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335ad41c1cde963664b13052bbcaf023 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_15 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ $O_1O_2 = 15$ $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. [asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\omega_1$",8*dir(110),SW); label("$\omega_2$",5*dir(70)+(15,0),SE); label("$O_1$",O1,W); label("$O_2$",O2,E); label("$B$",B,N+1/2*E); label("$A$",A,N+1/2*W); label("$C$",C,S+1/4*W); label("$D$",D,S+1/4*E); label("$15$",midpoint(O1--O2),N); label("$16$",midpoint(C--D),N); label("$2$",midpoint(A--B),S); label("$\Omega$",o.C+(o.r-1)*dir(270)); [/asy] | Let points $A'$ and $B'$ be the reflections of $A$ and $B,$ respectively, about the perpendicular bisector of $O_1 O_2.$ \[B'O_2 = BO_1 = O_1 P = O_1 C,\] \[A'O_1 = AO_2 = O_2 P = O_2 D.\] We establish the equality of the arcs and conclude that the corresponding chords are equal \[\overset{\Large\frown} {CO_1} + \overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} = \overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} =\overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {DO_2} +\overset{\Large\frown} {A'B'}\] \[\implies A'D = B'C = O_1 O_2 = 15.\] Similarly $A'C = B'D \implies \triangle A'CO_1 = \triangle B'DO_2.$
Ptolemy's theorem on $A'CDB'$ yields \[B'D \cdot A'C + A'B' \cdot CD = A'D \cdot B'C \implies\] \[B'D^2 + 2 \cdot 16 = 15^2 \implies B'D = A'C = \sqrt{193}.\] The area of the trapezoid $A'CDB'$ is equal to the area of an isosceles triangle with sides $A'D = B'C = 15$ and $A'B' + CD = 18.$
The height of this triangle is $\sqrt{15^2-9^2} = 12.$ The area of $A'CDB'$ is $108.$
\[\sin \angle B'CD = \frac{12}{15} = \frac{4}{5},\] \[\angle B'CD + \angle B'O_2 D = 180^o \implies \sin \angle B'O_2 D = \frac{4}{5}.\]
Denote $\angle B'O_2 D = 2\alpha.$ $\angle B'O_2 D > \frac{\pi}{2},$ hence $\cos \angle B'O_2 D = \cos 2\alpha = -\frac{3}{5}.$ \[\tan \alpha =\frac { \sin 2 \alpha}{1+\cos 2 \alpha} = \frac {4/5}{1 - 3/5}=2.\]
Semiperimeter of $\triangle B'O_2 D$ is $s = \frac {15 + \sqrt{193}}{2}.$
The distance from the vertex $O_2$ to the tangent points of the inscribed circle of the triangle $B'O_2 D$ is equal $s – B'D = \frac{15 – \sqrt{193}}{2}.$
The radius of the inscribed circle is $r = (s – B'D) \tan \alpha.$
The area of triangle $B'O_2 D$ is $[B'O_2 D] = sr = s (s – B'D) \tan \alpha = \frac {15^2 – 193}{2} = 16.$
The hexagon $ABO_1 CDO_2$ has the same area as hexagon $B'A'O_1 CDO_2.$
The area of hexagon $B'A'O_1 CDO_2$ is equal to the sum of the area of the trapezoid $A'CDB'$ and the areas of two equal triangles $B'O_2 D$ and $A'O_1 C,$ so the area of the hexagon $ABO_1 CDO_2$ is \[108 + 16 + 16 = \boxed{140}.\] | null | 140 |
78692c88b6b91edefd22a39ececabb65 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_1 | Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | For the next five races, Zou wins four and loses one. Let $W$ and $L$ denote a win and a loss, respectively. There are five possible outcome sequences for Zou:
We proceed with casework:
Case (1): Sequences #1-4, in which Zou does not lose the last race.
The probability that Zou loses a race is $\frac13,$ and the probability that Zou wins the next race is $\frac13.$ For each of the three other races, the probability that Zou wins is $\frac23.$
There are four sequences in this case. The probability of one such sequence is $\left(\frac13\right)^2\left(\frac23\right)^3.$
Case (2): Sequence #5, in which Zou loses the last race.
The probability that Zou loses a race is $\frac13.$ For each of the four other races, the probability that Zou wins is $\frac23.$
There is one sequence in this case. The probability is $\left(\frac13\right)^1\left(\frac23\right)^4.$
Answer
The requested probability is \[4\left(\frac13\right)^2\left(\frac23\right)^3+\left(\frac13\right)^1\left(\frac23\right)^4=\frac{32}{243}+\frac{16}{243}=\frac{48}{243}=\frac{16}{81},\] from which the answer is $16+81=\boxed{097}.$ | null | 097 |
78692c88b6b91edefd22a39ececabb65 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_1 | Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | We have $5$ cases, depending on which race Zou lost. Let $\text{W}$ denote a won race, and $\text{L}$ denote a lost race for Zou. The possible cases are $\text{WWWWL, WWWLW, WWLWW, WLWWW, LWWWW}$ . The first case has probability $\left(\frac{2}{3} \right)^4 \cdot \frac{1}{3} = \frac{16}{3^5}$ . The second case has probability $\left( \frac{2}{3} \right)^3 \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{8}{3^5}$ . The third has probability $\left( \frac{2}{3} \right)^2 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{8}{3^5}$ . The fourth has probability $\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^2 = \frac{8}{3^5}$ . Lastly, the fifth has probability $\frac{1}{3} \cdot \frac{1}{3} \cdot \left( \frac{2}{3} \right)^3 = \frac{8}{3^5}$ . Adding these up, the total probability is $\frac{16 + 8 \cdot 4}{3^5} = \frac{16 \cdot 3}{3^5} = \frac{16}{81}$ , so $m+n = \boxed{097}$ | null | 097 |
78692c88b6b91edefd22a39ececabb65 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_1 | Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Case 1: Zou loses the first race
In this case, Zou must win the rest of the races. Thus, our probability is $\frac{8}{243}$
Case 2: Zou loses the last race
There is only one possibility for this, so our probability is $\frac{16}{243}$
Case 3: Neither happens
There are three ways that this happens. Each has one loss that is not the last race. Therefore, the probability that one happens is $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{8}{243}$ . Thus, the total probability is $\frac{8}{243} \cdot 3 = \frac{24}{243}$
Adding these up, we get $\frac{48}{243} = \frac{16}{81}$ , so $16+81=\boxed{097}$ | null | 097 |
78692c88b6b91edefd22a39ececabb65 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_1 | Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Note that Zou wins one race. The probability that he wins the last race is $\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)=\frac{16}{243}.$ Now, if he doesn't win the last race, then there must be two races where the winner of the previous race loses. We can choose any $4$ of the middle races for Zou to win. So the probability for this case is $4\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^2=\frac{32}{243}.$ Thus, the answer is $\frac{16}{243}+\frac{32}{243}=\frac{16}{81}\implies\boxed{097}.$ | null | 097 |
b2069d0c22703a044f4314aa75ab3b98 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2 | In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy] | Again, let the intersection of $AE$ and $BC$ be $G$ . By AA similarity, $\triangle AFG \sim \triangle CDG$ with a $\frac{7}{3}$ ratio. Define $x$ as $\frac{[CDG]}{9}$ . Because of similar triangles, $[AFG] = 49x$ . Using $ABCD$ , the area of the parallelogram is $33-18x$ . Using $AECF$ , the area of the parallelogram is $63-98x$ . These equations are equal, so we can solve for $x$ and obtain $x = \frac{3}{8}$ . Thus, $18x = \frac{27}{4}$ , so the area of the parallelogram is $33 - \frac{27}{4} = \frac{105}{4}$
Finally, the answer is $105+4=\boxed{109}$ | null | 109 |
b2069d0c22703a044f4314aa75ab3b98 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2 | In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy] | Let $P = AD \cap FC$ , and $K = AE \cap BC$ . Also let $AP = x$
$CK$ also has to be $x$ by parallelogram properties. Then $PD$ and $BK$ must be $11-x$ because the sum of the segments has to be $11$
We can easily solve for $PC$ by the Pythagorean Theorem: \begin{align*} DC^2 + PD^2 &= PC^2\\ 9 + (11-x)^2 &= PC^2 \end{align*} It follows shortly that $PC = \sqrt{x^2-22x+30}$
Also, $FC = 9$ , and $FP + PC = 9$ . We can then say that $PC = \sqrt{x^2-22x+30}$ , so $FP = 9 - \sqrt{x^2-22x+30}$
Now we can apply the Pythagorean Theorem to $\triangle AFP$ \begin{align*} AF^2 + FP^2 = AP^2\\ 49 + \left(9 - \sqrt{x^2-22x+30}\right)^2 = x^2 \end{align*}
This simplifies (not-as-shortly) to $x = \dfrac{35}{4}$ . Now we have to solve for the area of $APCK$ . We know that the height is $3$ because the height of the parallelogram is the same as the height of the smaller rectangle.
From the area of a parallelogram (we know that the base is $\dfrac{35}{4}$ and the height is $3$ ), it is clear that the area is $\dfrac{105}{4}$ , giving an answer of $\boxed{109}$ | null | 109 |
b2069d0c22703a044f4314aa75ab3b98 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2 | In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy] | Suppose $B=(0,0).$ It follows that $A=(0,3),C=(11,0),$ and $D=(11,3).$
Since $AECF$ is a rectangle, we have $AE=FC=9$ and $EC=AF=7.$ The equation of the circle with center $A$ and radius $\overline{AE}$ is $x^2+(y-3)^2=81,$ and the equation of the circle with center $C$ and radius $\overline{CE}$ is $(x-11)^2+y^2=49.$
We now have a system of two equations with two variables. Expanding and rearranging respectively give \begin{align*} x^2+y^2-6y&=72, &(1) \\ x^2+y^2-22x&=-72. &(2) \end{align*} Subtracting $(2)$ from $(1),$ we obtain $22x-6y=144.$ Simplifying and rearranging produce \[x=\frac{3y+72}{11}. \hspace{34.5mm} (*)\] Substituting $(*)$ into $(1)$ gives \[\left(\frac{3y+72}{11}\right)^2+y^2-6y=72,\] which is a quadratic of $y.$ We clear fractions by multiplying both sides by $11^2=121,$ then solve by factoring: \begin{align*} \left(3y+72\right)^2+121y^2-726y&=8712 \\ \left(9y^2+432y+5184\right)+121y^2-726y&=8712 \\ 130y^2-294y-3528&=0 \\ 2(5y+21)(13y-84)&=0 \\ y&=-\frac{21}{5},\frac{84}{13}. \end{align*} Since $E$ is in Quadrant IV, we have $E=\left(\frac{3\left(-\frac{21}{5}\right)+72}{11},-\frac{21}{5}\right)=\left(\frac{27}{5},-\frac{21}{5}\right).$ It follows that the equation of $\overleftrightarrow{AE}$ is $y=-\frac{4}{3}x+3.$
Let $G$ be the intersection of $\overline{AD}$ and $\overline{FC},$ and $H$ be the intersection of $\overline{AE}$ and $\overline{BC}.$ Since $H$ is the $x$ -intercept of $\overleftrightarrow{AE},$ we get $H=\left(\frac94,0\right).$
By symmetry, quadrilateral $AGCH$ is a parallelogram. Its area is $HC\cdot AB=\left(11-\frac94\right)\cdot3=\frac{105}{4},$ from which the requested sum is $105+4=\boxed{109}.$ | null | 109 |
b2069d0c22703a044f4314aa75ab3b98 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2 | In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ [asy] pair A, B, C, D, E, F; A=(0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy] | Let the intersection of $AE$ and $BC$ be $G$ . It is useful to find $\tan(\angle DAE)$ , because $\tan(\angle DAE)=\frac{3}{BG}$ and $\frac{3}{\tan(\angle DAE)}=BG$ . From there, subtracting the areas of the two triangles from the larger rectangle, we get Area = $33-3BG=33-\frac{9}{\tan(\angle DAE)}$
let $\angle CAD = \alpha$ . Let $\angle CAE = \beta$ . Note, $\alpha+\beta=\angle DAE$
$\alpha=\tan^{-1}\left(\frac{3}{11}\right)$
$\beta=\tan^{-1}\left(\frac{7}{9}\right)$
$\tan(\angle DAE) = \tan\left(\tan^{-1}\left(\frac{3}{11}\right)+\tan^{-1}\left(\frac{7}{9}\right)\right) = \frac{\frac{3}{11}+\frac{7}{9}}{1-\frac{3}{11}\cdot\frac{7}{9}} = \frac{\frac{104}{99}}{\frac{78}{99}} = \frac{4}{3}$
$\mathrm{Area}=33-\frac{9}{\frac{4}{3}} = 33-\frac{27}{4 } = \frac{105}{4}$ . The answer is $105+4=\boxed{109}$ | null | 109 |
2f305d5e8d7d3021e31a2bfc3aa89294 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_3 | Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$ | We want to find the number of positive integers $n<1000$ which can be written in the form $n = 2^a - 2^b$ for some non-negative integers $a > b \ge 0$ (note that if $a=b$ , then $2^a-2^b = 0$ ). We first observe $a$ must be at most 10; if $a \ge 11$ , then $2^a - 2^b \ge 2^{10} > 1000$ . As $2^{10} = 1024 \approx 1000$ , we can first choose two different numbers $a > b$ from the set $\{0,1,2,\ldots,10\}$ in $\binom{11}{2}=55$ ways. This includes $(a,b) = (10,0)$ $(10,1)$ $(10,2)$ $(10,3)$ $(10,4)$ which are invalid as $2^a - 2^b > 1000$ in this case. For all other choices $a$ and $b$ , the value of $2^a - 2^b$ is less than 1000.
We claim that for all other choices of $a$ and $b$ , the values of $2^a - 2^b$ are pairwise distinct. More specifically, if $(a_1,b_1) \neq (a_2,b_2)$ where $10 \ge a_1 > b_1 \ge 0$ and $10 \ge a_2 > b_2 \ge 0$ , we must show that $2^{a_1}-2^{b_1} \neq 2^{a_2} - 2^{b_2}$ . Suppose otherwise for sake of contradiction; rearranging yields $2^{a_1}+2^{b_2} = 2^{a_2}+2^{b_1}$ . We use the fact that every positive integer has a unique binary representation:
If $a_1 \neq b_2$ then $\{a_1,b_2\} = \{a_2,b_1\}$ ; from here we can deduce either $a_1=a_2$ and $b_1=b_2$ (contradicting the assumption that $(a_1,b_1) \neq (a_2,b_2)$ , or $a_1=b_1$ and $a_2=b_2$ (contradicting the assumption $a_1>b_1$ and $a_2>b_2$ ).
If $a_1 = b_2$ then $2^{a_1}+2^{b_2} = 2 \times 2^{a_1}$ , and it follows that $a_1=a_2=b_1=b_2$ , also contradicting the assumption $(a_1,b_1) \neq (a_2,b_2)$ . Hence we obtain contradiction.*
Then there are $\binom{11}{2}-5$ choices for $(a,b)$ for which $2^a - 2^b$ is a positive integer less than 1000; by the above claim, each choice of $(a,b)$ results in a different positive integer $n$ . Then there are $55-5 = \boxed{050}$ integers which can be expressed as a difference of two powers of 2. | null | 050 |
2f305d5e8d7d3021e31a2bfc3aa89294 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_3 | Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$ | Case 1: When our answer is in the form $2^n-2^i$ , where $i$ is an integer such that $0\le i\le 4$
We start with the subcase where it is $2^n-2^0$ , for some integer $n$ where $n>0$ (this is because the case where $n=0$ yields $2^0-2^0=0$ , which doesn't work because it must be a positive integer.)
Note that $2^{10}=1024$ , and $2^9=512$ . Our answer needs to be less than $1000$ , so the maximum possible result (in this case) is $2^9-2^0$ . Our lowest result is $2^1-2^0$ . All the positive powers of two less than $1024$ work, so we have $9$ possibilities for this subcase. For subcases $i=1, i=2, i=3,$ and $i=4$ , we have $8, 7, 6,$ and $5$ possibilities, respectively.
Case 2: When our answer is in the form of $2^n-2^j$ , where $j$ is an integer such that $5\le j\le 9$
We can start with the subcase where $j=5$ . We notice that $2^5=32$ , and $2^{10}-2^5=992$ which is less than $1000$ , so the greatest result in this subcase is actually $2^{10}-2^5$ , and the lowest is $2^6-2^5$ . Thus, we have $5$ possibilities. For the other four subcases, we have $4, 3, 2,$ and $1$ possibilities, respectively.
Answer: We note that these are our only cases, as numbers in the form of $2^n-2^{10}$ and beyond are greater than $1000$
Thus, our result is $9+8+7+6+5+5+4+3+2+1=(9+8+7+6+5+4+3+2+1)+5=\boxed{50}$ . ~jehu26 | null | 50 |
2f305d5e8d7d3021e31a2bfc3aa89294 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_3 | Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$ | We look for all positive integers of the form $2^a-2^b<1000,$ where $0\leq b<a.$ Performing casework on $a,$ we can enumerate all possibilities in the table below: \[\begin{array}{c|c} & \\ [-2.25ex] \boldsymbol{a} & \boldsymbol{b} \\ \hline & \\ [-2ex] 1 & 0 \\ 2 & 0,1 \\ 3 & 0,1,2 \\ 4 & 0,1,2,3 \\ 5 & 0,1,2,3,4 \\ 6 & 0,1,2,3,4,5 \\ 7 & 0,1,2,3,4,5,6 \\ 8 & 0,1,2,3,4,5,6,7 \\ 9 & 0,1,2,3,4,5,6,7,8 \\ 10 & \xcancel{0},\xcancel{1},\xcancel{2},\xcancel{3},\xcancel{4},5,6,7,8,9 \\ [0.5ex] \end{array}\] As indicated by the X-marks, the ordered pairs $(a,b)=(10,0),(10,1),(10,2),(10,3),(10,4)$ generate $2^a-2^b>1000,$ which are invalid.
Note that each of the remaining ordered pairs generates one unique desired positive integer.
We prove this statement as follows:
Together, we have justified our claim in bold. The answer is \[1+2+3+4+5+6+7+8+9+5=\boxed{050}.\] | null | 050 |
2f305d5e8d7d3021e31a2bfc3aa89294 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_3 | Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$ | Because the difference is less than $1000$ , we can simply list out all numbers that satisfy $2^n < 1000$ . We get $0 \le n < 10$ , where n is an integer. Because the sequence $2^n$ is geometric, the difference of any two terms will be unique. $\binom{10}{2}$ will be the number of differences for $0\le n < 10$ . However, we also need to consider the case in which $n=10$ . With simple counting, we find that $5$ numbers: $(32, 64, 128, 256, 512)$ could be subtracted from $1024$ , which makes another 5 cases. There is no need to check for higher exponents since the lowest difference would be $2^{11} - 2^{10} = 1024$ , which exceeds $1000$
Thus, the final answer is $\binom{10}{2} + 5 = \boxed{050}.$ | null | 050 |
bf89c60f7dc1fdf2863e7c321461a63d | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_4 | Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile. | Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)$ all work, for a total of $29$ . Continuing this pattern until $21$ coins in the first pile, we have the sum \begin{align*} 31+29+28+26+25+\cdots+4+2+1 &= (31+28+25+22+\cdots+1)+(29+26+23+\cdots+2) \\ &= 176+155 \\ &= \boxed{331} | null | 331 |
bf89c60f7dc1fdf2863e7c321461a63d | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_4 | Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile. | We make an equation: $a+b+c=66,$ where $a<b<c.$ We don't have a clear solution, so we'll try complementary counting. First, let's find where $a\geq b\geq c.$ By stars and bars, we have $\dbinom{65}{2}=2080$ to assign positive integer solutions to $a + b + c = 66.$ Now we need to subtract off the cases where it doesn't satisfy the condition.
We start with $a = b.$ We can write that as $2b + c = 66.$ We can find there are 32 integer solutions to this equation. There are $32$ solutions for $b=c$ and $a = c$ by symmetry. We also need to add back $2$ because we subtracted $(a,b,c)=(22,22,22)$ $3$ times.
We then have to divide by $6$ because there are $3!=6$ ways to order $a, b,$ and $c.$ Therefore, we have $\dfrac{\dbinom{65}{2}-96+2}{6} = \dfrac{1986}{6} = \boxed{331}.$ | null | 331 |
bf89c60f7dc1fdf2863e7c321461a63d | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_4 | Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile. | Let the piles have $a, b$ and $c$ coins, with $0 < a < b < c$ . Then, let $b = a + k_1$ , and $c = b + k_2$ , such that each $k_i \geq 1$ . The sum is then $a + a+k_1 + a+k_1+k_2 = 66 \implies 3a+2k_1 + k_2 = 66$ . This is simply the number of positive solutions to the equation $3x+2y+z = 66$ . Now, we take cases on $a$
If $a = 1$ , then $2k_1 + k_2 = 63 \implies 1 \leq k_1 \leq 31$ . Each value of $k_1$ corresponds to a unique value of $k_2$ , so there are $31$ solutions in this case. Similarly, if $a = 2$ , then $2k_1 + k_2 = 60 \implies 1 \leq k_1 \leq 29$ , for a total of $29$ solutions in this case. If $a = 3$ , then $2k_1 + k_2 = 57 \implies 1 \leq k_1 \leq 28$ , for a total of $28$ solutions. In general, the number of solutions is just all the numbers that aren't a multiple of $3$ , that are less than or equal to $31$
We then add our cases to get \begin{align*} 1 + 2 + 4 + 5 + \cdots + 31 &= 1 + 2 + 3 + \cdots + 31 - 3(1 + 2 + 3 + \cdots + 10) \\ &= \frac{31(32)}{2} - 3(55) \\ &= 31 \cdot 16 - 165 \\ &= 496 - 165 \\ &= \boxed{331} as our answer. | null | 331 |
1bf5f347021dedb9c72017a0953cd0eb | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_5 | Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences. | Let the terms be $a-b$ $a$ , and $a+b$ . Then we want $(a-b)^2+a^2+(a+b)^2=ab^2$ , or $3a^2+2b^2=ab^2$ . Rearranging, we get $b^2=\frac{3a^2}{a-2}$ . Simplifying further, $b^2=3a+6+\frac{12}{a-2}$ . Looking at this second equation, since the right side must be an integer, $a-2$ must equal $\pm1, 2, 3, 4, 6, 12$ . Looking at the first equation, we see $a>2$ since $b^2$ is positive. This means we must test $a=3, 4, 5, 6, 8, 14$ . After testing these, we see that only $a=5$ and $a=14$ work which give $b=5$ and $b=7$ respectively. Thus the answer is $10+21=\boxed{031}$ .
~JHawk0224 | null | 031 |
1bf5f347021dedb9c72017a0953cd0eb | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_5 | Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences. | Let the common difference be $d$ and let the middle term be $x$ . Then, we have that the sequence is \[x-d,~x,~x+d.\] This means that the sum of the squares of the 3 terms of the sequence is \[(x-d)^2+x^2+(x+d)^2=x^2-2xd+d^2+x^2+x^2+2xd+d^2=3x^2+2d^2.\] We know that this must be equal to $xd^2,$ so we can write that \[3x^2+2d^2=xd^2,\] and it follows that \[3x^2-xd^2+2d^2=3x^2-\left(d^2\right)x+2d^2=0.\]
Now, we can treat $d$ as a constant and use the quadratic formula to get \[x=\frac{d^2\pm \sqrt{d^4-4(3)(2d^2)}}{6}.\] We can factor pull $d^2$ out of the square root to get \[x=\frac{d^2\pm d\sqrt{d^2-24}}{6}.\] Here, it is easy to figure out the values of $d$ . Let $\sqrt{d^2-24} = k$ , then $d^2-k^2=24$ which is $(d+k)(d-k)=24,$ note that $d$ $k$ are integers. Examining the parity, we find that $d+k$ and $d-k$ are of the same parity. Now, we solve by factoring. We can find that $d=5$ and $d=7$ are the only positive integer values of $d$ that make $\sqrt{d^2-24}$ a positive integer. $^{*}$ $d=5$ gives $x=5$ and $x=\frac{10}{3}$ , but we can ignore the latter. $d=7$ gives $x=14$ , as well as a fraction which we can ignore.
Since $d=5,~x=5$ and $d=7, x=14$ are the only two solutions and we want the sum of the third terms, our answer is $(5+5)+(7+14)=10+21=\boxed{031}$ . -BorealBear, minor edit by Kinglogic | null | 031 |
1bf5f347021dedb9c72017a0953cd0eb | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_5 | Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences. | As in Solution 1, write the three integers in the sequence as $a-d$ $a$ , and $a+d$
Then the sum of the squares of the three integers is $(a-d)^2+a^2+(a+d)^2 = 3a^2+2d^2$
Setting this equal to the middle term times the common difference squared, which is $ad^2$
and solving for $d^2$ we get:
$3a^2+2d^2 = ad^2 \implies ad^2-2d^2 = 3a^2 \implies d^2(a-2) = 3a^2 \implies d^2 = \frac{3a^2}{a-2}$
The numerator has to be positive, so the denominator has to be positive too for the sequence
to be strictly increasing; that is, $a>2$
For $\frac{3a^2}{a-2}$ to be a perfect square, $\frac{3}{a-2}$ must be a perfect square as well.
This means that $a-2$ is divisible by 3, and whatever left over is a perfect square.
We can express this as an equation: let the perfect square left over be $n^2$ . Then:
$3n^2 = a-2$ . Now when you divide the numerator and denominator by 3, you are left with
$d^2 = \frac{a^2}{n^2} \implies d = \frac{a}{n}$ . Because the sequence is of integers, d must also be an
integer, which means that $n$ must divide $a$
Taking the above equation we can solve for $a$ $3n^2 = a-2 \implies a = 3n^2+2$
This means that $3n^2+2$ is divisible by $n$ $3n^2$ is automatically divisible by $n$ , so
$2$ must be divisible by $n$ . Then $n$ must be either of $\{1,2\}$ . Plugging back into the
equation,
$n = 1 \implies a = 5 \implies d = 5$ , so $a+d = 5+5 = 10$
$n = 2 \implies a = 14 \implies d = 7$ , so $a+d = 14+7 = 21$
Finally, $10+21 = \boxed{031}$ | null | 031 |
1bf5f347021dedb9c72017a0953cd0eb | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_5 | Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences. | Following from previous solutions, we derive $3x^2+2a^2=xa^2.$ We divide both sides to get $3\left(\frac{x}{a}\right)^2+2=x.$ Since $x$ is an integer, $\frac{x}{a}$ must also be an integer, so we have $x=pa$ , for some factor $p$ . We then get $3p^2+2=pa.$ We then take this to modulo $p$ , getting $2\equiv 0 \pmod p.$ The only possibilities for $p$ are therefore 1 and 2. We plug these into $3p^2+2=pa$ , for $a=5$ and $x=5$ , giving us the sequence $0,5,10$ , or $2a=14$ and $x=14$ , for the sequence $7,14,21.$ $10+21 = \boxed{031}.$ | null | 031 |
1724982405561821e411f6c79caec021 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6 | Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$ | First scale down the whole cube by $12$ . Let point $P$ have coordinates $(x, y, z)$ , point $A$ have coordinates $(0, 0, 0)$ , and $s$ be the side length. Then we have the equations \begin{align*} (s-x)^2+y^2+z^2&=\left(5\sqrt{10}\right)^2, \\ x^2+(s-y)^2+z^2&=\left(5\sqrt{5}\right)^2, \\ x^2+y^2+(s-z)^2&=\left(10\sqrt{2}\right)^2, \\ (s-x)^2+(s-y)^2+(s-z)^2&=\left(3\sqrt{7}\right)^2. \end{align*} These simplify into \begin{align*} s^2+x^2+y^2+z^2-2sx&=250, \\ s^2+x^2+y^2+z^2-2sy&=125, \\ s^2+x^2+y^2+z^2-2sz&=200, \\ 3s^2-2s(x+y+z)+x^2+y^2+z^2&=63. \end{align*} Adding the first three equations together, we get $3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575$ .
Subtracting this from the fourth equation, we get $2(x^2+y^2+z^2)=512$ , so $x^2+y^2+z^2=256$ . This means $PA=16$ . However, we scaled down everything by $12$ so our answer is $16*12=\boxed{192}$ | null | 192 |
1724982405561821e411f6c79caec021 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6 | Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$ | Once the equations for the distance between point P and the vertices of the cube have been written, we can add the first, second, and third to receive, \[2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.\] Subtracting the fourth equation gives \begin{align*} 2(x^2 + y^2 + z^2) &= 575 - 63 \\ x^2 + y^2 + z^2 &= 256 \\ \sqrt{x^2 + y^2 + z^2} &= 16. \end{align*} Since point $A = (0,0,0), PA = 16$ , and since we scaled the answer is $16 \cdot 12 = \boxed{192}$ | null | 192 |
1724982405561821e411f6c79caec021 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6 | Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$ | Let $E$ be the vertex of the cube such that $ABED$ is a square.
Using the British Flag Theorem , we can easily show that \[PA^2 + PE^2 = PB^2 + PD^2\] and \[PA^2 + PG^2 = PC^2 + PE^2\] Hence, by adding the two equations together, we get $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$ . Substituting in the values we know, we get $2PA^2 + 7\cdot 36^2 =10\cdot60^2 + 5\cdot 60^2 + 2\cdot 120^2$
Thus, we can solve for $PA$ , which ends up being $\boxed{192}$ | null | 192 |
1724982405561821e411f6c79caec021 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_6 | Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ $CP=60\sqrt{5}$ $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$ | For all points $X$ in space, define the function $f:\mathbb{R}^{3}\rightarrow\mathbb{R}$ by $f(X)=PX^{2}-GX^{2}$ . Then $f$ is linear; let $O=\frac{2A+G}{3}$ be the center of $\triangle BCD$ . Then since $f$ is linear, \begin{align*} 3f(O)=f(B)+f(C)+f(D)&=2f(A)+f(G) \\ \left(PB^{2}-GB^{2}\right)+\left(PC^{2}-GC^{2}\right)+\left(PD^{2}-GD^{2}\right)&=2\left(PA^{2}-GA^{2}\right)+PG^{2} \\ \left(60\sqrt{10}\right)^{2}-2x^{2}+\left(60\sqrt{5}\right)^{2}-2x^{2}+\left(120\sqrt{2}\right)^{2}-2x^{2}&=2PA^{2}-2\cdot 3x^{2}+\left(36\sqrt{7}\right)^{2}, \end{align*} where $x$ denotes the side length of the cube. Thus \begin{align*} 36\text000+18\text000+28\text800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\ 82\text800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\ 73\text728&=2PA^{2} \\ 36\text864&=PA^{2} \\ PA&=\boxed{192} | null | 192 |
d607be777000c9cf5f4562a6f9c37fa8 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_7 | Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\] | It is trivial that the maximum value of $\sin \theta$ is $1$ , is achieved at $\theta = \frac{\pi}{2}+2k\pi$ for some integer $k$
This implies that $\sin(mx) = \sin(nx) = 1$ , and that $mx = \frac{\pi}{2}+2a\pi$ and $nx = \frac{\pi}{2}+2b\pi$ , for integers $a, b$
Taking their ratio, we have \[\frac{mx}{nx} = \frac{\frac{\pi}{2}+2a\pi}{\frac{\pi}{2}+2b\pi} \implies \frac{m}{n} = \frac{4a + 1}{4b + 1} \implies \frac{m}{4a + 1} = \frac{n}{4b + 1} = k.\] It remains to find all $m, n$ that satisfy this equation.
If $k = 1$ , then $m \equiv n \equiv 1 \pmod 4$ . This corresponds to choosing two elements from the set $\{1, 5, 9, 13, 17, 21, 25, 29\}$ . There are $\binom 82$ ways to do so.
If $k < 1$ , by multiplying $m$ and $n$ by the same constant $c = \frac{1}{k}$ , we have that $mc \equiv nc \equiv 1 \pmod 4$ . Then either $m \equiv n \equiv 1 \pmod 4$ , or $m \equiv n \equiv 3 \pmod 4$ . But the first case was already counted, so we don't need to consider that case. The other case corresponds to choosing two numbers from the set $\{3, 7, 11, 15, 19, 23, 27\}$ . There are $\binom 72$ ways here. (This argument seems to have a logical flaw)
Finally, if $k > 1$ , note that $k$ must be an integer. This means that $m, n$ belong to the set $\{k, 5k, 9k, \dots\}$ , or $\{3k, 7k, 11k, \dots\}$ . Taking casework on $k$ , we get the sets $\{2, 10, 18, 26\}, \{6, 14, 22, 30\}, \{4, 20\}, \{12, 28\}$ . Some sets have been omitted; this is because they were counted in the other cases already. This sums to $\binom 42 + \binom 42 + \binom 22 + \binom 22$
In total, there are $\binom 82 + \binom 72 + \binom 42 + \binom 42 + \binom 22 + \binom 22 = \boxed{063}$ pairs of $(m, n)$ | null | 063 |
d607be777000c9cf5f4562a6f9c37fa8 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_7 | Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\] | In order for $\sin(mx) + \sin(nx) = 2$ $\sin(mx) = \sin(nx) = 1$
This happens when $mx \equiv nx \equiv \frac{\pi}{2} ($ mod $2\pi).$
This means that $mx = \frac{\pi}{2} + 2\pi\alpha$ and $nx = \frac{\pi}{2} + 2\pi\beta$ for any integers $\alpha$ and $\beta$
As in Solution 1, take the ratio of the two equations: \[\frac{mx}{nx} = \frac{\frac{\pi}{2}+2\pi\alpha}{\frac{\pi}{2}+2\pi\beta} \implies \frac{m}{n} = \frac{\frac{1}{2}+2\alpha}{\frac{1}{2}+2\beta} \implies \frac{m}{n} = \frac{4\alpha+1}{4\beta+1}\]
Now notice that the numerator and denominator of $\frac{4\alpha+1}{4\beta+1}$ are both odd, which means that $m$ and $n$ have the same power of two (the powers of 2 cancel out).
Let the common power be $p$ : then $m = 2^p\cdot a$ , and $n = 2^p\cdot b$ where $a$ and $b$ are integers between 1 and 30.
We can now rewrite the equation: \[\frac{2^p\cdot a}{2^p\cdot b} = \frac{4\alpha+1}{4\beta+1} \implies \frac{a}{b} = \frac{4\alpha+1}{4\beta+1}\]
Now it is easy to tell that $a \equiv 1 ($ mod $4)$ and $b \equiv 1 ($ mod $4)$ . However, there is another case: that
$a \equiv 3 ($ mod $4)$ and $b \equiv 3 ($ mod $4)$ . This is because multiplying both $4\alpha+1$ and $4\beta+1$ by $-1$ will not change the fraction, but each congruence will be changed to $-1 ($ mod $4) \equiv 3 ($ mod $4)$
From the first set of congruences, we find that $a$ and $b$ can be two of $\{1, 5, 9, \ldots, 29\}$
From the second set of congruences, we find that $a$ and $b$ can be two of $\{3, 7, 11, \ldots, 27\}$
Now all we have to do is multiply by $2^p$ to get back to $m$ and $n$ .
Let’s organize the solutions in order of increasing values of $p$ , keeping in mind that $m$ and $n$ are bounded between 1 and 30.
For $p = 0$ we get $\{1, 5, 9, \ldots, 29\}, \{3, 7, 11, \ldots, 27\}$
For $p = 1$ we get $\{2, 10, 18, 26\}, \{6, 14, 22, 30\}$
For $p = 2$ we get $\{4, 20\}, \{12, 28\}$
Note that $16\mid{a}$ since $m$ will cancel out a factor of 4 from $a$ , and $\frac{a}{m}$ must contain a factor of 4. Again, $1-4X$ will never contribute a factor of 2. Simply inspecting, we see two feasible values for $a$ and $m$ such that $a+m\leq30$
If we increase the value of $p$ more, there will be less than two integers in our sets, so we are done there.
There are 8 numbers in the first set, 7 in the second, 4 in the third, 4 in the fourth, 2 in the fifth, and 2 in the sixth.
In each of these sets we can choose 2 numbers to be $m$ and $n$ and then assign them in increasing order. Thus there are:
\[\dbinom{8}{2}+\dbinom{7}{2}+\dbinom{4}{2}+\dbinom{4}{2}+\dbinom{2}{2}+\dbinom{2}{2} = 28+21+6+6+1+1 = \boxed{063}\] possible pairs $(m,n)$ that satisfy the conditions. | null | 063 |
d607be777000c9cf5f4562a6f9c37fa8 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_7 | Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\] | We know that the range of sine is between $-1$ and $1$ , inclusive.
Thus, the only way for the sum to be $2$ is for $\sin(mx)=\sin(nx)=1$
Note that $\sin(90+360k)=1$
Assuming $mx$ and $nx$ are both positive, $m$ and $n$ could be $1,5,9,13,17,21,25,29$ . There are $8$ ways, so $\dbinom{8}{2}$
If both are negative, $m$ and $n$ could be $3,7,11,15,19,23,27$ . There are $7$ ways, so $\dbinom{7}{2}$
However, the pair $(1,5)$ could also be $(2, 10)$ and so on. The same goes for some other pairs.
In total there are $14$ of these extra pairs.
The answer is $28+21+14 = \boxed{063}$ | null | 063 |
d607be777000c9cf5f4562a6f9c37fa8 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_7 | Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\] | The equation implies that $\sin(mx)=\sin(nx)=1$ . Therefore, we can write $mx$ as $2{\pi}k_1+\frac{\pi}{2}$ and $nx$ as $2{\pi}k_2+\frac{\pi}{2}$ for integers $k_1$ and $k_2$ . Then, $\frac{mx}{nx}=\frac{m}{n}=\frac{2k_1+\frac{1}{2}}{2k_2+\frac{1}{2}}$ . Cross multiplying, we get $m\cdot{(2k_2+\frac{1}{2})}=n\cdot{(2k_1+\frac{1}{2})} \Longrightarrow 4k_2m-4k_1n=n-m$ . Let $n-m=a$ so the equation becomes $4(m(k_2-k_1)+k_1a)=a$ . Let $k_2-k_1=X$ and $k_1=Y$ , then the equation becomes $a=4Ym+4Xa \Longrightarrow \frac{a(1-4X)}{m}=4Y$ . Note that $X$ and $Y$ can vary accordingly, and $4\mid{a}$ . Next, we do casework on $m\pmod{4}$
If $m\equiv 1\pmod{4}$
Once $a$ and $m$ are determined, $n$ is determined, so $a+m\leq30$ $a\in \{4,8,12,\dots,28\}$ and $m\in \{1,5,9,\dots,29\}$ . Therefore, there are $\sum_{i=1}^{7}{i}=28$ ways for this case such that $a+m\leq30$
If $m\equiv 3\pmod{4}$
$a\in \{4,8,12,\dots,28\}$ and $m\in \{3,7,11,\dots,27\}$ . Therefore, there are $\sum_{i=1}^{6}{i}=21$ ways such that $a+m\leq30$
If $m\equiv 2\pmod{4}$
Note that $8\mid{a}$ since $m$ in this case will have a factor of 2, which will cancel out a factor of 2 in $a$ , and we need the left hand side to divide 4. Also, $1-4X\equiv 1\pmod{4}$ so it is odd and will therefore never contribute a factor of 2. $a\in \{8,16,24\}$ and $m\in \{2,6,10,\dots,30\}$ . Following the condition $a+m\leq30$ , we conclude that there are $6+4+2=12$ ways for this case.
If $m\equiv 0\pmod{4}$
Adding all the cases up, we obtain $28+21+12+2=\boxed{063}$ | null | 063 |
e3fcb517f38b3cdb98bf957f18def378 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_8 | Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\] has $12$ distinct real solutions. | We take cases for the outermost absolute value, then rearrange: \[\left|20|x|-x^2\right|=c\pm21.\] Let $f(x)=\left|20|x|-x^2\right|.$ We rewrite $f(x)$ as a piecewise function without using absolute values: \[f(x) = \begin{cases} \left|-20x-x^2\right| & \mathrm{if} \ x \le 0 \begin{cases} 20x+x^2 & \mathrm{if} \ x\le-20 \\ -20x-x^2 & \mathrm{if} \ -20<x\leq0 \end{cases} \\ \left|20x-x^2\right| & \mathrm{if} \ x > 0 \begin{cases} 20x-x^2 & \mathrm{if} \ 0<x\leq20 \\ -20x+x^2 & \mathrm{if} \ x>20 \end{cases} \end{cases}.\] We graph $y=f(x)$ with all extremum points labeled, as shown below. The fact that $f(x)$ is an even function ( $f(x)=f(-x)$ holds for all real numbers $x,$ so the graph of $y=f(x)$ is symmetric about the $y$ -axis) should facilitate the process of graphing. [asy] /* Made by MRENTHUSIASM */ size(1200,300); real xMin = -65; real xMax = 65; real yMin = -50; real yMax = 125; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); real f(real x) { return abs(20*abs(x)-x^2); } real g(real x) { return 21; } real h(real x) { return -21; } draw(graph(f,-25,25),red,"$y=\left|20|x|-x^2\right|$"); draw(graph(g,-65,65),blue,"$y=\pm21$"); draw(graph(h,-65,65),blue); pair A[]; A[0] = (-20,0); A[1] = (-10,100); A[2] = (0,0); A[3] = (10,100); A[4] = (20,0); for(int i = 0; i <= 4; ++i) { dot(A[i],red+linewidth(4.5)); } label("$(-20,0)$",A[0],(-1.5,-1.5),red,UnFill); label("$(-10,100)$",A[1],(-1.5,1.5),red); label("$(0,0)$",A[2],(0,-1.5),red,UnFill); label("$(10,100)$",A[3],(1.5,1.5),red); label("$(20,0)$",A[4],(1.5,-1.5),red,UnFill); add(legend(),point(E),40E,UnFill); [/asy] Since $f(x)=c\pm21$ has $12$ distinct real solutions, it is clear that each case has $6$ distinct real solutions geometrically. We shift the graphs of $y=\pm21$ up $c$ units, where $c\geq0:$
Taking the intersection of these two cases gives $21<c<79,$ from which there are $79-21-1=\boxed{057}$ such integers $c.$ | null | 057 |
e3fcb517f38b3cdb98bf957f18def378 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_8 | Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\] has $12$ distinct real solutions. | Graph $y=|20|x|-x^2|$ (If you are having trouble, look at the description in the next two lines and/or the diagram in Solution 1). Notice that we want this to be equal to $c-21$ and $c+21$
We see that from left to right, the graph first dips from very positive to $0$ at $x=-20$ , then rebounds up to $100$ at $x=-10$ , then falls back down to $0$ at $x=0$
The positive $x$ are symmetric, so the graph re-ascends to $100$ at $x=10$ , falls back to $0$ at $x=10$ , and rises to arbitrarily large values afterwards.
Now we analyze the $y$ (varied by $c$ ) values. At $y=k<0$ , we will have no solutions, as the line $y=k$ will have no intersections with our graph.
At $y=0$ , we will have exactly $3$ solutions for the three zeroes.
At $y=n$ for any $n$ strictly between $0$ and $100$ , we will have exactly $6$ solutions.
At $y=100$ , we will have $4$ solutions, because local maxima are reached at $x= \pm 10$
At $y=m>100$ , we will have exactly $2$ solutions.
To get $12$ distinct solutions for $y=|20|x|-x^2|=c \pm 21$ , both $c +21$ and $c-21$ must produce $6$ solutions.
Thus $0<c-21$ and $c+21<100$ , so $c \in \{ 22, 23, \dots , 77, 78 \}$ is required.
It is easy to verify that all of these choices of $c$ produce $12$ distinct solutions (none overlap), so our answer is $\boxed{057}$ | null | 057 |
e3fcb517f38b3cdb98bf957f18def378 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_8 | Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\] has $12$ distinct real solutions. | Let $y = |x|.$ Then the equation becomes $\left|\left|20y-y^2\right|-c\right| = 21$ , or $\left|y^2-20y\right| = c \pm 21$ . Note that since $y = |x|$ $y$ is nonnegative, so we only care about nonnegative solutions in $y$ . Notice that each positive solution in $y$ gives two solutions in $x$ $x = \pm y$ ), whereas if $y = 0$ is a solution, this only gives one solution in $x$ $x = 0$ . Since the total number of solutions in $x$ is even, $y = 0$ must not be a solution. Hence, we require that $\left|y^2-20y\right| = c \pm 21$ has exactly $6$ positive solutions and is not solved by $y = 0.$
If $c < 21$ , then $c - 21$ is negative, and therefore cannot be the absolute value of $y^2 - 20y$ . This means the equation's only solutions are in $\left|y^2-20y\right| = c + 21$ . There is no way for this equation to have $6$ solutions, since the quadratic $y^2-20y$ can only take on each of the two values $\pm(c + 21)$ at most twice, yielding at most $4$ solutions. Hence, $c \ge 21$ $c$ also can't equal $21$ , since this would mean $y = 0$ would solve the equation. Hence, $c > 21.$
At this point, the equation $y^2-20y = c \pm 21$ will always have exactly $2$ positive solutions, since $y^2-20y$ takes on each positive value exactly once when $y$ is restricted to positive values (graph it to see this), and $c \pm 21$ are both positive. Therefore, we just need $y^2-20y = -(c \pm 21)$ to have the remaining $4$ solutions exactly. This means the horizontal lines at $-(c \pm 21)$ each intersect the parabola $y^2 - 20y$ in two places. This occurs when the two lines are above the parabola's vertex $(10,-100)$ . Hence we have \begin{align*} -(c + 21) &> -100 \\ c + 21 &< 100 \\ c &< 79. \end{align*} Hence, the integers $c$ satisfying the conditions are those satisfying $21 < c < 79.$ There are $\boxed{057}$ such integers. | null | 057 |
e3fcb517f38b3cdb98bf957f18def378 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_8 | Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\] has $12$ distinct real solutions. | Removing the absolute value bars from the equation successively, we get \begin{align*} \left|\left|20|x|-x^2\right|-c\right|&=21 \\ \left|20|x|-x^2\right|&= c \pm21 \\ 20|x|-x^2 &= \pm c \pm 21 \\ x^2 \pm 20x \pm c \pm21 &= 0. \end{align*} The discriminant of this equation is \[\sqrt{400-4(\pm c \pm 21)}.\] Equating the discriminant to $0$ , we see that there will be two distinct solutions to each of the possible quadratics above only in the interval $-79 < c < 79$ . However, the number of zeros the equation $ax^2+b|x|+k$ has is determined by where $ax^2+bx+k$ and $ax^2-bx+k$ intersect, namely at $(0,k)$ . When $k<0$ $a>0$ $ax^2+b|x|+k$ will have only $2$ solutions, and when $k>0$ $a>0$ , then there will be $4$ real solutions, if they exist at all.
In order to have $12$ solutions here, we thus need to ensure $-c+21<0$ , so that exactly $2$ out of the $4$ possible equations of the form $ax^2+b|x|+k$ given above have y-intercepts below $0$ and only $2$ real solutions, while the remaining $2$ equations have $4$ solutions. This occurs when $c>21$ , so our final bounds are $21<c<79$ , giving us $\boxed{057}$ valid values of $c$ | null | 057 |
8fde86b9be192eaa35b981f9a71f8206 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to $\overleftrightarrow{BC}, \overleftrightarrow{CD},$ and $\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\overline{AF}$ and $\overline{BD}.$
We set $AB=x$ and $AH=y,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); dot("$A$",A,1.5*NW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*S,linewidth(4)); dot("$G$",G,SE,linewidth(4)); dot("$H$",H,SE,linewidth(4)); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label("$10$",midpoint(A--G),1.5*(1,0)); label("$15$",midpoint(A--E),1.5*N); Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); label("$x$",midpoint(A--B),N); label("$y$",midpoint(A--H),W); [/asy] From here, we obtain $HF=18-y$ by segment subtraction, and $BG=\sqrt{x^2-10^2}$ and $HG=\sqrt{y^2-10^2}$ by the Pythagorean Theorem.
Since $\angle ABG$ and $\angle HAG$ are both complementary to $\angle AHB,$ we have $\angle ABG = \angle HAG,$ from which $\triangle ABG \sim \triangle HAG$ by AA. It follows that $\frac{BG}{AG}=\frac{AG}{HG},$ so $BG\cdot HG=AG^2,$ or \[\sqrt{x^2-10^2}\cdot\sqrt{y^2-10^2}=10^2. \hspace{10mm} (1)\] Since $\angle AHB = \angle FHD$ by vertical angles, we have $\triangle AHB \sim \triangle FHD$ by AA, with the ratio of similitude $\frac{AH}{FH}=\frac{BA}{DF}.$ It follows that $DF=BA\cdot\frac{FH}{AH}=x\cdot\frac{18-y}{y}.$
Since $\angle EBA = \angle ECD = \angle FDA$ by angle chasing, we have $\triangle EBA \sim \triangle FDA$ by AA, with the ratio of similitude $\frac{EA}{FA}=\frac{BA}{DA}.$ It follows that $DA=BA\cdot\frac{FA}{EA}=x\cdot\frac{18}{15}=\frac{6}{5}x.$
By the Pythagorean Theorem on right $\triangle ADF,$ we have $DF^2+AF^2=AD^2,$ or \[\left(x\cdot\frac{18-y}{y}\right)^2+18^2=\left(\frac{6}{5}x\right)^2. \hspace{7mm} (2)\] Solving this system of equations ( $(1)$ and $(2)$ ), we get $x=\frac{45\sqrt2}{4}$ and $y=\frac{90}{7},$ so $AB=x=\frac{45\sqrt2}{4}$ and $CD=AB+2DF=x+2\left(x\cdot\frac{18-y}{y}\right)=\frac{81\sqrt2}{4}.$ Finally, the area of $ABCD$ is \[K=\frac{AB+CD}{2}\cdot AF=\frac{567\sqrt2}{2},\] from which $\sqrt2 \cdot K=\boxed{567}.$ | null | 567 |
8fde86b9be192eaa35b981f9a71f8206 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | First, draw the diagram. Then, notice that since $ABCD$ is isosceles, $\Delta ABD \cong \Delta BAC$ , and the length of the altitude from $B$ to $AC$ is also $10$ . Let the foot of this altitude be $F$ , and let the foot of the altitude from $A$ to $BC$ be denoted as $E$ . Then, $\Delta BCF \sim \Delta ACE$ . So, $\frac{BC}{AC} = \frac{BF}{AE} = \frac{2}{3}$ . Now, notice that $[ABC] = \frac{10 \cdot AC} {2} = \frac{AB \cdot 18}{2} \implies AC = \frac{9 \cdot AB}{5}$ , where $[ABC]$ denotes the area of triangle $ABC$ . Letting $AB = x$ , this equality becomes $AC = \frac{9x}{5}$ . Also, from $\frac{BC}{AC} = \frac{2}{3}$ , we have $BC = \frac{6x}{5}$ . Now, by the Pythagorean theorem on triangles $ABF$ and $CBF$ , we have $AF = \sqrt{x^{2}-100}$ and $CF = \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$ . Notice that $AC = AF + CF$ , so $\frac{9x}{5} = \sqrt{x^{2}-100} + \sqrt{ \left( \frac{6x}{5} \right) ^{2}-100}$ . Squaring both sides of the equation once, moving $x^{2}-100$ and $\left( \frac{6x}{5} \right) ^{2}-100$ to the right, dividing both sides by $2$ , and squaring the equation once more, we are left with $\frac{32x^{4}}{25} = 324x^{2}$ . Dividing both sides by $x^{2}$ (since we know $x$ is positive), we are left with $\frac{32x^{2}}{25} = 324$ . Solving for $x$ gives us $x = \frac{45}{2\sqrt{2}}$
Now, let the foot of the perpendicular from $A$ to $CD$ be $G$ . Then let $DG = y$ . Let the foot of the perpendicular from $B$ to $CD$ be $H$ . Then, $CH$ is also equal to $y$ . Notice that $ABHG$ is a rectangle, so $GH = x$ . Now, we have $CG = GH + CH = x + y$ . By the Pythagorean theorem applied to $\Delta AGC$ , we have $(x+y)^{2}+18^{2}= \left( \frac{9x}{5} \right) ^{2}$ . We know that $\frac{9x}{5} = \frac{9}{5} \cdot \frac{45}{2\sqrt{2}} = \frac{81}{2\sqrt{2}}$ , so we can plug this into this equation. Solving for $x+y$ , we get $x+y=\frac{63}{2\sqrt{2}}$
Finally, to find $[ABCD]$ , we use the formula for the area of a trapezoid: $K = [ABCD] = \frac{b_{1}+b_{2}}{2} \cdot h = \frac{AB+CD}{2} \cdot 18 = \frac{x+(CG+DG)}{2} \cdot 18 = \frac{2x+2y}{2} \cdot 18 = (x+y) \cdot 18 = \frac{63}{2\sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}}$ . The problem asks us for $K \cdot \sqrt{2}$ , which comes out to be $\boxed{567}$ | null | 567 |
8fde86b9be192eaa35b981f9a71f8206 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Make $AE$ perpendicular to $BC$ $AG$ perpendicular to $BD$ $AF$ perpendicular $DC$
It's obvious that $\triangle{AEB} \sim \triangle{AFD}$ . Let $EB=5x; AB=5y; DF=6x; AD=6y$ . Then make $BQ$ perpendicular to $DC$ , it's easy to get $BQ=18$
Since $AB$ parallel to $DC$ $\angle{ABG}=\angle{BDQ}$ , so $\triangle{ABG} \sim \triangle{BDQ}$ . After drawing the altitude, it's obvious that $FQ=AB=5y$ , so $DQ=5y+6x$ . According to the property of similar triangles, $AG/BQ=BG/DQ$ . So, $\frac{5}{9}=\frac{GB}{(6x+5y)}$ , or $GB=\frac{(30x+25y)}{9}$
Now, we see the $\triangle AEB$ , pretty easy to find that $15^2+(5x)^2=(5y)^2$ , then we get $x^2+9=y^2$ , then express $y$ into $x$ form that $y=\sqrt{x^2+9}$ we put the length of $BG$ back to $\triangle AGB$ $BG^2+100=AB^2$ . So, \[\frac{[30x+25\sqrt{(x^2+9)}]^2}{81}+100=(5\sqrt{x^2+9})^2.\] After calculating, we can have a final equation of $x^2+9=\sqrt{x^2+9}\cdot3x$ . It's easy to find $x=\frac{3\sqrt{2}}{4}$ then $y=\frac{9\sqrt{2}}{4}$ . So, \[\sqrt{2}\cdot K = \sqrt{2}\cdot(5y+5y+6x+6x)\cdot9=\boxed{567}.\] ~bluesoul | null | 567 |
8fde86b9be192eaa35b981f9a71f8206 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Let the foot of the altitude from $A$ to $BC$ be $P$ , to $CD$ be $Q$ , and to $BD$ be $R$
Note that all isosceles trapezoids are cyclic quadrilaterals; thus, $A$ is on the circumcircle of $\triangle BCD$ and we have that $PRQ$ is the Simson Line from $A$ . As $\angle QAB = 90^\circ$ , we have that $\angle QAR = 90^\circ - \angle RAB =\angle ABR = \angle APR = \angle APQ$ , with the last equality coming from cyclic quadrilateral $APBR$ . Thus, $\triangle QAR \sim \triangle QPA$ and we have that $\frac{AQ}{AR} = \frac{PQ}{PA}$ or that $\frac{18}{10} = \frac{QP}{15}$ , which we can see gives us that $QP = 27$ . Further ratios using the same similar triangles gives that $QR = \frac{25}{3}$ and $RP = \frac{56}{3}$
We also see that quadrilaterals $APBR$ and $ARQD$ are both cyclic, with diameters of the circumcircles being $AB$ and $AD$ respectively. The intersection of the circumcircles are the points $A$ and $R$ , and we know $DRB$ and $QRP$ are both line segments passing through an intersection of the two circles with one endpoint on each circle. By Fact 5, we know then that there exists a spiral similarity with center $A$ taking $\triangle APQ$ to $\triangle APD$ . Because we know a lot about $\triangle APQ$ but very little about $\triangle APD$ and we would like to know more, we wish to find the ratio of similitude between the two triangles.
To do this, we use the one number we have for $\triangle APD$ : we know that the altitude from $A$ to $BD$ has length $10$ . As the two triangles are similar, if we can find the height from $A$ to $PQ$ , we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that $QP = 27$ . Using this, we can drop the altitude from $A$ to $QP$ and let it intersect $QP$ at $H$ . Then, let $QH = x$ and thus $HP=27-x$ . We then have by the Pythagorean Theorem on $\triangle AQH$ and $\triangle APH$ \begin{align*} 15^2 - x^2 &= 18^2 - (27-x)^2 \\ 225 - x^2 &= 324 - (x^2-54x+729) \\ 54x &= 630 \\ x &= \frac{35}{3}. \end{align*} Then, $RH = QH - QR = \frac{35}{3} - \frac{25}{3} = \frac{10}{3}$ . This gives us then from right triangle $\triangle ARH$ that $AH = \frac{20\sqrt{2}}{3}$ and thus the ratio of $\triangle APQ$ to $\triangle ABD$ is $\frac{3\sqrt{2}}{4}$ . From this, we see then that \[AB = AP \cdot \frac{3\sqrt{2}}{4} = 15 \cdot \frac{3\sqrt{2}}{4} = \frac{45\sqrt{2}}{4}\] and \[AD = AQ \cdot \frac{3\sqrt{2}}{4} = 18 \cdot \frac{3\sqrt{2}}{4} = \frac{27\sqrt{2}}{2}.\] The Pythagorean Theorem on $\triangle AQD$ then gives that \[QD = \sqrt{AD^2 - AQ^2} = \sqrt{\left(\frac{27\sqrt{2}}{2}\right)^2 - 18^2} = \sqrt{\frac{81}{2}} = \frac{9\sqrt{2}}{2}.\] Then, we have the height of trapezoid $ABCD$ is $AQ = 18$ , the top base is $AB = \frac{45\sqrt{2}}{4}$ , and the bottom base is $CD = \frac{45\sqrt{2}}{4} + 2\cdot\frac{9\sqrt{2}}{2}$ . From the equation of a trapezoid, $K = \frac{b_1+b_2}{2} \cdot h = \frac{63\sqrt{2}}{4} \cdot 18 = \frac{567\sqrt{2}}{2}$ , so the answer is $K\sqrt{2} = \boxed{567}$ | null | 567 |
8fde86b9be192eaa35b981f9a71f8206 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Let $E,F,$ and $G$ be the feet of the altitudes from $A$ to $BC,CD,$ and $DB$ , respectively.
Claim: We have $2$ pairs of similar right triangles: $\triangle AEB \sim \triangle AFD$ and $\triangle AGD \sim \triangle AEC$
Proof: Note that $ABCD$ is cyclic. We need one more angle, and we get this from this cyclic quadrilateral: \begin{align*} \angle ABE &= 180^\circ - \angle ABC =\angle ADC = \angle ADG, \\ \angle ADG &= \angle ADB =\angle ACB = \angle ACE. \hspace{20mm} \square \end{align*} Let $AD=a$ . We obtain from the similarities $AB = \frac{5a}{6}$ and $AC=BD=\frac{3a}{2}$
By Ptolemy, $\left(\frac{3a}{2}\right)^2 = a^2 + \frac{5a}{6} \cdot CD$ , so $\frac{5a^2}{4} = \frac{5a}{6} \cdot CD$
We obtain $CD=\frac{3a}{2}$ , so $DF=\frac{CD-AB}{2}=\frac{a}{3}$
Applying the Pythagorean theorem on $\triangle ADF$ , we get $324=a^2 - \frac{a^2}{9}=\frac{8a^2}{9}$
Thus, $a=\frac{27}{\sqrt{2}}$ , and $[ABCD]=\frac{AB+CD}{2} \cdot 18 = \frac{\frac{5a}{6} +\frac{9a}{6}}{2} \cdot 18 = 18 \cdot \frac{7}{6} \cdot \frac{27}{\sqrt{2}} = \frac{567}{\sqrt{2}}$ , yielding $\sqrt2\cdot[ABCD]=\boxed{567}$ | null | 567 |
8fde86b9be192eaa35b981f9a71f8206 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Let $AD=BC=a$ . Draw diagonal $AC$ and let $G$ be the foot of the perpendicular from $B$ to $AC$ $F$ be the foot of the perpendicular from $A$ to line $BC$ , and $H$ be the foot of the perpendicular from $A$ to $DC$
Note that $\triangle CBG\sim\triangle CAF$ , and we get that $\frac{10}{15}=\frac{a}{AC}$ . Therefore, $AC=\frac32 a$ . It then follows that $\triangle ABF\sim\triangle ADH$ . Using similar triangles, we can then find that $AB=\frac{5}{6}a$ . Using the Law of Cosines on $\triangle ABC$ , We can find that the $\cos\angle ABC=-\frac{1}{3}$ . Since $\angle ABF=\angle ADH$ , and each is supplementary to $\angle ABC$ , we know that the $\cos\angle ADH=\frac{1}{3}$ . It then follows that $a=\frac{27\sqrt{2}}{2}$ . Then it can be found that the area $K$ is $\frac{567\sqrt{2}}{2}$ . Multiplying this by $\sqrt{2}$ , the answer is $\boxed{567}$ | null | 567 |
8fde86b9be192eaa35b981f9a71f8206 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Draw the distances in terms of $B$ , as shown in the diagram. By similar triangles, $\triangle{AEC}\sim\triangle{BIC}$ . As a result, let $AB=u$ , then $BC=AD=\frac{6}{5}u$ and $2AC=3BC$ . The triangle $ABC$ is $6-5-9$ which $\cos(\angle{ABC})=-\frac{1}{3}$ . By angle subtraction, $\cos(180-\theta)=-\cos\theta$ . Therefore, $AB=\frac{45}{2\sqrt{2}}=\frac{45\sqrt{2}}{4}$ and $AD=BC=\frac{27}{\sqrt{2}}$ . By trapezoid area formula, the area of $ABCD$ is equal to $(AB+DF)\cdot 18=567\cdot \frac{\sqrt{2}}{2}$ which $\sqrt{2}\cdot k=\boxed{567}$ | null | 567 |
8fde86b9be192eaa35b981f9a71f8206 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | [asy] size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); filldraw(A--D--F--cycle,yellow,black+linewidth(1.5)); filldraw(A--B--E--cycle,yellow,black+linewidth(1.5)); dot("$A$",A,1.5*NW,linewidth(4)); dot("$B$",B,1.5*NE,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$D$",D,1.5*SW,linewidth(4)); dot(E,linewidth(4)); dot(F,linewidth(4)); dot(G,linewidth(4)); label("$E$",E,NE); label("$F$",F, S); label("$G$",G,SE); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label("$10$",midpoint(A--G),1.5*(1,0)); label("$15$",midpoint(A--E),1.5*N); label("$5x$",midpoint(A--B),S); label("$6x$",midpoint(A--D),1.5*(-1,0)); Label L = Label("$18$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); [/asy] Let the points formed by dropping altitudes from $A$ to the lines $BC$ $CD$ , and $BD$ be $E$ $F$ , and $G$ , respectively.
We have \[\triangle ABE \sim \triangle ADF \implies \frac{AD}{18} = \frac{AB}{15} \implies AD = \frac{6}{5}AB\] and \[BD\cdot10 = 2[ABD] = AB\cdot18 \implies BD = \frac{9}{5}AB.\] For convenience, let $AB = 5x$ . By Heron's formula on $\triangle ABD$ , we have sides $5x,6x,9x$ and semiperimeter $10x$ , so \[\sqrt{10x\cdot5x\cdot4x\cdot1x} = [ABD] = \frac{AB\cdot18}{2} = 45x \implies 10\sqrt{2}x^2 = 45x \implies x= \frac{45}{10\sqrt{2}},\] so $AB = 5x = \frac{45}{2\sqrt{2}}$
Then, \[BE = \sqrt{AB^2 - CA^2} = \sqrt{\left(\frac{45}{2\sqrt{2}}\right)^2 - 15^2} = \sqrt{\frac{225}{8}} = \frac{15}{2\sqrt{2}}\] and \[\triangle ABE \sim \triangle ADF \implies DF = \frac{6}{5}BE = \frac{6}{5}\cdot\frac{15}{2\sqrt{2}} = \frac{18}{2\sqrt{2}}.\] Finally, recalling that $ABCD$ is isosceles, \[K = [ABCD] = \frac{18}{2}(AB + (AB + 2DF)) = 18(AB + DF) = 18\left(\frac{45}{2\sqrt{2}} + \frac{18}{2\sqrt{2}}\right) = \frac{567}{\sqrt{2}},\] so $\sqrt{2}\cdot K = \boxed{567}$ | null | 567 |
8fde86b9be192eaa35b981f9a71f8206 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to ${BC}, {CD},$ and ${BD},$ respectively. $AE = 15, AF = 18, AG =10$ .
Denote by $G'$ the base of the perpendicular from $B$ to $AC, H$ be the base of the perpendicular from $C$ to $AB$ . Denote $\theta = \angle{CBH}.$ It is clear that \[BG' = AG, CH = AF, \triangle CBH \ =\triangle ADF,\] the area of $ABCD$ is equal to the area of the rectangle $AFCH.$
The problem is reduced to finding $AH$
In triangle $ABC$ all altitudes are known: \[AB : BC : AC = \frac{1}{CH}\ : \frac{1}{AE}\ : \frac{1}{BG'}\ =\] \[= \frac{1}{AF}\ : \frac{1}{AE}\ : \frac{1}{AG}\ = 5 : 6 : 9.\] We apply the Law of Cosines to $\triangle ABC$ and get $:$ \begin{align*} 2\cdot AB\cdot BC \cdot \cos\theta = AC^2 – AB^2 – BC^2, \end{align*} \begin{align*} 2\cdot 5\cdot 6\cdot \cos\theta = 60 \cos\theta = 9^2 – 5^2 – 6^2 = 20, \cos\theta =\frac{1}{3}. \end{align*} \begin{align*} BH = BC \cos\theta = \frac{BC}{3}.\end{align*} We apply the Pythagorean Law to $\triangle HBC$ and get $:$ \begin{align*} HC^2 = 18^2 = BC^2 – BH^2 = 9\cdot BH^2 – BH^2 = 8 BH^2.\end{align*} \begin{align*} BH = \frac{9}{\sqrt2}, AH = (\frac{5}{2} + 1)\cdot BH = \frac{63}{2\cdot \sqrt2}. \end{align*} Required area is \begin{align*} K = \frac{63}{2\cdot \sqrt{2}} \cdot 18 = \frac{567}{\sqrt{2}} \implies \sqrt{2} K=\boxed{567} | null | 567 |
8fde86b9be192eaa35b981f9a71f8206 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | Let $F$ be on $DC$ such that $AF \| DC$ . Let $G$ be on $BD$ such that $AG \| BD$
Let $m$ be the length of $AB$ . Let $n$ be the length of $AD$
The area of $\triangle ABD$ can be expressed in three ways: $\frac{1}{2}(15)(BC) = \frac{1}{2}(15)(n)$ $\frac{1}{2}(18)(m)$ , and $\frac{1}{2}(10)(BD)$
\[\frac{1}{2}(15)(n) = \frac{1}{2}(18)(m)\] \[15n = 18m\] \[5n = 6m\] \[n = \frac{6}{5}m\]
Now, $BD = BG + GD = \sqrt{m^2-100} + \sqrt{n^2-100}$ . We can substitute in $n = \frac{6}{5}m$ to get
$BD = \sqrt{m^2-100} + \sqrt{(\frac{6}{5}m)^2-100}$
We have \[\frac{1}{2}(10)\left(\sqrt{m^2-100} + \sqrt{(\frac{6}{5}m)^2-100}\right) = \frac{1}{2}(18)(m)\] After a fairly straightforward algebraic bash, we get $m = \frac{45\sqrt{2}}{4}$ , and $n = (\frac{6}{5})(\frac{45\sqrt{2}}{4}) = \frac{27\sqrt{2}}{2}$ . By the Pythagorean Theorem on $\triangle ADF$ $DF^2 = n^2 - 18^2 = \frac{729}{2} - 324 = \frac{81}{2}$ , and $DF = \frac{9\sqrt{2}}{2}$
Thus, $DC = 2DF + AB = 9\sqrt{2}+\frac{45\sqrt{2}}{4} = \frac{81\sqrt{2}}{4}$ . Therefore, $K = \frac{1}{2}(\frac{45\sqrt{2}}{4}+\frac{81\sqrt{2}}{4}) \cdot 18 = \frac{63\sqrt{2}}{2} \cdot 18 = \frac{567\sqrt{2}}{2}$ . The requested answer is $K \cdot \sqrt{2} = \boxed{567}$ | null | 567 |
d59c33e1b8af976abddb7209dbb49d68 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_10 | Consider the sequence $(a_k)_{k\ge 1}$ of positive rational numbers defined by $a_1 = \frac{2020}{2021}$ and for $k\ge 1$ , if $a_k = \frac{m}{n}$ for relatively prime positive integers $m$ and $n$ , then
\[a_{k+1} = \frac{m + 18}{n+19}.\] Determine the sum of all positive integers $j$ such that the rational number $a_j$ can be written in the form $\frac{t}{t+1}$ for some positive integer $t$ | We know that $a_{1}=\tfrac{t}{t+1}$ when $t=2020$ so $1$ is a possible value of $j$ . Note also that $a_{2}=\tfrac{2038}{2040}=\tfrac{1019}{1020}=\tfrac{t}{t+1}$ for $t=1019$ . Then $a_{2+q}=\tfrac{1019+18q}{1020+19q}$ unless $1019+18q$ and $1020+19q$ are not relatively prime which happens when $q+1$ divides $18q+1019$ (by the Euclidean Algorithm), or $q+1$ divides $1001$ . Thus, the least value of $q$ is $6$ and $j=2+6=8$ . We know $a_{8}=\tfrac{1019+108}{1020+114}=\tfrac{1127}{1134}=\tfrac{161}{162}$ . Now $a_{8+q}=\tfrac{161+18q}{162+19q}$ unless $18q+161$ and $19q+162$ are not relatively prime which happens the first time $q+1$ divides $18q+161$ or $q+1$ divides $143$ or $q=10$ , and $j=8+10=18$ . We have $a_{18}=\tfrac{161+180}{162+190}=\tfrac{341}{352}=\tfrac{31}{32}$ . Now $a_{18+q}=\tfrac{31+18q}{32+19q}$ unless $18q+31$ and $19q+32$ are not relatively prime. This happens the first time $q+1$ divides $18q+31$ implying $q+1$ divides $13$ , which is prime so $q=12$ and $j=18+12=30$ . We have $a_{30}=\tfrac{31+216}{32+228}=\tfrac{247}{260}=\tfrac{19}{20}$ . We have $a_{30+q}=\tfrac{18q+19}{19q+20}$ , which is always reduced by EA, so the sum of all $j$ is $1+2+8+18+30=\boxed{059}$ | null | 059 |
d59c33e1b8af976abddb7209dbb49d68 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_10 | Consider the sequence $(a_k)_{k\ge 1}$ of positive rational numbers defined by $a_1 = \frac{2020}{2021}$ and for $k\ge 1$ , if $a_k = \frac{m}{n}$ for relatively prime positive integers $m$ and $n$ , then
\[a_{k+1} = \frac{m + 18}{n+19}.\] Determine the sum of all positive integers $j$ such that the rational number $a_j$ can be written in the form $\frac{t}{t+1}$ for some positive integer $t$ | Let $a_{j_1}, a_{j_2}, a_{j_3}, \ldots, a_{j_u}$ be all terms in the form $\frac{t}{t+1},$ where $j_1<j_2<j_3<\cdots<j_u,$ and $t$ is some positive integer.
We wish to find $\sum_{i=1}^{u}{j_i}.$ Suppose $a_{j_i}=\frac{m}{m+1}$ for some positive integer $m.$
To find $\boldsymbol{a_{j_{i+1}},}$ we look for the smallest positive integer $\boldsymbol{k'}$ for which \[\boldsymbol{a_{j_{i+1}}=a_{j_i+k'}=\frac{m+18k'}{m+1+19k'}}\] is reducible:
If $\frac{m+18k'}{m+1+19k'}$ is reducible, then there exists a common factor $d>1$ for $m+18k'$ and $m+1+19k'.$ By the Euclidean Algorithm , we have \begin{align*} d\mid m+18k' \text{ and } d\mid m+1+19k' &\implies d\mid m+18k' \text{ and } d\mid k'+1 \\ &\implies d\mid m-18 \text{ and } d\mid k'+1. \end{align*} Since $m-18$ and $k'+1$ are not relatively prime, and $m$ is fixed, the smallest value of $k'$ such that $\frac{m+18k'}{m+1+19k'}$ is reducible occurs when $k'+1$ is the smallest prime factor of $m-18.$
We will prove that for such value of $\boldsymbol{k',}$ the number $\boldsymbol{a_{j_{i+1}}}$ can be written in the form $\boldsymbol{\frac{t}{t+1}:}$ \[a_{j_{i+1}}=a_{j_i+k'}=\frac{m+18k'}{m+1+19k'}=\frac{(m-18)+18(k'+1)}{(m-18)+19(k'+1)}=\frac{\frac{m-18}{k'+1}+18}{\frac{m-18}{k'+1}+19}, \hspace{10mm} (*)\] where $t=\frac{m-18}{k'+1}+18$ must be a positive integer.
We start with $m=2020$ and $a_{j_1}=a_1=\frac{2020}{2021},$ then find $a_{j_2}, a_{j_3}, \ldots, a_{j_u}$ by filling out the table below recursively: \[\begin{array}{c|c|c|c|c|c} & & & & & \\ [-2ex] \boldsymbol{i} & \boldsymbol{m} & \boldsymbol{m-18} & \boldsymbol{k'+1} & \boldsymbol{k'} & \boldsymbol{a_{j_{i+1}} \left(\textbf{by } (*)\right)} \\ [0.5ex] \hline & & & & & \\ [-1.5ex] 1 & 2020 & 2002 & 2 & 1 & \hspace{4.25mm} a_2 = \frac{1019}{1020} \\ [1ex] 2 & 1019 & 1001 & 7 & 6 & \hspace{2.75mm} a_8 = \frac{161}{162} \\ [1ex] 3 & 161 & 143 & 11 & 10 & a_{18} = \frac{31}{32} \\ [1ex] 4 & 31 & 13 & 13 & 12 & a_{30} = \frac{19}{20} \\ [1ex] 5 & 19 & 1 & \text{N/A} & \text{N/A} & \text{N/A} \\ [1ex] \end{array}\] As $\left(j_1,j_2,j_3,j_4,j_5\right)=(1,2,8,18,30),$ the answer is $\sum_{i=1}^{5}{j_i}=\boxed{059}.$ | null | 059 |
2ec6103043f7a195d120eb00176c5158 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11 | Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Note that $\cos(180^\circ-\theta)=-\cos\theta$ holds for all $\theta.$ We apply the Law of Cosines to $\triangle ABE, \triangle BCE, \triangle CDE,$ and $\triangle DAE,$ respectively: \begin{alignat*}{12} &&&AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\angle AEB&&=AB^2&&\quad\implies\quad AE^2+BE^2-2\cdot AE\cdot BE\cdot\cos\theta&&=4^2, \hspace{15mm} &(1\star) \\ &&&BE^2+CE^2-2\cdot BE\cdot CE\cdot\cos\angle BEC&&=BC^2&&\quad\implies\quad BE^2+CE^2+2\cdot BE\cdot CE\cdot\cos\theta&&=5^2, \hspace{15mm} &(2\star) \\ &&&CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\angle CED&&=CD^2&&\quad\implies\quad CE^2+DE^2-2\cdot CE\cdot DE\cdot\cos\theta&&=6^2, \hspace{15mm} &(3\star) \\ &&&DE^2+AE^2-2\cdot DE\cdot AE\cdot\cos\angle DEA&&=DA^2&&\quad\implies\quad DE^2+AE^2+2\cdot DE\cdot AE\cdot\cos\theta&&=7^2. \hspace{15mm} &(4\star) \\ \end{alignat*} We subtract $(1\star)+(3\star)$ from $(2\star)+(4\star):$ \begin{align*} 2\cdot AE\cdot BE\cdot\cos\theta+2\cdot BE\cdot CE\cdot\cos\theta+2\cdot CE\cdot DE\cdot\cos\theta+2\cdot DE\cdot AE\cdot\cos\theta&=22 \\ 2\cdot\cos\theta\cdot(\phantom{ }\underbrace{AE\cdot BE+BE\cdot CE+CE\cdot DE+DE\cdot AE}_{\text{Use the result from }\textbf{Remark}\text{.}}\phantom{ })&=22 \\ 2\cdot\cos\theta\cdot59&=22 \\ \cos\theta&=\frac{11}{59}. \end{align*} Finally, substituting this result into $(\bigstar)$ gives $22\cos\theta=\frac{242}{59},$ from which the answer is $242+59=\boxed{301}.$ | null | 301 |
2ec6103043f7a195d120eb00176c5158 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11 | Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Let the brackets denote areas.
We find $[ABCD]$ in two different ways:
Equating the expressions for $[ABCD],$ we have \[\frac12\cdot\sin\theta\cdot59=2\sqrt{210},\] so $\sin\theta=\frac{4\sqrt{210}}{59}.$ Since $0^\circ<\theta<90^\circ,$ we have $\cos\theta>0.$ It follows that \[\cos\theta=\sqrt{1-\sin^2\theta}=\frac{11}{59}.\] Finally, substituting this result into $(\bigstar)$ gives $22\cos\theta=\frac{242}{59},$ from which the answer is $242+59=\boxed{301}.$ | null | 301 |
2ec6103043f7a195d120eb00176c5158 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11 | Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length $\sqrt{4 \cdot 6 + 5 \cdot 7} = \sqrt{59}.$ [I don't believe this is correct... are the two diagonals of $ABCD$ necessarily congruent? -peace09]* WLOG we focus on diagonal $BD.$ To find the diagonal of the inner quadrilateral, we drop the altitude from $A$ and $C$ and calculate the length of $A_1C_1.$ Let $x$ be $A_1D$ (Thus $A_1B = \sqrt{59} - x.$ By Pythagorean theorem, we have \[49 - x^2 = 16 - \left(\sqrt{59} - x\right)^2 \implies 92 = 2\sqrt{59}x \implies x = \frac{46}{\sqrt{59}} = \frac{46\sqrt{59}}{59}.\] Now let $y$ be $C_1D.$ (thus making $C_1B = \sqrt{59} - y$ ). Similarly, we have \[36 - y^2 = 25 - \left(\sqrt{59} - y\right)^2 \implies 70 = 2\sqrt{59}y \implies y = \frac{35}{\sqrt{59}} = \frac{35\sqrt{59}}{59}.\] We see that $A_1C_1$ , the scaled down diagonal is just $x - y = \frac{11\sqrt{59}}{59},$ which is $\frac{\frac{11\sqrt{59}}{59}}{\sqrt{59}} = \frac{11}{59}$ times our original diagonal $BD,$ implying a scale factor of $\frac{11}{59}.$ Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply $\frac{11}{59} \cdot 22 = \frac{242}{59},$ making our answer $242+59 = \boxed{301}.$ | null | 301 |
2ec6103043f7a195d120eb00176c5158 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11 | Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | Solution
In accordance with Claim 1, the ratios of pairs of one-color segments are the same and equal to $\cos \theta,$ where $\theta$ is the acute angle between the diagonals. \begin{align*} s &= A'B' + B'C' + C'D' + D'A' \\ &= (AB + BC + CD + DA)\cos \theta \\ &= (a + b + c + d)\cos \theta \\ &= 22\cos \theta. \end{align*} In accordance with Claim 2, \begin{align*} 2(ac + bd)\cos \theta = |d^2 – c^2 + b^2 – a^2|.\end{align*} \[2 \cdot 59 \cos \theta = |13 + 9|.\] \[s = 22\cos \theta = \frac{22 \cdot 11}{59} = \frac{242}{59}.\] Therefore, the answer is $242+59=\boxed{301}.$ | null | 301 |
8772ef5daf80fff73216f98e0be66c41 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_12 | Let $A_1A_2A_3\ldots A_{12}$ be a dodecagon ( $12$ -gon). Three frogs initially sit at $A_4,A_8,$ and $A_{12}$ . At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Define the distance between two frogs as the number of sides between them that do not contain the third frog.
Let $E(a,b,c)$ denote the expected number of minutes until the frogs stop jumping, such that the distances between the frogs are $a,b,$ and $c$ (in either clockwise or counterclockwise order). Without the loss of generality, assume that $a\leq b\leq c.$
We wish to find $E(4,4,4).$ Note that:
We have the following system of equations: \begin{align*} E(4,4,4)&=1+\frac{2}{8}E(4,4,4)+\frac{6}{8}E(2,4,6), \\ E(2,4,6)&=1+\frac{4}{8}E(2,4,6)+\frac{1}{8}E(4,4,4)+\frac{1}{8}E(2,2,8), \\ E(2,2,8)&=1+\frac{2}{8}E(2,2,8)+\frac{2}{8}E(2,4,6). \end{align*} Rearranging and simplifying each equation, we get \begin{align*} E(4,4,4)&=\frac{4}{3}+E(2,4,6), &(1) \\ E(2,4,6)&=2+\frac{1}{4}E(4,4,4)+\frac{1}{4}E(2,2,8), &\hspace{12.75mm}(2) \\ E(2,2,8)&=\frac{4}{3}+\frac{1}{3}E(2,4,6). &(3) \end{align*} Substituting $(1)$ and $(3)$ into $(2),$ we obtain \[E(2,4,6)=2+\frac{1}{4}\left[\frac{4}{3}+E(2,4,6)\right]+\frac{1}{4}\left[\frac{4}{3}+\frac{1}{3}E(2,4,6)\right],\] from which $E(2,4,6)=4.$ Substituting this into $(1)$ gives $E(4,4,4)=\frac{16}{3}.$
Therefore, the answer is $16+3=\boxed{019}.$ | null | 019 |
8772ef5daf80fff73216f98e0be66c41 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_12 | Let $A_1A_2A_3\ldots A_{12}$ be a dodecagon ( $12$ -gon). Three frogs initially sit at $A_4,A_8,$ and $A_{12}$ . At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We can solve the problem by removing $1$ frog, and calculate the expected time for the remaining $2$ frogs. In the original problem, when the movement stops, $2$ of the $3$ frogs meet. Because the $3$ frogs cannot meet at one vertex, the probability that those two specific frogs meet is $\tfrac13$ . If the expected time for the two frog problem is $E'$ , then the expected time for the original problem is $\tfrac 13 E'$
The distance between the two frogs can only be $0$ $2$ $4$ $6$ . We use the distances as the states to draw the following Markov Chain . This Markov Chain is much simpler than that of Solution 1 Supplement in the Remark section.
\begin{align*} E(2) &= 1 + \frac12 \cdot E(2) + \frac14 \cdot E(4)\\ E(4) &= 1 + \frac14 \cdot E(2) + \frac12 \cdot E(4) + \frac14 \cdot E(6)\\ E(6) &= 1 + \frac12 \cdot E(4) + \frac12 \cdot E(6) \end{align*}
By solving the above system of equations, $E(4) = 16$ . The answer for the original problem is $\frac{16}{3}$ $16 + 3 = \boxed{019}$ | null | 019 |
750d3f543dbc68a1533a2d54636d80b6 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$ | Let $O_i$ and $r_i$ be the center and radius of $\omega_i$ , and let $O$ and $r$ be the center and radius of $\omega$
Since $\overline{AB}$ extends to an arc with arc $120^\circ$ , the distance from $O$ to $\overline{AB}$ is $r/2$ . Let $X=\overline{AB}\cap \overline{O_1O_2}$ . Consider $\triangle OO_1O_2$ . The line $\overline{AB}$ is perpendicular to $\overline{O_1O_2}$ and passes through $X$ . Let $H$ be the foot from $O$ to $\overline{O_1O_2}$ ; so $HX=r/2$ . We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$ . Let $O_1O_2=d$ [asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D("O",A,dir(A)); D("O_1",B,dir(B)); D("O_2",C,dir(C)); D("H",H,dir(270)); D("X",X,dir(225)); D("A",R1,dir(180)); D("B",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy] Since $X$ is on the radical axis of $\omega_1$ and $\omega_2$ , it has equal power with respect to both circles, so \[O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d}\] since $O_1X+O_2X=d$ . Now we can solve for $O_1X$ and $O_2X$ , and in particular, \begin{align*} O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\ O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}. \end{align*} We want to solve for $d$ . By the Pythagorean Theorem (twice): \begin{align*} &\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\ &\implies \left(d+r-\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_2)^2 = \left(d-r+\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_1)^2 \\ &\implies 2dr - 2(r_1^2-r_2^2)-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\ &\implies 4dr = 8rr_2-8rr_1 \\ &\implies d=2r_2-2r_1 \end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\boxed{672}$ | null | 672 |
750d3f543dbc68a1533a2d54636d80b6 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$ | Let $O_{1}$ $O_{2}$ , and $O$ be the centers of $\omega_{1}$ $\omega_{2}$ , and $\omega$ with $r_{1}$ $r_{2}$ , and $r$ their radii, respectively. Then, the distance from $O$ to the radical axis $\ell\equiv\overline{AB}$ of $\omega_{1}, \omega_{2}$ is equal to $\frac{1}{2}r$ . Let $x=O_{1}O_{2}$ and $O^{\prime}$ the orthogonal projection of $O$ onto line $\ell$ . Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by \[f(X)=\text{Pow}_{\omega_{1}}(X)-\text{Pow}_{\omega_{2}}(X).\] Then \begin{align*} f(O_{1})=-r_{1}^{2}-(x-r_{2})(x+r_{2})&=-x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O_{2})=(x-r_{1})(x+r_{1})-(-r_{2}^{2})&=x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O)=r(r+2r_{1})-r(r+2r_{2})&=2r(r_{1}-r_{2}), \\ f(O^{\prime})&=0. \end{align*} By linearity \[\frac{f(O_{2})-f(O_{1})}{f(O)-f(O^{\prime})}=\frac{O_{1}O_{2}}{OO^{\prime}}=\frac{x}{\tfrac{1}{2}r}=\frac{2x}{r}.\] Notice that $f(O_{2})-f(O_{1})=x^{2}-(-x^{2})=2x^{2}$ and $f(O)-f(O^{\prime})=2r(r_{1}-r_{2})$ , thus \begin{align*}\frac{2x^{2}}{2r(r_{1}-r_{2})}&=\frac{2x}{r}\end{align*} Dividing both sides by $\frac{2x}{r}$ (which is obviously nonzero as $x$ is nonzero) gives us \begin{align*}\frac{x}{2(r_{1}-r_{2})}&=1\end{align*} so $x=2(r_{1}-r_{2})$ . Since $r_{1}=961$ and $r_{2}=625$ , the answer is $x=2\cdot(961-625)=\boxed{672}$ | null | 672 |
750d3f543dbc68a1533a2d54636d80b6 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$ | Denote by $O_1$ $O_2$ , and $O$ the centers of $\omega_1$ $\omega_2$ , and $\omega$ , respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\omega_1$ and $\omega_2$ respectively, $r$ be the radius of $\omega$ , and $\ell$ the distance from $O$ to the line $AB$ . We claim that \[\dfrac{\ell}{r} = \dfrac{R_2-R_1}{d},\] where $d = O_1O_2$ . This solves the problem, for then the $\widehat{PQ} = 120^\circ$ condition implies $\tfrac{\ell}r = \cos 60^\circ = \tfrac{1}{2}$ , and then we can solve to get $d = \boxed{672}$ [asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label("$O_1$",origin,N); label("$O_2$",(d,0),N); label("$O$",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label("$X$",T,N,gray(0.6)); label("$Y$",foot(X[0],O1,O2),NE,gray(0.6)); label("$\ell$",(O+Tp)/2,S,gray(0.6)); [/asy] | null | 672 |
750d3f543dbc68a1533a2d54636d80b6 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$ | Suppose we label the points as shown below [asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot("$A$",A,dir(A-O2/2)); dot("$B$",B,dir(B-O2/2)); dot("$O_2$",O2,right+up); dot("$O_1$",O1,left+up); dot("$O$",O,dir(O-O2)); dot("$D$",D,dir(170)); dot("$E$",E,dir(E-O1)); dot("$X$",X,dir(X-D)); dot("$Y$",Y,dir(Y-D)); label("$R$",O--E,right+up,p); label("$R$",O--D,left+down,p); label("$2R$",(X+O)/2-(150,0),down,p); label("$961$",O1--D,2*(left+down),p); label("$625$",O2--E,2*(right+up),p); MA("",E,D,O1,100,fuchsia+linewidth(1)); MA("",X,D,O,100,fuchsia+linewidth(1)); MA("",Y,E,O2,100,orange+linewidth(1)); MA("",D,E,O,100,orange+linewidth(1)); [/asy] By radical axis, the tangents to $\omega$ at $D$ and $E$ intersect on $AB$ . Thus $PDQE$ is harmonic, so the tangents to $\omega$ at $P$ and $Q$ intersect at $X \in DE$ . Moreover, $OX \parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$ , and $OX = 2OP$ because $\angle POQ = 120^{\circ}$ . Thus \[O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}\] by similar triangles. | null | 672 |
750d3f543dbc68a1533a2d54636d80b6 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$ | Like in other solutions, let $O$ be the center of $\omega$ with $r$ its radius; also, let $O_{1}$ and $O_{2}$ be the centers of $\omega_{1}$ and $\omega_{2}$ with $R_{1}$ and $R_{2}$ their radii, respectively. Let line $OP$ intersect line $O_{1}O_{2}$ at $T$ , and let $u=TO_{2}$ $v=TO_{1}$ $x=PT$ , where the length $O_{1}O_{2}$ splits as $u+v$ . Because the lines $PQ$ and $O_{1}O_{2}$ are perpendicular, lines $OT$ and $O_{1}O_{2}$ meet at a $60^{\circ}$ angle.
Applying the Law of Cosines to $\triangle O_{2}PT$ $\triangle O_{1}PT$ $\triangle O_{2}OT$ , and $\triangle O_{1}OT$ gives \begin{align*}\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\ \triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\ \triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\ \triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\end{align*}
Adding the first and fourth equations, then subtracting the second and third equations gives us \[\left(O_{2}P^{2}-O_{1}P^{2}\right)+\left(R_{1}^{2}-R_{2}^{2}\right)+2r(R_{1}-R_{2})=r(u+v)\]
Since $P$ lies on the radical axis of $\omega_{1}$ and $\omega_{2}$ , the power of point $P$ with respect to either circle is \[O_{2}P^{2}-R_{2}^{2}=O_{1}P^{2}-R_{1}^{2}.\]
Hence $2r(R_{1}-R_{2})=r(u+v)$ which simplifies to \[u+v=2(R_{1}-R_{2}).\]
The requested distance \[O_{1}O_{2}=O_{1}T+O_{2}T=u+v\] is therefore equal to $2\cdot(961-625)=\boxed{672}$ | null | 672 |
750d3f543dbc68a1533a2d54636d80b6 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$ | Let circle $\omega$ tangent circles $\omega_1$ and $\omega_2,$ respectively at distinct points $C$ and $D$ . Let $O, O_1, O_2 (r, r_1, r_2)$ be the centers (the radii) of $\omega, \omega_1$ and $\omega_2,$ respectively. WLOG $r_1 < r_2.$ Let $F$ be the point of $OO_2$ such, that $OO_1 =OF.$ Let $M$ be the midpoint $FO_1, OE \perp AB, CT$ be the radical axes of $\omega_1$ and $\omega, T \in AB.$
Then $T$ is radical center of $\omega, \omega_1$ and $\omega_2, TD = CT.$
In $\triangle OFO_1 (OF = OO_1) OT$ is bisector of $\angle O, OM$ is median
$\hspace{10mm} \implies O,T,$ and $M$ are collinear.
\[\angle OCT = \angle ODT = \angle OET = 90^\circ \implies\]
$OCTDE$ is cyclic (in $\Omega), OT$ is diameter $\Omega.$ $O_1O_2 \perp AB, OM \perp FO_1 \implies \angle FO_1O_2 = \angle OTE$ $\angle OTE = \angle ODE$ as they intercept the same arc $\overset{\Large\frown}{OE}$ in $\Omega.$ \[OE \perp AB, O_1O_2 \perp AB \implies O_1 O_2 || OE \implies\] \[\angle OO_2O_1 = \angle O_2 OE \implies \triangle ODE \sim \triangle O_2 O_1 F \implies\] \[\frac {OE}{OD} = \frac {O_2F}{O_1O_2} \implies \cos \frac {120^\circ}{2} = \frac{r_2 + r - r_1 -r} {O_1O_2}\implies {O_1O_2}= 2|r_2 – r_1|.\]
Since $r_{1}=625$ and $r_{2}=961$ , the answer is $2\cdot|961-625|=\boxed{672}$ | null | 672 |
750d3f543dbc68a1533a2d54636d80b6 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$ | We are not given the radius of circle $w$ , but based on the problem statement, that radius isn't important. We can set $w$ to have radius infinity (solution 8), but if you didn't observe that, you could also set the radius to be $2r$ so that the line containing the center of $w$ , call it $W$ , and $w_2$ , call it $W_2$ , is perpendicular to the line containing the center of $w_1$ , call it $W_1$ and $w_2$ . Let $AB = 2h$ and $W_1W_2 = x$ . Also, let the projections of $W$ and $W_1$ onto line $AB$ be $X$ and $Y$ , respectively.
By Pythagorean Theorem on $\triangle WW_1W_2$ , we get \[x^2+(625+2x)^2=(961+2r)^2 \;(1)\] Note that since $\angle PWQ = 120$ $\angle PWX = 60$ . So, $WX = 2r/2 = r = W_1Y$ . We now get two more equations from Pythag: \[h^2+r^2 = 625^2 \; (2)\] \[h^2+(x-r)^2 = 961^2 \; (3)\] From subtracting $(2)$ and $(3)$ $x^2-2rx=961^2-625^2 \; (4)$ . Rearranging $(3)$ yields $x^2-1344r = 961^2+625^2$ . Plugging in our result from $(4)$ $x^2-1344r= x^2-2rx \implies 1344r = 2rx \implies x=\boxed{672}$ | null | 672 |
750d3f543dbc68a1533a2d54636d80b6 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_13 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$ | Let the circle $\omega$ be infinitely big (a line). Then for it to be split into an arc of $120^{\circ}$ $\overline{PQ}$ must intersect at a $60^{\circ}$ with line $\omega$
Notice the 30-60-90 triangle in the image. $O_1R = 961 - 625$
Thus, the distance between the centers of $\omega_1$ and $\omega_2$ is $2(961-625)=\boxed{672}$ | null | 672 |
891fbd11f453d2b468075929a7f4cfd8 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14 | problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object | We first claim that $n$ must be divisible by $42$ . Since $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$ , we can first consider the special case where $a$ is prime and $a \neq 0,1 \pmod{43}$ . By Dirichlet's Theorem (Refer to the Remark section.), such $a$ always exists.
Then $\sigma(a^n)-1 = \sum_{i=1}^n a^i = a\left(\frac{a^n - 1}{a-1}\right)$ . In order for this expression to be divisible by $2021=43\cdot 47$ , a necessary condition is $a^n - 1 \equiv 0 \pmod{43}$ . By Fermat's Little Theorem $a^{42} \equiv 1 \pmod{43}$ . Moreover, if $a$ is a primitive root modulo $43$ , then $\text{ord}_{43}(a) = 42$ , so $n$ must be divisible by $42$
By similar reasoning, $n$ must be divisible by $46$ , by considering $a \not\equiv 0,1 \pmod{47}$
We next claim that $n$ must be divisible by $43$ . By Dirichlet, let $a$ be a prime that is congruent to $1 \pmod{43}$ . Then $\sigma(a^n) \equiv n+1 \pmod{43}$ , so since $\sigma(a^n)-1$ is divisible by $43$ $n$ is a multiple of $43$
Alternatively, since $\left(\frac{a(a^n - 1^n)}{a-1}\right)$ must be divisible by $43,$ by LTE, we have $v_{43}(a)+v_{43}{(a-1)}+v_{43}{(n)}-v_{43}{(a-1)} \geq 1,$ which simplifies to $v_{43}(n) \geq 1,$ which implies the desired result.
Similarly, $n$ is a multiple of $47$
Lastly, we claim that if $n = \text{lcm}(42, 46, 43, 47)$ , then $\sigma(a^n) - 1$ is divisible by $2021$ for all positive integers $a$ . The claim is trivially true for $a=1$ so suppose $a>1$ . Let $a = p_1^{e_1}\ldots p_k^{e_k}$ be the prime factorization of $a$ . Since $\sigma(n)$ is multiplicative , we have \[\sigma(a^n) - 1 = \prod_{i=1}^k \sigma(p_i^{e_in}) - 1.\] We can show that $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ for all primes $p_i$ and integers $e_i \ge 1$ , so \[\sigma(p_i^{e_in}) = 1 + (p_i + p_i^2 + \ldots + p_i^n) + (p_i^{n+1} + \ldots + p_i^{2n}) + \ldots + (p_i^{n(e_i-1)+1} + \ldots + p_i^{e_in}),\] where each expression in parentheses contains $n$ terms. It is easy to verify that if $p_i = 43$ or $p_i = 47$ then $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ for this choice of $n$ , so suppose $p_i \not\equiv 0 \pmod{43}$ and $p_i \not\equiv 0 \pmod{47}$ . Each expression in parentheses equals $\frac{p_i^n - 1}{p_i - 1}$ multiplied by some power of $p_i$ . If $p_i \not\equiv 1 \pmod{43}$ , then FLT implies $p_i^n - 1 \equiv 0 \pmod{43}$ , and if $p_i \equiv 1 \pmod{43}$ , then $p_i + p_i^2 + \ldots + p_i^n \equiv 1 + 1 + \ldots + 1 \equiv 0 \pmod{43}$ (since $n$ is also a multiple of $43$ , by definition). Similarly, we can show $\sigma(p_i^{e_in}) \equiv 1 \pmod{47}$ , and a simple CRT argument shows $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ . Then $\sigma(a^n) \equiv 1^k \equiv 1 \pmod{2021}$
Then the prime factors of $n$ are $2,3,7,23,43,$ and $47,$ and the answer is $2+3+7+23+43+47 = \boxed{125}$ | null | 125 |
891fbd11f453d2b468075929a7f4cfd8 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14 | problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object | $n$ only needs to satisfy $\sigma(a^n)\equiv 1 \pmod{43}$ and $\sigma(a^n)\equiv 1 \pmod{47}$ for all $a$ . Let's work on the first requirement (mod 43) first. All $n$ works for $a=1$ . If $a>1$ , let $a$ 's prime factorization be $a=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ . The following three statements are the same:
We can show this by casework on $p$
Similar arguments for modulo $47$ lead to $46|n$ and $47|n$ . Therefore, we get $n=\operatorname{lcm}[42,43,46,47]$ . Following the last paragraph of Solution 1 gives the answer $\boxed{125}$ | null | 125 |
891fbd11f453d2b468075929a7f4cfd8 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14 | problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object | We perform casework on $a:$
Finally, the least such positive integer $n$ for all cases is \begin{align*} n&=\operatorname{lcm}(42,43,46,47) \\ &=\operatorname{lcm}(2\cdot3\cdot7,43,2\cdot23,47) \\ &=2\cdot3\cdot7\cdot23\cdot43\cdot47, \end{align*} so the sum of its prime factors is $2+3+7+23+43+47=\boxed{125}.$ | null | 125 |
891fbd11f453d2b468075929a7f4cfd8 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_14 | problem_id
891fbd11f453d2b468075929a7f4cfd8 For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8 Warning: This solution doesn't explain why $43...
Name: Text, dtype: object | Since the problem works for all positive integers $a$ , let's plug in $a=2$ and see what we get. Since $\sigma({2^n}) = 2^{n+1}-1,$ we have $2^{n+1} \equiv 2 \pmod{2021}.$ Simplifying using CRT and Fermat's Little Theorem , we get that $n \equiv 0 \pmod{42}$ and $n \equiv 0 \pmod{46}.$ Then, we can look at $a$ being a $1\pmod{43}$ prime and a $1\pmod{47}$ prime, just like in Solution 1, to find that $43$ and $47$ also divide $n$ . There don't seem to be any other odd "numbers" to check, so we can hopefully assume that the answer is the sum of the prime factors of $\text{lcm(42, 43, 46, 47)}.$ From here, follow solution 1 to get the final answer $\boxed{125}$ | null | 125 |
fff95c70430ec7f9488f6b5567fea357 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_15 | Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$ | Note that $y=x^2-k$ is an upward-opening parabola with the vertex at $(0,-k),$ and $x=2(y-20)^2-k$ is a rightward-opening parabola with the vertex at $(-k,20).$ We consider each condition separately:
Taking the intersection of Conditions 1 and 2 produces $5\leq k\leq280.$ Therefore, the answer is $5+280=\boxed{285}.$ | null | 285 |
fff95c70430ec7f9488f6b5567fea357 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_15 | Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$ | Make the translation $y \rightarrow y+20$ to obtain $20+y=x^2-k$ and $x=2y^2-k$ . Multiply the first equation by $2$ and sum, we see that $2(x^2+y^2)=3k+40+2y+x$ . Completing the square gives us $\left(y- \frac{1}{2}\right)^2+\left(x - \frac{1}{4}\right)^2 = \frac{325+24k}{16}$ ; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that $LHS \leq 21^2=441 \rightarrow 24k \leq 6731$ , so $k \leq 280$
For the lower bound, we need to ensure there are $4$ intersections to begin with. (Here I'm using the un-translated coordinates.) Draw up a graph, and realize that two intersections are guaranteed, on the so called "right branch" of $y=x^2-k$ . As we increase the value of $k$ , two more intersections appear on the "left branch":
$k=4$ does not work because the "leftmost" point of $x=2(y-20)^2-4$ is $(-4,20)$ which lies to the right of $\left(-\sqrt{24}, 20\right)$ , which is on the graph $y=x^2-4$ . While technically speaking this doesn't prove that there are no intersections (why?), drawing the graph should convince you that this is the case. Clearly, $k<4$ does not work.
$k=5$ does work because the two graphs intersect at $(-5,20)$ , and by drawing the graph, you realize this is not a tangent point and there is in fact another intersection nearby, due to slope. Therefore, the answer is $5+280=\boxed{285}$ | null | 285 |
fff95c70430ec7f9488f6b5567fea357 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_15 | Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$ | Claim
Let the axes of two parabolas be perpendicular, their focal parameters be $p_1$ and $p_2$ and the distances from the foci to the point of intersection of the axes be $x_2$ and $y_1$ . Suppose that these parabolas intersect at four points.
Then these points lie on the circle centered at point $(p_2, p_1)$ with radius $r = \sqrt{2(p_1^2 + p_2^2 + p_1 y_1 + p_2 x_2)}.$
Proof
Let's introduce a coordinate system with the center at the point of intersection of the axes. Let the first (red) parabola have axis $x = 0,$ focal parameter $p_1$ and focus at point $A(0, –y_1), y_1 > 0.$ Let second (blue) parabola have axis $y = 0,$ focal parameter $p_2$ and focus at point $B(–x_2,0), x_2 > 0.$ Let us denote the angle between the vector connecting the focus of the first parabola and its point and the positive direction of the ordinate axis $2\theta,$ its length $\rho_1(\theta),$ the angle between the vector connecting the focus of the second parabola and its point and the positive direction of the abscissa axis $2\phi,$ its length $\rho_2(\phi).$ Then \[\rho_1(\theta) = \frac{p_1}{1 - \cos(2\theta)}, \rho_2(\phi) = \frac{p_2}{1 - \cos(2\phi)}.\]
Abscissa of the point of intersection is \begin{align*} x =\rho_1 \sin(2\theta) = p_1\cot\theta = \rho_2 \cos (2\phi) - x_2 = \frac{p_2}{2} (\cot^2\phi - 1)- x_2,\end{align*} \begin{align*} x^2 = p_1^2 \cot ^2 \theta , 2 p_1\cot\theta = p_2 \cos^2 \phi - p_2 - 2x_2 .\end{align*} Ordinate of the point of intersection is \begin{align*} y =\rho_2 \sin 2\phi = p_2\cot\phi = \rho_1 \cos 2\theta - y_1 = \frac{p_1}{2} (\cot^2\theta - 1)- y_1,\end{align*} \begin{align*} y^2 = p_2^2 \cot ^2 \phi , 2 p_2\cot\phi = p_1 \cos^2 \theta - p_1 - 2y_1 .\end{align*} The square of the distance from point of intersection to the point $(p_2, p_1)$ is \begin{align*} r^2 = (x-p_2)^2 + (y-p_1)^2 = x^2 + y^2 - 2 p_1 y - 2 p_2 x + p_1^2 + p_2^2 .\end{align*} After simple transformations, we get $r^2 = 2(p_1^2 + p_2^2 + p_1 y_1 + p_2 x_2).$
Hence, any intersection point has the same distance $r$ from the point $(p_2, p_1).$
Solution
Parameters of the parabola $y = x^2 – k$ are $p_1 = \frac{1}{2}, y_1 = 20 + k – \frac{1}{2}.$
Parameters of the parabola $\frac{x}{2} = (y – 20)^2 – \frac{k}{2}$ are $p_2 = \frac{1}{4}, x_2 = k – \frac{1}{4} \implies r^2 = 20 + \frac{3k}{2}.$
If $r \le 21, k \le \frac{842}{3},$ then integer $k \le 280.$
The vertex of the second parabola is point $(– k,20)$ can be on the parabola $y = x^2 – k$ or below the point of the parabola with the same abscissa. So \[20 \ge (– k)^2 – k \implies 5 \le k \le 280.\] Therefore, the answer is $5+280=\boxed{285}$ | null | 285 |
85385e4568ecc54fa7380589bae6a42c | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1 | Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .) | Recall that the arithmetic mean of all the $n$ digit palindromes is just the average of the largest and smallest $n$ digit palindromes, and in this case the $2$ palindromes are $101$ and $999$ and $\frac{101+999}{2}=\boxed{550},$ which is the final answer. | null | 550 |
85385e4568ecc54fa7380589bae6a42c | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1 | Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .) | For any palindrome $\underline{ABA},$ note that $\underline{ABA}$ is $100A + 10B + A = 101A + 10B.$ The average for $A$ is $5$ since $A$ can be any of $1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$ The average for $B$ is $4.5$ since $B$ is either $0, 1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$ Therefore, the answer is $505 + 45 = \boxed{550}.$ | null | 550 |
85385e4568ecc54fa7380589bae6a42c | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1 | Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .) | For every three-digit palindrome $\underline{ABA}$ with $A\in\{1,2,3,4,5,6,7,8,9\}$ and $B\in\{0,1,2,3,4,5,6,7,8,9\},$ note that $\underline{(10-A)(9-B)(10-A)}$ must be another palindrome by symmetry. Therefore, we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to \begin{align*} \underline{ABA}+\underline{(10-A)(9-B)(10-A)}&=\left[100A+10B+A\right]+\left[100(10-A)+10(9-B)+(10-A)\right] \\ &=\left[100A+10B+A\right]+\left[1000-100A+90-10B+10-A\right] \\ &=1000+90+10 \\ &=1100. \end{align*} For instances: \begin{align*} 171+929&=1100, \\ 262+838&=1100, \\ 303+797&=1100, \\ 414+686&=1100, \\ 545+555&=1100, \end{align*} and so on.
From this symmetry, the arithmetic mean of all the three-digit palindromes is $\frac{1100}{2}=\boxed{550}.$ | null | 550 |
85385e4568ecc54fa7380589bae6a42c | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1 | Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .) | We notice that a three-digit palindrome looks like this: $\underline{aba}.$
And we know $a$ can be any digit from $1$ through $9,$ and $b$ can be any digit from $0$ through $9,$ so there are $9\times{10}=90$ three-digit palindromes.
We want to find the sum of these $90$ palindromes and divide it by $90$ to find the arithmetic mean.
How can we do that? Instead of adding the numbers up, we can break each palindrome into two parts: $101a+10b.$
Thus, all of these $90$ palindromes can be broken into this form.
Thus, the sum of these $90$ palindromes will be $101\times{(1+2+...+9)}\times{10}+10\times{(0+1+2+...+9)}\times{9},$ because each $a$ will be in $10$ different palindromes (since for each $a,$ there are $10$ choices for $b$ ). The same logic explains why we multiply by $9$ when computing the total sum of $b.$
We get a sum of $45\times{1100},$ but don't compute this! Divide this by $90$ and you will get $\boxed{550}.$ | null | 550 |
85385e4568ecc54fa7380589bae6a42c | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1 | Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .) | The possible values of the first and last digits each are $1, 2, ..., 8, 9$ with a sum of $45$ so the average value is $5.$ The middle digit can be any digit from $0$ to $9$ with a sum of $45,$ so the average value is $4.5.$ The average of all three-digit palindromes is $5\cdot 10^2+4.5\cdot 10+5=\boxed{550}.$ | null | 550 |
1a2b1c3fb5eef892c0b7c1af36df4057 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2 | Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$ $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$ . Find $AF$ [asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy] | By angle chasing, we conclude that $\triangle AGF$ is a $30^\circ\text{-}30^\circ\text{-}120^\circ$ triangle, and $\triangle BED$ is a $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangle.
Let $AF=x.$ It follows that $FG=x$ and $EB=FC=840-x.$ By the side-length ratios in $\triangle BED,$ we have $DE=\frac{840-x}{2}$ and $DB=\frac{840-x}{2}\cdot\sqrt3.$
Let the brackets denote areas. We have \[[AFG]=\frac12\cdot AF\cdot FG\cdot\sin{\angle AFG}=\frac12\cdot x\cdot x\cdot\sin{120^\circ}=\frac12\cdot x^2\cdot\frac{\sqrt3}{2}\] and \[[BED]=\frac12\cdot DE\cdot DB=\frac12\cdot\frac{840-x}{2}\cdot\left(\frac{840-x}{2}\cdot\sqrt3\right).\]
We set up and solve an equation for $x:$ \begin{align*} \frac{[AFG]}{[BED]}&=\frac89 \\ \frac{\frac12\cdot x^2\cdot\frac{\sqrt3}{2}}{\frac12\cdot\frac{840-x}{2}\cdot\left(\frac{840-x}{2}\cdot\sqrt3\right)}&=\frac89 \\ \frac{2x^2}{(840-x)^2}&=\frac89 \\ \frac{x^2}{(840-x)^2}&=\frac49. \end{align*} Since $0<x<840,$ it is clear that $\frac{x}{840-x}>0.$ Therefore, we take the positive square root for both sides: \begin{align*} \frac{x}{840-x}&=\frac23 \\ 3x&=1680-2x \\ 5x&=1680 \\ x&=\boxed{336} | null | 336 |
1a2b1c3fb5eef892c0b7c1af36df4057 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2 | Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$ $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$ . Find $AF$ [asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy] | We express the areas of $\triangle BED$ and $\triangle AFG$ in terms of $AF$ in order to solve for $AF.$
We let $x = AF.$ Because $\triangle AFG$ is isosceles and $\triangle AEF$ is equilateral, $AF = FG = EF = AE = x.$
Let the height of $\triangle ABC$ be $h$ and the height of $\triangle AEF$ be $h'.$ Then we have that $h = \frac{\sqrt{3}}{2}(840) = 420\sqrt{3}$ and $h' = \frac{\sqrt{3}}{2}(EF) = \frac{\sqrt{3}}{2}x.$
Now we can find $DB$ and $BE$ in terms of $x.$ $DB = h - h' = 420\sqrt{3} - \frac{\sqrt{3}}{2}x,$ $BE = AB - AE = 840 - x.$ Because we are given that $\angle DBC = 90,$ $\angle DBE = 30.$ This allows us to use the sin formula for triangle area: the area of $\triangle BED$ is $\frac{1}{2}(\sin 30)\left(420\sqrt{3} - \frac{\sqrt{3}}{2}x\right)(840-x).$ Similarly, because $\angle AFG = 120,$ the area of $\triangle AFG$ is $\frac{1}{2}(\sin 120)(x^2).$
Now we can make an equation: \begin{align*} \frac{\triangle AFG}{\triangle BED} &= \frac{8}{9} \\ \frac{\frac{1}{2}(\sin 120)(x^2)}{\frac{1}{2}(\sin 30)\left(420\sqrt{3} - \frac{\sqrt{3}}{2}x\right)(840-x)} &= \frac{8}{9} \\ \frac{x^2}{\left(420 - \frac{x}{2}\right)(840-x)} &= \frac{8}{9}. \end{align*} To make further calculations easier, we scale everything down by $420$ (while keeping the same variable names, so keep that in mind). \begin{align*} \frac{x^2}{\left(1-\frac{x}{2}\right)(2-x)} &= \frac{8}{9} \\ 8\left(1-\frac{x}{2}\right)(2-x) &= 9x^2 \\ 16-16x + 4x^2 &= 9x^2 \\ 5x^2 + 16x -16 &= 0 \\ (5x-4)(x+4) &= 0. \end{align*} Thus $x = \frac{4}{5}.$ Because we scaled down everything by $420,$ the actual value of $AF$ is $\frac{4}{5}(420) = \boxed{336}.$ | null | 336 |
1a2b1c3fb5eef892c0b7c1af36df4057 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2 | Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$ $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$ . Find $AF$ [asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy] | $\angle AFE = \angle AEF = \angle EAF = 60^{0} \Rightarrow \angle AFG = 120^{0}$ So, If $\Delta AFG$ is isosceles, it means that $AF = FG$
Let $AF = FG = AE = EF = x$
So, $[\Delta AFG] = \frac{1}{2} \cdot x^{2} \textup{sin} 120^{0} = \frac{\sqrt{3}}{4}x^{2}$
In $\Delta BED$ $BE = 840 - x$ , Hence $DE = \frac{840 - x}{2}$ (because $\textup{sin} 30^{0} = \frac{1}{2}$
Therefore, $[\Delta BED] = \frac{1}{2} (840 - x) \left (\frac{840-x}{2} \right) \textup{sin} 60^{0}$
So, $[\Delta BED] = \frac{\sqrt{3}}{4} (840 - x) \left (\frac{840-x}{2} \right) = \frac{\sqrt{3}}{8} (840 - x)^{2}$
Now, as we know that the ratio of the areas of $\Delta AFG$ and $\Delta BED$ is $8:9$
Substituting the values, we get
$\frac{\frac{\sqrt{3}}{4}x^{2}}{\frac{\sqrt{3}}{8} (840 - x)^{2}} = \frac{8}{9} \Rightarrow \left (\frac{x}{840 - x} \right)^{2} = \frac{4}{9}$ Hence, $\frac{x}{840 - x} = \frac{2}{3}$ . Solving this, we easily get $x = 336$
We have taken $AF = x$ , Hence, $AF = \boxed{336}$ | null | 336 |
1a2b1c3fb5eef892c0b7c1af36df4057 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2 | Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$ $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$ . Find $AF$ [asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy] | Since $\triangle AFG$ is isosceles, $AF = FG$ , and since $\triangle AEF$ is equilateral, $AF = EF$ . Thus, $EF = FG$ , and since these triangles share an altitude, they must have the same area.
Drop perpendiculars from $E$ and $F$ to line $BC$ ; call the meeting points $P$ and $Q$ , respectively. $\triangle BEP$ is clearly congruent to both $\triangle BED$ and $\triangle FQC$ , and thus each of these new triangles has the same area as $\triangle BED$ . But we can "slide" $\triangle BEP$ over to make it adjacent to $\triangle FQC$ , thus creating an equilateral triangle whose area has a ratio of $18:8$ when compared to $\triangle AEF$ (based on our conclusion from the first paragraph). Since these triangles are both equilateral, they are similar, and since the area ratio $18:8$ reduces to $9:4$ , the ratio of their sides must be $3:2$ . So, because $FC$ and $AF$ represent sides of these triangles, and they add to $840$ $AF$ must equal two-fifths of $840$ , or $\boxed{336}$ | null | 336 |
715443e32214a2f3f1fac0e7781e090d | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_3 | Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\] is divisible by $3$ | Since $3$ is one of the numbers, a product with a $3$ in it is automatically divisible by $3,$ so WLOG $x_3=3,$ we will multiply by $5$ afterward since any of $x_1, x_2, \ldots, x_5$ would be $3,$ after some cancelation we see that now all we need to find is the number of ways that $x_5x_1(x_4+x_2)$ is divisible by $3,$ since $x_5x_1$ is never divisible by $3,$ now we just need to find the number of ways $x_4+x_2$ is divisible by $3.$ Note that $x_2$ and $x_4$ can be $(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (4, 5),$ or $(5, 4).$ We have $2$ ways to designate $x_1$ and $x_5$ for a total of $8 \cdot 2 = 16.$ So the desired answer is $16 \cdot 5=\boxed{080}.$ | null | 080 |
715443e32214a2f3f1fac0e7781e090d | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_3 | Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\] is divisible by $3$ | The expression $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2$ has cyclic symmetry. Without the loss of generality, let $x_1=3.$ It follows that $\{x_2,x_3,x_4,x_5\}=\{1,2,4,5\}.$ We have:
We construct the following table for the case $x_1=3,$ with all values in modulo $3:$ \[\begin{array}{c||c|c|c|c|c||c} & & & & & & \\ [-2.5ex] \textbf{Row} & \boldsymbol{x_2} & \boldsymbol{x_3} & \boldsymbol{x_4} & \boldsymbol{x_5} & \boldsymbol{x_2x_3x_4 + x_3x_4x_5} & \textbf{Valid?} \\ [0.5ex] \hline & & & & & & \\ [-2ex] 1 & 1 & 1 & 2 & 2 & 0 & \checkmark \\ 2 & 1 & 2 & 1 & 2 & 0 & \checkmark \\ 3 & 1 & 2 & 2 & 1 & 2 & \\ 4 & 2 & 1 & 1 & 2 & 1 & \\ 5 & 2 & 1 & 2 & 1 & 0 & \checkmark \\ 6 & 2 & 2 & 1 & 1 & 0 & \checkmark \end{array}\] For Row 1, $(x_2,x_3)$ can be either $(1,4)$ or $(4,1),$ and $(x_4,x_5)$ can be either $(2,5)$ or $(5,2).$ By the Multiplication Principle, Row 1 produces $2\cdot2=4$ permutations. Similarly, Rows 2, 5, and 6 each produce $4$ permutations.
Together, we get $4\cdot4=16$ permutations for the case $x_1=3.$ By the cyclic symmetry, the cases $x_2=3, x_3=3, x_4=3,$ and $x_5=3$ all have the same count. Therefore, the total number of permutations $x_1, x_2, x_3, x_4, x_5$ is $16\cdot5=\boxed{080}.$ | null | 080 |
715443e32214a2f3f1fac0e7781e090d | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_3 | Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\] is divisible by $3$ | WLOG, let $x_{3} = 3$ So, the terms $x_{1}x_{2}x_{3}, x_{2}x_{3}x_{4},x_{3}x_{4}x_{5}$ are divisible by $3$
We are left with $x_{4}x_{5}x_{1}$ and $x_{5}x_{1}x_{2}$ .
We need $x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} \equiv 0 \pmod{3}$ .
The only way is when They are $(+1,-1)$ or $(-1, +1) \pmod{3}$
The numbers left with us are $1,2,4,5$ which are $+1,-1,+1,-1\pmod{3}$ respectively.
$+1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$ $= +1 \cdot +1 \cdot +1$ $\;\;\; OR \;\;\;+1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$ ) = $-1 \cdot -1 \cdot +1$
$-1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$ $= -1 \cdot -1 \cdot -1$ $\;\;\; OR \;\;\;-1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$ ) = $-1 \cdot +1 \cdot +1$
But, as we have just two $+1's$ and two $-1's$ .
Hence, We will have to take $+1 = +1 \cdot -1 \cdot -1$ and $-1 = -1 \cdot +1 \cdot +1$ .
Among these two, we have a $+1$ and $-1$ in common, i.e. $(x_{5}, x_{1}) = (+1, -1) or (-1, +1)$ (because $x_{1}$ and $x_{5}$ .
are common in $x_{4}x_{5}x_{1}$ and $x_{5}x_{1}x_{2}$ ).
So, $(x_{5}, x_{1}) \in {(1,2), (1,5), (4,2), (4,5), (2,1), (5,1), (2,4), (5,4)}$ i.e. $8$ values.
For each value of $(x_{5}, x_{1})$ we get $2$ values for $(x_{2}, x_{4})$ .
Hence, in total, we have $8 \times 2 = 16$ ways.
But any of the $x_{i} 's$ can be $3$ .
So, $16 \times 5 = \boxed{080}$ | null | 080 |
715443e32214a2f3f1fac0e7781e090d | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_3 | Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\] is divisible by $3$ | This is my first time doing a solution (feel free to edit it)
We have \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2.\]
We have $5$ numbers. Considering any $x$ as $3,$ we see that we are left with two terms that are not always divisible by $3,$ which means that already gives us 5 options.
Let's now consider $x_1 =3:$ We are left with \[3x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + 3x_4x_5 + 3x_5x_2.\] The two terms left over are \[x_2x_3x_4 + x_3x_4x_5 \equiv 0 \pmod{3}\] since we already have used $3$ the remaining numbers are $1,2,4,5.$
We now factor \begin{align*} (x_2 + x_5)(x_3x_4) &\equiv 0 \pmod{3} \\ (x_2 + x_5) &\equiv 0 \pmod{3} \end{align*} since $1,2,4,5$ are all not factors of $3.$
Now using the number $1,2,4,5,$ we take two to get a number divisible by $3$ for $(x_2 + x_5):$ \begin{align*} 1+5 &\equiv 0 \pmod{3}, \\ 4+2 &\equiv 0 \pmod{3}, \\ 4+5 &\equiv 0 \pmod{3}, \\ 1+2 &\equiv 0 \pmod{3}. \end{align*} We have $4$ possibilities from above. Since we can also have $5+1$ or $2+4,$ there are $4\cdot2=8$ possibilities in all.
Now using \[(x_2 + x_5)(x_3x_4) \equiv 0 \pmod{3},\] we have $(x_3x_4),$ which results in $8$ more possibilities of $2$ times more. So, we get $2\cdot2\cdot4=16.$
Remember that $3$ can be any of $5$ different variables. So, we multiply by $5$ to get the answer $\boxed{080}.$ | null | 080 |
c08e60a37cae51350d41e44ff0e3a680 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are complex conjugates. Let $z=m+\sqrt{n}\cdot i$ and $\overline{z}=m-\sqrt{n}\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\overline{z}.$
We know that \begin{align*} z+\overline{z}&=2m, & (1) \\ z\overline{z}&=m^2+n. & (2) \end{align*} Applying Vieta's Formulas to $x^3+ax+b,$ we have $-20+z+\overline{z}=0.$ Substituting $(1)$ into this equation, we get $m=10.$
Applying Vieta's Formulas to $x^3+cx^2+d,$ we have $-21z-21\overline{z}+z\overline{z}=0,$ or $-21(z+\overline{z})+z\overline{z}=0.$ Substituting $(1)$ and $(2)$ into this equation, we get $n=320.$
Finally, the answer is $m+n=\boxed{330}.$ | null | 330 |
c08e60a37cae51350d41e44ff0e3a680 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | $(-20)^{3} + (-20)a + b = 0$ , hence $-20a + b = 8000$
Also, $(-21)^{3} + c(-21)^{2} + d = 0$ , hence $441c + d = 9261$
$m + i \sqrt{n}$ satisfies both $\Rightarrow$ we can put it in both equations and equate to 0.
In the first equation, we get $(m + i \sqrt{n})^{3} + a(m + i \sqrt{n}) + b = 0$ Simplifying this further, we get $(m^{3} - 3mn + am + b) + i(3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n}) = 0$
Hence, $m^{3} - 3mn + am + b = 0$ and $3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n} = 0 \Rightarrow 3m^{2} - n + a = 0 \rightarrow (1)$
In the second equation, we get $(m + i \sqrt{n})^{3} + c(m + i \sqrt{n})^{2} + d = 0$ Simplifying this further, we get $(m^{3} + m^{2}c - nc - 3mn + d) + i(3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n}) = 0$
Hence, $m^{3} + m^{2}c - nc - 3mn + d = 0$ and $3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n} = 0 \Rightarrow 3m^{2} - n + 2mc = 0 \rightarrow (2)$
Comparing (1) and (2),
$a = 2mc$ and $am + b = m^{2}c - nc + d \rightarrow (3)$
$b = 8000 + 20a \Rightarrow b = 40mc + 8000$ $d = 9261 - 441c$
Substituting these in $(3)$ gives, $2m^{2}c + 8000 + 40mc = m^{2}c - nc + 9261 - 441c$
This simplifies to $m^{2}c + nc + 40mc + 441c = 1261 \Rightarrow c(m^{2} + n + 40m + 441) = 1261$
Hence, $c|1261 \Rightarrow c \in {1,13,97,1261}$
Consider case of $c = 1$
$c = 1 \Rightarrow d = 8820$ Also, $a = 2m, b = 8000 + 40m$
$am + b = m^{2} - n + 8820$ (because c = 1)
Also, $m^{2} + n + 40m = 820 \rightarrow (4)$ Also, Equation (2) gives $3m^{2} - n + 2m = 0 \rightarrow (5)$
Solving (4) and (5) simultaneously gives $m = 10, n = 320$
[AIME can not have more than one answer, so we can stop here ... Not suitable for Subjective exam]
Hence, $m + n = 10 + 320 = \boxed{330}$ | null | 330 |
c08e60a37cae51350d41e44ff0e3a680 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | start off by applying vieta's and you will find that $a=m^2+n-40m$ $b=20m^2+20n$ $c=21-2m$ and $d=21m^2+21n$ . After that, we have to use the fact that $-20$ and $-21$ are roots of $x^3+ax+b$ and $x^3+cx^2+d$ , respectively. Since we know that if you substitute the root of a function back into the function, the output is zero, therefore $(-20)^3-20(a)+b=0$ and $(-21)^3+c*(-21)^2+d=0$ and you can set these two equations equal to each other while also substituting the values of $a$ $b$ $c$ , and $d$ above to give you $21m^2+21n-1682m+8000=0$ , then you can rearrange the equation into $21n = -21m^2+1682m-8000$ . With this property, we know that $-21m^2+1682m-8000$ is divisible by $21$ therefore that means $1682m-8000=0(mod 21)$ which results in $2m-20=0(mod 21)$ which finally gives us m=10 mod 21. We can test the first obvious value of $m$ which is $10$ and we see that this works as we get $m=10$ and $n=320$ . That means your answer will be $m + n = 10 + 320 = \boxed{330}$ | null | 330 |
c08e60a37cae51350d41e44ff0e3a680 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | We note that $x^3 + ax + b = (x+20)P(x)$ and $x^3 + cx^2 + d = (x+21)Q(x)$ for some polynomials $P(x)$ and $Q(x)$
Through synthetic division (ignoring the remainder as we can set $b$ and $d$ to constant values such that the remainder is zero), $P(x) = x^2 - 20x + (400+a)$ , and $Q(x) = x^2 + (c-21)x + (441 - 21c)$
By the complex conjugate root theorem, we know that $P(x)$ and $Q(x)$ share the same roots, and they share the same leading coefficient, so $P(x) = Q(x)$
Therefore, $c-21 = -20$ and $441-21c = 400 + a$ . Solving the system of equations, we get $a = 20$ and $c = 1$ , so $P(x) = Q(x) = x^2 - 20x + 420$
Finally, by the quadratic formula, we have roots of $\frac{20 \pm \sqrt{400 - 1680}}{2} = 10 \pm \sqrt{320}i$ , so our final answer is $10 + 320 = \boxed{330}$ | null | 330 |
c08e60a37cae51350d41e44ff0e3a680 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | We plug -20 into the equation obtaining $(-20)^3-20a+b$ , likewise, plugging -21 into the second equation gets $(-21)^3+441c+d$
Both equations must have 3 solutions exactly, so the other two solutions must be $m + \sqrt{n} \cdot i$ and $m - \sqrt{n} \cdot i$
By Vieta's, the sum of the roots in the first equation is $0$ , so $m$ must be $10$
Next, using Vieta's theorem on the second equation, you get $x1x2+x2x3+x1x3 = 0$ . However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with -21, so $-21*(20)+(m^2+n)=0$
Given that $m$ is $10$ , then $n$ is equal to $320$
Therefore, the answer to the equation is $\boxed{330}$ | null | 330 |
c08e60a37cae51350d41e44ff0e3a680 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 | There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | Since $m+i\sqrt{n}$ is a common root and all the coefficients are real, $m-i\sqrt{n}$ must be a common root, too.
Now that we know all three roots of both polynomials, we can match coefficients (or more specifically, the zero coefficients).
First, however, the product of the two common roots is: \begin{align*} &&&(x-m-i\sqrt{n})(x-m+i\sqrt{n})\\ &=&&x^2-x(m+i\sqrt{n}+m-i\sqrt{n})+(m+i\sqrt{n})(m-i\sqrt{n})\\ &=&&x^2-2xm+(m^2-i^2n)\\ &=&&x^2-2xm+m^2+n \end{align*}
Now, let's equate the two forms of both the polynomials: \[x^3+ax+b=(x^2-2xm+m^2+n)(x+20)\] \[x^3+cx^2+d=(x^2-2xm+m^2+n)(x+21)\] Now we can match the zero coefficients. \[-2m+20=0\to m=10\text{ and}\] \[-42m+m^2+n=0\to-420+100+n=0\to n=320\text{.}\] Thus, $m+n=10+320=\boxed{330}$ | null | 330 |
2d72e62786ff7a51631df3405ac73aa4 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given $4$ and $10$ as the sides, so we know that the $3$ rd side is between $6$ and $14$ , exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the $3$ rd side. So the triangles' sides are between $6$ and $\sqrt{84}$ exclusive, and the larger bound is between $\sqrt{116}$ and $14$ , exclusive. The area of these triangles are from $0$ (straight line) to $2\sqrt{84}$ on the first "small bound" and the larger bound is between $0$ and $20$ $0 < s < 2\sqrt{84}$ is our first equation, and $0 < s < 20$ is our $2$ nd equation. Therefore, the area is between $\sqrt{336}$ and $\sqrt{400}$ , so our final answer is $\boxed{736}$ | null | 736 |
2d72e62786ff7a51631df3405ac73aa4 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | If $a,b,$ and $c$ are the side-lengths of an obtuse triangle with $a\leq b\leq c,$ then both of the following must be satisfied:
For one such obtuse triangle, let $4,10,$ and $x$ be its side-lengths and $K$ be its area. We apply casework to its longest side:
Case (1): The longest side has length $\boldsymbol{10,}$ so $\boldsymbol{0<x<10.}$
By the Triangle Inequality Theorem, we have $4+x>10,$ from which $x>6.$
By the Pythagorean Inequality Theorem, we have $4^2+x^2<10^2,$ from which $x<\sqrt{84}.$
Taking the intersection produces $6<x<\sqrt{84}$ for this case.
At $x=6,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\sqrt{84},$ the obtuse triangle degenerates into a right triangle with area $K=\frac12\cdot4\cdot\sqrt{84}=2\sqrt{84}.$ Together, we obtain $0<K<2\sqrt{84},$ or $K\in\left(0,2\sqrt{84}\right).$
Case (2): The longest side has length $\boldsymbol{x,}$ so $\boldsymbol{x\geq10.}$
By the Triangle Inequality Theorem, we have $4+10>x,$ from which $x<14.$
By the Pythagorean Inequality Theorem, we have $4^2+10^2<x^2,$ from which $x>\sqrt{116}.$
Taking the intersection produces $\sqrt{116}<x<14$ for this case.
At $x=14,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\sqrt{116},$ the obtuse triangle degenerates into a right triangle with area $K=\frac12\cdot4\cdot10=20.$ Together, we obtain $0<K<20,$ or $K\in\left(0,20\right).$
Answer
It is possible for noncongruent obtuse triangles to have the same area. Therefore, all such positive real numbers $s$ are in exactly one of $\left(0,2\sqrt{84}\right)$ or $\left(0,20\right).$ By the exclusive disjunction, the set of all such $s$ is \[[a,b)=\left(0,2\sqrt{84}\right)\oplus\left(0,20\right)=\left[2\sqrt{84},20\right),\] from which $a^2+b^2=\boxed{736}.$ | null | 736 |
2d72e62786ff7a51631df3405ac73aa4 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | We have the diagram below.
[asy] draw((0,0)--(1,2*sqrt(3))); draw((1,2*sqrt(3))--(10,0)); draw((10,0)--(0,0)); label("$A$",(0,0),SW); label("$B$",(1,2*sqrt(3)),N); label("$C$",(10,0),SE); label("$\theta$",(0,0),NE); label("$\alpha$",(1,2*sqrt(3)),SSE); label("$4$",(0,0)--(1,2*sqrt(3)),WNW); label("$10$",(0,0)--(10,0),S); [/asy]
We proceed by taking cases on the angles that can be obtuse, and finding the ranges for $s$ that they yield .
If angle $\theta$ is obtuse, then we have that $s \in (0,20)$ . This is because $s=20$ is attained at $\theta = 90^{\circ}$ , and the area of the triangle is strictly decreasing as $\theta$ increases beyond $90^{\circ}$ . This can be observed from \[s=\frac{1}{2}(4)(10)\sin\theta\] by noting that $\sin\theta$ is decreasing in $\theta \in (90^{\circ},180^{\circ})$
Then, we note that if $\alpha$ is obtuse, we have $s \in (0,4\sqrt{21})$ . This is because we get $x=\sqrt{10^2-4^2}=\sqrt{84}=2\sqrt{21}$ when $\alpha=90^{\circ}$ , yileding $s=4\sqrt{21}$ . Then, $s$ is decreasing as $\alpha$ increases by the same argument as before.
$\angle{ACB}$ cannot be obtuse since $AC>AB$
Now we have the intervals $s \in (0,20)$ and $s \in (0,4\sqrt{21})$ for the cases where $\theta$ and $\alpha$ are obtuse, respectively. We are looking for the $s$ that are in exactly one of these intervals, and because $4\sqrt{21}<20$ , the desired range is \[s\in [4\sqrt{21},20)\] giving \[a^2+b^2=\boxed{736}\Box\] | null | 736 |
2d72e62786ff7a51631df3405ac73aa4 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | Note: Archimedes15 Solution which I added an answer
here are two cases. Either the $4$ and $10$ are around an obtuse angle or the $4$ and $10$ are around an acute triangle. If they are around the obtuse angle, the area of that triangle is $<20$ as we have $\frac{1}{2} \cdot 40 \cdot \sin{\alpha}$ and $\sin$ is at most $1$ . Note that for the other case, the side lengths around the obtuse angle must be $4$ and $x$ where we have $16+x^2 < 100 \rightarrow x < 2\sqrt{21}$ . Using the same logic as the other case, the area is at most $4\sqrt{21}$ . Square and add $4\sqrt{21}$ and $20$ to get the right answer \[a^2+b^2= \boxed{736}\Box\] | null | 736 |
2d72e62786ff7a51631df3405ac73aa4 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | For $\triangle ABC,$ we fix $AB=10$ and $BC=4.$ Without the loss of generality, we consider $C$ on only one side of $\overline{AB}.$
As shown below, all locations for $C$ at which $\triangle ABC$ is an obtuse triangle are indicated in red, excluding the endpoints. [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C1 = intersectionpoints(D--D+(100,0),Arc(B,Q,P))[1]; C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C2)^^Arc(B,D,P),red); dot("$A$", A, 1.5*S, linewidth(4.5)); dot("$B$", B, 1.5*S, linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^D^^Q, linewidth(0.8), UnFill); Label L1 = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$4$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); label("$\angle C$ obtuse",(midpoint(Arc(B,D,P)).x,2),2.5*W,red); label("$\angle B$ obtuse",(midpoint(Arc(B,Q,C2)).x,2),5*E,red); [/asy] Note that:
For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$ [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C1)^^Arc(B,D,P),red); draw(Arc(B,C1,C2),green); draw((A.x,D.y)--(Q.x,D.y),dashed); dot("$A$", A, 1.5*S, linewidth(4.5)); dot("$B$", B, 1.5*S, linewidth(4.5)); dot("$D$", D, 1.5*dir(75), linewidth(0.8), UnFill); dot("$C_2$", C2, 1.5*N, linewidth(4.5)); dot("$C_1$", C1, 1.5*dir(C1-B), linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^Q, linewidth(0.8), UnFill); dot(C1, green+linewidth(4.5)); Label L1 = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$4$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] Let the brackets denote areas:
Finally, the set of all such $s$ is $[a,b)=\left[2\sqrt{84},20\right),$ from which $a^2+b^2=\boxed{736}.$ | null | 736 |
2d72e62786ff7a51631df3405ac73aa4 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | Let a triangle in $\tau(s)$ be $ABC$ , where $AB = 4$ and $BC = 10$ . We will proceed with two cases:
Case 1: $\angle ABC$ is obtuse. If $\angle ABC$ is obtuse, then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $(0,10)$ ; therefore, the area of the triangle will fall in the range of $(0, 20)$
Case 2: $\angle BAC$ is obtuse. Then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $\left(0, \sqrt{10^{2} - 4^{2}}\right)$ . Therefore, the area of the triangle will fall in the range of $\left(0, 2 \sqrt{84}\right)$
If $s < 2 \sqrt{84}$ , there will exist two types of triangles in $\tau(s)$ - one type with $\angle ABC$ obtuse; the other type with $\angle BAC$ obtuse. If $s \geq 2 \sqrt{84}$ , as we just found, $\angle BAC$ cannot be obtuse, so therefore, there is only one type of triangle - the one in which $\angle ABC$ is obtuse. Also, if $s > 20$ , no triangle exists with lengths $4$ and $10$ . Therefore, $s$ is in the range $\left[ 2 \sqrt{84}, 20\right)$ , so our answer is $\left(2 \sqrt{84}\right)^{2} + 20^{2} = \boxed{736}$ | null | 736 |
2d72e62786ff7a51631df3405ac73aa4 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | Let's rephrase the condition. It is required to find such values of the area of an obtuse triangle with sides $4$ and $10,$ when there is exactly one such obtuse triangle. In the diagram, $AB = 4, AC = 10.$
The largest area of triangle with sides $4$ and $10$ is $20$ for a right triangle with legs $4$ and $10$ $AC\perp AB$ ).
The diagram shows triangles with equal heights. The yellow triangle $ABC'$ has the longest side $BC',$ the blue triangle $ABC$ has the longest side $AC.$ If $BC\perp AB,$ then $BC = \sqrt {AC^2 – AB^2} = 2 \sqrt{21}$ the area is equal to $4\sqrt{21}.$ In the interval, the blue triangle $ABC$ is acute-angled, the yellow triangle $ABC'$ is obtuse-angled. Their heights and areas are equal. The condition is met.
If the area is less than $4\sqrt{21},$ both triangles are obtuse, not equal, so the condition is not met.
Therefore, $s$ is in the range $\left[ 4 \sqrt{21}, 20\right)$ , so answer is $\left(4 \sqrt{21}\right)^{2} + 20^{2} = \boxed{736}$ | null | 736 |
2d72e62786ff7a51631df3405ac73aa4 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 | For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ | If $4$ and $10$ are the shortest sides and $\angle C$ is the included angle, then the area is \[\frac{4\cdot10\cdot\sin\angle C}{2} = 20\sin\angle C.\] Because $0\leq\sin\angle C\leq1$ , the maximum value of $20\sin\angle C$ is $20$ , so $s\leq20$
If $4$ is a shortest side and $10$ is the longest side, the length of the other short side is $4\cos\angle C+2\sqrt{4\cos^2 \angle C+21}$ by law of cosines, and the area is $2\left(4\cos\angle C+2\sqrt{4\cos^2\angle C+21}\right)\sqrt{1-\cos\angle C}$ . Because $-1\le \cos\angle C\le 0$ , this is minimized if $\cos\angle C=0$ , where $s=4\sqrt{21}$
So, the answer is $20^2+\left(4\sqrt{21}\right)^2=\boxed{736}$ | null | 736 |
4e22b21ca71adb0a2ce336423d8cd2f0 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 | For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | By PIE, $|A|+|B|-|A \cap B| = |A \cup B|$ . Substituting into the equation and factoring, we get that $(|A| - |A \cap B|)(|B| - |A \cap B|) = 0$ , so therefore $A \subseteq B$ or $B \subseteq A$ . WLOG $A\subseteq B$ , then for each element there are $3$ possibilities, either it is in both $A$ and $B$ , it is in $B$ but not $A$ , or it is in neither $A$ nor $B$ . This gives us $3^{5}$ possibilities, and we multiply by $2$ since it could have also been the other way around. Now we need to subtract the overlaps where $A=B$ , and this case has $2^{5}=32$ ways that could happen. It is $32$ because each number could be in the subset or it could not be in the subset. So the final answer is $2\cdot 3^5 - 2^5 = \boxed{454}$ | null | 454 |
4e22b21ca71adb0a2ce336423d8cd2f0 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 | For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | We denote $\Omega = \left\{ 1 , 2 , 3 , 4 , 5 \right\}$ .
We denote $X = A \cap B$ $Y = A \backslash \left( A \cap B \right)$ $Z = B \backslash \left( A \cap B \right)$ $W = \Omega \backslash \left( A \cup B \right)$
Therefore, $X \cup Y \cup Z \cup W = \Omega$ and the intersection of any two out of sets $X$ $Y$ $Z$ $W$ is an empty set.
Therefore, $\left( X , Y , Z , W \right)$ is a partition of $\Omega$
Following from our definition of $X$ $Y$ $Z$ , we have $A \cup B = X \cup Y \cup Z$
Therefore, the equation
\[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]
can be equivalently written as
\[\left( | X | + | Y | \right) \left( | X | + | Z | \right) = | X | \left( | X | + | Y | + | Z | \right) .\]
This equality can be simplified as
\[| Y | \cdot | Z | = 0 .\]
Therefore, we have the following three cases: (1) $|Y| = 0$ and $|Z| \neq 0$ , (2) $|Z| = 0$ and $|Y| \neq 0$ , (3) $|Y| = |Z| = 0$ .
Next, we analyze each of these cases, separately.
Case 1: $|Y| = 0$ and $|Z| \neq 0$
In this case, to count the number of solutions, we do the complementary counting.
First, we count the number of solutions that satisfy $|Y| = 0$
Hence, each number in $\Omega$ falls into exactly one out of these three sets: $X$ $Z$ $W$ .
Following from the rule of product, the number of solutions is $3^5$
Second, we count the number of solutions that satisfy $|Y| = 0$ and $|Z| = 0$
Hence, each number in $\Omega$ falls into exactly one out of these two sets: $X$ $W$ .
Following from the rule of product, the number of solutions is $2^5$
Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy $|Y| = 0$ minus the number of solutions that satisfy $|Y| = 0$ and $|Z| = 0$ , i.e., $3^5 - 2^5$
Case 2: $|Z| = 0$ and $|Y| \neq 0$
This case is symmetric to Case 1. Therefore, the number of solutions in this case is the same as the number of solutions in Case 1, i.e., $3^5 - 2^5$
Case 3: $|Y| = 0$ and $|Z| = 0$
Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is $2^5$
By putting all cases together, following from the rule of sum, the total number of solutions is equal to
\begin{align*} \left( 3^5 - 2^5 \right) + \left( 3^5 - 2^5 \right) + 2^5 & = 2 \cdot 3^5 - 2^5 \\ & = \boxed{454} | null | 454 |
4e22b21ca71adb0a2ce336423d8cd2f0 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 | For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | The answer is \begin{align*} \sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] &= 2\sum_{k=0}^{5}\binom{5}{k}2^{5-k}-\sum_{k=0}^{5}\binom{5}{k} \\ &=2(2+1)^5-(1+1)^5 \\ &=2(243)-32 \\ &=\boxed{454} ~MRENTHUSIASM | null | 454 |
4e22b21ca71adb0a2ce336423d8cd2f0 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 | For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | The answer is \begin{align*} &\hspace{5.125mm}\sum_{k=0}^{5}\left[2\binom{5}{k}2^{5-k}-\binom{5}{k}\right] \\ &=\left[2\binom{5}{0}2^{5-0}-\binom{5}{0}\right] + \left[2\binom{5}{1}2^{5-1}-\binom{5}{1}\right] + \left[2\binom{5}{2}2^{5-2}-\binom{5}{2}\right] + \left[2\binom{5}{3}2^{5-3}-\binom{5}{3}\right] + \left[2\binom{5}{4}2^{5-4}-\binom{5}{4}\right] + \left[2\binom{5}{5}2^{5-5}-\binom{5}{5}\right] \\ &=\left[2\left(1\right)2^5-1\right] + \left[2\left(5\right)2^4-5\right] + \left[2\left(10\right)2^3-10\right] + \left[2\left(10\right)2^2-10\right] + \left[2\left(5\right)2^1-5\right] + \left[2\left(1\right)2^0-1\right] \\ &=63+155+150+70+15+1 \\ &=\boxed{454} ~MRENTHUSIASM | null | 454 |
4e22b21ca71adb0a2ce336423d8cd2f0 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 | For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\] | Proceed with Solution 1 to get $(|A| - |A \cap B|)(|B| - |A \cap B|) = 0$ . WLOG, assume $|A| = |A \cap B|$ . Thus, $A \subseteq B$
Since $A \subseteq B$ , if $|B| = n$ , there are $2^n$ possible sets $A$ , and there are also ${5 \choose n}$ ways of choosing such $B$
Therefore, the number of possible pairs of sets $(A, B)$ is
\[\sum_{k=0}^{5} 2^n {5 \choose n}\]
We can compute this manually since it's only from $k=0$ to $5$ , and computing gives us $243$ . We can double this result for $B \subseteq A$ , and we get $2(243) = 486$
However, we have double counted the cases where $A$ and $B$ are the same sets. There are $32$ possible such cases, so we subtract $32$ from $486$ to get $\boxed{454}$ | null | 454 |
bf3ffc78287fe32fa9ccb286343337b3 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7 | Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$ | From the fourth equation we get $d=\frac{30}{abc}.$ Substitute this into the third equation and you get $abc + \frac{30(ab + bc + ca)}{abc} = abc - \frac{120}{abc} = 14$ . Hence $(abc)^2 - 14(abc)-120 = 0$ . Solving, we get $abc = -6$ or $abc = 20$ . From the first and second equation, we get $ab + bc + ca = ab-3c = -4 \Longrightarrow ab = 3c-4$ . If $abc=-6$ , substituting we get $c(3c-4)=-6$ . If you try solving this you see that this does not have real solutions in $c$ , so $abc$ must be $20$ . So $d=\frac{3}{2}$ . Since $c(3c-4)=20$ $c=-2$ or $c=\frac{10}{3}$ . If $c=\frac{10}{3}$ , then the system $a+b=-3$ and $ab = 6$ does not give you real solutions. So $c=-2$ . Since you already know $d=\frac{3}{2}$ and $c=-2$ , so you can solve for $a$ and $b$ pretty easily and see that $a^{2}+b^{2}+c^{2}+d^{2}=\frac{141}{4}$ . So the answer is $\boxed{145}$ | null | 145 |
bf3ffc78287fe32fa9ccb286343337b3 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7 | Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$ | Note that $ab + bc + ca = -4$ can be rewritten as $ab + c(a+b) = -4$ . Hence, $ab = 3c - 4$
Rewriting $abc+bcd+cda+dab = 14$ , we get $ab(c+d) + cd(a+b) = 14$ .
Substitute $ab = 3c - 4$ and solving, we get \[3c^{2} - 4c - 4d - 14 = 0.\] We refer to this as Equation 1.
Note that $abcd = 30$ gives $(3c-4)cd = 30$ . So, $3c^{2}d - 4cd = 30$ , which implies $d(3c^{2} - 4c) = 30$ or \[3c^{2} - 4c = \frac{30}{d}.\] We refer to this as Equation 2.
Substituting Equation 2 into Equation 1 gives, $\frac{30}{d} - 4d - 14 = 0$
Solving this quadratic yields that $d \in \left\{-5, \frac{3}{2}\right\}$
Now we just try these two cases:
For $d = \frac{3}{2}$ substituting in Equation 1 gives a quadratic in $c$ which has roots $c \in \left\{\frac{10}{3}, -2\right\}$
Again trying cases, by letting $c = -2$ , we get $ab = 3c-4$ , Hence $ab = -10$ .
We know that $a + b = -3$ , Solving these we get $a = -5, b = 2$ or $a= 2, b = -5$ (doesn't matter due to symmetry in $a,b$ ).
So, this case yields solutions $(a,b,c,d) = \left(-5, 2 , -2, \frac{3}{2}\right)$
Similarly trying other three cases, we get no more solutions, Hence this is the solution for $(a,b,c,d)$
Finally, $a^{2} + b^{2} + c^{2} + d^{2} = 25 + 4 + 4 + \frac{9}{4} = \frac{141}{4} = \frac{m}{n}$
Therefore, $m + n = 141 + 4 = \boxed{145}$ | null | 145 |
bf3ffc78287fe32fa9ccb286343337b3 | https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7 | Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$ | For simplicity purposes, we number the given equations $(1),(2),(3),$ and $(4),$ in that order.
Rearranging $(2)$ and solving for $c,$ we have \begin{align*} ab+(a+b)c&=-4 \\ ab-3c&=-4 \\ c&=\frac{ab+4}{3}. \hspace{14mm} (5) \end{align*} Substituting $(5)$ into $(4)$ and solving for $d,$ we get \begin{align*} ab\left(\frac{ab+4}{3}\right)d&=30 \\ d&=\frac{90}{ab(ab+4)}. \hspace{5mm} (6) \end{align*} Substituting $(5)$ and $(6)$ into $(3)$ and simplifying, we rewrite the left side of $(3)$ in terms of $a$ and $b$ only: \begin{align*} ab\left[\frac{ab+4}{3}\right] + b\left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right] + \left[\frac{ab+4}{3}\right]\left[\frac{90}{ab(ab+4)}\right]a + \left[\frac{90}{ab(ab+4)}\right]ab &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{30}{a} + \frac{30}{b}}_{\text{Group them.}} + \frac{90}{ab+4} &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \frac{30(a+b)}{ab} + \frac{90}{ab+4} &= 14 \\ ab\left[\frac{ab+4}{3}\right] + \underbrace{\frac{-90}{ab} + \frac{90}{ab+4}}_{\text{Group them.}} &= 14 \\ ab\left[\frac{ab+4}{3}\right] - \frac{360}{ab(ab+4)}&=14. \end{align*} Let $t=ab(ab+4),$ from which \[\frac{t}{3}-\frac{360}{t}=14.\] Multiplying both sides by $3t,$ rearranging, and factoring give $(t+18)(t-60)=0.$ Substituting back and completing the squares produce \begin{align*} \left[ab(ab+4)+18\right]\left[ab(ab+4)-60\right]&=0 \\ \left[(ab)^2+4ab+18\right]\left[(ab)^2+4ab-60\right]&=0 \\ \underbrace{\left[(ab+2)^2+14\right]}_{ab+2=\pm\sqrt{14}i\implies ab\not\in\mathbb R}\underbrace{\left[(ab+2)^2-64\right]}_{ab+2=\pm8}&=0 \\ ab&=6,-10. \end{align*} If $ab=6,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x+6=0.$ Since the discriminant is negative, neither $a$ nor $b$ is a real number.
If $ab=-10,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x-10=0,$ or $(x+5)(x-2)=0,$ from which $\{a,b\}=\{-5,2\}.$ Substituting $ab=-10$ into $(5)$ and $(6),$ we obtain $c=-2$ and $d=\frac32,$ respectively. Together, we have \[a^2+b^2+c^2+d^2=\frac{141}{4},\] so the answer is $141+4=\boxed{145}.$ | null | 145 |
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