id int64 0 59 | problem stringlengths 133 794 | solution stringlengths 319 6.91k | answer stringlengths 2 3 | url stringlengths 77 79 |
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0 | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | Let $R(x)=P(x)+Q(x).$ Since the $x^2$-terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.
Note that
\begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \end{alignat*}
so the slope of $R(x)$ is $\frac{106-108}{20-16}=-\frac12.$
It follows ... | 116 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_1 |
1 | Three spheres with radii $11$, $13$, and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$, $B$, and $C$, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$. Find $AC^2$. | This solution refers to the Diagram section.
We let $\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$. We take a cross-section that contains $A$ and $B$, which contains these two spheres but not the third, as shown below:
Because the plane cu... | 756 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_10 |
2 | Let $ABCD$ be a parallelogram with $\angle BAD < 90^\circ.$ A circle tangent to sides $\overline{DA},$ $\overline{AB},$ and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ,$ as shown. Suppose that $AP=3,$ $PQ=9,$ and $QC=16.$ Then the area of $ABCD$ can be expressed in the form $... | Let's redraw the diagram, but extend some helpful lines.
We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \cdot (... | 150 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_11 |
3 | For any finite set $X$, let $| X |$ denote the number of elements in $X$. Define
\[S_n = \sum | A \cap B | ,\]
where the sum is taken over all ordered pairs $(A, B)$ such that $A$ and $B$ are subsets of $\left\{ 1 , 2 , 3, \cdots , n \right\}$ with $|A| = |B|$.
For example, $S_2 = 4$ because the sum is taken over the ... | Let's try out for small values of $n$ to get a feel for the problem. When $n=1, S_n$ is obviously $1$. The problem states that for $n=2, S_n$ is $4$. Let's try it out for $n=3$.
Let's perform casework on the number of elements in $A, B$.
$\textbf{Case 1:} |A| = |B| = 1$
In this case, the only possible equivalencies w... | 245 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_12 |
4 | Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.\overline{abcd},$ where at least one of the digits $a,$ $b,$ $c,$ or $d$ is nonzero. Let $N$ be the number of distinct numerators obtained when numbers in $S$ are written as fractions in lowest terms. For example, bot... | $0.\overline{abcd}=\frac{abcd}{9999} = \frac{x}{y}$, $9999=9\times 11\times 101$.
Then we need to find the number of positive integers $x$ that (with one of more $y$ such that $y|9999$) can meet the requirement $1 \leq {x}\cdot\frac{9999}{y} \leq 9999$.
Make cases by factors of $x$. (A venn diagram of cases would be ni... | 392 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_13 |
5 | Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive intege... | Steven Chen (www.professorchenedu.com)
We wish to solve the Diophantine equation $a^2+ab+b^2=3^2 \cdot 73^2$. It can be shown that $3|a$ and $3|b$, so we make the substitution $a=3x$ and $b=3y$ to obtain $x^2+xy+y^2=73^2$ as our new equation to solve for.
Notice that $r^2+r+1=(r-\omega)(r-{\omega}^2)$, where $\omega=e... | 459 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_14 |
6 | Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations:
\begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*}
Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ a... | First, let define a triangle with side lengths $\sqrt{2x}$, $\sqrt{2z}$, and $l$, with altitude from $l$'s equal to $\sqrt{xz}$. $l = \sqrt{2x - xz} + \sqrt{2z - xz}$, the left side of one equation in the problem.
Let $\theta$ be angle opposite the side with length $\sqrt{2x}$. Then the altitude has length $\sqrt{2z} ... | 033 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_15 |
7 | Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. | We are given that \[100a + 10b + c = 81b + 9c + a,\] which rearranges to \[99a = 71b + 8c.\]
Taking both sides modulo $71,$ we have
\begin{align*} 28a &\equiv 8c \pmod{71} \\ 7a &\equiv 2c \pmod{71}. \end{align*}
The only solution occurs at $(a,c)=(2,7),$ from which $b=2.$
Therefore, the requested three-digit positive ... | 227 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_2 |
8 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | We have the following diagram:
Let $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$, and $YZ$ be the height of $\triangle AZB$. As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$. We can apply this logic to triangles $BCQ$ and $XCQ$ as well, gi... | 242 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_3 |
9 | Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$ | We rewrite $w$ and $z$ in polar form:
\begin{align*} w &= e^{i\cdot\frac{\pi}{6}}, \\ z &= e^{i\cdot\frac{2\pi}{3}}. \end{align*}
The equation $i \cdot w^r = z^s$ becomes
\begin{align*} e^{i\cdot\frac{\pi}{2}} \cdot \left(e^{i\cdot\frac{\pi}{6}}\right)^r &= \left(e^{i\cdot\frac{2\pi}{3}}\right)^s \\ e^{i\left(\frac{\pi... | 834 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_4 |
10 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per ... | Define $m$ as the number of minutes they swim for.
Let their meeting point be $A$. Melanie is swimming against the current, so she must aim upstream from point $A$, to compensate for this; in particular, since she is swimming for $m$ minutes, the current will push her $14m$ meters downstream in that time, so she must a... | 550 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_5 |
11 | Find the number of ordered pairs of integers $(a, b)$ such that the sequence\[3, 4, 5, a, b, 30, 40, 50\]is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression. | Since $3,4,5,a$ and $3,4,5,b$ cannot be an arithmetic progression, $a$ or $b$ can never be $6$. Since $b, 30, 40, 50$ and $a, 30, 40, 50$ cannot be an arithmetic progression, $a$ and $b$ can never be $20$. Since $a < b$, there are ${24 - 2 \choose 2} = 231$ ways to choose $a$ and $b$ with these two restrictions in mind... | 228 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_6 |
12 | Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{1}{7\cdot8\cdot9}.$
If we minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f = 1.$ Note that $a \cdot b \cdot c \cdot d \... | 289 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_7 |
13 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six po... | We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\omega_A$ is the incircle of $\triangle AB'C'$.
[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); ... | 378 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_8 |
14 | Ellina has twelve blocks, two each of red ($\textbf{R}$), blue ($\textbf{B}$), yellow ($\textbf{Y}$), green ($\textbf{G}$), orange ($\textbf{O}$), and purple ($\textbf{P}$). Call an arrangement of blocks $\textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For example, the... | Consider this position chart: \[\textbf{1 2 3 4 5 6 7 8 9 10 11 12}\]
Since there has to be an even number of spaces between each pair of the same color, spots $1$, $3$, $5$, $7$, $9$, and $11$ contain some permutation of all $6$ colored balls. Likewise, so do the even spots, so the number of even configurations is $6!... | 247 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_I_Problems/Problem_9 |
15 | Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. | Let $x$ be the number of people at the party before the bus arrives. We know that $x\equiv 0\pmod {12}$, as $\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \equiv 0 \pmod{25}$, as $\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x + 5... | 154 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_1 |
16 | Find the remainder when\[\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}\]is divided by $1000$. | We first write the expression as a summation.
\begin{align*} \sum_{i=3}^{40} \binom{\binom{i}{2}}{2} & = \sum_{i=3}^{40} \binom{\frac{i \left( i - 1 \right)}{2}}{2} \\ & = \sum_{i=3}^{40} \frac{\frac{i \left( i - 1 \right)}{2} \left( \frac{i \left( i - 1 \right)}{2}- 1 \right)}{2} \\ & = \frac{1}{8} \sum_{i=3}^{40} i... | 004 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_10 |
17 | Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | [asy] defaultpen(fontsize(12)+0.6); size(300); pair A,B,C,D,M,H; real xb=71, xd=121; A=origin; D=(7,0); B=2*dir(xb); C=3*dir(xd)+D; M=(B+C)/2; H=foot(M,A,D); path c=CR(D,3); pair A1=bisectorpoint(D,A,B), D1=bisectorpoint(C,D,A), Bp=IP(CR(A,2),A--H), Cp=IP(CR(D,3),D--H); draw(B--A--D--C--B); draw(A--M--D^^M--H^^Bp--M... | 180 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_11 |
18 | Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that\[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\]Find the least possible value of $a+b.$ | Denote $P = \left( x , y \right)$.
Because $\frac{x^2}{a^2}+\frac{y^2}{a^2-16} = 1$, $P$ is on an ellipse whose center is $\left( 0 , 0 \right)$ and foci are $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$.
Hence, the sum of distance from $P$ to $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$ is equal to twice... | 023 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_12 |
19 | There is a polynomial $P(x)$ with integer coefficients such that\[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\]holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$. | Because $0 < x < 1$, we have
\begin{align*} P \left( x \right) & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \binom{6}{a} x^{2310a} \left( - 1 \right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\ & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\... | 220 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_13 |
20 | For positive integers $a$, $b$, and $c$ with $a < b < c$, consider collections of postage stamps in denominations $a$, $b$, and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, ... | Notice that we must have $a = 1$; otherwise $1$ cent stamp cannot be represented. At least $b-1$ numbers of $1$ cent stamps are needed to represent the values less than $b$. Using at most $c-1$ stamps of value $1$ and $b$, it can have all the values from $1$ to $c-1$ cents. Plus $\lfloor \frac{999}{c} \rfloor$ stamps o... | 188 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_14 |
21 | 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... | First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$. Let points $A'$ and $B'$ be the reflections of $A$ and $B$, respectively, about the perpendicular bisector of $\overline{O_1O_2}$. Then quadrilaterals $ABO_1O_2$ and $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. Furthermor... | 140 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_15 |
22 | Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probabili... | Let $A$ be Azar, $C$ be Carl, $J$ be Jon, and $S$ be Sergey. The $4$ circles represent the $4$ players, and the arrow is from the winner to the loser with the winning probability as the label.
[2022AIMEIIP2.png](https://artofproblemsolving.com/wiki/index.php/File:2022AIMEIIP2.png)
This problem can be solved by using $2... | 125 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_2 |
23 | A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | [2022 AIME II 3.png](https://artofproblemsolving.com/wiki/index.php/File:2022_AIME_II_3.png)
Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius $\frac{6}{\sqrt{2}} = 3\sqrt{2}$). Since all five vertices are on th... | 021 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_3 |
24 | There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that\[\log_{20x} (22x)=\log_{2x} (202x).\]The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | Define $a$ to be $\log_{20x} (22x) = \log_{2x} (202x)$, what we are looking for. Then, by the definition of the logarithm,
\[\begin{cases} (20x)^{a} &= 22x \\ (2x)^{a} &= 202x. \end{cases}\]
Dividing the first equation by the second equation gives us $10^a = \frac{11}{101}$, so by the definition of logs, $a = \log_{1... | 112 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_4 |
25 | Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points. | [isabelchen](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Isabelchen)
Note: This solution seems incorrect.
Although the answer is correct, solution 2 below is a more accurate way to approach this problem. I agree, I don't get how $a + 2 \leq 20$.
As above, we must deduce that the ... | 072 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_5 |
26 | Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$. Among all such $100$-tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find... | [isabelchen](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Isabelchen)
Define $s_N$ to be the sum of all the negatives, and $s_P$ to be the sum of all the positives.
Since the sum of the absolute values of all the numbers is $1$, $|s_N|+|s_P|=1$.
Since the sum of all the numbers is ... | 841 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_6 |
27 | A circle with radius $6$ is externally tangent to a circle with radius $24$. Find the area of the triangular region bounded by the three common tangent lines of these two circles. | Professor Rat's solution, added by @heheman and edited by @megahertz13 and @Yrock for $\LaTeX$.
First, we want to find $O_2D$. We know that $\angle O_1AD = \angle O_2BD = 90^{\circ}$, so by AA similarity, $\triangle O_1AD \sim \triangle O_2BD$. We want to find the length of $x$, and using the similar triangles, we writ... | 192 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_7 |
28 | Find the number of positive integers $n \le 600$ whose value can be uniquely determined when the values of $\left\lfloor \frac n4\right\rfloor$, $\left\lfloor\frac n5\right\rfloor$, and $\left\lfloor\frac n6\right\rfloor$ are given, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to the real n... | We need to find all numbers between $1$ and $600$ inclusive that are multiples of $4$, $5$, and/or $6$ which are also multiples of $4$, $5$, and/or $6$ when $1$ is added to them.
We begin by noting that the LCM of $4$, $5$, and $6$ is $60$. We can therefore simplify the problem by finding all such numbers described ab... | 080 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_8 |
29 | Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$, distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$, and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$. Additionally, when segments $\overline{A_iB_j}$ are drawn for a... | We can use recursion to solve this problem:
1. Fix 7 points on $\ell_A$, then put one point $B_1$ on $\ell_B$. Now, introduce a function $f(x)$ that indicates the number of regions created, where x is the number of points on $\ell_B$. For example, $f(1) = 6$ because there are 6 regions.
2. Now, put the second point $B... | 244 | https://artofproblemsolving.com/wiki/index.php/2022_AIME_II_Problems/Problem_9 |
30 | Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are $14!$ arrangements without restrictions.
First, there are $\binom{7}{5}$ ways to choose the man-woman diameters. Then, there are $10\cdot8\cdot6\cdot4\cdot2$ ways to place the fiv... | 191 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_1 |
31 | There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$. For that unique $a$, find $a+U$.
(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$.) | minor edit by KevinChen_Yay
We define $U' = \sum^{2023}_{n=1} {\frac{n^2-na}{5}}$. Since for any real number $x$, $\lfloor x \rfloor \le x \le \lfloor x \rfloor + 1$, we have $U \le U' \le U + 2023$. Now, since $-1000 \le U \le 1000$, we have $-1000 \le U' \le 3023$.
Now, we can solve for $U'$ in terms of $a$. We have:... | 944 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_10 |
32 | Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | Define $f(x)$ to be the number of subsets of $\{1, 2, 3, 4, \ldots x\}$ that have $0$ consecutive element pairs, and $f'(x)$ to be the number of subsets that have $1$ consecutive pair.
Using casework on where the consecutive element pair is, there is a unique consecutive element pair that satisfies the conditions. It i... | 235 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_11 |
33 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\a... | By Miquel's theorem, $P=(AEF)\cap(BFD)\cap(CDE)$ (intersection of circles). The law of cosines can be used to compute $DE=42$, $EF=35$, and $FD=13$. Toss the points on the coordinate plane; let $B=(-7, 0)$, $D=(0, 0)$, and $C=(48, 0)$, where we will find $\tan^{2}\left(\measuredangle CDP\right)$ with $P=(BFD)\cap(CDE)$... | 075 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_12 |
34 | Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$.
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$, where $m$ and $n$
are relatively prime positive integers. Find $m + n$. A parallelepiped is a sol... | Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Let one of the vertices be at the origin and the three adjacent vertices be $u$, $v$, and $w$. For one of the parallelepipeds, the three diagonals involving the origin have length $\sqrt {21}$. Hence, $(u+v)\cdot (u+v)=u\cdot u+v\cdot v+2u\cdot v=2... | 125 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_13 |
35 | The following analog clock has two hands that can move independently of each other.
Initially, both hands point to the number $12$. The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock face while the other hand does not move.
Let... | This problem is, in essence, the following: A $12\times12$ coordinate grid is placed on a (flat) torus; how many loops are there that pass through each point while only moving up or right? In other words, Felix the frog starts his journey at $(0,0)$ in the coordinate plane. Every second, he jumps either to the right... | 608 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_14 |
36 | Find the largest prime number $p<1000$ for which there exists a complex number $z$ satisfying
the real and imaginary part of $z$ are both integers;
$|z|=\sqrt{p},$ and
there exists a triangle whose three side lengths are $p,$ the real part of $z^{3},$ and the imaginary part of $z^{3}.$ | Assume that $z=a+bi$. Then,
\[z^3=(a^3-3ab^2)+(3a^2b-b^3)i\]Note that by the Triangle Inequality,
\[|(a^3-3ab^2)-(3a^2b-b^3)|<p\implies |a^3+b^3-3ab^2-3a^2b|<a^2+b^2\]Thus, we know
\[|a+b||a^2+b^2-4ab|<a^2+b^2\]Without loss of generality, assume $a>b$ (as otherwise, consider $i^3\overline z=b+ai$). If $|a/b|\geq 4$, th... | 349 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_15 |
37 | Positive real numbers $b \not= 1$ and $n$ satisfy the equations \[\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).\] The value of $n$ is $\frac{j}{k},$ where $j$ and $k$ are relatively prime positive integers. Find $j+k.$ | Denote $x = \log_b n$.
Hence, the system of equations given in the problem can be rewritten as
\begin{align*} \sqrt{x} & = \frac{1}{2} x , \\ bx & = 1 + x . \end{align*}
Solving the system gives $x = 4$ and $b = \frac{5}{4}$.
Therefore,
\[n = b^x = \frac{625}{256}.\]
Therefore, the answer is $625 + 256 = \boxed{881}$. | 881 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_2 |
38 | A plane contains $40$ lines, no $2$ of which are parallel. Suppose that there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lines intersect. Fin... | In this solution, let $\boldsymbol{n}$-line points be the points where exactly $n$ lines intersect. We wish to find the number of $2$-line points.
There are $\binom{40}{2}=780$ pairs of lines. Among them:
The $3$-line points account for $3\cdot\binom32=9$ pairs of lines.
The $4$-line points account for $4\cdot\binom42... | 607 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_3 |
39 | The sum of all positive integers $m$ such that $\frac{13!}{m}$ is a perfect square can be written as $2^a3^b5^c7^d11^e13^f,$ where $a,b,c,d,e,$ and $f$ are positive integers. Find $a+b+c+d+e+f.$ | We first rewrite $13!$ as a prime factorization, which is $2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.$
For the fraction to be a square, it needs each prime to be an even power. This means $m$ must contain $7\cdot11\cdot13$. Also, $m$ can contain any even power of $2$ up to $2^{10}$, any odd power of $3$ up to $3^{5}$,... | 012 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_4 |
40 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | [Ptolemy's theorem](https://artofproblemsolving.com/wiki/index.php/Ptolemy%27s_theorem) states that for cyclic quadrilateral $WXYZ$, $WX\cdot YZ + XY\cdot WZ = WY\cdot XZ$.
We may assume that $P$ is between $B$ and $C$. Let $PA = a$, $PB = b$, $PC = c$, $PD = d$, and $AB = s$. We have $a^2 + c^2 = AC^2 = 2s^2$, because... | 106 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_5 |
41 | Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. ... | We break the problem into stages, one for each card revealed, then further into cases based on the number of remaining unrevealed cards of each color. Since [expected value](https://artofproblemsolving.com/wiki/index.php/Expected_value) is linear, the expected value of the total number of correct card color guesses acr... | 051 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_6 |
42 | Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$. | $n$ can either be $0$ or $1$ (mod $2$).
Case 1: $n \equiv 0 \pmod{2}$
Then, $n \equiv 2 \pmod{4}$, which implies $n \equiv 1 \pmod{3}$ and $n \equiv 4 \pmod{6}$, and therefore $n \equiv 3 \pmod{5}$. Using [CRT](https://artofproblemsolving.com/wiki/index.php/Chinese_Remainder_Theorem), we obtain $n \equiv 58 \pmod{60}$,... | 049 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_7 |
43 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | This solution refers to the Diagram section.
Let $O$ be the incenter of $ABCD$ for which $\odot O$ is tangent to $\overline{DA},\overline{AB},$ and $\overline{BC}$ at $X,Y,$ and $Z,$ respectively. Moreover, suppose that $R,S,$ and $T$ are the feet of the perpendiculars from $P$ to $\overleftrightarrow{DA},\overleftrigh... | 125 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_8 |
44 | Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$ | Plugging $2$ and $m$ into $P(x)$ and equating them, we get $8+4a+2b+c = m^3+am^2+bm+c$. Rearranging, we have \[(m^3-8) + (m^2 - 4)a + (m-2)b = 0.\] Note that the value of $c$ won't matter as it can be anything in the provided range, giving a total of $41$ possible choices for $c.$ So what we just need to do is to just ... | 738 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_I_Problems/Problem_9 |
45 | The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples ... | In the arithmetic sequence, let $a$ be the first term and $d$ be the common difference, where $d>0.$ The sum of the first six terms is \[a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d) = 6a+15d.\]
We are given that
\begin{align*} 6a+15d &= 990, \\ 2a &= a+5d. \end{align*}
The second equation implies that $a=5d.$ Substituting this ... | 220 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_1 |
46 | Let $N$ be the number of ways to place the integers $1$ through $12$ in the $12$ cells of a $2 \times 6$ grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by $3.$ One way to do this is shown below. Find the number of positive integer divisors of $N.$
\[\be... | We replace the numbers which are 0 mod(3) to 0, 1 mod(3) to 1, and 2 mod(3) to 2.
Then, the problem is equivalent to arranging 4 0's,4 1's, and 4 2's into the grid (and then multiplying by $4!^3$ to account for replacing the remainder numbers with actual numbers) such that no 2 of the same numbers are adjacent.
Then, ... | 144 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_10 |
47 | Find the number of collections of $16$ distinct subsets of $\{1,2,3,4,5\}$ with the property that for any two subsets $X$ and $Y$ in the collection, $X \cap Y \not= \emptyset.$ | Denote by $\mathcal C$ a collection of 16 distinct subsets of $\left\{ 1, 2, 3, 4, 5 \right\}$.
Denote $N = \min \left\{ |S|: S \in \mathcal C \right\}$.
Case 1: $N = 0$.
This entails $\emptyset \in \mathcal C$.
Hence, for any other set $A \in \mathcal C$, we have $\emptyset \cap A = \emptyset$. This is infeasible.
Cas... | 081 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_11 |
48 | In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$... | Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Define $L_1$ to be the foot of the altitude from $A$ to $BC$. Furthermore, define $L_2$ to be the foot of the altitude from $Q$ to $BC$. From here, one can find $AL_1=12$, either using the 13-14-15 triangle or by calculating the area of $ABC$ two w... | 247 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_12 |
49 | Let $A$ be an acute angle such that $\tan A = 2 \cos A.$ Find the number of positive integers $n$ less than or equal to $1000$ such that $\sec^n A + \tan^n A$ is a positive integer whose units digit is $9.$ | Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
\[\tan A = 2 \cos A \implies \sin A = 2 \cos^2 A \implies \sin^2 A + \cos^2 A = 4 \cos^4 A + \cos^2 A = 1\]
\[\implies \cos^2 A = \frac {\sqrt {17} - 1}{8}.\]
\[c_n = \sec^n A + \tan^n A = \frac {1}{\cos^n A} + 2^n \cos^n A = (4\cos^2 A +1)^{\f... | 167 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_13 |
50 | A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangl... | Let's first view the cube from a direction perpendicular to $ABDC$, as illustrated above. Let $x$ be the cube's side length. Since $\triangle CHA \sim \triangle AGB$, we have
\[\frac{CA}{CH} = \frac{AB}{AG}.\]
We know $AB = x$, $AG = \sqrt{x^2-2^2}$, $AC = \sqrt{2}x$, $CH = 8$. Plug them into the above equation, we get... | 751 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_14 |
51 | For each positive integer $n$ let $a_n$ be the least positive integer multiple of $23$ such that $a_n \equiv 1 \pmod{2^n}.$ Find the number of positive integers $n$ less than or equal to $1000$ that satisfy $a_n = a_{n+1}.$ | [Bxiao31415](https://artofproblemsolving.com/wiki/index.php/User:Bxiao31415)
(small changes by bobjoebilly and [IraeVid13](https://artofproblemsolving.com/wiki/index.php/User:Iraevid13))
We first check that $\gcd(23, 2^n) = 1$ hence we are always seeking a unique modular inverse of $23$, $b_n$, such that $a_n \equiv 23... | 363 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_15 |
52 | Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than $1000$ that is a palindrome both when written in base ten and when written in base eight, such as $292 = 444_{\text{eight}}.$ | Assuming that such palindrome is greater than $777_8 = 511,$ we conclude that the palindrome has four digits when written in base $8.$ Let such palindrome be \[(\underline{ABBA})_8 = 512A + 64B + 8B + A = 513A + 72B.\]
It is clear that $A=1,$ so we repeatedly add $72$ to $513$ until we get palindromes less than $1000:$... | 585 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_2 |
53 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | This solution refers to the Diagram section.
Let $\angle PAB = \angle PBC = \angle PCA = \theta,$ from which $\angle PAC = 90^\circ-\theta,$ and $\angle APC = 90^\circ.$
Moreover, we have $\angle PBA = \angle PCB = 45^\circ-\theta,$ as shown below:
Note that $\triangle PAB \sim \triangle PBC$ by AA Similarity. The rat... | 250 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_3 |
54 | Let $x,y,$ and $z$ be real numbers satisfying the system of equations
\begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*}
Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | We first subtract the second equation from the first, noting that they both equal $60$.
\begin{align*} xy+4z-yz-4x&=0 \\ 4(z-x)-y(z-x)&=0 \\ (z-x)(4-y)&=0 \end{align*}
Case 1: Let $y=4$.
The first and third equations simplify to:
\begin{align*} x+z&=15 \\ xz&=44 \end{align*}
from which it is apparent that $x=4$ and $x... | 273 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_4 |
55 | Let $S$ be the set of all positive rational numbers $r$ such that when the two numbers $r$ and $55r$ are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of $S$ can be... | Denote $r = \frac{a}{b}$, where $\left( a, b \right) = 1$.
We have $55 r = \frac{55a}{b}$.
Suppose $\left( 55, b \right) = 1$, then the sum of the numerator and the denominator of $55r$ is $55a + b$.
This cannot be equal to the sum of the numerator and the denominator of $r$, $a + b$.
Therefore, $\left( 55, b \right) \... | 719 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_5 |
56 | Consider the L-shaped region formed by three unit squares joined at their sides, as shown below. Two points $A$ and $B$ are chosen independently and uniformly at random from inside the region. The probability that the midpoint of $\overline{AB}$ also lies inside this L-shaped region can be expressed as $\frac{m}{n},$ w... | We proceed by calculating the complement.
Note that the only configuration of the 2 points that makes the midpoint outside of the L shape is one point in the top square, and one in the right square. This occurs with $\frac{2}{3} \cdot \frac{1}{3}$ probability.
Let the topmost coordinate have value of: $(x_1,y_1+1)$, ... | 35 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_6 |
57 | Each vertex of a regular dodecagon ($12$-gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle. | Note that the condition is equivalent to stating that there are no 2 pairs of oppositely spaced vertices with the same color.
Case 1: There are no pairs. This yields $2$ options for each vertices 1-6, and the remaining vertices 7-12 are set, yielding $2^6=64$ cases.
Case 2: There is one pair. Again start with 2 options... | 928 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_7 |
58 | Let $\omega = \cos\frac{2\pi}{7} + i \cdot \sin\frac{2\pi}{7},$ where $i = \sqrt{-1}.$ Find the value of the product\[\prod_{k=0}^6 \left(\omega^{3k} + \omega^k + 1\right).\] | For any $k\in Z$, we have,
\begin{align*} & \left( \omega^{3k} + \omega^k + 1 \right) \left( \omega^{3\left( 7 - k \right)} + \omega^{\left( 7 - k \right)} + 1 \right) \\ & = \omega^{3 \cdot 7} + \omega^{2k + 7} + \omega^{3k} + \omega^{-2k + 3 \cdot 7} + \omega^7 + \omega^k + \omega^{3\left( 7 - k \right)} + \omega^{\l... | 024 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_8 |
59 | Circles $\omega_1$ and $\omega_2$ intersect at two points $P$ and $Q,$ and their common tangent line closer to $P$ intersects $\omega_1$ and $\omega_2$ at points $A$ and $B,$ respectively. The line parallel to $AB$ that passes through $P$ intersects $\omega_1$ and $\omega_2$ for the second time at points $X$ and $Y,$ r... | Mathkiddie
Notice that line $\overline{PQ}$ is the [radical axis](https://artofproblemsolving.com/wiki/index.php/Radical_axis) of circles $\omega_1$ and $\omega_2$. By the radical axis theorem, we know that the tangents of any point on line $\overline{PQ}$ to circles $\omega_1$ and $\omega_2$ are equal. Therefore, line... | 033 | https://artofproblemsolving.com/wiki/index.php/2023_AIME_II_Problems/Problem_9 |
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