloor$ odd-numbered apples, so $A_1 - P_1 = \\left\\lfloor \\frac{a}{2} \\right\\rfloor - \\left\\lfloor \\frac{b}{2} \\right\\rfloor - 1$. As $A_1 - P_1$ is invariant, it is impossible to attain the desired ending configuration. $\\blacksquare$\nClaim: If at least one of $a$ and $b$ is even, then the end goal of achieving $b$ pears followed by $a$ apples is possible.\nProof: Without loss of generality, assume $a$ is even. If only $b$ is even, then we can number the bowls in reverse, and swap apples with pears. We use two inductive arguments to show that $(a,b) = (2,b)$ is achievable for all $b\\ge 1$, then to show that $(a,b)$ is achievable for all even $a$ and all $b\\ge 1$.\nBase case: $(a,b) = (2,1)$. Then we can easily achieve the ending configuration in two moves, by moving the apple in bowl #1 and the pear in bowl #3 so that everything is in bowl #2, then finishing the next move.\nInductive step: suppose that for $a \\ge 2$ (even) apples and $b \\ge 1$ pears, that with $a$ apples and $b$ pears, the ending configuration is achievable. We will show two things: i) achievable with $a$ apples and $b+1$ pears, and ii) achievable with $a+2$ apples and $b$ pears.\ni) Apply the process on the $a$ apples and first $b$ pairs to get a configuration $bb\\ldots ba\\ldots ab$. Now we will \"swap\" the leftmost apple with the last pear by repeatedly applying the move on just these two fruits (as $a$ is even, the difference $i-j$ is even). This gives a solution for $a$ apples and $b+1$ pears. In particular, this shows that $(a,b) = (2,b)$ for all $b\\ge 1$ is achievable.\nii) To show $(a+2, b)$ is achievable, given that $(a,b)$ is achievable, apply the process on the last $a$ apples and $b$ pears to get the configuration $aabbb\\ldots baaa\\ldots a$. Then, because we have shown that 2 apples and $b$ pears is achievable in i), we can now reverse the first two apples and $b$ pears.\nThus for $ab$ even, the desired ending configuration is achievable. $\\blacksquare$\nCombining the above two claims, we see that this is possible if and only if $ab$ is even. -scrabbler94",
+        "Solution_2": "\"If\" part:\nNote that two opposite fruits can be switched if they have even distance. If one of $a$, $b$ is odd and the other is even, then switch $1$ with $a+b$, $2$ with $a+b-1$, $3$ with $a+b-2$... until all of one fruit is switched. If both are even, then if $a>b$, then switch $1$ with $a+1$, $2$ with $a+2$, $3$ with $a+3$... until all of one fruit is switched; if $a<b$ then switch $a$ with $a+b$, $a-1$ with $a+b-1$, $a-2$ with $a+b-2$... until all of one fruit is switched. Each of these processes achieve the end goal.\n\"Only if\" part:\nAssign each apple a value of $1$ and each pear a value of $-1$. At any given point in time, let $E$ be the sum of the values of the fruit in even-numbered bowls. Because $a$ and $b$ both are odd, at the beginning there are $\\frac{a-1}{2}$ apples in even bowls and $\\frac{b+1}{2}$ pears in even bowls, so at the beginning $E=\\frac{a-b}{2}-1$. After the end goal is achieved, there are $\\frac{a+1}{2}$ apples in even bowls and $\\frac{b-1}{2}$ pears in even bowls, so after the end goal is achieved $E=\\frac{a-b}{2}+1$. However, because two opposite fruits must have the same parity to move and will be the same parity after they move, we see that $E$ is invariant, which is a contradiction; hence, it is impossible for the end goal to be reached if $ab$ is odd.\nStormersyle",
+        "Solution_3": "We define a cycle to be when $2$ fruits in bowls $i$ and $j$ are swapped with eachother, provided that bowls $i$ contains an apple, bowl $j$ contains a pear, and the difference $i-j$ is even.\nWe claim that a cycle consists of only the valid moves described in the problem statement. \nProof: We can move the apple and pear one step at a time to eventually reach the other's bowl(i.e. the apple moves to $i+1$ and the pear moves to bowl $j-1$ at the beginning) using only valid moves. By a simple parity argument(if $k$ moves have already been done, $i+k-(j-k)=i+j+2k=0+2k=2k=0 \\textbf{mod 2}$), these moves are always valid, so we are done. $\\blacksquare$\nWe now split the \"if\" part of this proof into cases based on the parities of $a$ and $b$.\nCase 1: $a$ even, $b$ odd\nIn this case we can simply cycle the fruit in bowl $k+1$ with the fruit in bowl $a+b-k$ for all integer $k$ in the range $[0,b-1]$ to achieve our desired goal. The difference $a+b-k-k-1=a+b-1$ is even, so the move is valid.\nCase 2: $b$ even, $a$ odd\nUsing the same cycle as the previous case works here.\nCase 3: $a$ even, $b$ even\nHere, we can cycle the fruits in bowls $k$ and $a+b-k$ for all integer $k$ in the range $[1,b]$. Then, we finish by cycling the fruits in bowls $a$ and $a+b$. These are both valid moves by similar parity arguments as done in Case 1.\nSince we have exhausted all possible cases for the \"if\" part of this proof, we have completed the if part. We will proceed with the \"only if\" part.\nWe let the number of apples in odd positions be $A_1$, and the number of pears in odd positions be $A_2$. Obviously $A_1-A_2$ is invariant when using valid moves only. We can now use proof by contradiction(i.e. showing that this invariant does not occur when $a, b \\equiv 1 \\text{ mod } 2$ and our desired goal is achieved) to find that we have that $0=2$, which is obviously false, so we have completed the proof for the \"only if\" part.\nWe are now done, since we have completed our proofs for the \"if\" and \"only if\" parts. $\\blacksquare$\n~smartninja2000",
+        "Solution_4": "Begin by defining a $\\textit{swap}$ as a set of $i - j$ legal moves taking an apple in $i \\rightarrow j$ and the pear in $j \\rightarrow i$. Now we begin by proving if $ab$ is even, then the final configuration is possible.\nWLOG let $b$ be even. Then there are two cases to consider. $a < b, a \\ge b$ to provide valid constructions for. In the first case, consider the apple/pear pairs $(1,b+1), (2, b+2), \\dots, (i, b+i), \\dots, (a, b+a)$ and swap them. This results in the transformation, \n\nas desired.\nNow consider the case where $a \\ge b$. Also consider the pairing, $(a-b + 1, a + 1), (a - b + 2, a+2), \\dots, (a - b + i, a+i), \\dots, (a, a + b)$ and swap them. This results in the transformation, \n\nNow just consider the leftmost $a$ bowls. There are $a-b$ apples, and then $b$ pears. We have effectively reduced $(a,b) \\rightarrow (a-b, b)$. Reducing in this fashion recursively will lead to $(a \\pmod{b}, b)$ which can be constructed with the $a<b$ case. This finishes the construction for $a \\ge b$.\nNow assume $ab,a,b$ are odd. FTSOC, let this case satisfy the claim. Then, for each apple, define it's $\\textit{position}$ as the sum of the distance from it to all other pairs. So for apple in bowl $i \\le a$ originally, the position of $i$ is $\\sum_{j = a+1}^{b} j - i .$\nEvery time a legal move is made, the position of the apple decreases by $b - 1 + 2 = b+1$ (from the $b-1$ pears not part of the move and the decrease in $2$ by the corresponding pear). Also note that there needs to be exactly $ab$ moves in order to attain the final configurations. So the difference in sum of positions of the apples at the beginning and end is $ab(b+1)$. Mathematically,\n\nA symmetric argument for the position of pears gives $b = 1$. Clearly the $a=b=1$ does not have a solution, therefore we have contradiction. $\\square$\n~ Aaryabhatta1\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Let $\\mathbb Z$ be the set of all integers. Find all pairs of integers $(a,b)$ for which there exist functions $f:\\mathbb Z\\rightarrow\\mathbb Z$ and $g:\\mathbb Z\\rightarrow\\mathbb Z$ satisfying  for all integers $x$.",
+        "Solution_1": "We claim that the answer is $|a|=|b|$.\nProof:\n$f$ and $g$ are surjective because $x+a$ and $x+b$ can take on any integral value, and by evaluating the parentheses in different order, we find $f(g(f(x)))=f(x+b)=f(x)+a$ and $g(f(g(x)))=g(x+a)=g(x)+b$. We see that if $a=0$ then $g(x)=g(x)+b$ to $b=0$ as well, so similarly if $b=0$ then $a=0$, so now assume $a, b\\ne 0$.\nWe see that if $x=|b|n$ then $f(x)\\equiv f(0) \\pmod{|a|}$, if $x=|b|n+1$ then $f(x)\\equiv f(1)\\pmod{|a|}$, if $x=|b|n+2$ then $f(x)\\equiv f(2)\\pmod{|a|}$... if $x=|b|(n+1)-1$ then $f(x)\\equiv f(|b|-1)\\pmod{|a|}$. This means that the $b$-element collection $\\left\\{f(0), f(1), f(2), ... ,f(|b|-1)\\right\\}$ contains all $|a|$ residues mod $|a|$ since $f$ is surjective, so $|b|\\ge |a|$. Doing the same to $g$ yields that $|a|\\ge |b|$, so this means that only $|a|=|b|$ can work.\nFor $a=b$ let $f(x)=x$ and $g(x)=x+a$, and for $a=-b$ let $f(x)=-x$ and $g(x)=-x-a$, so  $|a|=|b|$ does work and are the only solutions, as desired.\n-Stormersyle",
+        "Solution_2": "We claim that $f$ and $g$ exist if and only if $|a|=|b|$.\nOnly If:\nFor some fixed $j$, let $f(j)=k$.\nIf $b=0$, then $g(k)=j$. Suppose $a\\ne 0$. Then $f(j)=f(g(k))=k+a\\ne k$, a contradiction. Thus, $a=0$. Similarly, if $a=0$, then $b=0$, satisfying $|a|=|b|$.\nOtherwise, $a,b\\ne 0$. We know that $g(k)=g(f(j))=j+b$, $f(j+b)=f(g(k))=k+a$, $g(k+a)=j+2b$, and so on: $f(j+nb)=k+na$ and $g(k+na)=j+(n+1)b$ for $n\\ge 0$.\nConsider the value of $g(k-a)$. Suppose $g(k-a)=j'\\ne j$. Then $f(j')=k$ and $g(f(j'))=j+b\\ne j'+b$, a contradiction. Thus, $g(k-a)=j$. We repeat with $f(j-b)$. Suppose $f(j-b)=k'-b\\ne k-b$. Then $g(k'-b)=j$ and $f(g(k'-b))=k\\ne k'$, a contradition. Thus, $f(j-b)=k-b$. Continuing, $g(k-2a)=j-a$, and so on: $f(j+nb)=k+na$ and $g(k+na)=j+(n+1)b$ now for all $n$.\nThis defines $f(x)$ for all $x\\equiv j\\pmod{|b|}$ and $g(x)$ for all $x\\equiv f(j)\\pmod{|a|}$.\nThis means that $x\\equiv j\\pmod{|b|}\\implies f(x)\\equiv f(j)\\pmod{|a|}$, and $y\\equiv f(j)\\pmod{|a|}\\implies g(y)\\equiv j\\pmod{|b|}$ which implies $f(x)\\equiv f(j)\\pmod{|a|}\\implies x\\equiv j\\pmod{|b|}$.\nAs a result, $f(x)$ maps each residue mod $|b|$ to a unique residue mod $|a|$, so $|a|\\ge|b|$. Similarly, $g(x)$ maps each residue mod $|a|$ to a unique residue mod $|b|$, so $|b|\\ge|a|$. Therefore, $|a|=|b|$.\nIf:\n$|a|=|b|$ means that either $a=b$ or $a=-b$. $f(x)=x,g(x)=x+a$ works for the former and $f(x)=-x,g(x)=-x-a$ works for the latter, and we are done.\n~emerald_block\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2+BC^2=AB^2$. The diagonals of $ABCD$ intersect at $E$. Let $P$ be a point on side $\\overline{AB}$ satisfying $\\angle APD=\\angle BPC$. Show that line $PE$ bisects $\\overline{CD}$.",
+        "Solution_1": "Let $PE \\cap DC = M$. Also, let $N$ be the midpoint of $AB$.\nNote that only one point $P$ satisfies the given angle condition. With this in mind, construct $P'$ with the following properties:\n$AP' \\cdot AB = AD^2 \\quad \\text{and} \\quad BP' \\cdot AB = CD^2$\nClaim: $P = P'$\nProof:\nThe conditions imply the similarities $ADP \\sim ABD$ and $BCP \\sim BAC$ whence $\\measuredangle APD = \\measuredangle BDA = \\measuredangle BCA = \\measuredangle CPB$ as desired. $\\square$\nClaim: $PE$ is a symmedian in $AEB$\nProof:\nWe have\n\nas desired. $\\blacksquare$\nSince $P$ is the isogonal conjugate of $N$, $\\measuredangle PEA = \\measuredangle MEC = \\measuredangle BEN$. However $\\measuredangle MEC = \\measuredangle BEN$ implies that $M$ is the midpoint of $CD$ from similar triangles, so we are done. $\\blacksquare$\n~sriraamster",
+        "Solution_2": "By monotonicity, we can see that the point $P$ is unique. Therefore, if we find another point $P'$ with all the same properties as $P$, then $P = P'$\nPart 1) Let $N$ be a point on $\\overline{AB}$ such that $AN\\cdot AB = AD^2$, and $BN \\cdot AB = BC^2$. Obviously $N$ exists because adding the two equations gives $AN\\cdot AB + BN \\cdot AB = AD^2 + BC^2 = AB^2$, which is the problem statement. Notice that converse PoP givesTherefore, $\\angle AND = \\angle ADB = \\angle ACB = \\angle CNB$, so $N$ does indeed satisfy all the conditions $P$ does, so $N = P$. Hence, $\\bigtriangleup ADP \\sim \\bigtriangleup ABD$ and $\\bigtriangleup CBP \\sim\\bigtriangleup ABC$.\nPart 2) Define $G$ as the midpoint of $CD$. Furthermore, create a point $X$ such that $DX || EC$ and $CX || ED$. Obviously $XCED$ must be a parallelogram. Now we set up for Jacobi's. The problem already gives us that $\\angle APD = \\angle CPB$, which is good for starters. Furthermore, $\\bigtriangleup ADP \\sim \\bigtriangleup ABD$ tells us thatThis gives us our second needed angle equivalence. Lastly, $\\bigtriangleup CBP \\sim\\bigtriangleup ABC$ will givewhich is our last necessary angle equivalence to apply Jacobi's. Finally, applying Jacobi's tells us that $AC$, $BD$, and $XP$ are concurrent $\\implies$ $X$, $E$, $P$ collinear. Additionally, since parallelogram diagonals bisect each other, $X$, $G$, and $E$ are collinear, so finally we obtain that $P$, $E$, and $G$ are collinear, as desired.\n-jj_ca888\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions."
+    },
+    {
+        "Problem": "$(*)$ Let $ABC$ be a triangle with $\\angle ABC$ obtuse. The $A$-excircle is a circle in the exterior of $\\triangle ABC$ that is tangent to side $\\overline{BC}$ of the triangle and tangent to the extensions of the other two sides. Let $E$, $F$ be the feet of the altitudes from $B$ and $C$ to lines $AC$ and $AB$, respectively. Can line $EF$ be tangent to the $A$-excircle?",
+        "Solution_1": "Instead of trying to find a synthetic way to describe $EF$ being tangent to the $A$-excircle (very hard), we instead consider the foot of the perpendicular from the $A$-excircle to $EF$, hoping to force something via the length of the perpendicular.  It would be nice if there were an easier way to describe $EF$, something more closely related to the $A$-excircle; as we are considering perpendicularity, if we could generate a line parallel to $EF$, that would be good.\nSo we recall that it is well known that triangle $AEF$ is similar to $ABC$.  This motivates reflecting $BC$ over the angle bisector at $A$ to obtain $B'C'$, which is parallel to $EF$ for obvious reasons.\nFurthermore, as reflection preserves intersection, $B'C'$ is tangent to the reflection of the $A$-excircle over the $A$-angle bisector.  But it is well-known that the $A$-excenter lies on the $A$-angle bisector, so the $A$-excircle must be preserved under reflection over the $A$-excircle.  Thus $B'C'$ is tangent to the $A$-excircle.Yet for all lines parallel to $EF$, there are only two lines tangent to the $A$-excircle, and only one possibility for $EF$,  so $EF = B'C'$.\nThus as $ABB'$ is isoceles,  contradiction. -alifenix-",
+        "Solution_2": "The answer is no.\nSuppose otherwise. Consider the reflection over the bisector of $\\angle BAC$. This swaps rays $AB$ and $AC$; suppose $E$ and $F$ are sent to $E'$ and $F'$. Note that the $A$-excircle is fixed, so line $E'F'$ must also be tangent to the $A$-excircle.\nSince $BEFC$ is cyclic, we obtain $\\measuredangle ECB = \\measuredangle EFB = \\measuredangle EF'E'$, so $\\overline{E'F'} \\parallel \\overline{BC}$. However, as $\\overline{EF}$ is a chord in the circle with diameter $\\overline{BC}$, $EF \\le BC$.\nIf $EF < BC$ then $E'F' < BC$ too, so then $\\overline{E'F'}$ lies inside $\\triangle ABC$ and cannot be tangent to the excircle.\nThe remaining case is when $EF = BC$. In this case, $\\overline{EF}$ is also a diameter, so $BECF$ is a rectangle. In particular $\\overline{BE} \\parallel \\overline{CF}$. However, by the existence of the orthocenter, the lines $BE$ and $CF$ must intersect, contradiction.",
+        "Solution_3": "The answer is $\\boxed{\\text{no}}$.\nSuppose for the sake of contradiction that it is possible for $EF$ to be tangent to the $A$-excircle. Call the tangency point $T$, and let $S_1, S_2$ denote the contact points of $AB, AC$ with the $A$-excircle, respectively. Let $s$ denote the semiperimeter of $ABC$. By equal tangents, we have It is also well known that $AS_1 = AS_2 = \\frac{s}{2}$, so It is well known (by an easy angle chase) that $\\triangle AEF \\sim \\triangle ABC$, so we must have the ratio of similitude is $2$. In particular, This results in which is absurd since $\\triangle BEC$ is a right triangle. We reached a contradiction, so we are done. $\\blacksquare$ ~ Mathscienceclass",
+        "Solution_4": "We claim that the answer is no. We proceed with contradiction. Suppose that $EF$ is indeed tangent to the a-excenter. Define the point of tangency to be $X$. Let $G$ to be the intersection of the a-excircle with the extension of $AB$ and $H$ to be the intersection of the a-excircle with the extension of $AC$. Define $I_a$ to be the a-excenter. It is a well-known fact that $I_a$ lies on the angle bisector of $\\angle BAC$. \nFor convenience, let $AB = c, AC = b, BC = a$, $\\angle CAB = A, \\angle CBA = B, \\angle ACB = C$.\nNotice that by Power of a Point:\n\n\nTherefore, adding, we see that:\n\nIt would be rather nice if we could re-write $FG$. Indeed, we can:\n\nand similarly for $FH$:\n\nWe now seek to re-write $FE$. Using the law of cosines:\n\nTherefore, \n\nPutting this all together, we see that:\n\n\nNow, we seek to write $\\tan \\left(\\frac{a}{2} \\right)$ in terms of the side lengths of the triangle. Notice that:\n\n\nwhere $s$ is the semi-perimeter. We get that:\n\nUsing the fact that $\\cos^2 \\left(\\frac{A}{2} \\right) + \\sin^2 \\left(\\frac{A}{2} \\right) = 1$, we have that:\n\nTherefore, \n\nReturning to our original problem:\n\nIt is a well-known fact that $r_a = \\sqrt{\\frac{s(s-b)(s-c)}{s-a}}$, so:\n which implies that:\n\nwhich is a contradiction. Hence, our original assumption that $EF$ is tangent to the a-excircle is incorrect. $\\blacksquare$\n~AopsUser101"
+    },
+    {
+        "Problem": "Let $n \\geq 2$ be an integer. Carl has $n$ books arranged on a bookshelf.\nEach book has a height and a width. No two books have the same height, and no two\nbooks have the same width.\nInitially, the books are arranged in increasing order of height from left to right. In a\nmove, Carl picks any two adjacent books where the left book is wider and shorter than\nthe right book, and swaps their locations. Carl does this repeatedly until no further\nmoves are possible.\nProve that regardless of how Carl makes his moves, he must stop after a finite number\nof moves, and when he does stop, the books are sorted in increasing order of width\nfrom left to right.",
+        "Solution": "Without loss of generality, let the books have widths and heights of 1, 2, 3,..., and n. First we will show that Carl can only make finitely many moves.\nNote that two books can never be swapped more than once with each other. This is because after a first swap, the wider book will be to the left of the other, so they can never be swapped back. Thus there can be at most ${n\\choose 2}$ moves.\nNow we will show that the books are in increasing order of width when Carl is done making moves. Assume otherwise for the sake of contradiction. Now let $f(k)$ be the width of the book in the kth spot from the left. Since the books are not in increasing order of width, there must be i such that $f(i)>f(i+1)$, for $1\\leq i\\leq n-1$.\nClaim: The ith book must be shorter than the $(i+1)$th book.\nProof: Assume otherwise for contradiction. Then the ith book must have started to the right of the $(i+1)$th book, since the ith book is taller. However, the ith book is now to the left of the $(i+1)$th book, so they must have been swapped at some point. But since $f(i)>f(i+1)$, the ith book is also wider. But the ith book started to the right, so since it was wider it cannot have been swapped with the $(i+1)$th book, so we arrive at a contradiction. $\\blacksquare$\nFrom the claim, the ith book must be shorter than the $(i+1)$th book. But since the ith book is also wider by definition (as $f(i)>f(i+1)$), Carl can swap those two books, so he still has legal moves and we arrive at a contradiction, so we are done. $\\blacksquare$\n~MortemEtInteritum\n"
+    },
+    {
+        "Problem": "Let $\\omega$ be the incircle of a fixed equilateral triangle $ABC$. Let $\\ell$ be a variable line that is tangent to $\\omega$ and meets the interior of segments $BC$ and $CA$ at points $P$ and $Q$, respectively. A point $R$ is chosen such that $PR = PA$ and $QR = QB$. Find all possible locations of the point $R$, over all choices of $\\ell$.",
+        "Solution_1": "Call a point good if it is a possible location for $R$.\nLet the incircle of $\\triangle ABC$ touch $BC$ at $D$, $AC$ at $E$, and $\\ell$ at $T$. Also, let the center of the incircle be $I$. Clearly, a point is good iff it lies on the circle containing $A$ with center $P$ as well as the circle containing $B$ with center $Q$. Call these circles $\\omega_1$ and $\\omega_2$, respectively.\nNote that point $T$ can only lie on minor arc $\\overarc{DE}$ (excluding the endpoints).\nClaim: A point $X$ is good iff $XT\\perp PQ$ and $XT=AD=BE$.\nProof: WLOG, let $X$ be on the same side of $PQ$ as $I$. Then we have that $T, I, X$ are collinear. In particular, we have $\\angle XTP=90^{\\circ}$. Then $TX=AD$, $PT=PD$, $\\angle XTP=\\angle ADP$, so that $\\triangle ADP\\cong \\triangle XTP\\rightarrow PA=PX$. Similarly, $QB=QX$, so $X$ is good. Then $X'$, the reflection of $X$ across $T$, is also good. But $\\omega_1$ and $\\omega_2$ have at most two intersections, so $X$ and $X'$ must be these intersections, and since a point is good iff it lies on both circles, we are done.\nNow, we know that $R, I, T$ are collinear. Then we have two cases:\nCase 1: $R$ and $I$ lie on the same side of $PQ$. Then we have $RI+IT=AI+ID=BI+IE$, so that $RI=AI=BI$. Then we have that $R, A, B$ lie on a circle with center $I$. Note that because $T$ lies on $\\overarc{DE}$, $R$ must lie on $\\overarc{AB}$. So, one of the solutions is $\\boxed{R \\text{ in arc } AB\\text{ (excluding the endpoints)}}$.\nCase 2: $R$ and $I$ lie on opposite sides of $PQ$. Then extend $ID$ out from $D$ to $X$ such that $DX=TR$, and extend $IE$ out from $E$ to $Y$ such that $EY=TR$. Then we see that $IR=IX=IY$, so that $\\boxed{R \\text{ lies on arc } XY\\text{ (excluding the endpoints)}}$.",
+        "Solution_2": "We claim that $R$ can lie on minor arc $AB$ of the circumcircle of triangle ABC, and it can also lie on the dilation of this arc about the center of triangle $ABC$ with a factor of $-2.$\nLet $D, E,$ and $F$ be the feet of the angle bisectors from points $A, B,$ and $C$ respectively. Trivially, $DEF$ is also the medial triangle, orthic triangle, and contact triangle (ABC is equilateral).\nLet $I$ be the incenter of ABC. Trivially, $I$ is the centroid, orthocenter, and circumcenter of ABC (ABC is equilateral). Also, $AD=BE=CF=3r$ where $r$ is the radius of circle $\\omega$ (This is trivial). $T$ is the point of tangency of $\\omega$ and segment $\\overline{PQ}$.\nR has to lie on the intersection of circles $\\omega1$(center P, radius PA) and $\\omega2$(center Q, radius QB), and for each choice of P, there exist two locations for R. The location that we claim to lie on the minor arc AB of the circumcircle of ABC shall be denoted M, and the other location shall be denoted N.\nDefine triangle XYZ to be the homothety of triangle ABC about I with a factor of -2.\nCritical claim: M, T, I, and N are collinear.\nProof:\nFirst we shall prove that T lies on MN using phantom points.\nLet the intersection of MN and PQ be denoted as K. We shall prove that K and T are the same point.\nLet $PT = p$ and $QT = q$. Because of the equal tangent theorem, $PD=PT=p$ and $QE=QT=q$. Hence, by the pythagorean theorem (recall $AD=BE=3r$), $PA^2 = 9r^2 + p^2$ and $QB^2 = 9r^2 + q^2$. Since PN = PA and QN = QB, then $PN^2 = 9r^2 + p^2$ and $QN^2 = 9r^2 + q^2$.\nPQ is the perpendicular bisector of MN because MN is the radical axis of $\\omega1$ and $\\omega2$. Hence, M is the reflection of N across K. Also, NK is the altitude of triangle PNQ, so $PK^2-QK^2 = PN^2-QN^2 = p^2 - q^2$ by using the pythagorean theorem and earlier expressions for $PN^2$ and $QN^2$. However, $PK+QK=PQ=p+q$. Now, we have a system of equations to solve for PK and QK in terms of p and q.\nDividing the first equation by the second (we can do this because p+q is always nonzero), we get $PK-QK=p-q$. Combining this with our PK+QK result, we get $PK = p$ and $QK = q$. However, $PT = p$ and $QT = q$, and only one point can exist on PQ for which this result holds true. As a result, K and T are the same point, otherwise it is a contradiction. Hence M, T, and N are collinear.\n$IT \\parallel MN$. This is because both MN and IT are perpendicular to PQ (IT is perpendicular to PQ because PQ is a tangent with point of tangency T). However, both lines share point T, as discussed earlier. Hence, IT and MN are the same line, and M, T, I, and N are collinear.\nIn fact, from our earlier results from the lengths of PN and PT, we can use the pythagorean theorem to get that $NT = 3r$, a result that is always true and independent of P and Q! Also, because M is the reflection of N over K (which is the same as T), $MT = 3r$ also. However, T varies based on P and Q. On the other hand, $IT = r$ and M, T, I and N are collinear. Remembering our earlier definitions of M and N, we get that $MI = 2r$ and $IN = 4r$, with M on the opposite side of N and T from I. Hence, M can be taken to N with a homothety about I with a factor of -2, and T can be taken to M with a homothety about I with a factor of -2. Since, trivially, the circumradius of ABC is 2r (ABC is equilateral), it seems like M can lie anywhere on the circumcircle of ABC.\nHowever, we must take into account the restrictions on P and Q. This limits T to only minor arc DE on the incircle of ABC, hence, because of our earlier homothety statement, M is restricted to minor arc AB on the circumcircle of ABC.  Because of our homothety statement about N, N has to lie on minor arc XY on the circumcircle of triangle XYZ.\nBecause we defined both M and N to be possible locations for R, $\\fbox{R can only lie on minor arc AB of the circumcircle of triangle ABC, and also on minor arc XY of the circumcircle of triangle XYZ}$.\n-QED\n-Solution by thanosaops"
+    },
+    {
+        "Problem": "An empty $2020 \\times 2020 \\times 2020$ cube is given, and a $2020 \\times 2020$ grid of square unit cells is drawn  on each of its six faces. A beam is a $1 \\times 1 \\times 2020$ rectangular prism. Several beams are placed inside the cube subject to the following conditions:\n- The two $1 \\times 1$ faces of each beam coincide with unit cells lying on opposite faces of the cube. (Hence, there are $3 \\cdot {2020}^2$ possible positions for a beam.)\n- No two beams have intersecting interiors.\n- The interiors of each of the four $1 \\times 2020$ faces of each beam touch either a face of the cube or the interior of the face of another beam.\nWhat is the smallest positive number of beams that can be placed to satisfy these conditions?",
+        "Solution": "Place the cube in the xyz-coordinate, with the positive x-axis pointing forward, the positive y-axis pointing right, and the positive z-axis pointing up. Let the position of a unit cube be $(x, y, z)$ if it is centered at $(x, y, z)$. Place the $2020 \\times 2020 \\times 2020$ cube so that the edges are parallel to the axes, and two of its corners are at $(1, 1, 1)$ and $(2020, 2020, 2020)$. Now call a beam z-oriented if its endpoints differ only in z-coordinates, and similarly call it x-oriented or y-oriented if the endpoints differ in x- or y-coordinates, respectively.\nWe claim that the answer is $\\boxed{3030}$. First we will prove 3030 suffices. Place the first beam so its endpoints lie at $(1, 1, 1)$, and $(2020, 1, 1)$. Place the 2nd beam so its endpoints lie at $(1, 1, 2)$ and $(1, 2020, 2)$. Place the 3rd beam so its endpoints lie at $(2, 2, 1)$ and $(2, 2, 2020)$. Now it is clear that the only faces among these three beams that are not touching the faces of the $2020 \\times 2020 \\times 2020$ cube of the face of another beam are the top face of beam 2, and the front and right faces of beam 3.\nNow place the 4th beam from $(1, 3, 3)$ to $(2020, 3, 3)$, 2 units up and right of beam 1. Then place beam 5 from $(3, 1, 4)$ to $(3, 2020, 4)$, 2 units forward and up of beam 2. Place beam 6 from $(4, 4, 1)$ to $(4, 4, 2020)$, 2 unites forward and right of beam 3. Note that these 3 beams will touch the top face of beam 2, and the front and right faces of beam 3. Now the only faces that are not touching another face are the top of beam 5, and the front and right of beam 6.\nWe can repeat this process 1010 times, placing the beams in groups of 3 so that beam $3n+1$ is 2 units up and right of beam $3n-2$, beam $3n+2$ is 2 units forward and up of beam $3n-1$, and beam $3n+3$ is 2 units forward and right of beam $3n$, for all $1\\leq n \\leq 1009$. After placing each set of 3 as such, the faces of the previous beams will all satisfy condition 3. And for the last set of 3 (that is, beams 3028, 3029, and 3030), their faces will satisfy condition 3 as well, since the top face of beam 3029 and the front/right faces of beam 3030 will touch the cube (again, note that all other faces will be touching another beam).\nThus, we have shown that 3030 beams suffices, so we will now show that 3030 beams is the minimum. Let there be $A$ x-oriented beams, $B$ y-oriented beams, and $C$ z-oriented beams. Now look each $2020 \\times 1 \\times 2020$ slice of the cube, parallel to the xz-plane. Note there are 2020 such slices. There are two cases. Either $B=2020^2$, or at least one of A or C is nonzero. In the first case, we clearly have at least $2020^2>3030$ beams as desired. In the second case, we must have at least one beam in each $2020 \\times 1 \\times 2020$ slice, in order to satisfy condition 3. If there is some slice without a beam, then there will be some x- or z-oriented beam with no beams touching its top or bottom face. Additionally, notice that only x- or z-oriented beams can fit in a $2020 \\times 1 \\times 2020$ slice, so $A+C\\geq 2020$. Similarly, we must have $A=2020^2$, $C=2020^2$, or both $A+B\\geq 2020$ and $B+C\\geq 2020$. In every case, the number of beams, $A+B+C$, will be at least 3030 as desired.$\\blacksquare$\n~MortemEtInteritum"
+    },
+    {
+        "Problem": "Let $ABCD$ be a convex quadrilateral inscribed in a circle and satisfying $DA < AB = BC < CD$. Points $E$ and $F$ are chosen on sides $CD$ and $AB$ such that $BE \\perp AC$ and $EF \\parallel BC$. Prove that $FB = FD$.",
+        "Solution": "Let $G$ be the intersection of $AE$ and $(ABCD)$ and $H$ be the intersection of $DF$ and $(ABCD)$.\nClaim: $GH || FE || BC$\nBy Pascal's on $GDCBAH$, we see that the intersection of $GH$ and $BC$, $E$, and $F$ are collinear. Since $FE || BC$, we know that $HG || BC$ as well. $\\blacksquare$\nNote that since all cyclic trapezoids are isosceles, $HB = GC$. Since $AB = BC$ and $EB \\perp AC$, we know that $EA = EC$, from which we have that $DGCA$ is an isosceles trapezoid and $DA = GC$. It follows that $DA = GC = HB$, so $BHAD$ is an isosceles trapezoid, from which $FB = FD$, as desired. $\\blacksquare$",
+        "Solution_2": "Let $G=\\overline{FE}\\cap\\overline{BC}$, and let $G=\\overline{AC}\\cap\\overline{BE}$. Now let $x=\\angle ACE$ and $y=\\angle BCA$.\nFrom $BA=BC$ and $\\overline{BE}\\perp \\overline{AC}$, we have $AE=EC$ so $\\angle EAC =\\angle ECA = x$. From cyclic quadrilateral ABCD, $\\angle ABD = \\angle ACD = x$. Since $BA=BC$, $\\angle BCA = \\angle BAC = y$.\nNow from cyclic quadrilateral ABC and $\\overline{FE}\\parallel \\overline{BC}$ we have $\\angle FAC = \\angle BAC = \\pi - \\angle BCD = \\pi - \\angle FED$. Thus F, A, D, and E are concyclic, and $\\angle DFG = \\angle DAE = \\angle DAC - \\angle EAC = \\angle DBC - x$ Let this be statement 1.\nNow since $\\overline{AH}\\perp \\overline {BH}$, triangle ABC gives us $\\angle BAH + \\angle ABG = \\frac{\\pi}{2}$. Thus $y+x+\\angle GBE=\\frac{\\pi}{2}$, or $\\angle GBE = \\frac{\\pi}{2}-x-y$.\nRight triangle BHC gives $\\angle HBC = \\frac{\\pi}{2}-y$, and $\\overline{BC}\\parallel \\overline{FE}$ implies $\\angle BEG=\\angle HBC = \\frac{\\pi}{2}-y.$\nNow triangle BGE gives $\\angle BGE = \\pi - \\angle BEG - \\angle GBE = \\pi - (\\frac{\\pi}{2}-y)-(\\frac{\\pi}{2}-x-y)=x+2y$. But $\\angle FGB = \\angle BGE$, so $\\angle FGB=x+2y$. Using triangle FGD and statement 1 gives\nThus, $\\angle FDB = \\angle FBD$, so $\\boxed{FB=FD}$ as desired.$\\blacksquare$\n~MortemEtInteritum",
+        "Solution_3_.28Angle-Chasing.29": "Proving that $FB=FD$ is equivalent to proving that $\\angle FBD= \\angle FDB$. Note that $\\angle FBD=\\angle ACD$ because quadrilateral $ABCD$ is cyclic. Also note that $\\angle BAC=\\angle ACB$ because $AB=BC$. $AE=EC$, which follows from the facts that $BE \\perp AC$ and $AB=AC$, implies that $\\angle CAE= \\angle ACE= \\angle ACD= \\angle FBD$. Thus, we would like to prove that triangle $FBD$ is similar to triangle $AEC$. In order for this to be true, then $\\angle BFD$ must equal $\\angle AEC$ which implies that $\\angle AFD$ must equal $\\angle AED$. In order for this to be true, then quadrilateral $AFED$ must be cyclic. Using the fact that $EF \\parallel BC$, we get that $\\angle AFE= \\angle ABC$, and that $\\angle FED= \\angle BCE$, and thus we have proved that quadrilateral $AFED$ is cyclic. Therefore, triangle $FDB$ is similar to isosceles triangle $AEC$ from AA and thus $FB=FD$.\n-xXINs1c1veXx",
+        "Solution_4": "BE is perpendicular bisector of AC, so $\\angle ACE = \\angle EAC$. FE is parallel to BC and ABCD is cyclic, so AFED is also cyclic. $\\angle AFD = \\angle AED = 2 \\angle ACE = 2 \\angle FBD$. Hence, $\\angle FBD = \\angle BDF$, $FD = FB$.\nMathdummy",
+        "Solution_5": "Let $G$ be on $AB$ such that $GD \\perp DB$, and $H = GD \\cap EB$. Then $\\angle{ADB} = \\angle{EDB} = 90^{\\circ} - \\angle{ABE} \\implies \\triangle{DAE}$ is the orthic triangle of $\\triangle{HGB}$. Thus, $F$ is the midpoint of $GB$ and lies on the $\\perp$ bisector of $DB$.",
+        "Solution_6": "Let $FE$ meet $AC$ at $J$, $BE$ meet $AC$ at $S$, connect $AE, SD$.\nDenote that $\\angle{BCA}=\\alpha; AB=BC, \\angle{BAC}=\\angle{BCA}=\\alpha$, since $EF$ is parallel to $BC$, $\\angle{AJF}=\\angle{ACB}=\\alpha$. $\\angle{AJF}$and $\\angle{EJS}$ are vertical angle, so they are equal to each other. \n$BE\\bot{AC}$,$\\angle{JES}=90^{\\circ}-\\alpha$, since $\\angle{EFB}=\\angle{AJF}+\\angle{FAJ}=2\\alpha$, we can express $\\angle{FBE}=180^{\\circ}-2\\alpha-(90^{\\circ}-\\alpha)=90^{\\circ}-\\alpha= \\angle{FEB}$, leads to $FE=FB$\nNotice that quadrilateral $AFED$ is a cyclic quadrilateral since $\\angle{ADE}+\\angle{AFE}=\\angle{ADE}+\\angle{ABC}=180^{\\circ}$.\nAssume $\\angle{ECA}=\\beta$, $\\triangle{AES}$ is congruent to $\\triangle{CES}$ since $AS=AS,\\angle{ASE}=\\angle{BSE}, SE=SE(SAS)$, so we can get $\\angle{EAS}=\\beta$ \nLet the circumcircle of $AFED$ meets $AC$ at $Q$\nNow notice that $\\widehat{QE}=\\widehat{QE}, \\angle{QAE}=\\angle{QDE}=\\beta$; similarly, $\\widehat{FQ}=\\widehat{FQ}; \\angle{FDQ}=\\angle{FAQ}=\\alpha$.\n$\\angle{FDE}=\\alpha+\\beta; \\angle{FED}=\\angle{BCD}=\\alpha+\\beta$, it leads to $FD=FE$.\nsince $FE=FB;FD=FE, DF=BF$ as desired\n~bluesoul"
+    },
+    {
+        "Problem": "Suppose that $(a_1,b_1),$ $(a_2,b_2),$ $\\dots,$ $(a_{100},b_{100})$ are distinct ordered pairs of nonnegative integers. Let $N$ denote the number of pairs of integers $(i,j)$ satisfying $1\\leq i<j\\leq 100$ and $|a_ib_j-a_jb_i|=1$. Determine the largest possible value of $N$ over all possible choices of the $100$ ordered pairs.",
+        "Solution": "Call the pair $(i, j)$ good if $1\\leq i < j \\leq 100$ and $|a_ib_j-a_jb_i|=1$. Note that we can reorder the pairs $(a_1, b_1), (a_2, b_2), \\ldots, (a_{100}, b_{100})$ without changing the number of good pairs. Thus, we can reorder them so that $a_1\\leq a_2\\leq\\ldots\\leq a_{100}$. Furthermore, reorder them so that if $a_i=a_j$ for some $i<j$, then $b_i<b_j$.\nNow I claim the maximum value of $N$ is $\\boxed{197}$. First, we will show $N\\leq 197$.\nThe key claim is that for any fixed integer $m$, $1\\leq m \\leq 100$, then there are at most 2 values of $i$ with $1\\leq i<m$ with the pair $(i, m)$ good. We will prove this with contradiction.\nAssume there are 3 integers $i$, $j$, and $k$ that are less than $m$, such that $(i, m)$, $(j, m)$, and $(k, m)$ are all good. Note that $gcd(a_m, b_m)$ divides $a_ib_m-a_mb_i=\\pm 1$, so $gcd(a_m, b_m)=1$. From $a_ib_m-a_mb_i=\\pm 1$, we have \nSimilarly,\nFrom $gcd(a_m, b_m)=1$ we have\nBut since $i$, $j$, and $k$ are less than $m$, from our reordering we have $0\\leq a_i<a_m$, $0\\leq a_j<a_m$, and $0\\leq a_k<a_m$. Thus our congruence gives us that $a_j=a_i$ or $a_j=a_m-a_i$, and similar for $a_k$. Thus at least two of $a_i, a_j, a_k$ are equal. WLOG let them be $a_i$ and $a_j$. Now we have 4 cases.\nCase 1: $a_m>2$\nWe have $a_ib_m-a_mb_i=\\pm 1$ and $a_jb_m-a_mb_j=\\pm 1$, so since $a_i=a_j$ we can subtract the two equations to get $a_mb_i-a_mb_j$ equals $\\pm 2, 0$. But $a_m>2$, so $|a_m(b_i-b_j)|\\leq 2$ implies that $b_i=b_j$. But all pairs are distinct and $a_i=a_j$, so this is a contradiction.\nCase 2: $a_m=2$\nWe have 2 subcases.\nSubcase 2a: $a_i\\equiv a_j\\equiv a_k\\equiv 1 \\mod a_m$\nClearly we must have $a_i=a_j=a_k=1$ in this case, as all 3 variables are less than $a_m=2$. Then $1\\cdot b_m-a_mb_i=\\pm 1$, $1\\cdot b_m-a_mb_j=\\pm 1$, and $1\\cdot b_m-a_mb_k=\\pm 1$. So at least 2 of $b_i$, $b_j$, and $b_k$ must be equal, but this contradicts the fact that all pairs $(a_x, b_x)$ are distinct.\nSubcase 2b: $a_i\\equiv a_j\\equiv a_k\\equiv 0 \\mod a_m$\n$a_i$ must be even in this case, but since $a_m$ is also even then $a_ib_m-a_mb_i$ must be even, which contradicts $|a_ib_m-a_mb_i|=1$.\nCase 3: $a_m=1$\nWe must have $a_i=0,1$, and similarly for $a_j$ and $a_k$. Thus at least 2 of the 3 must be equal. Now we have 2 subcases again.\nSubcase 3a: 2 of the 3 are equal to 0\nWLOG let $a_i=a_j=0$. Then $a_ib_m-a_mb_i=\\pm 1$ and $a_m=1$ so $b_i=\\pm 1$. Since $b_i$ is nonnegative, $b_i=1$. Similarly $b_j=1$, but all pairs are distinct so we have a contradiction.\nSubcase 3b: 2 of the 3 are equal to 1\nWLOG let $a_i=a_j=0$. Then $a_ib_m-a_mb_i=\\pm 1$, so $b_m-b_i=\\pm 1$. But $i<m$ and $a_i=a_m=1$, so by our ordering of the pairs we have $b_i<b_m$ so $b_m-b_i=1$. But similarly $b_m-b_j=1$, so $b_i=b_j$, which is a contradiction to all pairs being distinct.\nCase 4: $a_m=0$\nWe must have $a_i=0$, since $0\\leq a_i\\leq a_m=0$. Thus $a_ib_m-a_mb_i=0$, but $|a_ib_m-a_mb_i|=1$, contradiction.$\\blacksquare$\nThus our claim is proved. Now for $m=1$, there is clearly no $i$ with $1\\leq i < m \\leq 100$ and $(i, m)$ good. For $m=2$, there is at most one value of i with $1\\leq i < m \\leq 100$ and $(i, m)$ good (namely, this is when $i=1$). By our claim, for all $2<m\\leq 100$ there are at most 2 values of $i$ with $1\\leq i < m \\leq 100$ and $(i, m)$ good, so adding these up gives at most $2\\cdot 98+1=197$ good pairs. Now we will show that 197 good pairs is achievable.\nThe sequence $(1,1), (1,2), (2,3),\\ldots ,(99,100)$ has 197 good pairs - namely, $(1, n)$ is good for all $1<n\\leq 100$, and $(i, i+1)$ is good for all $1<i<100$. This adds up for a total of 197 good pairs, so we are done. $\\blacksquare$\n~MortemEtInteritum",
+        "Solution_2": "We claim the answer is $2n-3$ (so $197$), where $100$ is to be replaced with $n$.\nThe problem essentially counts unordered pairs $(P,Q)$ of lattice points so that $POQ$ has area $\\frac{1}{2}$ due to Shoelace ($O$ is the origin).\nFor a construction, take the points $(0,1),(1,1),(1,2), \\dots (1,n-1)$.\nWe now show by induction that $2n-3$ is the maximum, where $n=2$ obviously holds as a base case.\nTo reduce to the inductive hypothesis of $n-1$ points from $n$ points, we are allowed to consider the point with the greatest sum of coordinates $R$.\nThe key claim is that $R$ is part of at most $2$ pairs (triangles) with area $\\frac{1}{2}$.\nBefore we proceed, note that $R$ must have relatively prime coordinates $(m,n)$, or else any triangles found can be reduced down by a greatest common factor along $OR$, impossible since $\\frac{1}{2}$ is the minimum positive area.\nCall the hypothetical point(s) $X$. With triangle $XOR$ with area $\\frac{1}{2}$, we know the length of $OR$, so indeed the locus of $X$ is two lines evenly spaced next to $OR$ each parallel to $OR$ (before we consider the lattice condition).\nWhen we add in the lattice condition, we see that each of these two lines contain at most one valid $X$. This is because if two lattice points with non-negative integers lied on one of these lines, one would have to be a multiple of $m$ more than the other on the $x$-coordinates and a multiple of $n$ more on the $y$-coordinates. This contradicts the coordinate sum maximality of $R$.\nThus, by deleting $R$, we lose at most two pairs $(X,R)$, completing the induction and the problem."
+    },
+    {
+        "Problem": "Let $\\mathbb{N}$ denote the set of positive integers. Find all functions $f : \\mathbb{N} \\rightarrow \\mathbb{N}$ such that for positive integers $a$ and $b,$",
+        "Solution": "I claim that the only function $f$ that satisfies the constraints outlined within the problem is the function $f(n) = 1$ for all positive integers $n$.\nWe will proceed with strong induction. The base case is simple, as plugging $a=1$ into the second equation given within the problem gives $f(1)=f(1)^2$. Since $f(n)$ can only return a positive integer value, we have that $f(1)=1$.\nNow we proceed with the inductive step. If the next number $n$ is either a perfect square or can be represented as a sum of two perfect squares, then obviously $f(n)=1$, as it is either the product of two $f$-values that are both equal to $1$ from the inductive assumption or is the square of an $f$-value that is equal to $1$, again due to the inductive assumption. Otherwise, we can use the Sum of Two Squares Theorem, which tells us that $n$ has at least one prime in its prime factorization that is $3(\\mod 4)$ and is raised to an odd power.\nLemma 1: Given that $a^2+b^2=c^2$ and $f(b)=f(c)=1$, we then have $f(a)=1$.\nProof: Note that the first condition in the problem tells us that $f(a^2+b^2)=f(a)f(b)$, or $f(c^2)=f(a)f(b)$. Using the second condition gives us $f(c)^2=f(a)f(b)$. Plugging in the values of $f(b)$ and $f(c)$ gives us that $f(a)=1$.\nNow we will attempt to repeatedly remove prime factors that are $3(mod 4)$ and taken to an odd power, and we will move from the largest prime down to the smallest prime that satisfies above conditions. The prime factors will be removed by constructing a Pythagorean Triple with the prime being the smallest leg in the form of $p,\\frac{p^2-1}{2},\\frac{p^2+1}{2}$(this will always work as $p\\neq 2$(not 3 mod 4), and this method works via Lemma 1). We will then prove the ending numbers that we achieve via removing all the $(\\text{mod } 4)$ primes(which I will refer to as \"tips\") are equal to 1 b/c they can be expressed as the sum of two squares or is a perfect square(Sum of Two Squares Theorem). For example, if we took the number $21$, we would first aim to remove the $7$ by splitting it into $24$ and $25$, so $21$ would become $72$ and $75$. $75$ is divisible by $3$ to an odd power, so we transform it into $100$ and $125$. These two don't have any divisors that are $3 (\\text{mod } 4)$ raised to an odd power so we leave it alone, as sum of two squares will work on them or they are perfect squares. $72$ does have a prime factor that is $3(\\text{mod } 4)$, but it is an even power so we leave it alone. In this case, the tips are $72$, $100$, and $125$. However, now we need to prove two key facts: using this Pythagorean Triple Method will never generate another $3(\\text{mod } 4)$ prime that is bigger than the current one we are working on or create more primes or keep the same number of primes in the tips' prime factorizations., as otherwise it could cause an infinite cycle, and also we must prove when we use Sum of Two Squares/Perfect Square given in the second condition on the tips the square root of the square(s) used will never be greater than or equal to $n$.\nThe first claim can be proved rather simply. Note that $\\frac{p^2-1}{2}$ can have no prime factors greater than or equal to $p$, as it can be factored as $(p-1)(\\frac{p+1}{2})$, which are both less than $p$ and are integers($p+1$ must be an integer due to $p$ must being odd(not equal to $2$)). For $\\frac{p^2+1}{2}$, we can prove something a bit more general.\nLemma 2: For any positive integer $n$, all prime factors of $n^2+1$ must be $1(\\text{mod } 4)$.\nProof: Note that if $n^2+1\\equiv 0(\\text{mod } p)$, then we also must have $(n^2+1)(n^2-1)=(n^4-1)\\equiv 0(\\text{mod } p)$, or $n^4 \\equiv 1(\\text{mod } p)$. Now we can apply Fermat's Little Theorem to obtain $n^{p-1}\\equiv 1(\\text{mod } p)$. Note that since $n$ and $n^2$ are obviously not $1(\\text{mod } p)$, as $n^2\\equiv -1 (\\text{mod } p)$, we have that $p-1$ is a multiple of $4$, or $p \\equiv 1(\\text{mod } 4)$.\nWe can simply apply Lemma 2 to $p^2+1$ as $2$ is obviously not $3(\\text{mod } 4)$, and this means that a prime factor $3(\\text{mod } 4)$ will never be generated from this term. This completes the first of our two claims.\nNow we proceed to the second of our two claims. Note that every time we use the method on $n$ based on prime $p$, we will multiply by around $\\frac{p}{2}$, clearly less than $p$(if we multiply by $p$ we would get $p^2$ which is clearly greater than $\\frac{p^2+1}{2}$). We will never have to use the same prime twice in our method, so at max in the end we will multiply $n$ by the product of a little less than all $3(mod 4)$ primes that divide it, which is less than $n$ itself for $n$ greater than or equal to $3$(the smallest 3 mod 4 prime), meaning that the largest number that we must use two squares on that is generated by out method is less than $n^2$. We need to prove that the two squares that sum to this are both less than $n$, which is quite trivial, as they are less than $\\sqrt{n^2}$, which obviously means it must be less than $n$. This proves the second of our claims.\nThis completes the second case of the inductive step, and therefore completes both the induction and the problem.\n~Solution by hyxue",
+        "Solution_2_.28Taken_from_Twitch_Solves_ISL.29": "The Answer is $\\boxed{f\\equiv 1}$ which works but we want to prove that it's the only one. Claim: If $f(a)=f(b)=1$ and a>b, then $f(a^2-b^2)= f(2ab)= 1$. Proof: We can write $1=f(a)=f(b)= f(a^2+b^2)= \\sqrt{f((a^2+b^2)^2)} = \\sqrt{f((a^2-b^2)^2+ (2ab)^2)} = \\sqrt{f(a^2-b^2)f(2ab)}$. We set it to $a=b=1$ and we get that $f(1)=f(2)=1$. Easily by Induction it shows f(n)=1. We take if $n=2k$ take $(u,v)=(k,1)$ which $2uv=n$. If $n=2k+1$, just take $(u,v)= (k+1,k)$ which $u^2-v^2=n$. Thus the only answer is $\\boxed{f\\equiv 1}$ and we are done.",
+        "Solution_3_.28clear_Solution_2.29": "The answer is $\\boxed{f\\equiv 1}$, which works.\nTo show it is necessary, we first get $f(1)=f(1)^2$, so $f(1)=1$.\nThen, we get $f(2)=f(1^2 + 1^2)=f(1)^2 =1$.\nThe following claim finishes the problem via induction:\nClaim: If $f(n)=f(n-1)=1$ for $n \\ge 2$, we have $f(2n)=f(2n-1)=1$.\nProof: Note that $f(n^2+1)=f(n)f(1)=1$, so  implies $f(2n)=1$.\nNote that $f(2n^2 -2n+1)=f(n)f(n-1)=1$, so  implies $f(2n-1)=1$. $\\square$\nThus, the only solution is $\\boxed{f\\equiv 1}$."
+    },
+    {
+        "Problem": "Rectangles $BCC_1B_2,$ $CAA_1C_2,$ and $ABB_1A_2$ are erected outside an acute triangle $ABC.$ Suppose thatProve that lines $B_1C_2,$ $C_1A_2,$ and $A_1B_2$ are concurrent.",
+        "Solution": "We first claim that the three circles $(BCC_1B_2),$ $(CAA_1C_2),$ and $(ABB_1A_2)$ share a common intersection.\nLet the second intersection of $(BCC_1B_2)$ and $(CAA_1C_2)$ be $X$. Then \n\nwhich implies that $AA_1C_2CX$ is cyclic as desired.\nNow we show that $X$ is the intersection of $B_1C_2,$ $C_1A_2,$ and $A_1B_2.$ Note that $\\angle C_1XB = \\angle BXA_2 = 90^\\circ,$ so $A_2, X, C_1$ are collinear. Similarly, $B_1, X, C_2$ and $A_1, X, B_2$ are collinear, so the three lines concur and we are done.\n~Leonard_my_dude"
+    },
+    {
+        "Problem": "An equilateral triangle $\\Delta$ of side length $L>0$ is given. Suppose that $n$ equilateral triangles with side length 1 and with non-overlapping interiors are drawn inside $\\Delta$, such that each unit equilateral triangle has sides parallel to $\\Delta$, but with opposite orientation. (An example with $n=2$ is drawn below.)\n\nProve that",
+        "Solution": "I will use the word \"center\" to refer to the centroid of any equilateral triangle.\nConsider the center of each triangle we place and the surrounding area that is \"denied\" placement for any centers. With a little bit of testing and sliding the equilateral triangle along each side of the center triangle, it is easy to see that this region forms a hexagon of side length $1$ where any other center being in this hexagon would mean that the two equilateral triangles at those centers would overlap.\nLet us define a region for each center as its \"personal space\", where its personal space intersecting with any other center's personal space results in the two equilateral triangles intersecting. Given the above fact, it is easy to see that this personal space region encapsulates a hexagon of sidelength $\\frac{1}{2}$ centered at each center by simply scaling the denied region by a factor of $\\frac{1}{2}$. The area of each personal space would simply be $6(\\frac{\\sqrt{3}}{4}(\\frac{1}{2})^2)=\\frac{3\\sqrt{3}}{8}$. Since each personal space cannot intersect, the maximum amount of equilateral triangles would be bounded by $\\frac{\\text{Area of Big Triangle}}{\\text{Area of individual personal space}}=\\frac{L^2\\frac{\\sqrt{3}}{4}}{\\frac{3\\sqrt{3}}{8}}=\\frac{2}{3}L^2$, which completes the problem. (Unrigorous solution that I hope someone else can improve on ~ hyxue)"
+    },
+    {
+        "Problem": "Carina has three pins, labeled $A, B$, and $C$, respectively, located at the origin of the coordinate plane. In a move, Carina may move a pin to an adjacent lattice point at distance $1$ away. What is the least number of moves that Carina can make in order for triangle $ABC$ to have area 2021?\n(A lattice point is a point $(x, y)$ in the coordinate plane where $x$ and $y$ are both integers, not necessarily positive.)",
+        "Solution_1": "The answer is $128$, achievable by $A=(10,0), B=(0,-63), C=(-54,1)$. We now show the bound.\nWe first do the following optimizations:\n-if you have a point goes both left and right, we may obviously delete both of these moves and decrease the number of moves by $2$.\n-if all of $A,B,C$ lie on one side of the plane, for example $y>0$, we shift them all down, decreasing the number of moves by $3$, until one of the points is on $y=0$ for the first time.\nNow we may assume that $A=(a,d)$, $B=(b,-e)$, $C=(-c,f)$ where $a,b,c,d,e,f \\geq 0$. Note we may still shift all $A,B,C$ down by $1$ if $d,f>0$, decreasing the number of moves by $1$, until one of $d,f$ is on $y=0$ for the first time. So we may assume one of $(a,b)$ and $(d,f)$ is $0$, by symmetry. In particular, by shoelace the answer to 2021 JMO Problem 4 is the minimum of the answers to the following problems:\nCase 1 (where $a=d=0$) if $wx-yz=4042$, find the minimum possible value of $w+x+y+z$.\nCase 2 (else) $wy+xy+xz=(w+x)(y+z)-wz=4042$, find the minimum possible value of $w+x+y+z$.\nNote that $(m+n)^2=4mn+(m-n)^2$ so if $m+n$ is fixed then $mn$ is maximized exactly when $|m-n|$ is minimized. In particular, if $m+n \\leq 127$ then $mn-op \\leq mn \\leq 63*64 = 4032 <4042$ as desired.\n~Lcz"
+    },
+    {
+        "Problem": "For which positive integers $m$ does there exist an infinite arithmetic sequence of integers $a_1,a_2,\\cdots$ and an infinite geometric sequence of integers $g_1,g_2,\\cdots$ satisfying the following properties?\n$\\bullet$ $a_n-g_n$ is divisible by $m$ for all integers $n>1$;\n$\\bullet$ $a_2-a_1$ is not divisible by $m$.",
+        "Solution_1": "We claim that $m$ satisfies the given conditions if and only if $m$ is a perfect square.\nTo begin, we let the common difference of $\\{a_n\\}$ be $d$ and the common ratio of $\\{g_n\\}$ be $r$. Then, rewriting the conditions modulo $m$ gives:\nCondition $(1)$ holds if no consecutive terms in $a_i$ are equivalent modulo $m$, which is the same thing as never having consecutive, equal, terms, in $a_i\\pmod{m}$. By Condition $(2)$, this is also the same as never having equal, consecutive, terms in $g_i\\pmod{m}$:\n\nAlso, Condition $(2)$ holds if\nRestating, $(1),(2)\\quad \\textrm{if} \\quad(3),(4)$, and the conditions $g_{l-1}(r-1)\\not\\equiv 0\\pmod{m}$ and $g_{l-1}(r-1)^2\\equiv0\\pmod{m}$ hold if and only if $m$ is a perfect square.\n[will finish that step here]"
+    },
+    {
+        "Problem": "Let $a$ and $b$ be positive integers. The cells of an $(a + b + 1)\\times (a + b + 1)$ grid are colored amber and bronze such that there are at least $a^2+ab-b$ amber cells and at least $b^2+ab-a$ bronze cells. Prove that it is possible to choose $a$ amber cells and $b$ bronze cells such that no two of the $a+b$ chosen cells lie in the same row or column.",
+        "Solution_1:_Pigeonhole": "Let $n=a+b+1$. There are $n!$ ways to select $n$ cells such that no two are in the same row or column. Each such selection can be specified by $p_i=(p_{i1},p_{i2},\\cdots,p_{in})$, a permutation of $1,2,3,\\cdots,n$, such that in the $k^\\text{th}$ row, the cell in column $p_{ik}$ is selected. Let $A(p_i)$ be the number of amber cells in the selection $p_i$. We just need to prove there exists a $p_i$ such that\n\nUsing contradiction, assuming no such $p_i$ exists.\nIn calculating $\\sum A(p_i)$, each amber cell is counted $(n-1)!$ times. We have\n\n\nWLOG, let $A(p_1)=\\max(A(p_i))$. With the contradiction assumption\n\nSimilarly, let $A(p_2)=\\min(A(p_i))$.\nOne can always move from $p_1$ to $p_2$ via a path in $p$ space\n\nsuch that in each step two numbers in the starting permutation are swapped to get the next permutation. For example if\n\none such path is \n\nThen the value of $A$ along the path changes by at most 2 in each step. \nOn the other hand since $A(p_i)\\neq a, a+1$, there is a gap of at least width 3 in the range of $A$. A path starting above the gap and ending below the gap with max step size two should not exist. Contradiction.\nTherefore there must be an $i$ such that either $A(p_i)=a$ or $A(p_i)=a+1$.\n(Y)",
+        "Solution_2": "We claim that it is possible to choose $a$ amber cells such that no two lie in the same row or column.\nNumber the rows and columns from $1$ to $a+b+1.$ In cell $(x,y),$ write the number $x+y\\pmod{a+b+1}.$  Then two same-numbered cells will be on different rows and columns.\nThere are $a+b+1$ different numbers and $\\geq a^2+ab-b = (a-1)(a+b+1)+1$ amber cells. Using Pigeonhole, we have $a+b+1$ holes, and cells in the same hole are in different rows and columns, and $\\geq(a-1)(a+b+1)+1$ pigeons, so there exists $a$ amber cells such that no two lie in the same row or column. Similarly, there exists b bronze cells such that no two lie in the same row or column.\nChoose those $a$ amber cells and $b$ bronze cells. If no two chosen cells lie in the same row or column then we are done. Otherwise, since there are $a+b+1$ rows and columns, there exists a row and a column with no chosen cell. If their intersection is an amber cell, choose that cells and unchoose an amber cell that is in the same row or column as a bronze cell. If the intersection is a bronze cell, choose that cell and unchoose a chosen bronze cell that is in the same row or column as an amber cell. Repeating this process, we can obtain $a$ amber cells and $b$ bronze cells such that no two lie in the same row or column.",
+        "Solution_3:_Graph_Theoretic": "We can prove this statement by using the Hall's marriage theorem. Let us define a bipartite graph $G = (A \\cup B, E)$, where $A$ is the set of rows of the grid, $B$ is the set of columns of the grid, and $E$ is the set of edges between $A$ and $B$. We say that a row $a \\in A$ is connected to a column $b \\in B$ if and only if the cell in row $a$ and column $b$ is colored amber.\nNow, we need to show that the conditions of Hall's marriage theorem are satisfied. That is, for any subset $S \\subseteq A$, the number of columns connected to $S$ is at least $|S|$. To prove this, let $S$ be any subset of $A$, and let $T$ be the set of columns connected to $S$. Then, the number of amber cells in the rows of $S$ is at least $a|S|$, and the number of amber cells in the rows not in $S$ is at least $b(a+b+1)-a|S|$. Therefore, the total number of amber cells is at least $a|S| + b(a+b+1)-a|S| = ab + b(a+b)$, which is at least $a^2+ab-b$ by assumption. Hence, the number of columns connected to $S$ is at least $b$.\nSimilarly, we can define a bipartite graph $G' = (A \\cup B', E')$, where $B'$ is the set of rows of the grid, and an edge between a row $a$ and a column $b' \\in B'$ exists if and only if the cell in row $a$ and column $b'$ is colored bronze. We can show that the conditions of Hall's marriage theorem are also satisfied for $G'$.\nTherefore, by Hall's marriage theorem, there exist matchings $M$ and $M'$ in $G$ and $G'$, respectively, such that each row and column in the grid is incident to at most one edge in $M$ or $M'$. Let $X$ be the set of cells in the grid that are covered by $M$, and $Y$ be the set of cells in the grid that are covered by $M'$. Then, $|X| = |Y| = a+b$, since each matching covers a total of $a+b$ cells. Moreover, no two cells in $X$ or $Y$ lie in the same row or column, since each row and column is incident to at most one edge in $M$ or $M'$.\nFinally, we can choose $a$ amber cells and $b$ bronze cells from $X$ and $Y$, respectively, such that no two of the chosen cells lie in the same row or column. This is because $|X| = |Y| = a+b$, and there are at least $a^2+ab-b$ amber cells in the grid and at least $b^2+ab-a$ bronze cells in the grid. Therefore, there are at least $a^2+ab-b$ cells colored amber in $X$, and at least $b^2+ab-a$ cells colored bronze in $Y$. This implies that there are at least $a$ amber cells in $X$ and at least $b$ bronze cells in $Y$, so we can choose a subset of $a$ amber cells from $X$ and a subset of $b$ bronze cells from $Y$, such that no two of the chosen cells lie in the same row or column."
+    },
+    {
+        "Problem": "Let $b\\geq2$ and $w\\geq2$ be fixed integers, and $n=b+w$. Given are $2b$ identical black rods and $2w$ identical white rods, each of side length $1$.\nWe assemble a regular $2n$-gon using these rods so that parallel sides are the same color. Then, a convex $2b$-gon $B$ is formed by translating the black rods, and a convex $2w$-gon $W$ is formed by translating the white rods. An example of one way of doing the assembly when $b=3$ and $w=2$ is shown below, as well as the resulting polygons $B$ and $W$.\n\nProve that the difference of the areas of $B$ and $W$ depends only on the numbers $b$ and $w$, and not on how the $2n$-gon was assembled.\n",
+        "Solution_1": "First notice that the black rods and the white rods form polygons iff in the original $2n$-gon, if a side is a color $x$, then the side that is parallel to that side in the original $2n$-gon is also the color $x$.\nWe can prove that the difference in areas is only affected by the values of $b$ and $w$ by showing that for any valid arrangement of $2b$ rods and $2w$ rods, we may switch any two adjacent black and white rods(and their \"parallel pairs\"), and end up with the same area difference.\n\nIn the figure above (click to expand), after the switch, we can see that after removing the mutually congruent parts, we are left with two parallelograms from each color. Let $x$, $\\alpha{}$, $y$, and $\\beta{}$ be defined as shown. Notice that if we angle chase, the sides of the other parallelogram are the same, but if the angles of the original $2n$-gon all have measure $2k$, the angles of the new parallelograms are $\\alpha{}+180-2k$ and $\\beta{}+180-2k$, as shown. We must prove that the differences between the areas are the same.\nUsing area formulas, the change in the difference of areas is $x\\sin{\\alpha{}}-y\\sin{\\beta{}}+y\\sin{(2k-\\beta{})}-x\\sin{(2k-\\alpha{})}$, which is equal to $x(\\sin{k}(2\\cos^2{k})-2\\sin{k}\\cos{k}\\cos{\\alpha{}})-y(\\sin{k}(2\\cos^2{k}))+y(\\sin{k}\\cos{k}\\cos{\\beta{}})$, or $2\\cos{k}(x\\sin{(\\alpha{}-k)}+y\\sin{(k-\\beta{}}))$. Since $\\cos{k}$ is not $0$ because $n\\geq{}3$, we are left with proving that $x\\sin{(\\alpha{}-k)}+y\\sin{(k-\\beta{}})=0$.\nNow we rotate the polygon so that the vertex between the two sides that we switched is at the point $(0,0)$, the angle bisector of that vertex is $y=0$, and the black side is in the positive $y$-direction. Now think of all the sides as vectors, all pointing in the clockwise direction of the $2n$-gon.\nNotice the part labeled $a$ in the black polygons. We have that the vector labeled $x$ is really just the sum of all of the vectors in the part labeled $a$ - or all the vectors in the $2n$-gon that are in the positive $y$-direction excluding the one that was interchanged. Also notice that the angle of this vector $x$ has a signed angle of $\\alpha{}-k$ with $y=0$ and has length $x$ - meaning that the vertical displacement of the vector $x$ from $y=0$ is equal to $x\\sin{(\\alpha{}-k)}$! Similarly, we get that the vertical displacement of the vector $y$ is equivalent to $y\\sin{(k-\\beta{}})$.\nAdding these two together, we get that $x\\sin{(\\alpha{}-k)}+y\\sin{(k-\\beta{}})$ is simply the vertical displacement of the sum of the vectors $x$ and $y$. Since the sum of the vectors $x$ and $y$ is equivalent to the sum of the vectors in the positive half of the polygon minus the sum of the black vector that would be switched with the white vector(the leftmost vector in the positive half of the polygon) and the rightmost vector in the positive half(which is the parallel pair of the white vector that would be interchanged later), and we know that this sum happens to have a vertical displacement of $0$, along with the fact that the positive half of the polygon summed together also has a vertical displacement of $0$, we get that the total vertical displacement is $0$, meaning that $x\\sin{(\\alpha{}-k)}+y\\sin{(k-\\beta{}})=0$, and we are done.\n~by @peppapig_\n",
+        "Solution_2": "Pick a pair of parallel sides of the regular $2n$-gon and flip their color. Let those two lines be the horizontal. The area of the polygon formed by rods of the original color decreases by a parallelogram with height equal to the sum of the vertical heights of the rods of that color divided by $2$ and base length $1$. Similarly, the area of the polygon formed by rods of the new color increases by the sum of the vertical heights of the rods of the new color divided by $2$. The sum of the heights of all the rod is equal to twice the height of the $2n$-gon, so the difference between the areas of $B$ and $W$ changes by the height of the $2n$-gon, which is fixed when $n$ is fixed.\nIf we pair two pairs different-colored pairs of parallel sides of the $2n$-gon and flip the colors of both pairs, then $b$ and $w$ and the difference in the areas of $B$ and $W$ will remain unchanged. Thus, the difference in the areas of $B$ and $W$ depends only on $b$ and $w$."
+    },
+    {
+        "Problem": "Let $ABCD$ be a rhombus, and let $K$ and $L$ be points such that $K$ lies inside the rhombus, $L$ lies outside the rhombus, and $KA=KB=LC=LD$. Prove that there exist points $X$ and $Y$ on lines $AC$ and $BD$ such that $KXLY$ is also a rhombus.",
+        "Solution": "(Image of the solution is here [1])\nLet's draw ($\\ell$) perpendicular bisector of $\\overline{KL}$. Let $X, Y$ be intersections of $\\ell$ with $AC$ and $BD$, respectively. $KXLY$ is a kite. Let $O$ mid-point of $\\overline{KL}$. Let $M$ mid-point of $\\overline{BD}$ (and also $M$ is mid-point of $\\overline{AC}$). $X, O, Y$ are on the line $\\ell$.\n$BK=DL$, $BX=XD$, $XK=XL$ and so $\\triangle BXK \\cong \\triangle DXL$ (side-side-side). By spiral similarity, $\\triangle BXD \\sim\\triangle KXL$. Hence, we get\nSimilarly, $AK =CL$, $YK = YL$, $YA=YC$ and so $\\triangle AKY \\cong \\triangle CYL$ (side-side-side). From spiral similarity, $\\triangle YKL\\sim \\triangle YAC$. Thus,\n\nIf we can show that $a=b$, then the kite $KXLY$ will be a rhombus.\nBy spiral similarities, $\\dfrac{BX}{BD} = \\dfrac{XK}{KL}$ and $\\dfrac{YA}{AC} = \\dfrac{YK}{KL}$. Then, $KL = \\dfrac{BD \\cdot XK}{BX} = \\dfrac{AC \\cdot YK}{YA}$.\n$\\dfrac{XK}{YK} = \\dfrac{AC \\cdot BX}{AY \\cdot BD}$. Then, $\\dfrac{YK}{XK} = \\dfrac{(BD/2) \\cdot AY}{(AC/2) \\cdot BX} = \\dfrac{\\sin b}{\\sin a}$. Also, in the right triangles $\\triangle KXO$ and $\\triangle KYO$,  $\\dfrac{YK}{XK} = \\dfrac{OK/\\cos a}{OK/\\cos b} = \\dfrac{\\cos b}{\\cos a}$. Therefore,\n$\\sin a \\cos b = \\sin b \\cos a \\implies \\sin a \\cos b - \\sin b \\cos a = 0 \\implies \\sin(a-b) = 0$ and we get $a=b$.\n(Lokman GÖKÇE)"
+    },
+    {
+        "Problem": "Find all pairs of primes $(p,q)$ for which $p-q$ and $pq-q$ are both perfect squares.",
+        "Solution_1": "We first consider the case where one of $p,q$ is even. If $p=2$, $p-q=0$ and $pq-q=2$ which doesn't satisfy the problem restraints. If $q=2$, we can set $p-2=x^2$ and $2p-2=y^2$ giving us $p=y^2-x^2=(y+x)(y-x)$. This forces $y-x=1$ so $p=2x+1\\rightarrow 2x+1=x^2+2 \\rightarrow x=1$ giving us the solution $(p,q)=(3,2)$.\nNow assume that $p,q$ are both odd primes. Set $p-q=x^2$ and $pq-q=y^2$ so $(pq-q)-(p-q)=y^2-x^2 \\rightarrow p(q-1)$ $=(y+x)(y-x)$. Since $y+x>y-x$, $p | (x+y)$. Note that $q-1$ is an even integer and since $y+x$ and $y-x$ have the same parity, they both must be even. Therefore, $x+y=pk$ for some positive even integer $k$. On the other hand, $p>p-q=x^2 \\rightarrow p>x$ and $p^2-p>pq-q=y^2 \\rightarrow p>y$. Therefore, $2p>x+y$ so $x+y=p$, giving us a contradiction.\nTherefore, the only solution to this problem is $(p,q)=(3,2)$.\n~BennettHuang"
+    },
+    {
+        "Problem": "Find all triples of positive integers $(x,y,z)$ that satisfy the equation\n\n",
+        "Solution_1": "We claim that the only solutions are $(2,3,3)$ and its permutations.\nFactoring the above squares and canceling the terms gives you:\n$8(xyz)^2 + 2(x^2 +y^2 + z^2) = 4((xy)^2 + (yz)^2 + (zx)^2) + 2024$\nJumping on the coefficients in front of the $x^2$, $y^2$, $z^2$ terms, we factor into:\n$(2x^2 - 1)(2y^2 - 1)(2z^2 - 1) = 2023$\nRealizing that the only factors of 2023 that could be expressed as $(2x^2 - 1)$ are $1$, $7$, and $17$, we simply find that the only solutions are $(2,3,3)$ by inspection.\n-Max\nAlternatively, a more obvious factorization is:\n$2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023$\n$(\\sqrt{2}x+\\sqrt{2}y+\\sqrt{2}z+2\\sqrt{2}xyz)^2-(2xy+2yz+2zx+1)^2=2023$\n$(2\\sqrt{2}xyz+2xy+2yz+2zx+\\sqrt{2}x+\\sqrt{2}y+\\sqrt{2}z+1)(2\\sqrt{2}xyz-2xy-2yz-2zx+\\sqrt{2}x+\\sqrt{2}y+\\sqrt{2}z-1)=2023$\n$(\\sqrt{2}x+1)(\\sqrt{2}y+1)(\\sqrt{2}z+1)(\\sqrt{2}x-1)(\\sqrt{2}y-1)(\\sqrt{2}z-1)=2023$\n$(2x^2-1)(2y^2-1)(2z^2-1)=2023$\nProceed as above. ~eevee9406"
+    },
+    {
+        "Problem": "(Holden Mui) In an acute triangle $ABC$, let $M$ be the midpoint of $\\overline{BC}$. Let $P$ be the foot of the perpendicular from $C$ to $AM$. Suppose that the circumcircle of triangle $ABP$ intersects line $BC$ at two distinct points $B$ and $Q$. Let $N$ be the midpoint of $\\overline{AQ}$. Prove that $NB=NC$.",
+        "Solution_1": "The condition is solved only if $\\triangle{NBC}$ is isosceles, which in turn only happens if $\\overline{MN}$ is perpendicular to $\\overline{BC}$.\nNow, draw the altitude from $A$ to $\\overline{BC}$, and call that point $X$. Because of the Midline Theorem, the only way that this condition is met is if $\\triangle{AXQ} \\sim \\triangle{NMQ}$, or if $\\overline{XM}=\\overline{MQ}$.\nBy $AA$ similarity, $\\triangle{AXM} \\sim \\triangle{CPM}$. Using similarity ratios, we get that $\\frac{\\overline{AM}}{\\overline{XM}}=\\frac{\\overline{CM}}{\\overline{PM}}$. Rearranging, we get that $\\overline{AM} \\cdot \\overline{MP}=\\overline{XM} \\cdot \\overline{MC}$. This implies that $AXPC$ is cyclic.\nNow we start using Power of a Point. We get that $\\overline{BX} \\cdot \\overline {XQ}= \\overline{AM} \\cdot \\overline{MP}$, and $\\overline{AM} \\cdot \\overline{MP}=\\overline{XM} \\cdot \\overline{MC}$ from before. This leads us to get that $\\overline{BX} \\cdot \\overline {XQ}=\\overline{XM} \\cdot \\overline{MC}$.\nNow we assign variables to the values of the segments. Let $\\overline{BX}=a, \\overline{XM}=b, \\overline{MQ}=c,$ and $\\overline{QC}=d$. The equation from above gets us that $(a+b)c=b(c+d)$. As $a+b=c+d$ from the problem statements, this gets us that $b=c$ and $\\overline{XC}=\\overline{CQ}$, and we are done.\n-dragoon and rhydon516 (:",
+        "Solution_2": "Let $D$ be the foot of the altitude from $A$ onto $BC$. We want to show that $DM=MQ$ for obvious reasons.\nNotice that $ADPC$ is cyclic and that $M$ lies on the radical axis of $(ABPQ)$ and $(ADPC)$. By Power of a Point, $(CM)(DM)=(BM)(MQ)$. As $BM=CM$, we have $DM=MQ$, as desired.\n- Leo.Euler",
+        "Solution_3_.28Less_technical_bary.29": "We are going to use barycentric coordinates on $\\triangle ABC$. Let $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$, and $a=BC$, $b=CA$, $c=AB$. We have $M=\\left(0,\\frac{1}{2},\\frac{1}{2}\\right)$ and $P=(x:1:1)$ so $\\overrightarrow{CP}=\\left(\\frac{x}{x+2},\\frac{1}{x+2},\\frac{1}{x+2}-1\\right)$ and $\\overrightarrow{AM}=\\left(-1,\\frac{1}{2},\\frac{1}{2}\\right)$. Since $\\overleftrightarrow{CP}\\perp\\overleftrightarrow{AM}$, it follows that\n\nSolving this gives\n\nso\n\nThe equation for $(ABP)$ is\n\nPlugging in $A$ and $B$ gives $u=v=0$. Plugging in $P$ gives\n\nso\n\nNow let $Q=(0,t,1-t)$ where\n\nso $Q=\\left(0,\\frac{w}{a^2},1-\\frac{w}{a^2}\\right)$. It follows that $N=\\left(\\frac{1}{2},\\frac{w}{2a^2},1-\\frac{w}{2a^2}\\right)$. It suffices to prove that $\\overleftrightarrow{ON}\\perp\\overleftrightarrow{BC}$. Setting $\\overrightarrow{O}=0$, we get $\\overrightarrow{N}=\\left(\\frac{1}{2},\\frac{w}{2a^2},1-\\frac{w}{2a^2}\\right)$. Furthermore we have $\\overrightarrow{CB}=(0,1,-1)$ so it suffices to prove that\n\nwhich is valid. $\\square$\n~KevinYang2.71",
+        "Solution_4_.28Less_bashy_bary.29": "We employ barycentric coordinates. Set $AMC$ as the reference triangle with $A = (1, 0, 0)$, $M = (0, 1, 0)$, and $C = (0, 0, 1)$. We immediately have,\n\nSince it passes through $A$, for some $v, w$, the equation of circle $(ABP)$ is,\n\nPlugging in $P$,\n\n\nPlugging in $B$,\n\n\nIn conclusion the circle has formula,\n\n$Q$ is the second intersection of circle $(ABP)$ with $\\overline{CM}$. We let $Q = (0, 1 - t, t)$ for some $t \\neq -1$. Plugging this in,\n\nWe claim that $t = -\\frac{S_B}{a^2}$ is the other solution. \n\n\nFactoring out the $\\frac{3a^2 - b^2 + c^2}{2a^2}$, this is clearly true.\nWe also check that, these are not the same value.\n\n\n\nThe triangle is acute, so this is impossible.\nSince we had a quadratic in $t$ with at most two solutions, the second intersection $Q$ is indeed,\n\nTherefore,\n~ Daniel Ge"
+    },
+    {
+        "Problem": "Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a maximal grid-aligned configuration on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$.",
+        "Solution": "To start off, we put the initial non-covered square in a corner (marked by the shaded square). Let's consider what happens when our first domino slides over the empty square. We will call such a move where we slide a domino over the uncovered square a \"step\":\n\nWhen the vertically-oriented domino above the shaded square moved down to cover the shaded square, it had uncovered a new square two squares above the original one. Next, we consider the second another direction the uncovered square can move to:\n\nIn this case, the horizontally-oriented domino besides the shaded square moved to cover the previous shaded square. The new uncovered square, as a result, would be either two squares to the left, or to the right of the previous shaded square.\nSo, to summarize, during each \"move\", the uncovered square can move either two steps up, down, left, or right. We can make a path for which the shaded square takes:\n\nWe put the original shaded square was in the corner, so, the possible number of squares that cound be uncovered is $\\left(\\dfrac{n+1}{2}\\right)^2 = \\frac{n^2+2n+1}{4}$, all of which can be connected by a single path. Let those $\\frac{n^2+2n+1}{4}$ belong to a set $S$ of squares.\nIf the initial uncovered square was not an element of $S$, then it could either be $1$ or $\\sqrt{2}$ away from a square in $S$. If the initial square was a distance of $1$ from a element of $S$, then there would be $\\left(\\dfrac{n+1}{2}\\right)\\left(\\dfrac{n-1}{2}\\right)$ possible uncovered squares:\n\nIf the initial square was a distance of $\\sqrt{2}$ away from a square in $S$, then there would be $\\left(\\dfrac{n-1}{2}\\right)^2$ possible squares that could be uncovered:\n\nSo, that means the maximal number of squares that can be uncovered is $\\frac{n^2+2n+1}{4}$, thus making a maxmial of $k(C)=\\boxed{\\frac{n^2+2n+1}{4}}$ achievable configurations from $C$ ($C$'s uncovered square must belong in $S$).\n~ sml1809"
+    },
+    {
+        "Problem": "Two players, $B$ and $R$, play the following game on an infinite grid of unit squares, all initially colored white. The players take turns starting with $B$. On $B$'s turn, $B$ selects one white unit square and colors it blue. On $R$'s turn, $R$ selects two white unit squares and colors them red. The players alternate until $B$ decides to end the game. At this point, $B$ gets a score, given by the number of unit squares in the largest (in terms of area) simple polygon containing only blue unit squares. What is the largest score $B$ can guarantee?\n(A simple polygon is a polygon (not necessarily convex) that does not intersect itself and has no holes.)",
+        "Solution": "It is clear that $B$ can guarantee a score of $4$ squares. We will show that $R$ has a strategy to limit blue to $4$ squares, thus solving the problem.\nPartition the grid into 2x2 squares. Red's strategy is as follows:\n- If $B$ plays in a 2x2 square, play the two adjacent squares to $B$'s square that are not in the 2x2 square.\n- If one (or both) of these moves are blocked, instead play a square a megaparsec away from the rest of the moves. This move can only benefit you and will not change the outcome of the game.\nBy induction, it is clear that no two blue squares that are adjacent are not in the same 2x2 square. Thus, we conclude that $R$ has limited blue to a maximum score of $2^2 = 4$, and the proof is complete. $\\square$\n~mathboy100"
+    },
+    {
+        "Problem": "A positive integer $a$ is selected, and some positive integers are written on a board. Alice and Bob play the following game. On Alice's turn, she must replace some integer $n$ on the board with $n+a$, and on Bob's turn he must replace some even integer $n$ on the board with $n/2$. Alice goes first and they alternate turns. If on his turn Bob has no valid moves, the game ends.\nAfter analyzing the integers on the board, Bob realizes that, regardless of what moves Alice makes, he will be able to force the game to end eventually. Show that, in fact, for this value of $a$ and these integers on the board, the game is guaranteed to end regardless of Alice's or Bob's moves.",
+        "Solution": "We claim that the game will always end if and only if $\\nu_{2}(n) < \\nu_{2} (a)$ for all $n$ that are in the list of positive integers.\nFirst, we will prove the if direction. Notice that if Alice adds $a$ to $n,$ since we have  for all integers $k>0,$ eventually Bob will decrease $\\nu_{2} (n)$ by $1,$ and Alice will not be able to change $\\nu_{2} (n),$ and so $\\nu_{2}(n)$ will eventually become $0.$\nSimilarly, all of the other numbers on the list will have the same fate as $n,$ and so, no matter what Bob or Alice do, the game will end.\nNow, we complete the only if direction, i.e. if $\\nu_{2} (n) \\ge \\nu_{2} (a),$ where $n$ is one of the numbers in the list, we will prove that Alice can keep the game going forever.\nNotice that Bob can only decrease $\\nu_{2} (n)$ by $1$ at a time, and so if $\\nu_{2} (n) \\le \\nu_{2} (a),$ we need $\\nu_{2} (n) = \\nu_{2} (a)$ at some point. But, then, if this is true, take $n = 2^{k}(2m+1),$ and $a = 2^{k}(2\\ell+1),$ and notice that \n so\n$\\nu_{2}(a+n) > \\nu_{2} (n).$\nThus, if Bob gets $\\nu_{2} (n)$ equal to $\\nu_{2} (a),$ Alice can simply add $a$ to $n,$ and then $\\nu_{2} (n) > \\nu_{2} (a)$ again. Thus, Alice can keep the game going forever, and hence we are done.\n~Mr.Sharkman"
+    },
+    {
+        "Problem": "Isosceles triangle $ABC$, with $AB=AC$, is inscribed in circle $\\omega$. Let $D$ be an arbitrary point inside $BC$ such that $BD\\neq DC$. Ray $AD$ intersects $\\omega$ again at $E$ (other than $A$). Point $F$ (other than $E$) is chosen on $\\omega$ such that $\\angle DFE = 90^\\circ$. Line $FE$ intersects rays $AB$ and $AC$ at points $X$ and $Y$, respectively. Prove that $\\angle XDE = \\angle EDY$.",
+        "Solution": "All angle and side length names are defined as in the figures below. Figure 1 is the diagram of the problem while Figure 2 is the diagram of the Ratio Lemma. Do note that the point names defined in the Ratio Lemma are not necessarily the same defined points on Figure 1.\nFirst, we claim the Ratio Lemma:\n\nWe prove this as follows:\nBy the Law of Sines of $\\triangle{ABP},$ we get:\n\nBy the Law of Sines of $\\triangle{ACP},$ we get:\n\nDividing each of the corresponding equations, we get:\n\nNoting that $\\sin \\theta_1= \\sin(180^\\circ - \\theta_1’)=\\sin \\theta_1’,$ we have:\nWe transition into our problem.\nBy Power of a Point on circle $\\omega,$ we have:\n\n\nBy power of a point on circle $\\omega_1,$ we have:\n\n\nUsing the transitive property, we have:\n\n\nThus, by reverse Power of a Point, we have that quadrilaterals $ADKB$ and $ACLD$ are cyclic.\nNow, we move to angle chasing.\nIt is from here in which we focus on quadrilateral $AKEL.$\nLet $\\angle KDE=\\beta$ and $\\angle EDL=\\gamma.$ Now realize that all we want to show, as given by the problem statement, is $\\beta=\\gamma.$\nThus, $\\angle{KED}=90^\\circ-\\beta, \\angle{DEL}=90^\\circ-\\gamma.$\nWe also have:\n\\begin{align*}\n\\angle LAD&=180^\\circ-(\\angle ADL+\\alpha)\\\\\n&=180^\\circ-((180^\\circ-\\gamma)+\\alpha))\\\\\n&=\\gamma-\\alpha.\n\\end{align*}\nSimilarly, $\\angle{KAD}=\\beta-\\alpha.$\nCalling back to what we want to show - if we want to show that $\\beta=\\gamma,$ it suffices to show that $\\angle KAD=\\angle LAD$ and $\\angle KED=\\angle LED.$ But this is equivalent to showing $\\triangle{AKE}\\sim\\triangle{ALE}.$\nHowever, applying our Ratio Lemma to $\\triangle{AKE},$ we have:\n\nApplying our Ratio Lemma to $\\triangle{ALE},$ we have:\n\nFrom the transitive property, we have:\n\nNow, realize that we already have a shared side $AE$ and equivalent angles $\\angle{ALE}=90^\\circ+\\alpha=\\angle{AKE}.$ Thus, we establish a stronger bound between these two triangles - a congruence.\nThus, these two triangles are similar, and our proof is complete.\nmathboy282\n\n(cred. to v_Enhance diagram)"
+    },
+    {
+        "Problem": "Let $ABCD$ be a cyclic quadrilateral with $AB = 7$ and $CD = 8$. Points $P$ and $Q$ are selected on segment $AB$ such that $AP = BQ = 3$. Points $R$ and $S$ are selected on segment $CD$ such that $CR = DS = 2$. Prove that $PQRS$ is a cyclic quadrilateral.",
+        "Solution_1": "First, let $E$ and $F$ be the midpoints of $AB$ and $CD$, respectively. It is clear that $AE=BE=3.5$, $PE=QE=0.5$, $DF=CF=4$, and $SF=RF=2$. Also, let $O$ be the circumcenter of $ABCD$.\n\nBy properties of cyclic quadrilaterals, we know that the circumcenter of a cyclic quadrilateral is the intersection of its sides' perpendicular bisectors. This implies that $OE\\perp AB$ and $OF\\perp CD$. Since $E$ and $F$ are also bisectors of $PQ$ and $RS$, respectively, if $PQRS$ is indeed a cyclic quadrilateral, then its circumcenter is also at $O$. Thus, it suffices to show that $OP=OQ=OR=OS$.\nNotice that $PE=QE$, $EO=EO$, and $\\angle QEO=\\angle PEO=90^\\circ$. By SAS congruency, $\\Delta QOE\\cong\\Delta POE\\implies QO=PO$. Similarly, we find that $\\Delta SOF\\cong\\Delta ROF$ and $OS=OR$. We now need only to show that these two pairs are equal to each other.\nDraw the segments connecting $O$ to $B$, $Q$, $C$, and $R$.\n\nAlso, let $r$ be the circumradius of $ABCD$. This means that $AO=BO=CO=DO=r$. Recall that $\\angle BEO=90^\\circ$ and $\\angle CFO=90^\\circ$. Notice the several right triangles in our figure.\nLet us apply Pythagorean Theorem on $\\Delta BEO$. We can see that $EO^2+EB^2=BO^2\\implies EO^2+3.5^2=r^2\\implies EO=\\sqrt{r^2-12.25}.$\nLet us again apply Pythagorean Theorem on $\\Delta QEO$. We can see that $QE^2+EO^2=QO^2\\implies0.5^2+r^2-12.25=QO^2\\implies QO=\\sqrt{r^2-12}.$\nLet us apply Pythagorean Theorem on $\\Delta CFO$. We get $CF^2+OF^2=OC^2\\implies4^2+OF^2=r^2\\implies OF=\\sqrt{r^2-16}$.\nWe finally apply Pythagorean Theorem on $\\Delta RFO$. This becomes $OF^2+FR^2=OR^2\\implies r^2-16+2^2=OR^2\\implies OR=\\sqrt{r^2-12}$.\nThis is the same expression as we got for $QO$. Thus, $OQ=OR$, and recalling that $OQ=OP$ and $OR=OS$, we have shown that $OP=OQ=OR=OS$. We are done. QED\n~Technodoggo",
+        "Solution_2": "We can consider two cases: $AB \\parallel CD$ or $AB \\nparallel CD.$ The first case is trivial, as $PQ \\parallel RS$ and we are done due to symmetry. For the second case, WLOG, assume that $A$ and $C$ are located on $XB$ and $XD$ respectively. Extend $AB$ and $CD$ to a point $X,$ and by Power of a Point, we have  which may be written as  or  We can translate this to  so  and therefore by the Converse of Power of a Point $PQRS$ is cyclic, and we are done.\n- spectraldragon8"
+    },
+    {
+        "Problem": "Find all functions $f:\\mathbb{R}\\rightarrow\\mathbb{R}$ that satisfy\n\nfor all $x,y\\in\\mathbb{R}$.",
+        "Solution_1": "I will denote the original equation $f(x^2-y)+2yf(x)=f(f(x))+f(y)$ as OE.\nI claim that the only solutions are $f(x) = -x^2, f(x) = 0,$ and $f(x) = x^2.$\nLemma 1: $f(0) = 0.$\nProof of Lemma 1:\nWe prove this by contradiction. Assume $f(0) = k \\neq 0.$\nBy letting $x=y=0$ in the OE, we have\nIf we let $x = 0$ and $y = k^2$ in the OE, we have\n and if we let $x = k$ and $y = k^2$ in the OE, we get\n\nHowever, upon substituting $x = k$ and $y = -k^2$ in the OE, this implies\n\nThis means $k = 0,$ but we assumed $k \\neq 0,$ contradiction, which proves the Lemma.\nSubstitute $y = 0$ in the OE to obtain\n\nand let $y = x^2$ in the OE to get\nThus we can write $f(x) = kx^2$ for some $k.$ By $f(x^2) = f^2(x),$ we have  so $k = -1, 0, 1,$ yielding the solutions\n- spectraldragon8"
+    }
+]
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