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1fece03f90c5f4895e9a75d8a686cfd4 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_11 | Let $R$ be the set of all possible remainders when a number of the form $2^n$ $n$ a nonnegative integer, is divided by 1000. Let $S$ be the sum of the elements in $R$ . Find the remainder when $S$ is divided by 1000. | If we write out the remainders when powers of $2$ are divided by $1000$ , we must eventually write a number we have already written down. After this happens, we will fall into a cycle, and thus, nothing new will be written down.
The answer extraction of the problem is equivalent to asking for $1+2+4 + \dots + 2^n \pmod{1000}$ , where $2^{n+1}$ is the first number whose remainder when divided by $1000$ is a repeat. This expression is equivalent to $2^{n+1}-1 \pmod{1000}$ . So our answer will just be the first repeated remainder minus $1$
(Everything up to this point has been perfectly rigorous; now, we will destroy this.)
Note that $1$ $2$ and $4$ cannot be the first repeated remainders, due to the $2$ $4$ and $8$ divisibility rules, respectively (i.e. $2^0$ $2^1$ and $2^2$ are the only powers of $2$ that leave a remainder of $1$ $2$ and $4$ , respectively, when divided by $1000$ .) There is no reason why $8$ could not be the first repeated remainder, so it probably is. Thus, our answer is $8-1 = \boxed{7}$ . (In fact, one can quite easily show that if $8$ reappears at all in the sequence, it must be the first repeated remainder.) | null | 7 |
53d186d84f7afecd5cae87d8b491a24b | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_12 | Six men and some number of women stand in a line in random order. Let $p$ be the probability that a group of at least four men stand together in the line, given that every man stands next to at least one other man. Find the least number of women in the line such that $p$ does not exceed 1 percent. | Let $n$ be the number of women present, and let _ be some positive number of women between groups of men. Since the problem states that every man stands next to another man, there cannot be isolated men. Thus, there are five cases to consider, where $(k)$ refers to a consecutive group of $k$ men:
For the first case, we can place the three groups of men in between women. We can think of the groups of men as dividers splitting up the $n$ women. Since there are $n+1$ possible places to insert the dividers, and we need to choose any three of these locations, we have $\dbinom{n+1}{3}$ ways.
The second, third, and fourth cases are like the first, only that we need to insert two dividers among the $n+1$ possible locations. Each gives us $\dbinom{n+1}{2}$ ways, for a total of $3\dbinom{n+1}{2}$ ways.
The last case gives us $\dbinom{n+1}{1}=n+1$ ways.
Therefore, the total number of possible ways where there are no isolated men is
\[\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1).\]
The total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or
\[2\dbinom{n+1}{2}+(n+1).\]
Thus, we want to find the minimum possible value of $n$ where $n$ is a positive integer such that
\[\dfrac{2\dbinom{n+1}{2}+(n+1)}{\dbinom{n+1}{3}+3\dbinom{n+1}{2}+(n+1)}\le\dfrac{1}{100}.\]
After simplification, we arrive at \[\dfrac{6(n+1)}{n^2+8n+6}\le\dfrac{1}{100}.\]
Simplifying again, we see that we seek the smallest positive integer value of $n$ such that $n(n-592)\ge594$ . Clearly $n>592$ , or the left side will not even be positive; we quickly see that $n=593$ is too small but $n=\boxed{594}$ satisfies the inequality. | null | 594 |
3e55e8905408ae010ba4dcc4a76f3610 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_13 | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$ . The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r-\sqrt{s}}{t}$ , where $r$ $s$ , and $t$ are positive integers. Find $r+s+t$ | Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$ , where $a^2+b^2+c^2=1$ . The distance from a point $(X,Y,Z)$ to a plane with equation $Ax+By+Cz+D=0$ is \[\frac{AX+BY+CZ+D}{\sqrt{A^2+B^2+C^2}},\] so the (directed) distance from any point $(x,y,z)$ to the plane is $ax+by+cz+d$ . So, by looking at the three vertices, we have $10a+d=10, 10b+d=11, 10c+d=12$ , and by rearranging and summing, \[(10-d)^2+(11-d)^2+(12-d)^2= 100\cdot(a^2+b^2+c^2)=100.\]
Solving the equation is easier if we substitute $11-d=y$ , to get $3y^2+2=100$ , or $y=\sqrt {98/3}$ . The distance from the origin to the plane is simply $d$ , which is equal to $11-\sqrt{98/3} =(33-\sqrt{294})/3$ , so $33+294+3=\boxed{330}$ | null | 330 |
3e55e8905408ae010ba4dcc4a76f3610 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_13 | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$ . The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r-\sqrt{s}}{t}$ , where $r$ $s$ , and $t$ are positive integers. Find $r+s+t$ | Let the vertices with distance $10,11,12$ be $B,C,D$ , respectively. An equilateral triangle $\triangle BCD$ is formed with side length $10\sqrt{2}$ . We care only about the $z$ coordinate: $B=10,C=11,D=12$ . It is well known that the centroid of a triangle is the average of the coordinates of its three vertices, so $\text{centroid}=(10+11+12)/3=11$ . Designate the midpoint of $BD$ as $M$ . Notice that median $CM$ is parallel to the plane because the $\text{centroid}$ and vertex $C$ have the same $z$ coordinate, $11$ , and the median contains $C$ and the $\text{centroid}$ . We seek the angle $\theta$ of the line: $(1)$ through the centroid $(2)$ perpendicular to the plane formed by $\triangle BCD$ $(3)$ with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular $\textit{in slope}$ to $BD$ . Since $BD$ makes a $2-14-10\sqrt{2}$ right triangle, the orthogonal line makes the same right triangle rotated $90^\circ$ . Therefore, $\sin\theta=\frac{14}{10\sqrt{2}}=\frac{7\sqrt{2}}{10}$
It is also known that the centroid of $\triangle BCD$ is a third of the way between vertex $A$ and $H$ , the vertex farthest from the plane. Since $AH$ is a diagonal of the cube, $AH=10\sqrt{3}$ . So the distance from the $\text{centroid}$ to $A$ is $10/\sqrt{3}$ . So, the $\Delta z$ from $A$ to the centroid is $\frac{10}{\sqrt{3}}\sin\theta=\frac{10}{\sqrt{3}}\left(\frac{7\sqrt{2}}{10}\right)=\frac{7\sqrt{6}}{3}$
Thus the distance from $A$ to the plane is $11-\frac{7\sqrt{6}}{3}=\frac{33-7\sqrt{6}}{3}=\frac{33-\sqrt{294}}{3}$ , and $33+294+3=\boxed{330}$ | null | 330 |
5ab115ffe4227351e131d3828d0131e3 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_14 | Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$ $M_3$ $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ $\overline{A_3 A_4}$ $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively. For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ $R_3 \perp R_5$ $R_5 \perp R_7$ , and $R_7 \perp R_1$ . Pairs of rays $R_1$ and $R_3$ $R_3$ and $R_5$ $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ $B_3$ $B_5$ $B_7$ respectively. If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers. Find $m + n$ | We use coordinates. Let the octagon have side length $2$ and center $(0, 0)$ . Then all of its vertices have the form $(\pm 1, \pm\left(1+\sqrt{2}\right))$ or $(\pm\left(1+\sqrt{2}\right), \pm 1)$
By symmetry, $B_{1}B_{3}B_{5}B_{7}$ is a square. Thus lines $\overleftrightarrow{B_{1}B_{3}}$ and $\overleftrightarrow{B_{5}B_{7}}$ are parallel, and its side length is the distance between these two lines. However, this is given to be the side length of the octagon, or $2$
Suppose the common slope of the lines is $m$ and let $m=\tan\theta$ . Then, we want to find \[\cos 2\left(90-\theta\right)=2\cos^{2}\left(90-\theta\right)-1=2\sin^{2}\theta-1.\]
It can easily be seen that the equations of the lines are \begin{align*} B_{1}B_{3}: y-mx+m\left(1+\sqrt{2}\right)=0 \\ B_{5}B_{7}: y-mx-m\left(1+\sqrt{2}\right)=0.\end{align*} By the distance between parallel lines formula , a corollary of the point to line distance formula , the distance between these two lines is \[\frac{|c_{2}-c_{1}|}{\sqrt{a^{2}+b^{2}}}=\frac{2m\left(1+\sqrt{2}\right)}{\sqrt{m^{2}+1}}.\] Since we want this to equal $2$ , we have \begin{align*}\frac{2m\left(1+\sqrt{2}\right)}{\sqrt{m^{2}+1}}&=2 \\ 4m^{2}\left(3+2\sqrt{2}\right)&=4m^{2}+4 \\ \left(12+8\sqrt{2}\right)m^{2}&=4m^{2}+4 \\ \left(8+8\sqrt{2}\right)m^{2}&=4 \\ m^{2}&=\frac{4}{8+8\sqrt{2}} \\ \Rightarrow m^{2}=\tan^{2}\theta=\frac{\sin^{2}\theta}{\cos^{2}\theta}&=\frac{1}{2+2\sqrt{2}}.\end{align*} Since $\sin^{2}\theta+\cos^{2}\theta=1,$ we have $\sin^{2}\theta=\frac{1}{3+2\sqrt{2}}$ . Thus \[2\sin^{2}\theta-1=\frac{2}{3+2\sqrt{2}}-1=\frac{-1-2\sqrt{2}}{3+2\sqrt{2}}=\frac{\left(-1-2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}=\frac{5-4\sqrt{2}}{1}=5-\sqrt{32}.\] The answer is $\boxed{037}$ | null | 037 |
5ab115ffe4227351e131d3828d0131e3 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_14 | Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$ $M_3$ $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ $\overline{A_3 A_4}$ $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively. For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ $R_3 \perp R_5$ $R_5 \perp R_7$ , and $R_7 \perp R_1$ . Pairs of rays $R_1$ and $R_3$ $R_3$ and $R_5$ $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ $B_3$ $B_5$ $B_7$ respectively. If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers. Find $m + n$ | Let $\theta=\angle M_1 M_3 B_1$ . Thus we have that $\cos 2 \angle A_3 M_3 B_1=\cos \left(2\theta + \frac{\pi}{2} \right)=-\sin2\theta$
Since $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ is a regular octagon and $B_1 B_3 = A_1 A_2$ , let $k=A_1 A_2 = A_2 A_3 = B_1 B_3$
Extend $\overline{A_1 A_2}$ and $\overline{A_3 A_4}$ until they intersect. Denote their intersection as $I_1$ . Through similar triangles & the $45-45-90$ triangles formed, we find that $M_1 M_3=\frac{k}{2}(2+\sqrt2)$
We also have that $\triangle M_7 B_7 M_1 =\triangle M_1 B_1 M_3$ through ASA congruence ( $\angle B_7 M_7 M_1 =\angle B_1 M_1 M_3$ $M_7 M_1 = M_1 M_3$ $\angle B_7 M_1 M_7 =\angle B_1 M_3 M_1$ ). Therefore, we may let $n=M_1 B_7 = M_3 B_1$
Thus, we have that $\sin\theta=\frac{n-k}{\frac{k}{2}(2+\sqrt2)}$ and that $\cos\theta=\frac{n}{\frac{k}{2}(2+\sqrt2)}$ . Therefore $\cos\theta-\sin\theta=\frac{k}{\frac{k}{2}(2+\sqrt2)}=\frac{2}{2+\sqrt2}=2-\sqrt2$
Squaring gives that $\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 6-4\sqrt2$ and consequently that $-2\sin\theta\cos\theta = 5-4\sqrt2 = -\sin2\theta$ through the identities $\sin^2\theta + \cos^2\theta = 1$ and $\sin2\theta = 2\sin\theta\cos\theta$
Thus we have that $\cos 2 \angle A_3 M_3 B_1=5-4\sqrt2=5-\sqrt{32}$ . Therefore $m+n=5+32=\boxed{037}$ | null | 037 |
5ab115ffe4227351e131d3828d0131e3 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_14 | Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$ $M_3$ $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ $\overline{A_3 A_4}$ $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively. For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ $R_3 \perp R_5$ $R_5 \perp R_7$ , and $R_7 \perp R_1$ . Pairs of rays $R_1$ and $R_3$ $R_3$ and $R_5$ $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ $B_3$ $B_5$ $B_7$ respectively. If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers. Find $m + n$ | Let $A_1A_2 = 2$ . Then $B_1$ and $B_3$ are the projections of $M_1$ and $M_5$ onto the line $B_1B_3$ , so $2=B_1B_3=-M_1M_5\cos x$ , where $x = \angle A_3M_3B_1$ . Then since $M_1M_5 = 2+2\sqrt{2}, \cos x = \dfrac{-2}{2+2\sqrt{2}}= 1-\sqrt{2}$ $\cos 2x = 2\cos^2 x -1 = 5 - 4\sqrt{2} = 5-\sqrt{32}$ , and $m+n=\boxed{037}$ | null | 037 |
5ab115ffe4227351e131d3828d0131e3 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_14 | Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$ $M_3$ $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ $\overline{A_3 A_4}$ $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively. For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ $R_3 \perp R_5$ $R_5 \perp R_7$ , and $R_7 \perp R_1$ . Pairs of rays $R_1$ and $R_3$ $R_3$ and $R_5$ $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ $B_3$ $B_5$ $B_7$ respectively. If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers. Find $m + n$ | Notice that $R_3$ and $R_7$ are parallel ( $B_1B_3B_5B_7$ is a square by symmetry and since the rays are perpendicular) and $B_1B_3=B_3B_5=s=$ the distance between the parallel rays. If the regular hexagon as a side length of $s$ , then $M_3M_7$ has a length of $s+s\sqrt{2}$ . Let $X$ be on $R_3$ such that $M_7X$ is perpendicular to $M_3X$ , and $\phi=\angle M_7M_3X$ . The distance between $R_3$ and $R_7$ is $s=M_7X$ , so $\sin\phi=\frac{s}{s+s\sqrt{2}}=\frac{1}{1+\sqrt{2}}$
Since we are considering a regular hexagon, $M_3$ is directly opposite to $M_7$ and $\angle A_3M_3B_1=90 ^\circ +\phi$ . All that's left is to calculate $\cos 2\angle A_3M_3B_1=\cos^2(90^\circ+\phi)-\sin^2(90^\circ+\phi)=\sin^2\phi-\cos^2\phi$ . By drawing a right triangle or using the Pythagorean identity, $\cos^2\phi=\frac{2+2\sqrt2}{3+2\sqrt2}$ and $\cos 2\angle A_3M_3B_1=\frac{-1-2\sqrt2}{3+2\sqrt2}=5-4\sqrt2=5-\sqrt{32}$ , so $m+n=\boxed{037}$ | null | 037 |
5ab115ffe4227351e131d3828d0131e3 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_14 | Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$ $M_3$ $M_5$ , and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$ $\overline{A_3 A_4}$ $\overline{A_5 A_6}$ , and $\overline{A_7 A_8}$ , respectively. For $i = 1, 3, 5, 7$ , ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$ $R_3 \perp R_5$ $R_5 \perp R_7$ , and $R_7 \perp R_1$ . Pairs of rays $R_1$ and $R_3$ $R_3$ and $R_5$ $R_5$ and $R_7$ , and $R_7$ and $R_1$ meet at $B_1$ $B_3$ $B_5$ $B_7$ respectively. If $B_1 B_3 = A_1 A_2$ , then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$ , where $m$ and $n$ are positive integers. Find $m + n$ | Assume that $A_1A_2=1.$ Denote the center $O$ , and the midpoint of $B_1$ and $B_3$ as $B_2$ . Then we have that \[\cos\angle A_3M_3B_1=\cos(\angle A_3M_3O+\angle OM_3B_1)=-\sin(\angle OM_3B_1)=-\frac{OB_2}{OM_3}=-\frac{1/2}{1/2+\sqrt2/2}=-\frac{1}{\sqrt2+1}=1-\sqrt2.\] Thus, by the cosine double-angle theorem, \[\cos2\angle A_3M_3B_1=2(1-\sqrt2)^2-1=5-\sqrt{32},\] so $m+n=\boxed{037}$ | null | 037 |
3a0839ddaf5e056fbf6c89e7e0dbc12a | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_15 | For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ $b$ , and $c$ . Find $|a| + |b| + |c|$ | From Vieta's formulas, we know that $a+b+c = 0$ , and $ab+bc+ac = -2011$ . Thus $a = -(b+c)$ . All three of $a$ $b$ , and $c$ are non-zero: say, if $a=0$ , then $b=-c=\pm\sqrt{2011}$ (which is not an integer). $\textsc{wlog}$ , let $|a| \ge |b| \ge |c|$ . If $a > 0$ , then $b,c < 0$ and if $a < 0$ , then $b,c > 0$ . We have \[-2011=ab+bc+ac = a(b+c)+bc = -a^2+bc\] Thus $a^2 = 2011 + bc$ . We know that $b$ $c$ have the same sign. So $|a| \ge 45 = \lceil \sqrt{2011} \rceil$
Also, if we fix $a$ $b+c$ is fixed, so $bc$ is maximized when $b = c$ . Hence, \[2011 = a^2 - bc > \tfrac{3}{4}a^2 \qquad \Longrightarrow \qquad a ^2 < \tfrac{4}{3}\cdot 2011 = 2681+\tfrac{1}{3}\] So $|a| \le 51$ . Thus we have bounded $a$ as $45\le |a| \le 51$ , i.e. $45\le |b+c| \le 51$ since $a=-(b+c)$ . Let's analyze $bc=(b+c)^2-2011$ . Here is a table:
We can tell we don't need to bother with $45$
$105 = (3)(5)(7)$ , So $46$ won't work. $198/47 > 4$
$198$ is not divisible by $5$ $198/6 = 33$ , which is too small to get $47$
$293/48 > 6$ $293$ is not divisible by $7$ or $8$ or $9$ , we can clearly tell that $10$ is too much.
Hence, $|a| = 49$ $a^2 -2011 = 390$ $b = 39$ $c = 10$
Answer: $\boxed{098}$ | null | 098 |
3a0839ddaf5e056fbf6c89e7e0dbc12a | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_15 | For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ $b$ , and $c$ . Find $|a| + |b| + |c|$ | Starting off like the previous solution, we know that $a + b + c = 0$ , and $ab + bc + ac = -2011$
Therefore, $c = -b-a$
Substituting, $ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011$
Factoring the perfect square, we get: $ab-(b+a)^2=-2011$ or $(b+a)^2-ab=2011$
Therefore, a sum ( $a+b$ ) squared minus a product ( $ab$ ) gives $2011$ ..
We can guess and check different $a+b$ ’s starting with $45$ since $44^2 < 2011$
$45^2 = 2025$ therefore $ab = 2025-2011 = 14$
Since no factors of $14$ can sum to $45$ $1+14$ being the largest sum), a + b cannot equal $45$
$46^2 = 2116$ making $ab = 105 = 3 * 5 * 7$
$5 * 7 + 3 < 46$ and $3 * 5 * 7 > 46$ so $46$ cannot work either.
We can continue to do this until we reach $49$
$49^2 = 2401$ making $ab = 390 = 2 * 3 * 5* 13$
$3 * 13 + 2* 5 = 49$ , so one root is $10$ and another is $39$ . The roots sum to zero, so the last root must be $-49$
$|-49|+10+39 = \boxed{098}$ | null | 098 |
3a0839ddaf5e056fbf6c89e7e0dbc12a | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_15 | For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ $b$ , and $c$ . Find $|a| + |b| + |c|$ | Let us first note the obvious that is derived from Vieta's formulas: $a+b+c=0, ab+bc+ac=-2011$ . Now, due to the first equation, let us say that $a+b=-c$ , meaning that $a,b>0$ and $c<0$ . Now, since both $a$ and $b$ are greater than 0, their absolute values are both equal to $a$ and $b$ , respectively. Since $c$ is less than 0, it equals $-a-b$ . Therefore, $|c|=|-a-b|=a+b$ , meaning $|a|+|b|+|c|=2(a+b)$ . We now apply Newton's sums to get that $a^2+b^2+ab=2011$ ,or $(a+b)^2-ab=2011$ . Solving, we find that $49^2-390$ satisfies this, meaning $a+b=49$ , so $2(a+b)=\boxed{098}$ | null | 098 |
3a0839ddaf5e056fbf6c89e7e0dbc12a | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_15 | For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ $b$ , and $c$ . Find $|a| + |b| + |c|$ | We have
As a result, we have
$a+b+c=0$
$ab+bc+ac=-2011$
$abc=-m$
So, $a=-b-c$
As a result, $ab+bc+ac=(-b-c)b+(-b-c)c+bc=-b^2-c^2-bc=-2011$
Solve $b=\frac {-c+\sqrt{c^2-4(c^2-2011)}}{2}$ and $\Delta =8044-3c^2=k^2$ , where $k$ is an integer
Cause $89<\sqrt{8044}<90$
So, after we tried for $2$ times, we get $k=88$ and $c=10$
then $b=39$ $a=-b-c=-49$
As a result, $|a|+|b|+|c|=10+39+49=\boxed{098}$ | null | 098 |
3a0839ddaf5e056fbf6c89e7e0dbc12a | https://artofproblemsolving.com/wiki/index.php/2011_AIME_I_Problems/Problem_15 | For some integer $m$ , the polynomial $x^3 - 2011x + m$ has the three integer roots $a$ $b$ , and $c$ . Find $|a| + |b| + |c|$ | First, derive the equations $a=-b-c$ and $ab+bc+ca=-2011\implies b^2+bc+c^2=2011$ . Since the product is negative, $a$ is negative, and $b$ and $c$ positive. Now, a simple mod 3 testing of all cases shows that $b\equiv \{1,2\} \pmod{3}$ , and $c$ has the repective value. We can choose $b$ not congruent to 0, make sure you see why. Now, we bash on values of $b$ , testing the quadratic function to see if $c$ is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for $b=10$ $c=39, -49$ . Choosing $c$ positive we get $a=-49$ , so $|a|+|b|+|c|=10+29+39=\boxed{098}$ ~firebolt360 | null | 098 |
061f745ce5e23bbab2501911de3dff19 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_1 | Gary purchased a large beverage, but only drank $m/n$ of it, where $m$ and $n$ are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only $2/9$ as much beverage. Find $m+n$ | Let $x$ be the fraction consumed, then $(1-x)$ is the fraction wasted. We have $\frac{1}{2} - 2x =\frac{2}{9} (1-x)$ , or $9 - 36x = 4 - 4x$ , or $32x = 5$ or $x = 5/32$ . Therefore, $m + n = 5 + 32 = \boxed{037}$ | null | 037 |
b0c18c1657d82ad08385156c9588ae71 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_3 | The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle | Another way to solve this problem would be to use exterior angles. Exterior angles of any polygon add up to $360^{\circ}$ . Since there are $18$ exterior angles in an 18-gon, the average measure of an exterior angles is $\frac{360}{18}=20^\circ$ . We know from the problem that since the exterior angles must be in an arithmetic sequence, the median and average of them is $20$ . Since there are even number of exterior angles, the middle two must be $19^\circ$ and $21^\circ$ , and the difference between terms must be $2$ . Check to make sure the smallest exterior angle is greater than $0$ $19-2(8)=19-16=3^\circ$ . It is, so the greatest exterior angle is $21+2(8)=21+16=37^\circ$ and the smallest interior angle is $180-37=\boxed{143}$ | null | 143 |
b0c18c1657d82ad08385156c9588ae71 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_3 | The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle | The sum of the angles in a 18-gon is $(18-2) \cdot 180^\circ = 2880 ^\circ.$ Because the angles are in an arithmetic sequence, we can also write the sum of the angles as $a+(a+d)+(a+2d)+\dots+(a+17d)=18a+153d,$ where $a$ is the smallest angle and $d$ is the common difference. Since these two are equal, we know that $18a+153d = 2880 ^\circ,$ or $2a+17d = 320^\circ.$ The smallest value of $d$ that satisfies this is $d=2,$ so $a=143.$ Other values of $d$ and $a$ satisfy that equation, but if we tried any of them the last angle would be greater than $180,$ so the only value of $a$ that works is $a=\boxed{143}$ | null | 143 |
b0c18c1657d82ad08385156c9588ae71 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_3 | The degree measures of the angles in a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle | Each individual angle in a $18$ -gon is $\frac {(18-2) \cdot 180^\circ}{18} = 160^\circ$ . Since no angle in a convex polygon can be larger than $180^\circ$ , the smallest angle possible is in the set $159, 161, 157, 163, 155, 165, 153, 167, 151, 169, 149, 171, 147, 173, 145, 175, 143, 177$
Our smallest possible angle is $\boxed{143}$ | null | 143 |
216a3e14cf7880de4246a48350cf18b9 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_4 | In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] pointpen = black; pathpen = linewidth(0.7); pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C), D2 = IP(D--D+P-B, A--C); D(MP("A",D(A))--MP("B",D(B),N)--MP("C",D(C))--cycle); D(A--MP("D",D(D),NE)--MP("D'",D(D2))); D(B--MP("P",D(P))); D(MP("M",M,NW)); MP("20",(B+D)/2,ENE); MP("11",(C+D)/2,ENE); [/asy] Let $D'$ be on $\overline{AC}$ such that $BP \parallel DD'$ . It follows that $\triangle BPC \sim \triangle DD'C$ , so \[\frac{PC}{D'C} = 1 + \frac{BD}{DC} = 1 + \frac{AB}{AC} = \frac{31}{11}\] by the Angle Bisector Theorem . Similarly, we see by the Midline Theorem that $AP = PD'$ . Thus, \[\frac{CP}{PA} = \frac{1}{\frac{PD'}{PC}} = \frac{1}{1 - \frac{D'C}{PC}} = \frac{31}{20},\] and $m+n = \boxed{51}$ | null | 51 |
216a3e14cf7880de4246a48350cf18b9 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_4 | In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Assign mass points as follows: by Angle-Bisector Theorem, $BD / DC = 20/11$ , so we assign $m(B) = 11, m(C) = 20, m(D) = 31$ . Since $AM = MD$ , then $m(A) = 31$ , and $\frac{CP}{PA} = \frac{m(A) }{ m(C)} = \frac{31}{20}$ , so $m+n = \boxed{51}$ | null | 51 |
216a3e14cf7880de4246a48350cf18b9 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_4 | In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | By Menelaus' Theorem on $\triangle ACD$ with transversal $PB$ \[1 = \frac{CP}{PA} \cdot \frac{AM}{MD} \cdot \frac{DB}{CB} = \frac{CP}{PA} \cdot \left(\frac{1}{1+\frac{AC}{AB}}\right) \quad \Longrightarrow \quad \frac{CP}{PA} = \frac{31}{20}.\] So $m+n = \boxed{051}$ | null | 051 |
216a3e14cf7880de4246a48350cf18b9 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_4 | In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We will use barycentric coordinates. Let $A = (1, 0, 0)$ $B = (0, 1, 0)$ $C = (0, 0, 1)$ . By the Angle Bisector Theorem $D = [0:11:20] = \left(0, \frac{11}{31}, \frac{20}{31}\right)$ . Since $M$ is the midpoint of $AD$ $M = \frac{A + D}{2} = \left(\frac{1}{2}, \frac{11}{62}, \frac{10}{31}\right)$ . Therefore, the equation for line BM is $20x = 31z$ . Let $P = (x, 0, 1-x)$ . Using the equation for $BM$ , we get \[20x = 31(1-x)\] \[x = \frac{31}{51}\] Therefore, $\frac{CP}{PA} = \frac{1-x}{x} = \frac{31}{20}$ so the answer is $\boxed{051}$ | null | 051 |
216a3e14cf7880de4246a48350cf18b9 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_4 | In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $DC=x$ . Then by the Angle Bisector Theorem, $BD=\frac{20}{11}x$ . By the Ratio Lemma, we have that $\frac{PC}{AP}=\frac{\frac{31}{11}x\sin\angle PBC}{20\sin\angle ABP}.$ Notice that $[\triangle BAM]=[\triangle BMD]$ since their bases have the same length and they share a height. By the sin area formula, we have that \[\frac{1}{2}\cdot20\cdot BM\cdot \sin\angle ABP=\frac{1}{2}\cdot \frac{20}{11}x\cdot BM\cdot\sin\angle PBC.\] Simplifying, we get that $\frac{\sin\angle PBC}{\sin\angle ABP}=\frac{11}{x}.$ Plugging this into what we got from the Ratio Lemma, we have that $\frac{PC}{AP}=\frac{31}{20}\implies\boxed{051.}$ | null | 051. |
216a3e14cf7880de4246a48350cf18b9 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_4 | In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | First, we will find $\frac{MP}{BP}$ . By Menelaus on $\triangle BDM$ and the line $AC$ , we have \[\frac{BC}{CD}\cdot\frac{DA}{AM}\cdot\frac{MP}{PB}=1\implies \frac{62MP}{11BP}=1\implies \frac{MP}{BP}=\frac{11}{62}.\] This implies that $\frac{MB}{BP}=1-\frac{MP}{BP}=\frac{51}{62}$ . Then, by Menelaus on $\triangle AMP$ and line $BC$ , we have \[\frac{AD}{DM}\cdot\frac{MB}{BP}\cdot\frac{PC}{CA}=1\implies \frac{PC}{CA}=\frac{31}{51}.\] Therefore, $\frac{PC}{AP}=\frac{31}{51-31}=\frac{31}{20}.$ The answer is $\boxed{051}$ . -brainiacmaniac31 | null | 051 |
216a3e14cf7880de4246a48350cf18b9 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_4 | In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Assume $ABC$ is a right triangle at $A$ . Line $AD = x$ and $BC = \tfrac{-11}{20}x + 11$ . These two lines intersect at $D$ which have coordinates $(\frac{220}{31},\frac{220}{31})$ and thus $M$ has coordinates $(\frac{110}{31},\frac{110}{31})$ . Thus, the line $BM = \tfrac{11}{51} \cdot (20-x)$ . When $x = 0$ $P$ has $y$ coordinate equal to $\frac{11\cdot20}{51} \frac{AP + CP}{AP} = 1 + \frac{CP}{AP}$ $\tfrac{51}{20} = 1 + \frac{CP}{AP},$ which equals ${\tfrac{31}{20}},$ giving an answer of $\boxed{51}.$ | null | 51 |
216a3e14cf7880de4246a48350cf18b9 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_4 | In triangle $ABC$ $AB=20$ and $AC=11$ . The angle bisector of $\angle A$ intersects $BC$ at point $D$ , and point $M$ is the midpoint of $AD$ . Let $P$ be the point of the intersection of $AC$ and $BM$ . The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We start by using Menelaus' theorem on $\triangle ABD$ and $EC$ .
So, we see that $\frac{BC}{DC}\cdot\frac{DM}{AM}\cdot\frac{AE}{EB}=1$ .
By Angle Bisector theorem, $\frac{BC}{DC}=\frac{31}{11}$ , and therefore after plugging in our values we get $\frac{AE}{EB}=\frac{11}{31}$ .
Then, by Ceva's on the whole figure, we have $\frac{CP}{PA}\cdot\frac{AE}{EB}\cdot\frac{BD}{DC}=1$ .
Plugging in our values, we get $\frac{CP}{PA}=\frac{31}{20}$ , giving an answer of $\boxed{51}$ .
~ESAOPS | null | 51 |
9a81dfca356aca414667ac8387491bbe | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_5 | The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms. | Since the sum of the first $2011$ terms is $200$ , and the sum of the first $4022$ terms is $380$ , the sum of the second $2011$ terms is $180$ .
This is decreasing from the first 2011, so the common ratio is less than one.
Because it is a geometric sequence and the sum of the first 2011 terms is $200$ , second $2011$ is $180$ , the ratio of the second $2011$ terms to the first $2011$ terms is $\frac{9}{10}$ . Following the same pattern, the sum of the third $2011$ terms is $\frac{9}{10}*180 = 162$
Thus, $200+180+162=542$ , so the sum of the first $6033$ terms is $\boxed{542}$ | null | 542 |
9a81dfca356aca414667ac8387491bbe | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_5 | The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms. | Solution by e_power_pi_times_i
The sum of the first $2011$ terms can be written as $\dfrac{a_1(1-k^{2011})}{1-k}$ , and the first $4022$ terms can be written as $\dfrac{a_1(1-k^{4022})}{1-k}$ . Dividing these equations, we get $\dfrac{1-k^{2011}}{1-k^{4022}} = \dfrac{10}{19}$ . Noticing that $k^{4022}$ is just the square of $k^{2011}$ , we substitute $x = k^{2011}$ , so $\dfrac{1}{x+1} = \dfrac{10}{19}$ . That means that $k^{2011} = \dfrac{9}{10}$ . Since the sum of the first $6033$ terms can be written as $\dfrac{a_1(1-k^{6033})}{1-k}$ , dividing gives $\dfrac{1-k^{2011}}{1-k^{6033}}$ . Since $k^{6033} = \dfrac{729}{1000}$ , plugging all the values in gives $\boxed{542}$ | null | 542 |
9a81dfca356aca414667ac8387491bbe | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_5 | The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms. | The sum of the first 2011 terms of the sequence is expressible as $a_1 + a_1r + a_1r^2 + a_1r^3$ .... until $a_1r^{2010}$ . The sum of the 2011 terms following the first 2011 is expressible as $a_1r^{2011} + a_1r^{2012} + a_1r^{2013}$ .... until $a_1r^{4021}$ . Notice that the latter sum of terms can be expressed as $(r^{2011})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})$ . We also know that the latter sum of terms can be obtained by subtracting 200 from 380, which then means that $r^{2011} = 9/10$ . The terms from 4023 to 6033 can be expressed as $(r^{4022})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})$ , which is equivalent to $((9/10)^2)(200) = 162$ . Adding 380 and 162 gives the answer of $\boxed{542}$ | null | 542 |
d958b3940e914e22e5733aa2e31b5d3a | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_6 | Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$ , and $a+d>b+c$ . How many interesting ordered quadruples are there? | We first start out when the value of $a=1$
Doing casework, we discover that $d=5,6,7,8,9,10$ . We quickly find a pattern.
Now, doing this for the rest of the values of $a$ and $d$ , we see that the answer is simply:
$(1)+(2)+(1+3)+(2+4)+(1+3+5)+(2+4+6)+(1)+(2)+(1+3)+(2+4)$ $+(1+3+5)+(1)+(2)+(1+3)+(2+4)+(1)+(2)+(1+3)+(1)+(2)+(1)=\boxed{080}$ | null | 080 |
d958b3940e914e22e5733aa2e31b5d3a | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_6 | Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$ , and $a+d>b+c$ . How many interesting ordered quadruples are there? | Notice that if $a+d>b+c$ , then $(11-a)+(11-d)<(11-b)+(11-c)$ , so there is a bijection between the number of ordered quadruples with $a+d>b+c$ and the number of ordered quadruples with $a+d<b+c$
Quick counting gives that the number of ordered quadruples with $a+d=b+c$ is 50. To count this, consider our numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ . Notice that if, for example, $a+d=b+c=8$ , that the average of $a,d$ and $b,c$ must both be $4$ . In this way, there is a symmetry for this case, centered at $4$ . If instead, say, $a+d=b+c=7$ , an odd number, then there is symmetry with $(a,d);(b,c)$ about $3.5$ . Further, the number of cases for each of these centers of symmetry correspond to a triangular number. Eg centered at $2.5,3,8,8.5$ , there is $1$ case for each and so on, until centered at $5.5$ , there are $10$ possible cases. Adding these all, we have $2(1+3+6)+10=50$
Thus the answer is $\frac{\binom{10}{4}-50}{2} = \boxed{080}.$ | null | 080 |
d958b3940e914e22e5733aa2e31b5d3a | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_6 | Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$ , and $a+d>b+c$ . How many interesting ordered quadruples are there? | Think about a,b,c,and d as distinct objects, that we must place in 4 of 10 spaces. However, in only 1 of 24 of these combinations, will the placement of these objects satisfy the condition in the problem. So we know the total number of ordered quadruples is $(10*9*8*7/24)=210$
Next, intuitively, the number of quadruples where $a+d>b+c$ is equal to the number of quadruples where $a+d<b+c$ . So we need to find the number of quadruples where the two quantities are equal. To do this, all we have to do is consider the cases when $a-d$ ranges from 3 to 9. It would seem natural that a range of 3 would produce 1 option, and a range of 4 would produce 2 options. However, since b and c cannot be equal, a range of 3 or 4 produces 1 option each, a range of 5 or 6 produces 2 options each, a range of 7 or 8 produces 3 options each, and a range of 9 will produce 4 options. In addition, a range of n has 10-n options for combinations of a and d. Multiplying the number of combinations of a and d by the corresponding number of options for b and c gives us 50 total quadruplets where $a+d=b+c$
So the answer will be $\frac{210-50}{2} = \boxed{080}.$ | null | 080 |
d958b3940e914e22e5733aa2e31b5d3a | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_6 | Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$ , and $a+d>b+c$ . How many interesting ordered quadruples are there? | Let $b = a+x$ and $c = a+x+y$ and $d = a+x+y+z$ for positive integers $x,y,z.$ In order to satisfy the other condition we need $z > x$ so we let $z = x+k.$ Now the only other condition we need to satisfy so $a+2x+y+k \le 10.$ This condition can be transformed into $a+2x+y+k+b = 11$ for positive $a,x,y,k,b.$ Now we use generating functions to finish. We find the generating function of the whole expression is $(x + x^2 + \cdots)^4 \cdot (x^2+x^4 + \cdots)$ and we are looking for the $x^{11}$ coefficient. This simplifies to finding the $x^5$ coefficient of $(1+x+\cdots)^4 \cdot (1+x^2+\cdots) =\frac{1}{(1-x)^4} \cdot\frac{1}{1+x}.$ Now this expression simplifies to \[\left(\binom{4}{4}+\binom{5}{4} + \cdots +\binom{4+k}{4}x^k\right)(1-x+x^2-x^3 \cdots).\] The $x^5$ coefficient ends up to be $\binom{9}{4} -\binom{8}{4} +\binom{7}{4} -\binom{6}{4} +\binom{5}{4} -\binom{4}{4} = 126 - 70 + 35 - 15 + 5 - 1 = \boxed{080}.$ | null | 080 |
d958b3940e914e22e5733aa2e31b5d3a | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_6 | Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$ , and $a+d>b+c$ . How many interesting ordered quadruples are there? | First, let $a=1$ and $d=10$ . If $b=2$ , then $c$ can be from $3$ to $8$ . If $b=3$ , then $4$ to $7$ . If $b=4$ , then $c$ is between $5$ and $6$ . We find a pattern that whenever $b$ increases by $1$ , when $a$ and $d$ are stationary, then the possible values of $c$ decrease by 2, unless it gets to zero or negative, in which case that case ends. Counting up, we have $6+4+2=12$ different possibilities when $a=1$ and $d=10$ . For $a=1$ and $d=9$ $b=2$ , then $c$ can be from $3$ to $7$ . If $b=3$ , then $c$ can be from $4$ to $6$ , and so on. Notice that the possible values for each case of $b$ gets one less than if $d$ were one greater, unless that number is zero, in which it stays zero. We then use this pattern to find all the values: $12+9+6+4+2+1+9+6+4+2+1+6+4+2+1+4+2+1+2+1+1 \Rightarrow \\ 12\cdot1+9\cdot2+6\cdot3+4\cdot4+2\cdot5+1\cdot6= \\ 12+18+18+16+10+6=\boxed{080}.$ | null | 080 |
d958b3940e914e22e5733aa2e31b5d3a | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_6 | Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$ , and $a+d>b+c$ . How many interesting ordered quadruples are there? | Rearranging the equation obtains $b-a<d-c$ . Let $a-0=e_0$ $b-a=e_3+e_1$ $c-b=e_2$ $d-c=e_3$ $11-d=e_4$ . Add up all of these newly defined equations to obtain $e_0+e_1+e_2+2e_3+e_4=11$ . Note that since all $e_n$ were defined to be $e_n\ge1$ , to form our stars and bars argument we can let $d_n+1=e_n$ for all $n$ . Then we obtain $d_0+d_1+d_2+2d_3+d_4=5$ where $d_n$ is nonnegative. Now, we can move the $2d_3$ term to the other side and perform casework.
If $2d_3=0$ : 5 objects for 4 variables -> $\binom{8}{3}$
If $2d_3=2$ : 3 objects for 4 variables -> $\binom{6}{3}$
If $2d_3=4$ : 1 object for 4 variables -> $\binom{4}{3}$
Adding all of these cases up, we get $56+20+4=\boxed{080}$ as our requested answer. | null | 080 |
e2e267bb7089705d403c39a83d2c8874 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_7 | Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal to the number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let $m$ be the maximum number of red marbles for which such an arrangement is possible, and let $N$ be the number of ways he can arrange the $m+5$ marbles to satisfy the requirement. Find the remainder when $N$ is divided by $1000$ | We are limited by the number of marbles whose right hand neighbor is not the same color as the marble. By surrounding every green marble with red marbles - RGRGRGRGRGR. That's 10 "not the same colors" and 0 "same colors." Now, for every red marble we add, we will add one "same color" pair and keep all 10 "not the same color" pairs. It follows that we can add 10 more red marbles for a total of $m = 16$ . We can place those ten marbles in any of 6 "boxes": To the left of the first green marble, to the right of the first but left of the second, etc. up until to the right of the last. This is a stars-and-bars problem, the solution of which can be found as $\binom{n+k}{k}$ where n is the number of stars and k is the number of bars. There are 10 stars (The unassigned Rs, since each "box" must contain at least one, are not counted here) and 5 "bars," the green marbles. So the answer is $\binom{15}{5} = 3003$ , take the remainder when divided by 1000 to get the answer: $\boxed{003}$ | null | 003 |
e2e267bb7089705d403c39a83d2c8874 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_7 | Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal to the number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let $m$ be the maximum number of red marbles for which such an arrangement is possible, and let $N$ be the number of ways he can arrange the $m+5$ marbles to satisfy the requirement. Find the remainder when $N$ is divided by $1000$ | Begin with the above solution's reasoning to find that the ideal sequence is RGRGRGRGRGR + 10 Rs. To count the number of arrangements where the Gs are separate, group a G and an R together (GR) which will be rearranged/counted as one. However this also introduces the case of having GRGR.... which results in too few number of marbles whose right-hand neighbor is different, so we fix an R in the beginning of our sequence which also removes it from our counting calculation. Finally we are looking at the arrangement of 5 (GR)s + 10 Rs, all indistinguishable, which is $\frac{15!}{5!10!}=3003\Longrightarrow\boxed{003}$ | null | 003 |
9711a0692e81af62cc3248e80c114354 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_8 | Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$ . For each $j$ , let $w_j$ be one of $z_j$ or $i z_j$ . Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$ | The twelve dots above represent the $12$ roots of the equation $z^{12}-2^{36}=0$ . If we write $z=a+bi$ , then the real part of $z$ is $a$ and the real part of $iz$ is $-b$ . The blue dots represent those roots $z$ for which the real part of $z$ is greater than the real part of $iz$ , and the red dots represent those roots $z$ for which the real part of $iz$ is greater than the real part of $z$ . Now, the sum of the real parts of the blue dots is easily seen to be $8+16\cos\frac{\pi}{6}=8+8\sqrt{3}$ and the negative of the sum of the imaginary parts of the red dots is easily seen to also be $8+8\sqrt{3}$ . Hence our desired sum is $16+16\sqrt{3}=16+\sqrt{768}$ , giving the answer $\boxed{784}.$ | null | 784 |
9711a0692e81af62cc3248e80c114354 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_8 | Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$ . For each $j$ , let $w_j$ be one of $z_j$ or $i z_j$ . Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$ | The equation $z^{12}-2^{36}=0$ can be factored as follows: \[(z^6-2^{18})(z^6+2^{18})=0\] \[(z^2-2^6)(z^2+2^6)({(z^2+2^6)}^2-z^2\cdot2^6)({(z^2-2^6)}^2+z^2\cdot2^6)=0\] \[(z^2-2^6)(z^2+2^6)(z^2+2^6-z\cdot2^3)(z^2+2^6+z\cdot2^3) (z^2-2^6-iz\cdot2^3)(z^2-2^6+i z\cdot2^3)=0\]
Since this is a 12th degree equation, there are 12 roots. Also, since each term in the equation is even, the positive or negative value of each root is another root. That would mean there are 6 roots that can be multiplied by $-1$ and since we have 6 factors, that’s 1 root per factor. We just need to solve for $z$ in each factor and pick whether or not to multiply by $i$ and $-1$ for each one depending on the one that yields the highest real value. After that process, we get $8+8+2((4\sqrt{3}+4)+(4\sqrt{3}-4))$ Adding the values up yields $16+16\sqrt{3}$ , or $16+\sqrt{768}$ , and $16+768=\boxed{784}$ | null | 784 |
9711a0692e81af62cc3248e80c114354 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_8 | Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$ . For each $j$ , let $w_j$ be one of $z_j$ or $i z_j$ . Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$ | Clearly, the roots are: $2^3*(\cos{\frac{k\pi}{12}}+i\sin{\frac{k\pi}{12}}), k\in{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}$
Now, realize for $z=a+bi$ $\operatorname{Re}(iz)=-b$ $\operatorname{Re}(z)=a$ $\operatorname{Re}(z)<\operatorname{Re}(iz)$ is true when $a<-b$
This means:
When $a>0$ $b<-a<0$
When $a<0$ $0<b<-a$
For the 12 roots of the polynomial in the original equation, $8\cos{k\pi/12}=\operatorname{Re}(z), k\in{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}$ $-8\sin{k\pi/12}=\operatorname{Im}(iz), k\in{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}$
So $8\cos{k\pi/12}<-8\sin{k\pi/12}$ $-\cos{k\pi/12}>\sin{k\pi/12}$ . This can be easily true for roots that are in the 3rd quadrant in the complex plane.
This cannot be true for roots in the 1st quadrant because that would yield a negative number bigger than a positive one.
Consider the roots in the 2nd and 4th quadrants. Calculate the roots, choose, and then add the ones up. You will get $\boxed{784}$ | null | 784 |
9711a0692e81af62cc3248e80c114354 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_8 | Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$ . For each $j$ , let $w_j$ be one of $z_j$ or $i z_j$ . Then the maximum possible value of the real part of $\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$ | Use De Moivre's Theorem to brute force all the roots out. Then choose the greater value of $\operatorname{Re}(z), \operatorname{Re}(iz)$ . After adding everything up, you get $\boxed{784}$ | null | 784 |
70d0742a906089237154f42059ebb500 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_10 | circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$ . The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m + n$ is divided by 1000. | We begin as in the first solution. Once we see that $\triangle EOF$ has side lengths $12$ $20$ , and $24$ , we can compute its area with Heron's formula:
\[K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}.\]
Thus, the circumradius of triangle $\triangle EOF$ is $R = \frac{abc}{4K} = \frac{45}{\sqrt{14}}$ . Looking at $EPFO$ , we see that $\angle OEP = \angle OFP = 90^\circ$ , which makes it a cyclic quadrilateral. This means $\triangle EOF$ 's circumcircle and $EPFO$ 's inscribed circle are the same.
Since $EPFO$ is cyclic with diameter $OP$ , we have $OP = 2R = \frac{90}{\sqrt{14}}$ , so $OP^2 = \frac{4050}{7}$ and the answer is $\boxed{057}$ | null | 057 |
70d0742a906089237154f42059ebb500 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_10 | circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$ . The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m + n$ is divided by 1000. | Let $OP=x$
Proceed as the first solution in finding that quadrilateral $EPFO$ has side lengths $OE=20$ $OF=24$ $EP=\sqrt{x^2-20^2}$ , and $PF=\sqrt{x^2-24^2}$ , and diagonals $OP=x$ and $EF=12$
We note that quadrilateral $EPFO$ is cyclic and use Ptolemy's theorem to solve for $x$
\[20\cdot \sqrt{x^2-24^2} + 12\cdot x = 24\cdot \sqrt{x^2-20^2}\]
Solving, we have $x^2=\frac{4050}{7}$ so the answer is $\boxed{057}$ | null | 057 |
70d0742a906089237154f42059ebb500 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_10 | circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$ . The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m + n$ is divided by 1000. | Let $M$ be the midpoint of $AB$ and $N$ of $CD$ . As $\angle OMP = \angle ONP$ , quadrilateral $OMPN$ is cyclic with diameter $OP$ . By Cyclic quadrilaterals note that $\angle MPO = \angle MNO$
The area of $\triangle MNP$ can be computed by Herons as \[[MNO] = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{28\cdot 16\cdot 8\cdot 4} = 32\sqrt{14}.\] The area is also $\frac{1}{2}ON \cdot MN \sin{\angle MNO}$ . Therefore, \begin{align*} \sin{\angle MNO} &= \frac{2[MNO]}{ON \cdot MN} \\ &= \frac{2}{9}\sqrt{14} \\ \sin{\angle MNO} &= \frac{OM}{OP} \\ &= \frac{2}{9}\sqrt{14} \\ OP &= \frac{90\sqrt{14}}{14} \\ OP^2 &= \frac{4050}{7} \implies \boxed{057} | null | 057 |
70d0742a906089237154f42059ebb500 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_10 | circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$ . The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m + n$ is divided by 1000. | Define $M$ and $N$ as the midpoints of $AB$ and $CD$ , respectively. Because $\angle OMP = \angle ONP = 90^{\circ}$ , we have that $ONPM$ is a cyclic quadrilateral. Hence, $\angle PNM = \angle POM.$ Then, let these two angles be denoted as $\alpha$ .
Now, assume WLOG that $PD = x < 7$ and $PB = y < 15$ (We can do this because one of $PD$ or $PC$ must be less than 7, and similarly for $PB$ and $PA$ ). Then, by Power of a Point on P with respect to the circle with center $O$ , we have that \[(14-x)x = (30-y)y\] \[(7-x)^{2}+176=(15-y)^{2}.\] Then, let $z = (7-x)^{2}$ . From Law of Cosines on $\triangle NMP$ , we have that \[\textrm{cos } \angle MNP = \frac{NP^{2}+MN^{2}-MP^{2}}{2 \cdot NP \cdot MN}\] \[\textrm{cos } \alpha = \frac{(7-x)^{2} + 12^{2} - (14-x)^{2}}{24 \cdot (7-x)}.\] Plugging in $z$ in gives \[\textrm{cos } \alpha = \frac{-32}{24 \cdot \sqrt{z}}\] \[\textrm{cos } \alpha = \frac{-4}{3\sqrt{z}}\] \[\textrm{cos }^{2} \alpha = \frac{16}{9z}.\] Hence, \[\textrm{tan }^{2} \alpha = \frac{\frac{9z-16}{9z}}{\frac{16}{9z}} = \frac{9z-16}{16}.\] Then, we also know that \[\textrm{tan } \alpha = \textrm{tan } \angle MOP = \frac{MP}{OM} = \frac{14-y}{20}.\] Squaring this, we get \[\textrm{tan }^{2} \alpha = \frac{z+176}{400}.\] Equating our expressions for $z$ , we get $\frac{z+176}{400} = \frac{9z-16}{16}.$ Solving gives us that $z = \frac{18}{7}$ .
Since $\angle ONP = 90^{\circ}$ , from the Pythagorean Theorem, $OP^{2} = ON^{2}+PN^{2} = 25^{2}-7^{2} + z = 576+z = \frac{4050}{7}$ ,
and thus the answer is $4050+7 = 4057$ , which when divided by a thousand leaves a remainder of $\boxed{57}.$ | null | 57 |
70d0742a906089237154f42059ebb500 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_10 | circle with center $O$ has radius 25. Chord $\overline{AB}$ of length 30 and chord $\overline{CD}$ of length 14 intersect at point $P$ . The distance between the midpoints of the two chords is 12. The quantity $OP^2$ can be represented as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find the remainder when $m + n$ is divided by 1000. | 2011 AIMEII Problem 10 CASE 2.png
Let $E$ and $F$ be the midpoints of $\overline{AB}$ and $\overline{CD}$ , respectively, such that $\overline{BE}$ intersects $\overline{CF}$
Since $E$ and $F$ are midpoints, $BE = 15$ and $CF = 7$
$B$ and $C$ are located on the circumference of the circle, so $OB = OC = 25$
Since $\overline{OE}\perp \overline{AB}$ and $\overline{OF}\perp \overline{CD}$ $OE = \sqrt{OB^2-BE^2}=20$ and $OF = \sqrt{OC^2-OF^2}=24$
With law of cosines, $\cos \angle EOF = \frac{OE^2+OF^2-EF^2}{2\cdot OE\cdot OF} = \frac{13}{15}$
Since $EF < OF$ $\angle EOF$ is acute angle. $\sin \angle EOF = \sqrt{1-\cos^2 \angle EOF} = \frac{\sqrt{56}}{15}$ and $\tan \angle EOF = \frac{\sqrt{56}}{13}$
Let $\overline{OF}$ line be $x$ axis.
Line $\overline{DC}$ equation is $x = OF$
Since line $\overline{AB}$ passes point $E$ and perpendicular to $\overline{OD}$ , its equation is $y - E_y = -\frac{1}{\tan \angle EOF} (x - E_x)$
where $E_x = OE\cos{\angle EOF}$ $E_y = OE\sin{\angle EOF}$
Since $P$ is the intersection of $\overline{AB}$ and $\overline{CD}$
$P_x = OF = 24$
$P_y = E_y -\frac{1}{\tan \angle EOF} (OF - E_x) = - \frac{3\sqrt{14}}{7}$ (Negative means point $P$ is between point $F$ and $C$
$OP^2 = P_x^2 + P_y^2 = \frac{4050}{7}$ and the answer is $\boxed{057}$ | null | 057 |
f95378aa464431f4f196ad1cb21550f9 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_11 | Let $M_n$ be the $n \times n$ matrix with entries as follows: for $1 \le i \le n$ $m_{i,i} = 10$ ; for $1 \le i \le n - 1$ $m_{i+1,i} = m_{i,i+1} = 3$ ; all other entries in $M_n$ are zero. Let $D_n$ be the determinant of matrix $M_n$ . Then $\sum_{n=1}^{\infty} \frac{1}{8D_n+1}$ can be represented as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$
Note: The determinant of the $1 \times 1$ matrix $[a]$ is $a$ , and the determinant of the $2 \times 2$ matrix $\left[ {\begin{array}{cc} a & b \\ c & d \\ \end{array} } \right] = ad - bc$ ; for $n \ge 2$ , the determinant of an $n \times n$ matrix with first row or first column $a_1$ $a_2$ $a_3$ $\dots$ $a_n$ is equal to $a_1C_1 - a_2C_2 + a_3C_3 - \dots + (-1)^{n+1}a_nC_n$ , where $C_i$ is the determinant of the $(n - 1) \times (n - 1)$ matrix formed by eliminating the row and column containing $a_i$ | \[D_{1}=\begin{vmatrix} 10 \end{vmatrix} = 10, \quad D_{2}=\begin{vmatrix} 10 & 3 \\ 3 & 10 \\ \end{vmatrix} =(10)(10) - (3)(3) = 91, \quad D_{3}=\begin{vmatrix} 10 & 3 & 0 \\ 3 & 10 & 3 \\ 0 & 3 & 10 \\ \end{vmatrix}.\] Using the expansionary/recursive definition of determinants (also stated in the problem):
$D_{3}=\left| {\begin{array}{ccc} 10 & 3 & 0 \\ 3 & 10 & 3 \\ 0 & 3 & 10 \\ \end{array} } \right|=10\left| {\begin{array}{cc} 10 & 3 \\ 3 & 10 \\ \end{array} } \right| - 3\left| {\begin{array}{cc} 3 & 3 \\ 0 & 10 \\ \end{array} } \right| + 0\left| {\begin{array}{cc} 3 & 10 \\ 0 & 3 \\ \end{array} } \right| = 10D_{2} - 9D_{1} = 820$
This pattern repeats because the first element in the first row of $M_{n}$ is always 10, the second element is always 3, and the rest are always 0. The ten element directly expands to $10D_{n-1}$ . The three element expands to 3 times the determinant of the the matrix formed from omitting the second column and first row from the original matrix. Call this matrix $X_{n}$ $X_{n}$ has a first column entirely of zeros except for the first element, which is a three. A property of matrices is that the determinant can be expanded over the rows instead of the columns (still using the recursive definition as given in the problem), and the determinant found will still be the same. Thus, expanding over this first column yields $3D_{n-2} + 0(\text{other things})=3D_{n-2}$ . Thus, the $3 \det(X_{n})$ expression turns into $9D_{n-2}$ . Thus, the equation $D_{n}=10D_{n-1}-9D_{n-2}$ holds for all n > 2.
This equation can be rewritten as $D_{n}=10(D_{n-1}-D_{n-2}) + D_{n-2}$ . This version of the equation involves the difference of successive terms of a recursive sequence. Calculating $D_{0}$ backwards from the recursive formula and $D_{4}$ from the formula yields $D_{0}=1, D_{4}=7381$ . Examining the differences between successive terms, a pattern emerges. $D_{0}=1=9^{0}$ $D_{1}-D_{0}=10-1=9=9^{1}$ $D_{2}-D_{1}=91-10=81=9^{2}$ $D_{3}-D_{2}=820-91=729=9^{3}$ $D_{4}-D_{3}=7381-820=6561=9^{4}$ .
Thus, $D_{n}=D_{0} + 9^{1}+9^{2}+ . . . +9^{n}=\sum_{i=0}^{n}9^{i}=\frac{(1)(9^{n+1}-1)}{9-1}=\frac{9^{n+1}-1}{8}$
Thus, the desired sum is $\sum_{n=1}^{\infty}\frac{1}{8\frac{9^{n+1}-1}{8}+1}=\sum_{n=1}^{\infty}\frac{1}{9^{n+1}-1+1} = \sum_{n=1}^{\infty}\frac{1}{9^{n+1}}$
This is an infinite geometric series with first term $\frac{1}{81}$ and common ratio $\frac{1}{9}$ . Thus, the sum is $\frac{\frac{1}{81}}{1-\frac{1}{9}}=\frac{\frac{1}{81}}{\frac{8}{9}}=\frac{9}{(81)(8)}=\frac{1}{(9)(8)}=\frac{1}{72}$
Thus, $p + q = 1 + 72 = \boxed{073}$ | null | 073 |
043b37552b2e47757f9c3560f3a946b8 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_12 | Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | We use complementary counting.
We can order the $9$ people around a circle in $\frac{9!}{9} = 8!$ ways. Now we count when there is at least one delegate surrounded by people from only his/her country.
Let the countries be $A,B,C$ . Suppose that the group $XXX$ (for some country $X$ ) appears. To account for circular over counting we fix this group at the top. There are $6!$ ways to arrange the rest of the delegates and $3!$ ways to arrange inside the group. Since there are three countries this group can belong to, the total is $6!*3!*3$
But notice that when the group $XXX$ AND $YYY$ both appear is over counted. So, fix one group at the top. for the last country, we can insert $0,1,2,3$ people in between the two groups. Since we can choose two countries in $\binom{3}{2} = 3$ ways, the total is $3!*3!*3!*4*3$ ways.
Unfortunately, there is over count in the over count. If each country has their delegates together, this case must be added back. Think of these three groups as a whole, and there are $3!/3 = 2$ ways of arranging. In each group there is $3!$ ways of arrangements. So, $3!*3!*3!*2$ is the total for this case.
We now find the complementary probability total. This is \[\frac{6!*3!*3 - 3!*3!*3!*4*3 + 3!*3!*3!*2}{8!} = \frac{15}{56}\] so the actual probability is $1-\frac{15}{56} = \frac{41}{56}$ for an answer of $\boxed{97}$ | null | 97 |
043b37552b2e47757f9c3560f3a946b8 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_12 | Nine delegates, three each from three different countries, randomly select chairs at a round table that seats nine people. Let the probability that each delegate sits next to at least one delegate from another country be $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | $1680= \binom{9}{3} \binom{6}{3}$ is the total.
$540 = 3* 9* \binom{6}{3}$ is the case where one country has three people in a row. (Three is for selection of country and nine is for rotation of the seats.)
$108 = 3* 9* 4$ is the case where two countries has three people in a row. (Three and nine are for the same reasons above and four is for putting three people in a row for the remaining six seats.)
$18$ is the case where three countries has three people in a row.
$\frac{1680-540+108-18}{1680}=\frac{41}{56}$ so the answer is $\boxed{97}$ | null | 97 |
f1546942461b4c7f971d7e572d052e43 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_13 | Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ | This is a combination of Solutions 1 and 2.
First, draw $O_1P,O_2P,BP,DP$ . Then, observe that $\angle BAP=45$ implies that $\angle BO_1P=90$ . So, $\triangle BO_1P$ is a $45-90-45$ triangle. Similarly, observe that $DO_2P$ is too. So, a rotation of $\angle O_1PO_2$ to $\angle BPO_2$ adds $45$ degrees. Then, moving to $BPD$ subtracts $45$ degrees. Hence, $\angle BPD=120$ . Let the intersection of $BD$ and $AC$ be $Q$ . Then $BQP$ is a $30-90-60$ triangle, hence $QP=\frac{6\sqrt{2}}{\sqrt{3}}$ (We know that $BQ$ is $6\sqrt{2}$ ), or $QP=2\sqrt{6}$ Finally, $AP=QP+AQ=2\sqrt{6}+6\sqrt{2}=\sqrt{24}+\sqrt{72} \Rightarrow \boxed{096}$ | null | 096 |
f1546942461b4c7f971d7e572d052e43 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_13 | Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ | Use vectors. In an $xy$ plane, let $(-s,0)$ be $A$ $(0,s)$ be $B$ $(s,0)$ be $C$ $(0,-s)$ be $D$ , and $(p,0)$ be P, where $s=|AB|/\sqrt{2}=6\sqrt{2}$ . It remains to find $p$
The line $y=-x$ is the perpendicular bisector of $AB$ and $CD$ , so $O_1$ and $O_2$ lies on the line. Now compute the perpendicular bisector of $AP$ . The center has coordinate $(\frac{p-s}{2},0)$ , and the segment is part of the $x$ -axis, so the perpendicular bisector has equation $x=\frac{p-s}{2}$ . Since $O_1$ is the circumcenter of triangle $ABP$ , it lies on the perpendicular bisector of both $AB$ and $AP$ , so \[O_1=(\frac{p-s}{2},-\frac{p-s}{2})\] Similarly, \[O_2=(\frac{p+s}{2},-\frac{p+s}{2})\] The relation $\angle O_1PO_2=120^\circ$ can now be written using dot product as \[\vec{PO_1}\cdot\vec{PO_2}=|\vec{PO_1}|\cdot|\vec{PO_2}|\cos 120^\circ=-\frac{1}{2}|\vec{PO_1}|\cdot|\vec{PO_2}|\] Computation of both sides yields \[\frac{p^2-s^2}{p^2+s^2}=-\frac{1}{2}\] Solve for $p$ gives $p=s/\sqrt{3}=2\sqrt{6}$ , so $AP=s+p=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}$ . The answer is 72+24 $\Rightarrow\boxed{096}$ | null | 096 |
f1546942461b4c7f971d7e572d052e43 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_13 | Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ | Translate $\triangle{ABP}$ so that the image of $AB$ coincides $DC$ . Let the image of $P$ be $P’$
$\angle{DPC}=\angle{CPB}$ by symmetry, and $\angle{APB}=\angle{DP’C}$ because translation preserves angles. Thus $\angle{DP’C}+\angle{CPD}=\angle{CPB}+\angle{APB}=180^\circ$ . Therefore, quadrilateral $CPDP’$ is cyclic. Thus the image of $O_1$ coincides with $O_2$
$O_1P$ is parallel to $O_2P’$ so $\angle{P’O_2P}=\angle{O_1PO_2}=120^\circ$ , so $\angle{PDP’}=60^\circ$ and $\angle{PDC}=15^\circ$ , thus $\angle{ADP}=75^{\circ}$
Let $M$ be the foot of the perpendicular from $D$ to $AC$ . Then $\triangle{AMD}$ is a 45-45-90 triangle and $\triangle{DMP}$ is a 30-60-90 triangle. Thus
$AM=6\sqrt{2}$ and $MP=\frac{6\sqrt{2}}{\sqrt{3}}$
This gives us $AP=AM+MP=\sqrt{72}+\sqrt{24}$ , and the answer is $72+24=\boxed{096}.$ | null | 096 |
f1546942461b4c7f971d7e572d052e43 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_13 | Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ | Reflect $O_1$ across $AP$ to $O_1'$ . By symmetry $O_1’$ is the circumcenter of $\triangle{ADP}$
$\angle{DO_1’P}$ $2*\angle{DAP} = 90^\circ$ , so $\angle{O_1’PD}=45^\circ$
similarly $\angle{DO_2P}$ $2*\angle{DCP} = 90^\circ$ , so $\angle{O_2PD}=45^\circ$
Therefore $\angle{O_1’PO_2}=90^\circ$ , so that $\angle{O_1’PO_1} =120^\circ - 90^\circ = 30^\circ$
By symmetry, $\angle{O_1'PA} = \angle{APO_1} = 0.5*\angle{O_1’PO_1} = 15^\circ$
Therefore, since $O_1’$ is the circumcenter of $\triangle{ADP}$ $\angle{ADP}$ $0.5*(180^\circ - 2*\angle{O_1'PA}) = 75^\circ$
Therefore $\angle{APD} = 180^\circ - 45^\circ - 75^\circ = 60^\circ$
Using sine rule in $\triangle{ADP}$ $AP = (12 * \sin 75^\circ) / \sin 60^\circ =\sqrt{72}+\sqrt{24}$ , and the answer is $72+24=\boxed{096}.$ | null | 096 |
f1546942461b4c7f971d7e572d052e43 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_13 | Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ | Why not use coordinates? After all, 45 degrees is rather friendly in terms of ordered-pair representation! We can set $A=(0, 12)$ $B=(12,12)$ $C=(12, 0)$ $D=(0, 0)$ . Let this $P=(a, 12-a)$ for some $a$
We also know that the circumcenter is the intersection of all perpendicular bisectors of sides, but two will suffice also due to this property. Therefore, we see that $O_{1}$ is the intersection of $x=6$ and, knowing the midpoint of $AP$ to be $(\frac{a}{2}, \frac{12-a}{2})$ and thus the equation to be $y=x+(12-a)$ , we get $(6, 18-a)$ . Likewise for $O_{2}$ it's $(6, 6-a)$ . Now what do we see? $O_{1}P=O_{2}P$ (just look at the coordinates)! So both of those distances are $4\sqrt{3}$ . Solving for $a$ we get it to be $6+2\sqrt{3}$ , since $AP>CP$ . Multiply by $\sqrt{2}$ because we are looking for $AP$ to get the answer of $\boxed{096}$ | null | 096 |
f1546942461b4c7f971d7e572d052e43 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_13 | Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ | Let $\angle APD = \theta$ . Then $\angle ADP = 180^{\circ}-45^{\circ}-\theta=135^{\circ}-\theta \implies \angle PDC=\theta-45^{\circ}$ .
Realize that because $O_1$ is a circumcenter, $\angle DO_1P=\angle DCP=45^{\circ} \implies \angle DPO_1=\frac{180^{\circ}-\angle DO_1P}{2}=45^{\circ}$ . Then $\angle O_2PA=75^{\circ}-\theta \implies \angle AOP = 180^{\circ}-2\angle O_2PA = 2\theta+30^{\circ} \implies \angle ABP=\theta + 15^{\circ} \implies \angle PBC=75-\theta$ .
Now, because $P$ lies on diagonal $AC$ $\triangle PDC \cong \triangle PBC \implies \angle PDC = \angle PBC \implies \theta-45^{\circ}=75^{\circ}-\theta \implies \theta = 60^{\circ}$ .
To finish, we look at $\triangle ADP$ . Drop a perpendicular from $D$ to $AP$ at $E$ . Then $\triangle ADE$ is a $45-45-90$ and $\triangle PDE$ is a $30-60-90$ . Therefore, $DE=EA=6\sqrt{2}, EP=2\sqrt{6}$ , so $AP=AE+EP=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24} \implies \boxed{096}$ $\blacksquare$ ~msc | null | 096 |
f1546942461b4c7f971d7e572d052e43 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_13 | Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ | Both $O_1$ and $O_2$ lie on the perpendicular bisector of $AB$
Claim: $O_1O_2=12$ and $O_1P=O_2P$
Proof. Translate $O_1$ and $P$ $12$ units down, and let their images be $O_1'$ and $P'$ , respectively. Note that $\triangle ABP\cong\triangle DCP'$ . Additionally, \[\angle CP'D = \angle BPA = 180^{\circ} - \angle BPC = 180^{\circ} - \angle CPD,\] so $CPDP'$ is cyclic. This means $O_1'$ and $O_2$ coincide, so $O_1O_2=12$ . This also means the circumradii of both triangles are equal, so $O_1P=O_2P$ $\blacksquare$
Let the perpendicular from $P$ intersect $O_1O_2$ at $X$ and $AD$ at $Y$ . Since $\triangle O_1XP$ is 30-60-90, $XP=\frac{6}{\sqrt{3}} = 2\sqrt3$ . Since $YX=6$ $PY=6+2\sqrt3$ , so $AP=6\sqrt2+2\sqrt6 = \sqrt{72}+\sqrt{24} \implies\boxed{96}$ | null | 96 |
f1546942461b4c7f971d7e572d052e43 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_13 | Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$ . Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$ , then $AP = \sqrt{a} + \sqrt{b}$ , where $a$ and $b$ are positive integers. Find $a + b$ | \[BP = DP, \angle PAB =\angle PCD = 45^\circ \implies O_{1}P =O_{2}P\] by the Law of Sines.
\[AB= CD, O_{1}A = O_{2}D = O_{1}B = O_{2}C \implies \triangle AO_{1}B = \triangle DO_{2}C.\] They have translational symmetry. The translation vector is \[\vec {AD} \implies O_{1}O_{2} = AD.\]
\[\triangle PO_{1}O_{2} = \triangle O_{1}AB \implies \angle AO_{1}B = 120^\circ \implies\] \[\angle APB = 60^\circ \implies \angle ABP = 180^\circ – 60^\circ – 45^\circ = 75^\circ.\] By the Law of Sines \[AP = \frac {AB \cdot \sin 75^\circ}{\sin 60^\circ} = \frac {AB \cdot \sin (30^\circ + 45^\circ)} {\sin 60^\circ}\] \[AP = AB \cdot (\sin 45^\circ + \cos 45^\circ \cdot \tan 30^\circ),\] \[AP = \frac {AB}{\sqrt{2}} (1 + \frac {1}{\sqrt{3}}) = 6 \sqrt {2} + 2 \sqrt {6} \implies \boxed{096}\] [email protected], vvsss | null | 096 |
54d86e5356d1c7bcb909a07ad10e56b0 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_14 | There are $N$ permutations $(a_{1}, a_{2}, ... , a_{30})$ of $1, 2, \ldots, 30$ such that for $m \in \left\{{2, 3, 5}\right\}$ $m$ divides $a_{n+m} - a_{n}$ for all integers $n$ with $1 \leq n < n+m \leq 30$ . Find the remainder when $N$ is divided by $1000$ | Be wary of "position" versus "number" in the solution!
Each POSITION in the 30-position permutation is uniquely defined by an ordered triple $(i, j, k)$ . The $n$ th position is defined by this ordered triple where $i$ is $n \mod 2$ $j$ is $n \mod 3$ , and $k$ is $n \mod 5$ . There are 2 choices for $i$ , 3 for $j$ , and 5 for $k$ , yielding $2 \cdot 3 \cdot 5=30$ possible triples. Because the least common multiple of 2, 3, and 5 is 30, none of these triples are repeated and all are used. By the conditions of the problem, if $i$ is the same in two different triples, then the two numbers in these positions must be equivalent mod 2. If $j$ is the same, then the two numbers must be equivalent $\mod 3$ , and if $k$ is the same, the two numbers must be equivalent mod 5. Take care to note that that doesn't mean that the number 1 has to have $1 \mod 2$ ! It's that the POSITION which NUMBER 1 occupies has $1 \mod 2$
The ordered triple (or position) in which 1 can be placed has 2 options for i, 3 for j, and 5 for k, resulting in 30 different positions of placement.
The ordered triple where 2 can be placed in is somewhat constrained by the placement of 1. Because 1 is not equivalent to 2 in terms of mod 2, 3, or 5, the i, j, and k in their ordered triples must be different. Thus, for the number 2, there are (2-1) choices for i, (3-1) choices for j, and (5-1) choices for k. Thus, there are 1*2*4=8 possible placements for the number two once the number one is placed.
Because 3 is equivalent to 1 mod 2, it must have the same i as the ordered triple of 1. Because 3 is not equivalent to 1 or 2 in terms of mod 3 or 5, it must have different j and k values. Thus, there is 1 choice for i, (2-1) choices for j, and (4-1) choices for k, for a total of $1\cdot 1 \cdot 3=3$ choices for the placement of 3.
As above, 4 is even, so it must have the same value of i as 2. It is also 1 mod 3, so it must have the same j value of 1. 4 is not equivalent to 1, 2, or 3 mod 5, so it must have a different k value than that of 1, 2, and 3. Thus, there is 1 choice for i, 1 choice for j, and (3-1) choices for k, yielding a total of $1 \cdot 1 \cdot 2=2$ possible placements for 4.
5 is odd and is equivalent to 2 mod 3, so it must have the same i value as 1 and the same j value of 2. 5 is not equivalent to 1, 2, 3, or 4 mod 5, so it must have a different k value from 1, 2, 3, and 4. However, 4 different values of k are held by these four numbers, so 5 must hold the one remaining value. Thus, only one possible triple is found for the placement of 5.
All numbers from 6 to 30 are also fixed in this manner. All values of i, j, and k have been used, so every one of these numbers will have a unique triple for placement, as above with the number five. Thus, after 1, 2, 3, and 4 have been placed, the rest of the permutation is fixed.
Thus, $N = 30 \cdot 8 \cdot 3 \cdot 2=30 \cdot 48=1440$ . Thus, the remainder when $N$ is divided by $1000$ is $\boxed{440}.$ | null | 440 |
54d86e5356d1c7bcb909a07ad10e56b0 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_14 | There are $N$ permutations $(a_{1}, a_{2}, ... , a_{30})$ of $1, 2, \ldots, 30$ such that for $m \in \left\{{2, 3, 5}\right\}$ $m$ divides $a_{n+m} - a_{n}$ for all integers $n$ with $1 \leq n < n+m \leq 30$ . Find the remainder when $N$ is divided by $1000$ | Note that $30=2\cdot 3\cdot 5$ . Since $\gcd(2, 3, 5)=1$ , by CRT, for each value $k=0\ldots 29$ modulo $30$ there exists a unique ordered triple of values $(a, b, c)$ such that $k\equiv a\pmod{2}$ $k\equiv b\pmod{3}$ , and $k\equiv c\pmod{5}$ . Therefore, we can independently assign the residues modulo $2, 3, 5$ , so $N=2!\cdot 3!\cdot 5!=1440$ , and the answer is $\boxed{440}$ | null | 440 |
54d86e5356d1c7bcb909a07ad10e56b0 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_14 | There are $N$ permutations $(a_{1}, a_{2}, ... , a_{30})$ of $1, 2, \ldots, 30$ such that for $m \in \left\{{2, 3, 5}\right\}$ $m$ divides $a_{n+m} - a_{n}$ for all integers $n$ with $1 \leq n < n+m \leq 30$ . Find the remainder when $N$ is divided by $1000$ | First let's look at the situation when $m$ is equal to 2. It isn't too difficult to see the given conditions are satisfied iff the sequences $S_1 = a_1,a_3,a_5,a_7...$ and $S_2 =a_2,a_4,a_6...$ each are assigned either $\equiv 0 \pmod 2$ or $\equiv 1 \pmod 2$ . Another way to say this is each element in the sequence $a_1,a_3,a_5,a_7...$ would be the same mod 2, and similarly for the other sequence. There are 2! = 2 ways to assign the mods to the sequences.
Now when $m$ is equal to 3, the sequences are $T_1 = a_1, a_4,a_7..$ $T_2 = a_2,a_5,a_8,a_11...$ , and $T_3 = a_3,a_6,a_9...$ . Again, for each sequence, all of its elements are congruent either $0$ $1$ , or $2$ mod $3$ . There are $3! = 6$ ways to assign the mods to the sequences.
Finally do the same thing for $m = 5$ . There are $5!$ ways. In total there are $2 * 6*120 = 1440$ and the answer is $\boxed{440}$ | null | 440 |
84a4a4f6da98dadf28ccf5d105c54a90 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_15 | Let $P(x) = x^2 - 3x - 9$ . A real number $x$ is chosen at random from the interval $5 \le x \le 15$ . The probability that $\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$ $b$ $c$ $d$ , and $e$ are positive integers. Find $a + b + c + d + e$ | Make the substitution $y=2x-3$ , so $P(x)=\frac{y^2-45}{4}.$ We're looking for solutions to \[\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=\sqrt{\frac{\lfloor{y\rfloor}^2-45}{4}}\] with the new bounds $y\in{[7,27]}$ . Since the left side is an integer, it must be that $\frac{\lfloor{y\rfloor}^2-45}{4}$ is a perfect square. For simplicity, write $\lfloor{y\rfloor}=a$ and \[a^2-45=4b^2\implies{(a-2b)(a+2b)=45}.\] Since $a-2b<a+2b$ , it must be that $(a-2b,a+2b)=(1,45),(3,15),(5,9)$ , which gives solutions $(23,11),(9,3),(7,1)$ , respectively. But this gives us three cases to check:
Case 1: $\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=11$
In this case, we have \[11\leq{\sqrt{\frac{y^2-45}{4}}}<12\implies{y\in{[23,\sqrt{621})}}.\] Case 2: $\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=3$
In this case, we have \[3\leq{\sqrt{\frac{y^2-45}{4}}}<4\implies{y\in{[9,\sqrt{109})}}.\] Case 3: $\bigg\lfloor{\sqrt{\frac{y^2-45}{4}}\bigg\rfloor}=1$
In this case, we have \[1\leq{\sqrt{\frac{y^2-45}{4}}}<2\implies{y\in{[7,\sqrt{61})}}.\] To finish, the total length of the interval from which we choose $y$ is $27-7=20$ . The total length of the success intervals is \[(\sqrt{61}-7)+(\sqrt{109}-9)+(\sqrt{621}-23)=\sqrt{61}+\sqrt{109}+\sqrt{621}-39,\] which means the probability is \[\frac{\sqrt{61}+\sqrt{109}+\sqrt{621}-39}{20}.\] The requested sum is $\boxed{850}$ | null | 850 |
84a4a4f6da98dadf28ccf5d105c54a90 | https://artofproblemsolving.com/wiki/index.php/2011_AIME_II_Problems/Problem_15 | Let $P(x) = x^2 - 3x - 9$ . A real number $x$ is chosen at random from the interval $5 \le x \le 15$ . The probability that $\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$ $b$ $c$ $d$ , and $e$ are positive integers. Find $a + b + c + d + e$ | Note that all the "bounds" have to be less than the number+1, otherwise it wouldn't fit the answer format. Therefore, the answer is $\frac{3*3+\sqrt{9+4(4+9)}-10+\sqrt{9+4(16+9)}-12+\sqrt{9+4(144+9)}}{20} \implies \boxed{850}$ | null | 850 |
e957f9fb3fb04d6a5a2f62eb00055298 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_1 | Maya lists all the positive divisors of $2010^2$ . She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square . The probability $p$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | $2010^2 = 2^2\cdot3^2\cdot5^2\cdot67^2$ . Thus there are $(2+1)^4$ divisors, $(1+1)^4$ of which are squares (the exponent of each prime factor must either be $0$ or $2$ ). Therefore the probability is \[\frac {2\cdot2^4\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.\] | null | 107 |
e957f9fb3fb04d6a5a2f62eb00055298 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_1 | Maya lists all the positive divisors of $2010^2$ . She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square . The probability $p$ can be expressed in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | The prime factorization of $2010^2$ is $67^2\cdot3^2\cdot2^2\cdot5^2$ . Therefore, the number of divisors of $2010^2$ is $3^4$ or $81$ $16$ of which are perfect squares. The number of ways we can choose $1$ perfect square from the two distinct divisors is $\binom{16}{1}\binom{81-16}{1}$ . The total number of ways to pick two divisors is $\binom{81}{2}$
Thus, the probability is \[\frac {\binom{16}{1}\binom{81-16}{1}}{\binom{81}{2}} = \frac {16\cdot65}{81\cdot40} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.\] | null | 107 |
c1bbe1d5d493bfb740f5e1b79ff6b716 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_3 | Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$ | Substitute $y = \frac34x$ into $x^y = y^x$ and solve. \[x^{\frac34x} = \left(\frac34x\right)^x\] \[x^{\frac34x} = \left(\frac34\right)^x \cdot x^x\] \[x^{-\frac14x} = \left(\frac34\right)^x\] \[x^{-\frac14} = \frac34\] \[x = \frac{256}{81}\] \[y = \frac34x = \frac{192}{81}\] \[x + y = \frac{448}{81}\] \[448 + 81 = \boxed{529}\] | null | 529 |
c1bbe1d5d493bfb740f5e1b79ff6b716 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_3 | Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$ | We solve in general using $c$ instead of $3/4$ . Substituting $y = cx$ , we have:
Dividing by $x^x$ , we get $(x^x)^{c - 1} = c^x$
Taking the $x$ th root, $x^{c - 1} = c$ , or $x = c^{1/(c - 1)}$
In the case $c = \frac34$ $x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}$ $y = \frac {64}{27}$ $x + y = \frac {256 + 192}{81} = \frac {448}{81}$ , yielding an answer of $448 + 81 = \boxed{529}$ | null | 529 |
c1bbe1d5d493bfb740f5e1b79ff6b716 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_3 | Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$ | Taking the logarithm base $x$ of both sides, we arrive with: \[y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \left(\frac{3}{4}x\right) = \frac{3}{4}\] Now we proceed by the logarithm rule $\log(ab)=\log a + \log b$ . The equation becomes: \[\log_x \frac{3}{4} + \log_x x = \frac{3}{4}\] \[\Longleftrightarrow \log_x \frac{3}{4} + 1 = \frac{3}{4}\] \[\Longleftrightarrow \log_x \frac{3}{4} = -\frac{1}{4}\] \[\Longleftrightarrow x^{-\frac{1}{4}} = \frac{3}{4}\] \[\Longleftrightarrow \frac{1}{x^{\frac{1}{4}}} = \frac{3}{4}\] \[\Longleftrightarrow x^{\frac{1}{4}} = \frac{4}{3}\] \[\Longleftrightarrow \sqrt[4]{x} = \frac{4}{3}\] \[\Longleftrightarrow x = \left(\frac{4}{3}\right)^4=\frac{256}{81}\] Then find $y$ as in solution 3, and we get $\boxed{529}$ | null | 529 |
0a8fcc7c2f43dc66cbbae885b7baeeb1 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_4 | Jackie and Phil have two fair coins and a third coin that comes up heads with probability $\frac47$ . Jackie flips the three coins, and then Phil flips the three coins. Let $\frac {m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | This can be solved quickly and easily with generating functions
Let $x^n$ represent flipping $n$ heads.
The generating functions for these coins are $(1+x)$ $(1+x)$ ,and $(3+4x)$ in order.
The product is $3+10x+11x^2+4x^3$ . ( $ax^n$ means there are $a$ ways to get $n$ heads, eg there are $10$ ways to get $1$ head, and therefore $2$ tails, here.)
The sum of the coefficients squared (total number of possible outcomes, squared because the event is occurring twice) is $(4 + 11 + 10 + 3)^2 = 28^2 = 784$ and the sum of the squares of each coefficient (the sum of the number of ways that each coefficient can be chosen by the two people) is $4^2 + 11^2 + 10^2 + 3^2=246$ .
The probability is then $\frac{4^2 + 11^2 + 10^2 + 3^2}{28^2} = \frac{246}{784} = \frac{123}{392}$ .
(Notice the relationship between the addends of the numerator here and the cases in the following solution.)
$123 + 392 = \boxed{515}$ | null | 515 |
116d74d1ec4bb3d1428bf0e7e9d515d5 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_5 | Positive integers $a$ $b$ $c$ , and $d$ satisfy $a > b > c > d$ $a + b + c + d = 2010$ , and $a^2 - b^2 + c^2 - d^2 = 2010$ . Find the number of possible values of $a$ | Using the difference of squares $2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \ge a + b + c + d = 2010$ , where equality must hold so $b = a - 1$ and $d = c - 1$ . Then we see $a = 1004$ is maximal and $a = 504$ is minimal, so the answer is $\boxed{501}$ | null | 501 |
116d74d1ec4bb3d1428bf0e7e9d515d5 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_5 | Positive integers $a$ $b$ $c$ , and $d$ satisfy $a > b > c > d$ $a + b + c + d = 2010$ , and $a^2 - b^2 + c^2 - d^2 = 2010$ . Find the number of possible values of $a$ | Since $a+b$ must be greater than $1005$ , it follows that the only possible value for $a-b$ is $1$ (otherwise the quantity $a^2 - b^2$ would be greater than $2010$ ). Therefore the only possible ordered pairs for $(a,b)$ are $(504, 503)$ $(505, 504)$ , ... , $(1004, 1003)$ , so $a$ has $\boxed{501}$ possible values. | null | 501 |
67625137316c4c72cb7b75f0d977577b | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | Let $Q(x) = x^2 - 2x + 2$ $R(x) = 2x^2 - 4x + 3$ Completing the square , we have $Q(x) = (x-1)^2 + 1$ , and $R(x) = 2(x-1)^2 + 1$ , so it follows that $P(x) \ge Q(x) \ge 1$ for all $x$ (by the Trivial Inequality ).
Also, $1 = Q(1) \le P(1) \le R(1) = 1$ , so $P(1) = 1$ , and $P$ obtains its minimum at the point $(1,1)$ . Then $P(x)$ must be of the form $c(x-1)^2 + 1$ for some constant $c$ ; substituting $P(11) = 181$ yields $c = \frac 95$ . Finally, $P(16) = \frac 95 \cdot (16 - 1)^2 + 1 = \boxed{406}$ | null | 406 |
67625137316c4c72cb7b75f0d977577b | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | It can be seen that the function $P(x)$ must be in the form $P(x) = ax^2 - 2ax + c$ for some real $a$ and $c$ . This is because the derivative of $P(x)$ is $2ax - 2a$ , and a global minimum occurs only at $x = 1$ (in addition, because of this derivative, the vertex of any quadratic polynomial occurs at $\frac{-b}{2a}$ ). Substituting $(1,1)$ and $(11, 181)$ we obtain two equations:
Solving, we get $a = \frac{9}{5}$ and $c = \frac{14}{5}$ , so $P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}$ . Therefore, $P(16) = \boxed{406}$ | null | 406 |
67625137316c4c72cb7b75f0d977577b | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | Let $y = x^2 - 2x + 2$ ; note that $2y - 1 = 2x^2 - 4x + 3$ . Setting $y = 2y - 1$ , we find that equality holds when $y = 1$ and therefore when $x^2 - 2x + 2 = 1$ ; this is true iff $x = 1$ , so $P(1) = 1$
Let $Q(x) = P(x) - x$ ; clearly $Q(1) = 0$ , so we can write $Q(x) = (x - 1)Q'(x)$ , where $Q'(x)$ is some linear function. Plug $Q(x)$ into the given inequality:
$x^2 - 3x + 2 \le Q(x) \le 2x^2 - 5x + 3$
$(x - 1)(x - 2) \le (x - 1)Q'(x) \le (x - 1)(2x - 3)$ , and thus
$x - 2 \le Q'(x) \le 2x - 3$
For all $x > 1$ ; note that the inequality signs are flipped if $x < 1$ , and that the division is invalid for $x = 1$ . However,
$\lim_{x \to 1} x - 2 = \lim_{x \to 1} 2x - 3 = -1$
and thus by the sandwich theorem $\lim_{x \to 1} Q'(x) = -1$ ; by the definition of a continuous function, $Q'(1) = -1$ . Also, $Q(11) = 170$ , so $Q'(11) = 170/(11-1) = 17$ ; plugging in and solving, $Q'(x) = (9/5)(x - 1) - 1$ . Thus $Q(16) = 390$ , and so $P(16) = \boxed{406}$ | null | 406 |
67625137316c4c72cb7b75f0d977577b | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | Let $Q(x) = P(x) - (x^2-2x+2)$ , then $0\le Q(x) \le (x-1)^2$ (note this is derived from the given inequality chain). Therefore, $0\le Q(x+1) \le x^2 \Rightarrow Q(x+1) = Ax^2$ for some real value A.
$Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}$
$Q(16)=15^2A=180 \Rightarrow P(16)-(16^2-32+2) = 180 \Rightarrow P(16)=180+226= \boxed{406}$ | null | 406 |
67625137316c4c72cb7b75f0d977577b | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | Let $P(x) = ax^2 + bx + c$ . Plugging in $x = 1$ to the expressions on both sides of the inequality, we see that $a + b + c = 1$ . We see from the problem statement that $121a + 11b + c = 181$ . Since we know the vertex of $P(x)$ lies at $x = 1$ , by symmetry we get $81a -9b + c = 181$ as well. Since we now have three equations, we can solve this trivial system and get our answer of $\boxed{406}$ | null | 406 |
67625137316c4c72cb7b75f0d977577b | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | Similar to Solution 5, let $P(x) = ax^2 + bx + c$ . Note that $(1,1)$ is a vertex of the polynomial. Additionally, this means that $b = -2a$ (since $\frac{-b}{2a}$ is the minimum $x$ point). Thus, we have $P(x) = ax^2 - 2ax + c$ . Therefore $a - 2a + c = 1$ . Moreover, $99a + c = 181$ . And so our polynomial is $\frac{9}{5}x^2 - \frac{18}{5}x + \frac{14}{5}$ . Plug in $x = 16$ to get $\boxed{406}$ | null | 406 |
67625137316c4c72cb7b75f0d977577b | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_6 | Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$ , and suppose $P(11) = 181$ . Find $P(16)$ | Very similar to Solution 6, start by noticing that $P(x)$ is between two polynomials, we try to set them equal to find a point where the two polynomials intersect, meaning that $P(x)$ would also have to intersect that point (it must be between the two graphs). Setting $x^2 - 2x + 2 = 2x^2 - 4x + 3$ , we find that $x = 1$ . Note that both of these graphs have the same vertex (at $x = 1$ ), and so $P(x)$ must also have the same vertex $(1, 1)$ . Setting $P(x) = ax^2 - 2ax + a + 1$ (this is where we have a vertex at $(1, 1)$ ), we plug in $11$ and find that $a = 1.8$ . Evaluating $1.8x^2 - 3.6x + 2.8$ when $x = 16$ (our intended goal), we find that $P(16) = \boxed{406}$ | null | 406 |
8e853659e3b64adce178129f181ab503 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_8 | For a real number $a$ , let $\lfloor a \rfloor$ denote the greatest integer less than or equal to $a$ . Let $\mathcal{R}$ denote the region in the coordinate plane consisting of points $(x,y)$ such that $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$ . The region $\mathcal{R}$ is completely contained in a disk of radius $r$ (a disk is the union of a circle and its interior). The minimum value of $r$ can be written as $\frac {\sqrt {m}}{n}$ , where $m$ and $n$ are integers and $m$ is not divisible by the square of any prime. Find $m + n$ | The desired region consists of 12 boxes, whose lower-left corners are integers solutions of $x^2 + y^2 = 25$ , namely $(\pm5,0), (0,\pm5), (\pm3,\pm4), (\pm4,\pm3).$ Since the points themselves are symmetric about $(0,0)$ , the boxes are symmetric about $\left(\frac12,\frac12\right)$ . The distance from $\left(\frac12,\frac12\right)$ to the furthest point on a box that lays on an axis, for instance $(6,1)$ , is $\sqrt {\frac {11}2^2 + \frac12^2} = \sqrt {\frac {122}4}.$ The distance from $\left(\frac12,\frac12\right)$ to the furthest point on a box in the middle of a quadrant, for instance $(5,4)$ , is $\sqrt {\frac92^2 + \frac72^2} = \sqrt {\frac {130}4}.$ The latter is the larger, and is $\frac {\sqrt {130}}2$ , giving an answer of $130 + 2 = \boxed{132}$ | null | 132 |
2c6adc2842a8e365a8f81e01f4735d15 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_9 | Let $(a,b,c)$ be the real solution of the system of equations $x^3 - xyz = 2$ $y^3 - xyz = 6$ $z^3 - xyz = 20$ . The greatest possible value of $a^3 + b^3 + c^3$ can be written in the form $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | This is almost the same as solution 1. Note $a^3 + b^3 + c^3 = 28 + 3abc$ . Next, let $k = a^3$ . Note that $b = \sqrt [3]{k + 4}$ and $c = \sqrt [3]{k + 18}$ , so we have $28 + 3\sqrt [3]{k(k+4)(k+18)} = 28+3abc=a^3+b^3+c^3=3k+22$ . Move 28 over, divide both sides by 3, then cube to get $k^3-6k^2+12k-8 = k^3+22k^2+18k$ . The $k^3$ terms cancel out, so solve the quadratic to get $k = -2, -\frac{1}{7}$ . We maximize $abc$ by choosing $k = -\frac{1}{7}$ , which gives us $a^3+b^3+c^3 = 3k + 22 = \frac{151}{7}$ . Thus, our answer is $151+7=\boxed{158}$ | null | 158 |
d83e02b06c3cd4d9fb39460d596a337f | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_10 | Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$ | Note that $a_2\cdot 10^2 + a_0$ is the base $100$ representation of any number from $0$ to $9999$ , and similarly $10(a_3\cdot 10^2 + a_1)$ is ten times the base $100$ representation of any number from $0$ to $9999$ . Thus, the number of solutions is just the number of solutions to $2010 = 10a+b$ where $0\le a, b\le 9999$ , which is equal to $\boxed{202}$ as $a$ can range from $0$ to $201$ | null | 202 |
d83e02b06c3cd4d9fb39460d596a337f | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_10 | Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$ | Note that $a_0 \equiv 2010\ (\textrm{mod}\ 10)$ and $a_1 \equiv 2010 - a_0\ (\textrm{mod}\ 100)$ . It's easy to see that exactly 10 values in $0 \leq a_0 \leq 99$ that satisfy our first congruence. Similarly, there are 10 possible values of $a_1$ for each choice of $a_0$ . Thus, there are $10 \times 10 = 100$ possible choices for $a_0$ and $a_1$ . We next note that if $a_0$ and $a_1$ are chosen, then a valid value of $a_3$ determines $a_2$ , so we dive into some simple casework:
Our answer is thus $6 + 0 + 196 = \boxed{202}$ | null | 202 |
d83e02b06c3cd4d9fb39460d596a337f | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_10 | Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$ | We immediately see that $a_3$ can only be $0$ $1$ or $2$ . We also note that the maximum possible value for $10a_1 + a_0$ is $990 + 99 = 1089$ . We then split into cases:
Case 1: $a_3 = 0$ .
We try to find possible values of $a_2$ . We plug in $a_3 = 0$ and $10a_1 + a_0 = 1089$ to our initial equation, which gives us $2010 = 0 + 100a_2 + 1089$ . Thus $a_2 \geq 10$ . We also see that $a_2 \leq 20$ . We now take these values of $a_2$ and find the number of pairs $(a_1, a_0)$ that work. If $a_2 = 10$ $10a_1 + a_0 = 1010$ . We see that there are $8$ possible pairs in this case. Using the same logic, there are $10$ ways for $a_2 = 11, 12 \ldots 19$ . For $a_2 = 20$ , we get the equation $10a_1 + a_0 = 10$ , for 2 ways. Thus, for $a_3 = 0$ , there are $8 + 10 \cdot 9 + 2 = 100$ ways.
Case 2: $a_3 = 1$ .
This case is almost identical to the one above, except $0 \leq a_2 \leq 10$ . We also get 100 ways.
Case 3: $a_3 = 2$ .
If $a_3 = 2$ , our initial equation becomes $100a_2 + 10a_1 + a_0 = 10$ . It is obvious that $a_2 = 0$ , and we are left with $10a_1 + a_0 = 10$ . We saw above that there are $2$ ways.
Totaling everything, we get that there are $100 + 100 + 2 = \boxed{202}$ ways. | null | 202 |
d83e02b06c3cd4d9fb39460d596a337f | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_10 | Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$ | We will represent the problem using generating functions. Consider the generating function \[f(x) = (1+x^{1000}+x^{2000}+\cdots+x^{99000})(1+x^{100}+x^{200}+\cdots+x^{9900})(1+x^{10}+x^{20}+\cdots+x^{990})(1+x+x^2+\cdots+x^{99})\] where the first factor represents $a_3$ , the second factor $a_2$ , and so forth. We want to find the coefficient of $x^{2010}$ in the expansion of $f(x)$ . Now rewriting each factor using the geometric series yields \[f(x) = \frac{\cancel{x^{100}-1}}{x-1} \cdot \frac{\cancel{x^{1000}-1}}{x^{10}-1} \cdot \frac{x^{10000}-1}{\cancel{x^{100}-1}} \cdot \frac{x^{100000}-1}{\cancel{x^{1000}-1}}=\frac{x^{10000}-1}{x-1} \cdot \frac{x^{100000}-1}{x^{10}-1} = (1+x+x^2+\cdots + x^{9999})(1+x^{10}+x^{20}+\cdots+x^{99990})\] The coefficient of $x^{2010}$ in this is simply $\boxed{202}$ , as we can choose any of the first 202 terms from the second factor and pair it with exactly one term in the first factor. | null | 202 |
d83e02b06c3cd4d9fb39460d596a337f | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_10 | Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$ , where the $a_i$ 's are integers, and $0 \le a_i \le 99$ . An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$ . Find $N$ | First note that $a_3$ has to be a single-digit number( $0$ $1$ , or $2$ to be exact), and that $a_1$ has to be a two-digit multiple of ten.
Then, $a_3$ $a_2$ $a_1$ and $a_0$ can be represented as follows: \begin{align*} a_3 = a \\ a_2 = 10b+c \\ a_1= 10d+e \\ a_0 = 10f \end{align*} , where $a$ $b$ $c$ $d$ $e$ , and $f$ are all(not necessarily nonzero) digits.
Now, we can write our given equation as follows: \begin{align*} 2010 = 1000(a+b) + 100(c+d) + 10(e+f) \\ 201 = 100(a+b) + 10(c+d) + (e+f) \\ 201 = (100a + 10c + e) + (100b + d + f) \end{align*} Now, each integer between $0$ and $201$ inclusive can be represented in exactly one way as $100a + 10c + e$ , and this corresponds with one unique $100b + d + f$ , so it remains to count the number of integers between $0$ and $201$ inclusive. This is easily counted to be $\boxed{202}$ $\blacksquare$ | null | 202 |
8e46c0a543513bc87d4c50f7fb742ecc | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_11 | Let $\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \le 10$ and $3y - x \ge 15$ . When $\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$ , the volume of the resulting solid is $\frac {m\pi}{n\sqrt {p}}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime , and $p$ is not divisible by the square of any prime. Find $m + n + p$ | The inequalities are equivalent to $y \ge x/3 + 5, y \le 10 - |x - 8|$ . We can set them equal to find the two points of intersection, $x/3 + 5 = 10 - |x - 8| \Longrightarrow |x - 8| = 5 - x/3$ . This implies that one of $x - 8, 8 - x = 5 - x/3$ , from which we find that $(x,y) = \left(\frac 92, \frac {13}2\right), \left(\frac{39}{4}, \frac{33}{4}\right)$ . The region $\mathcal{R}$ is a triangle , as shown above. When revolved about the line $y = x/3+5$ , the resulting solid is the union of two right cones that share the same base and axis.
Let $h_1,h_2$ denote the height of the left and right cones, respectively (so $h_1 > h_2$ ), and let $r$ denote their common radius. The volume of a cone is given by $\frac 13 Bh$ ; since both cones share the same base, then the desired volume is $\frac 13 \cdot \pi r^2 \cdot (h_1 + h_2)$ . The distance from the point $(8,10)$ to the line $x - 3y + 15 = 0$ is given by $\left|\frac{(8) - 3(10) + 15}{\sqrt{1^2 + (-3)^2}}\right| = \frac{7}{\sqrt{10}}$ . The distance between $\left(\frac 92, \frac {13}2\right)$ and $\left(\frac{39}{4}, \frac{33}{4}\right)$ is given by $h_1 + h_2 = \sqrt{\left(\frac{18}{4} - \frac{39}{4}\right)^2 + \left(\frac{26}{4} - \frac{33}{4}\right)^2} = \frac{7\sqrt{10}}{4}$ . Thus, the answer is $\frac{343\sqrt{10}\pi}{120} = \frac{343\pi}{12\sqrt{10}} \Longrightarrow 343 + 12 + 10 = \boxed{365}$ . (Note to MAA: Is it a coincidence that this is the number of days in a non-leap year?) | null | 365 |
b2ff9fc1f8995e6a982ecbedcdb21cf6 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_12 | Let $m \ge 3$ be an integer and let $S = \{3,4,5,\ldots,m\}$ . Find the smallest value of $m$ such that for every partition of $S$ into two subsets, at least one of the subsets contains integers $a$ $b$ , and $c$ (not necessarily distinct) such that $ab = c$
Note : a partition of $S$ is a pair of sets $A$ $B$ such that $A \cap B = \emptyset$ $A \cup B = S$ | We claim that $243$ is the minimal value of $m$ . Let the two partitioned sets be $A$ and $B$ ; we will try to partition $3, 9, 27, 81,$ and $243$ such that the $ab=c$ condition is not satisfied. Without loss of generality , we place $3$ in $A$ . Then $9$ must be placed in $B$ , so $81$ must be placed in $A$ , and $27$ must be placed in $B$ . Then $243$ cannot be placed in any set, so we know $m$ is less than or equal to $243$
For $m \le 242$ , we can partition $S$ into $S \cap \{3, 4, 5, 6, 7, 8, 81, 82, 83, 84 ... 242\}$ and $S \cap \{9, 10, 11 ... 80\}$ , and in neither set are there values where $ab=c$ (since $8 < (3\text{ to }8)^2 < 81$ and $81^2>242$ and $(9\text{ to }80)^2 > 80$ ). Thus $m = \boxed{243}$ | null | 243 |
80c47714cd4a12b36aaeb9ef5d5bbd95 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_13 | Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$ , and segment $CD$ at distinct points $N$ $U$ , and $T$ , respectively. Line $\ell$ divides region $\mathcal{R}$ into two regions with areas in the ratio $1: 2$ . Suppose that $AU = 84$ $AN = 126$ , and $UB = 168$ . Then $DA$ can be represented as $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ | The center of the semicircle is also the midpoint of $AB$ . Let this point be O. Let $h$ be the length of $AD$
Rescale everything by 42, so $AU = 2, AN = 3, UB = 4$ . Then $AB = 6$ so $OA = OB = 3$
Since $ON$ is a radius of the semicircle, $ON = 3$ . Thus $OAN$ is an equilateral triangle.
Let $X$ $Y$ , and $Z$ be the areas of triangle $OUN$ , sector $ONB$ , and trapezoid $UBCT$ respectively.
$X = \frac {1}{2}(UO)(NO)\sin{O} = \frac {1}{2}(1)(3)\sin{60^\circ} = \frac {3}{4}\sqrt {3}$
$Y = \frac {1}{3}\pi(3)^2 = 3\pi$
To find $Z$ we have to find the length of $TC$ . Project $T$ and $N$ onto $AB$ to get points $T'$ and $N'$ . Notice that $UNN'$ and $TUT'$ are similar. Thus:
$\frac {TT'}{UT'} = \frac {UN'}{NN'} \implies \frac {TT'}{h} = \frac {1/2}{3\sqrt {3}/2} \implies TT' = \frac {\sqrt {3}}{9}h$
Then $TC = T'C - T'T = UB - TT' = 4 - \frac {\sqrt {3}}{9}h$ . So:
$Z = \frac {1}{2}(BU + TC)(CB) = \frac {1}{2}\left(8 - \frac {\sqrt {3}}{9}h\right)h = 4h - \frac {\sqrt {3}}{18}h^2$
Let $L$ be the area of the side of line $l$ containing regions $X, Y, Z$ . Then
$L = X + Y + Z = \frac {3}{4}\sqrt {3} + 3\pi + 4h - \frac {\sqrt {3}}{18}h^2$
Obviously, the $L$ is greater than the area on the other side of line $l$ . This other area is equal to the total area minus $L$ . Thus:
$\frac {2}{1} = \frac {L}{6h + \frac {9}{2}{\pi} - L} \implies 12h + 9\pi = 3L$
Now just solve for $h$
\begin{align*} 12h + 9\pi & = \frac {9}{4}\sqrt {3} + 9\pi + 12h - \frac {\sqrt {3}}{6}h^2 \\ 0 & = \frac {9}{4}\sqrt {3} - \frac {\sqrt {3}}{6}h^2 \\ h^2 & = \frac {9}{4}(6) \\ h & = \frac {3}{2}\sqrt {6} \end{align*}
Don't forget to un-rescale at the end to get $AD = \frac {3}{2}\sqrt {6} \cdot 42 = 63\sqrt {6}$
Finally, the answer is $63 + 6 = \boxed{069}$ | null | 069 |
80c47714cd4a12b36aaeb9ef5d5bbd95 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_13 | Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$ , and segment $CD$ at distinct points $N$ $U$ , and $T$ , respectively. Line $\ell$ divides region $\mathcal{R}$ into two regions with areas in the ratio $1: 2$ . Suppose that $AU = 84$ $AN = 126$ , and $UB = 168$ . Then $DA$ can be represented as $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ | Let $O$ be the center of the semicircle. It follows that $AU + UO = AN = NO = 126$ , so triangle $ANO$ is equilateral
Let $Y$ be the foot of the altitude from $N$ , such that $NY = 63\sqrt{3}$ and $NU = 21$
Finally, denote $DT = a$ , and $AD = x$ . Extend $U$ to point $Z$ so that $Z$ is on $CD$ and $UZ$ is perpendicular to $CD$ . It then follows that $ZT = a-84$ . Since $NYU$ and $UZT$ are similar
$\frac {x}{a-84} = \frac {63\sqrt{3}}{21} = 3\sqrt{3}$
Given that line $NT$ divides $R$ into a ratio of $1:2$ , we can also say that
$(x)(\frac{84+a}{2}) + \frac {126^2\pi}{6} - (63)(21)(\sqrt{3}) = (\frac{1}{3})(252x + \frac{126^2\pi}{2})$
where the first term is the area of trapezoid $AUTD$ , the second and third terms denote the areas of $\frac{1}{6}$ a full circle, and the area of $NUO$ , respectively, and the fourth term on the right side of the equation is equal to $R$ . Cancelling out the $\frac{126^2\pi}{6}$ on both sides, we obtain
$(x)(\frac{84+a}{2}) - \frac{252x}{3} = (63)(21)(\sqrt{3})$
By adding and collecting like terms, $\frac{3ax - 252x}{6} = (63)(21)(\sqrt{3})$
$\frac{(3x)(a-84)}{6} = (63)(21)(\sqrt{3})$
Since $a - 84 = \frac{x}{3\sqrt{3}}$
$\frac {(3x)(\frac{x}{3\sqrt{3}})}{6} = (63)(21)(\sqrt{3})$
$\frac {x^2}{\sqrt{3}} = (63)(126)(\sqrt{3})$
$x^2 = (63)(126)(3) = (2)(3^5)(7^2)$
$x = AD = (7)(3^2)(\sqrt{6}) = 63\sqrt{6}$ , so the answer is $\boxed{069}.$ | null | 069 |
80c47714cd4a12b36aaeb9ef5d5bbd95 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_13 | Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$ , and segment $CD$ at distinct points $N$ $U$ , and $T$ , respectively. Line $\ell$ divides region $\mathcal{R}$ into two regions with areas in the ratio $1: 2$ . Suppose that $AU = 84$ $AN = 126$ , and $UB = 168$ . Then $DA$ can be represented as $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ | Note that the total area of $\mathcal{R}$ is $252DA + \frac {126^2 \pi}{2}$ and thus one of the regions has area $84DA + \frac {126^2 \pi}{6}$
As in the above solutions we discover that $\angle AON = 60^\circ$ , thus sector $ANO$ of the semicircle has $\frac{1}{3}$ of the semicircle's area.
Similarly, dropping the $N'T'$ perpendicular we observe that $[AN'T'D] = 84DA$ , which is $\frac{1}{3}$ of the total rectangle.
Denoting the region to the left of $\overline {NT}$ as $\alpha$ and to the right as $\beta$ , it becomes clear that if $[\triangle UT'T] = [\triangle NUO]$ then the regions will have the desired ratio.
Using the 30-60-90 triangle, the slope of $NT$ , is ${-3\sqrt{3}}$ , and thus $[\triangle UT'T] = \frac {DA^2}{6\sqrt{3}}$
$[NUO]$ is most easily found by $\frac{absin(c)}{2}$ $[\triangle NUO] = \frac {126*42 * \frac {\sqrt{3}}{2}}{2}$
Equating, $\frac {126*42 * \frac {\sqrt{3}}{2}}{2} = \frac {DA^2}{6\sqrt{3}}$
Solving, $63 * 21 * 3 * 6 = DA^2$
$DA = 63 \sqrt{6} \longrightarrow \boxed{069}$ | null | 069 |
80c47714cd4a12b36aaeb9ef5d5bbd95 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_13 | Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$ , and segment $CD$ at distinct points $N$ $U$ , and $T$ , respectively. Line $\ell$ divides region $\mathcal{R}$ into two regions with areas in the ratio $1: 2$ . Suppose that $AU = 84$ $AN = 126$ , and $UB = 168$ . Then $DA$ can be represented as $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ | Like above solutions, note that $ANO$ is equilateral with side length $126,$ where $O$ is the midpoint of $AB.$ Then, if we let $DA=a$ and set origin at $D=(0,0),$ we get $N=(63,a+63\sqrt{3}), U=(84,a).$ Line $NU$ is then $y-a=\sqrt{27}(x-84),$ so it intersects $CA,$ the $x$ -axis, at $x=(a/\sqrt{27}+84),$ giving us point $T.$ Now the area of region $R$ is $252a+\pi(126)^2 / 2,$ so one third of that is $84a+\pi(126)^2 / 6.$
The area of the smaller piece of $R$ is $[AUTD] + [ANU] + [\text{lune} AN] = \frac{1}{2} \cdot a(84+\frac{a}{\sqrt{27}}+84)+\frac{1}{2} \cdot 84 \cdot 63 \sqrt{3}+ \frac{\pi (126)^2}{6}-\frac{1}{2} \cdot 126 \cdot 63 \sqrt{2}$ $=\frac{a^2}{2\sqrt{27}}+84a-21\cdot 63\sqrt{2} + \frac{\pi(126)^2}{6}.$ Setting this equal to $84a+\pi(126)^2 / 6$ and canceling the $84a + \pi(126)^2$ yields $\frac{a^2}{2\sqrt{27}}=21 \cdot 63 \sqrt{3},$ so $a = 63 \sqrt{6}$ and the anser is $\boxed{069}.$ ~ rzlng | null | 069 |
80c47714cd4a12b36aaeb9ef5d5bbd95 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_13 | Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$ , and segment $CD$ at distinct points $N$ $U$ , and $T$ , respectively. Line $\ell$ divides region $\mathcal{R}$ into two regions with areas in the ratio $1: 2$ . Suppose that $AU = 84$ $AN = 126$ , and $UB = 168$ . Then $DA$ can be represented as $m\sqrt {n}$ , where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n$ | Once we establish that $\Delta ANO$ is equilateral, we have \[[{\rm Sector } BON] = 2[{\rm Sector } AON], [BCT'U]=2[ADT'U]\] \[\Rightarrow [\overset{\large\frown}{NB} CT'UO]=2[\overset{\large\frown}{NA} DT'UO]\] On the other hand, \[[\overset{\large\frown}{NB} CT]=2[\overset{\large\frown}{NA} DT]\] Therefore, $[UT'T]=[NUO]$
Now, $UO=42, NU=21 \Rightarrow [UT'T]=[NUO]=2[NN'U]$ . Also $\Delta UT'T \sim \Delta NN'U$ . Therefore, \[DA=UT'=\sqrt{2} NN'=\sqrt{2} \left(\frac{\sqrt{3}}{2}\cdot 126\right)=63\sqrt{6}\longrightarrow \boxed{069}\] | null | 069 |
65a9b2c38f45db09022ecc223852fda1 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_14 | For each positive integer $n$ , let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$ . Find the largest value of $n$ for which $f(n) \le 300$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ | Observe that $f$ is strictly increasing in $n$ . We realize that we need $100$ terms to add up to around $300$ , so we need some sequence of $2$ s, $3$ s, and then $4$ s.
It follows that $n \approx 100$ (alternatively, use binary search to get to this, with $n\le 1000$ ). Manually checking shows that $f(109) = 300$ and $f(110) > 300$ . Thus, our answer is $\boxed{109}$ | null | 109 |
65a9b2c38f45db09022ecc223852fda1 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_14 | For each positive integer $n$ , let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$ . Find the largest value of $n$ for which $f(n) \le 300$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ | Because we want the value for which $f(n)=300$ , the average value of the 100 terms of the sequence should be around $3$ . For the value of $\lfloor \log_{10} (kn) \rfloor$ to be $3$ $1000 \le kn < 10000$ . We want kn to be around the middle of that range, and for k to be in the middle of 0 and 100, let $k=50$ , so $50n=\frac{10000+1000}{2}=\frac{11000}{2}=5500$ , and $n = 110$ $f(110) = 301$ , so we want to lower $n$ . Testing $109$ yields $300$ , so our answer is still $\boxed{109}$ | null | 109 |
65a9b2c38f45db09022ecc223852fda1 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_14 | For each positive integer $n$ , let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$ . Find the largest value of $n$ for which $f(n) \le 300$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ | For any $n$ where the sum is close to $300$ , all the terms in the sum must be equal to $2$ $3$ or $4$ . Let $M$ be the number of terms less than or equal to $3$ and $N$ be the number of terms equal to $2$ (also counted in $M$ ). With this definition of $M$ and $N$ the total will be $400 - M - N \le 300$ , from which $M + N \ge 100$ . Now $M+1$ is the smallest integer $k$ for which $\log_{10}(kn) \ge 4$ or $kn \ge 10000$ , thus \[M = \left\lfloor\frac{9999}{n}\right\rfloor.\] Similarly, \[N = \left\lfloor\frac{999}{n}\right\rfloor = \left\lfloor\frac{M}{10}\right\rfloor.\]
Therefore, \[M + \left\lfloor \frac{M}{10} \right\rfloor \ge 100 \implies M \ge \left\lceil\frac{1000}{11}\right\rceil = 91 \implies n \le \left\lfloor\frac{9999}{91}\right\rfloor = 109.\] Since we want the largest possible $n$ , the answer is $\boxed{109}$ | null | 109 |
65a9b2c38f45db09022ecc223852fda1 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_14 | For each positive integer $n$ , let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$ . Find the largest value of $n$ for which $f(n) \le 300$
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$ | Since we're working with base- $10$ logarithms, we can start by testing out $n$ 's that are powers of $10$ . For $n = 1$ , the terms in the sum are $\lfloor \log_{10} (1)\rfloor, \lfloor \log_{10} (2)\rfloor, \lfloor \log_{10} (3) \rfloor , . . . , \lfloor \log_{10} (100) \rfloor$ . For numbers $1$ $9$ $\lfloor \log_{10} (kn) \rfloor = 0$ . Then we have $90$ numbers, namely $10$ $99$ , for which $\lfloor \log_{10} (kn) \rfloor = 1$ . The last number we have is $100$ , which gives us $\lfloor \log_{10} (kn) \rfloor = 2$ . This sum gives us only $90 + 2 = 92$ , which is much too low. However, applying the same counting technique for $n = 10$ , our sum comes out to be $9 + 180 + 3 = 192$ , since there are $9$ terms for which $\lfloor \log_{10} (kn) \rfloor = 1$ $90$ terms for which $\lfloor \log_{10} (kn) \rfloor = 2$ , and one term for which $\lfloor \log_{10} (kn) \rfloor = 3$ . So we go up one more power of $10$ and get $18 + 270 + 4 = 292$ , which is very close to what we are looking for.
Now we only have to bump up the value of $n$ a bit and check our sum. Each increase in $n$ by $1$ actually increases the value of our sum by $1$ as well (except for $n = 101$ ), because whenever a $4$ is added to the sum, a $3$ is taken away. It doesn't take long to check and see that the value of $n$ we're looking for is $\boxed{109}$ , which corresponds to a sum of exactly $300$ | null | 109 |
f57e23a3227324264a5d93cb24506175 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_15 | In $\triangle{ABC}$ with $AB = 12$ $BC = 13$ , and $AC = 15$ , let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii . Then $\frac{AM}{CM} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Let $AM = x$ , then $CM = 15 - x$ . Also let $BM = d$ Clearly, $\frac {[ABM]}{[CBM]} = \frac {x}{15 - x}$ . We can also express each area by the rs formula. Then $\frac {[ABM]}{[CBM]} = \frac {p(ABM)}{p(CBM)} = \frac {12 + d + x}{28 + d - x}$ . Equating and cross-multiplying yields $25x + 2dx = 15d + 180$ or $d = \frac {25x - 180}{15 - 2x}.$ Note that for $d$ to be positive, we must have $7.2 < x < 7.5$
By Stewart's Theorem , we have $12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)$ or $432 = 3d^2 + 40x - 3x^2.$ Brute forcing by plugging in our previous result for $d$ , we have $432 = \frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.$ Clearing the fraction and gathering like terms, we get $0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.$
Aside: Since $x$ must be rational in order for our answer to be in the desired form, we can use the Rational Root Theorem to reveal that $6x$ is an integer because we can divide the polynomial by $2$ . The only such $x$ in the above-stated range is $\frac {22}3$
Legitimately solving that quartic, note that $x = 0$ and $x = 15$ should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get $0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).$ The only solution in the desired range is thus $\frac {22}3$ . Then $CM = \frac {23}3$ , and our desired ratio $\frac {AM}{CM} = \frac {22}{23}$ , giving us an answer of $\boxed{045}$ | null | 045 |
f57e23a3227324264a5d93cb24506175 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_15 | In $\triangle{ABC}$ with $AB = 12$ $BC = 13$ , and $AC = 15$ , let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii . Then $\frac{AM}{CM} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Let $AM = 2x$ and $BM = 2y$ so $CM = 15 - 2x$ . Let the incenters of $\triangle ABM$ and $\triangle BCM$ be $I_1$ and $I_2$ respectively, and their equal inradii be $r$ . From $r = \sqrt {(s - a)(s - b)(s - c)/s}$ , we find that
\begin{align*}r^2 & = \frac {(x + y - 6)( - x + y + 6)(x - y + 6)}{x + y + 6} & (1) \\ & = \frac {( - x + y + 1)(x + y - 1)( - x - y + 14)}{ - x + y + 14}. & (2) \end{align*}
Let the incircle of $\triangle ABM$ meet $AM$ at $P$ and the incircle of $\triangle BCM$ meet $CM$ at $Q$ . Then note that $I_1 P Q I_2$ is a rectangle. Also, $\angle I_1 M I_2$ is right because $MI_1$ and $MI_2$ are the angle bisectors of $\angle AMB$ and $\angle CMB$ respectively and $\angle AMB + \angle CMB = 180^\circ$ . By properties of tangents to circles $MP = (MA + MB - AB)/2 = x + y - 6$ and $MQ = (MB + MC - BC)/2 = - x + y + 1$ . Now notice that the altitude of $M$ to $I_1 I_2$ is of length $r$ , so by similar triangles we find that $r^2 = MP \cdot MQ = (x + y - 6)( - x + y + 1)$ (3). Equating (3) with (1) and (2) separately yields
\begin{align*} 2y^2 - 30 = 2xy + 5x - 7y \\ 2y^2 - 70 = - 2xy - 5x + 7y, \end{align*}
and adding these we have
\[4y^2 - 100 = 0\implies y = 5\implies x = 11/3 \\ \implies AM/MC = (22/3)/(15 - 22/3) = 22/23 \implies \boxed{045}.\] | null | 045 |
f57e23a3227324264a5d93cb24506175 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_15 | In $\triangle{ABC}$ with $AB = 12$ $BC = 13$ , and $AC = 15$ , let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii . Then $\frac{AM}{CM} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Let the incircle of $ABM$ hit $AM$ $AB$ $BM$ at $X_{1},Y_{1},Z_{1}$ , and let the incircle of $CBM$ hit $MC$ $BC$ $BM$ at $X_{2},Y_{2},Z_{2}$ . Draw the incircle of $ABC$ , and let it be tangent to $AC$ at $X$ . Observe that we have a homothety centered at A sending the incircle of $ABM$ to that of $ABC$ , and one centered at $C$ taking the incircle of $BCM$ to that of $ABC$ . These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is $AX_{1}/AX=CX_{2}/CX$
By standard computations, $AX=\dfrac{AB+AC-BC}{2}=7$ and $CX=\dfrac{BC+AC-AB}{2}=8$ . Now, let $AX_{1}=7x$ and $CX_{2}=8x$ . We will now go around and chase lengths. Observe that $BY_{1}=BA-AY_{1}=BA-AX_{1}=12-7x$ . Then, $BZ_{1}=12-7x$ . We also have $CY_{2}=CX_{2}=8x$ , so $BY_{2}=13-8x$ and $BZ_{2}=13-8x$
Observe now that $X_{1}M+MX_{2}=AC-15x=15(1-x)$ . Also, $X_{1}M-MX_{2}=MZ_{1}-MZ_{2}=BZ_{2}-BZ_{1}=BY_{2}-BY_{1}=(1-x)$ . Solving, we get $X_{1}M=8-8x$ and $MX_{2}=7-7x$ (as a side note, note that $AX_{1}+MX_{2}=X_{1}M+X_{2}C$ , a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).
Now, we get $BM=BZ_{2}+Z_{2}M=BZ_{2}+MX_{2}=20-15x$ . To finish, we will compute area ratios. $\dfrac{[ABM]}{[CBM]}=\dfrac{AM}{MC}=\dfrac{8-x}{7+x}$ . Also, since their inradii are equal, we get $\dfrac{[ABM]}{[CBM]}=\dfrac{40-16x}{40-14x}$ . Equating and cross multiplying yields the quadratic $3x^{2}-8x+4=0$ , so $x=2/3,2$ . However, observe that $AX_{1}+CX_{2}=15x<15$ , so we take $x=2/3$ . Our ratio is therefore $\dfrac{8-2/3}{7+2/3}=\dfrac{22}{23}$ , giving the answer $\boxed{045}$ | null | 045 |
f57e23a3227324264a5d93cb24506175 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_15 | In $\triangle{ABC}$ with $AB = 12$ $BC = 13$ , and $AC = 15$ , let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii . Then $\frac{AM}{CM} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Let the common inradius equal r, $BM = x$ $AM = y$ $MC = z$
From the prespective of $\triangle{ABM}$ and $\triangle{BMC}$ we get:
$S_{ABM} = rs = r \cdot (\frac{12+x+y}{2})$ $S_{BMC} = rs = r \cdot (\frac{13+x+z}{2})$
Add two triangles up, we get $\triangle{ABC}$
$S_{ABC} = S_{ABM} + S_{BMC} = r \cdot \frac{25+2x+y+z}{2}$
Since $y + z = 15$ , we get:
By drawing an altitude from $I_1$ down to a point $H_1$ and from $I_2$ to $H_2$ , we can get:
$r \cdot cot(\frac{\angle A}{2}) =r \cdot A H_1 = r \cdot \frac{12+y-x}{2}$ and
$r \cdot cot(\frac{\angle C}{2}) = r \cdot H_2 C = r \cdot \frac{13+z-x}{2}$
Adding these up, we get:
$r \cdot (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) = \frac{25+y+z-2x}{2} = \frac{40-2x}{2} = 20-x$
Now, we have 2 values equal to r, we can set them equal to each other:
If we let R denote the incircle of ABC, note:
AC = $(cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R = 15$ and
$S_{ABC} = \frac{12+13+15}{2} \cdot R = 20 \cdot R$
By cross multiplying the equation above, we get:
We can find out x:
Now, we can find ratio of y and z:
The answer is $\boxed{045}$ | null | 045 |
f57e23a3227324264a5d93cb24506175 | https://artofproblemsolving.com/wiki/index.php/2010_AIME_I_Problems/Problem_15 | In $\triangle{ABC}$ with $AB = 12$ $BC = 13$ , and $AC = 15$ , let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii . Then $\frac{AM}{CM} = \frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p + q$ | Let $CM=x, AM=rx, BM=d$ $x+rx=15\Rightarrow x=\frac{15}{1+r}$
Similar to Solution 1, we have \[r=\frac{[AMB]}{[CMB]}=\frac{12+rx+d}{13+x+d} \Rightarrow d=\frac{13r-12}{1-r}\] as well as \[12^2\cdot x + 13^2 rx=15x\cdot rx+15d^2 (\text{via Stewart's Theorem})\] \[\frac{(12^2 + 13^2r) \cdot 15}{1+r} - \frac{15r\cdot 15^2}{(1+r)^2}=\frac{15(13r-12)^2}{(1-r)^2}\] \[\frac{169r^2+88r+144}{(1+r)^2}=\frac{(13r-12)^2}{(1-r)^2}\] \[(169r^2+88r+144)((r^2+1)-2r)=(169r^2-312r+144)((r^2+1)+2r)\] \[(r^2+1)(400r)=2r(338r^2-224r+288)\] \[100(r^2+1)=169r^2-112r+144 \Rightarrow 69r^2-112r+44=(23r-22)(3r-2)=0\]
Since $d=\frac{13r-12}{1-r}>0$ , we have $r=\frac{22}{23} \longrightarrow \boxed{045}$ | null | 045 |
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