problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
For integers $0\le a\le n$ , let $f(n,a)$ denote the number of coefficients in the expansion of $(x+1)^a(x+2)^{n-a}$ that is divisible by $3.$ For example, $(x+1)^3(x+2)^1=x^4+5x^3+9x^2+7x+2$ , so $f(4,3)=1$ . For each positive integer $n$ , let $F(n)$ be the minimum of $f(n,0),f(n,1),\ldots ,f(n,n)$ .
(1... | Let $g(n,a)$ be the number of coefficients in the expansion of $(x+1)^a (x-1)^{n-a}$ that are NOT divisible by 3, and let $G(n)=\max\limits_{a=0}^n g(n,a)$ . We know $g(n,a)=(n+1)-f(n,a)$ Part a) I will show there exists $a$ such that $f(n,a)=\frac{n-1}{3}$ $n=4,7$ both work.
Rephrasing the problem, $G(n)=... | [
"After trying serveral cases, I believe it should be (n-1)/3 for both parts, not (n+1)/3.",
"Sketch:\nPart a) If n satisfies, then 3n+4 also satisfies.\nPart b) Induction. We have f(3n+1,3m+2)=3f(n-1,m)+2, f(3n+2,3m+3)=3f(n,m+1) and f(3m+3,3n+1)=f(n+1,m+1)+2f(n,m), then we will prove a stronger statement: There... | [
"origin:aops",
"2022 Contests",
"2022 China National Olympiad"
] | {
"answer_score": 106,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 China National Olympiad/2742835.json"
} |
Let $ABC$ be an acute triangle with $\angle ACB>2 \angle ABC$ . Let $I$ be the incenter of $ABC$ , $K$ is the reflection of $I$ in line $BC$ . Let line $BA$ and $KC$ intersect at $D$ . The line through $B$ parallel to $CI$ intersects the minor arc $BC$ on the circumcircle of $ABC$ at $E(E \neq ... | Let $CI$ meet circumcircle at $T$ .
Claim $: FTIB$ is parallelogram.
Proof $:$ Note that $TI || BF$ and $BF = CE = BT = TI$ (it's well-known that $TB = TI = TA$ ).
Claim $: TAF$ and $BFK$ are congruent.
Proof $:$ Note that $TF = BI = BK$ and $TA = TI = FB$ and $\angle FBK = \angle FBI + \angle IB... | [
"<blockquote>It is confusing why they placed two easy geometry in a single test (P1 and P4). Maybe an inequality would have been better. Once you take out the midpoint of the arc, this is done easily.</blockquote>\n\nD'KBF is cyclic, not D'CBF",
"This problem is not that easy if you attend the test. All four stud... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 62,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2808537.json"
} |
Let $C=\{ z \in \mathbb{C} : |z|=1 \}$ be the unit circle on the complex plane. Let $z_1, z_2, \ldots, z_{240} \in C$ (not necessarily different) be $240$ complex numbers, satisfying the following two conditions:
(1) For any open arc $\Gamma$ of length $\pi$ on $C$ , there are at most $200$ of $j ~(1 \le ... | Here is my <details><summary>solution</summary>Let the $240$ plurals be $e^{\theta_1 i},e^{\theta_2 i}, \cdots ,e^{\theta_{240}i},0 \leqslant \theta_1 \leqslant \theta_2 \leqslant \cdots \leqslant \theta_{240} \leqslant 2 \pi ,z_k=e^{\theta_k i}$ . And let $\omega_k=z_k+ $ $z_{k+40}+z_{k+80}+z_{k+120}+z_{k+160}+z_... | [
"Just notice that $z_1+z_2+...+z_{240}=\\int_0^{2\\pi}F(\\theta)e^{i\\theta}d\\theta$ ,where F( $\\theta$ ) indicates the number of $z_i$ s in the arc $(\\theta-\\frac{\\pi}{6},\\theta+\\frac{\\pi}{6})$ This intergal equals $\\frac{1}{6} \\int_0^{2\\pi}(F(\\theta)e^{i\\theta}+F(\\theta+\\frac{\\pi}{3})e^{i(\\th... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 46,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2808538.json"
} |
Let $m$ be a positive integer, and $A_1, A_2, \ldots, A_m$ (not necessarily different) be $m$ subsets of a finite set $A$ . It is known that for any nonempty subset $I$ of $\{1, 2 \ldots, m \}$ ,
\[ \Big| \bigcup_{i \in I} A_i \Big| \ge |I|+1. \]
Show that the elements of $A$ can be colored black and white,... | <details><summary>Solution</summary>Consider a bipartite graph $G(S,T,E)$ with $S=\{A_1,\cdots,A_m\}$ representing the sets, $T=\bigcup_{j=1}^m A_j$ representing the elements and $E$ representing the edges. Connect an edge between $A_j$ and $b\in T$ if and only if $b\in A_j$ . It is equivalent to prove tha... | [
"<details><summary>I think this works?</summary>Consider a bipartite graph $(S,T)$ with $S=\\{A_1,\\cdots,A_m\\}$ representing the sets and $T=\\cup_{j=1}^m A_j$ representing the elements. Connect an edge between $A_j$ and $b_i$ if $b_i\\in A_j$ . WLOG, the graph is connected, since we can solve for each... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 306,
"boxed": false,
"end_of_proof": true,
"n_reply": 17,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2808539.json"
} |
In a cyclic convex hexagon $ABCDEF$ , $AB$ and $DC$ intersect at $G$ , $AF$ and $DE$ intersect at $H$ . Let $M, N$ be the circumcenters of $BCG$ and $EFH$ , respectively. Prove that the $BE$ , $CF$ and $MN$ are concurrent. | Let $T = BE \cap CF$ . Consider the inversion $\Phi$ at $T$ fixing the circle. Angle chase shows $\angle BH^*C = 180^\circ - \angle BGC$ (where $H^* = \Phi(H)$ ), i.e. $H \in \odot(BCG)$ . So $\Phi$ just swaps $\odot(EFH)$ and $\odot(BCG)$ , implying points $N,T,M$ are collinear. $\blacksquare$ [asy]
s... | [
"<details><summary>Solution</summary>Let $S$ be the point of intersection of $BE$ and $CF$ . We need to prove that $M,N,S$ are collinear.\nBy Pascal on $AFCDEF$ , we have that $G,H,S$ are collinear.\nNow an easy angle chase yields\n\\[\\angle AHN=90^\\circ-\\angle HEF=90^\\circ-\\angle DAF\\]\nso that $M... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 212,
"boxed": false,
"end_of_proof": false,
"n_reply": 16,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2808540.json"
} |
Let $p$ be a prime, $A$ is an infinite set of integers. Prove that there is a subset $B$ of $A$ with $2p-2$ elements, such that the arithmetic mean of any pairwise distinct $p$ elements in $B$ does not belong to $A$ . | Nice problem!
We will denote by $\mathbb{N}$ the set of positive integers. In which follows, we will assume for simplicity that $A \cap \mathbb{N}$ is infinite, since the same argues would work anyways. It is sufficient to find a subset $B \subseteq { (A\cap\mathbb{N}) }$ and we will replace $A$ by $A\cap \ma... | [
"Here\nhttps://artofproblemsolving.com/community/u800085h2807854p24762139",
"After given a hint from my classmate, I got a solution as follows.\n<details><summary>Click to expand</summary>Let $S_{i}( \\alpha )= \\{x \\in A : x \\equiv i( \\bmod p^\\alpha)\\}$ .If there exist $i,j$ and $ \\alpha $ such that ... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 294,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2808541.json"
} |
Let $a, b, c, p, q, r$ be positive integers with $p, q, r \ge 2$ . Denote
\[Q=\{(x, y, z)\in \mathbb{Z}^3 : 0 \le x \le a, 0 \le y \le b , 0 \le z \le c \}. \]
Initially, some pieces are put on the each point in $Q$ , with a total of $M$ pieces. Then, one can perform the following three types of operations repeat... | Note: This seemingly easy problem is actually very hard. This is only a partial solution. I have yet to work out other details.
<details><summary>Solution if at least one of pqr is large...</summary>The answer is $p^aq^br^c$ . Let $f(x,y,z)$ be the number of pieces on the lattice point $(x,y,z)$ . For sets $X,Y,... | [
"Very nice problem!\nThere is my solution,i hope that there are no mistakes in it.\n<details><summary>my solution</summary>Answer is $p^{a}q^{b}r^{c}$ ,to the lowerbound place $p^{a}q^{b}r^{c}-1$ pieces in (a;b;c).\nNow we will proof that if we have $p^{a}q^{b}r^{c}$ pieces we can make a piece placed in (0;0;0... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 136,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2808544.json"
} |
Find all pairs of positive integers $(m, n)$ , such that in a $m \times n$ table (with $m+1$ horizontal lines and $n+1$ vertical lines), a diagonal can be drawn in some unit squares (some unit squares may have no diagonals drawn, but two diagonals cannot be both drawn in a unit square), so that the obtained grap... | The answer is $m=n$ only. Making the degree of all vertices even is the equivalent and sufficient condition to create an eulerian path.
Construction: let S be the set of points on an edge but not a corner. Connect each pair with a line of slope 1.
Proof of optimality: we prove a critical claim: let S be the set of ... | [
"<details><summary>My sol (BEWARE: AWFUL WORDING AND NO LATEX)</summary>CLAIM: m = n are the only integers such it is possible for the table to have an Eulerian cycle\nIt is well known that it is equivalent for every point to have an even degree for it to have an Eulerian cycle. \nAt m = n, we have a construction (... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 30,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2811451.json"
} |
Given a non-right triangle $ABC$ with $BC>AC>AB$ . Two points $P_1 \neq P_2$ on the plane satisfy that, for $i=1,2$ , if $AP_i, BP_i$ and $CP_i$ intersect the circumcircle of the triangle $ABC$ at $D_i, E_i$ , and $F_i$ , respectively, then $D_iE_i \perp D_iF_i$ and $D_iE_i = D_iF_i \neq 0$ . Let the ... | Great problem! My solution uses angle chasing to find interesting carectization of the points, and then the solution is quite natural!
*<span style="color:#00f">Key Lemma:</span> Let $K$ be intersection of the tangents from $B$ and $C$ to $(ABC)$ and then let $\omega$ be the circle with center $K$ passing... | [
"This is not hard. Notice that O,P1and P2 are collinear and OP1*OP2=R^2. Then it is easy by orthopole.",
"<blockquote>Interesting but easy problem.\nWe will use the following steps to finish the problem.\n<details><summary>Lemma 1</summary>**Lemma 1.** Let $P$ be a point lie on $(ABC)$ , $l$ be the Simson li... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 268,
"boxed": false,
"end_of_proof": true,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2811452.json"
} |
Let $a_1, a_2, \ldots, a_n$ be $n$ positive integers that are not divisible by each other, i.e. for any $i \neq j$ , $a_i$ is not divisible by $a_j$ . Show that
\[ a_1+a_2+\cdots+a_n \ge 1.1n^2-2n. \]
*Note:* A proof of the inequality when $n$ is sufficient large will be awarded points depending on your resu... | Partial solution with Desmos and Justanaccount that proves if the statement is true for all $n<C$ then it must be true for all $n$ by strong induction on $n$ . Because my solution seems a bit sketchy, if there are any mistakes in my solution, please point it out. Thank you!
<details><summary>Solution</summary>Let... | [
"This isn't a complete solution. Its a very stupid bounding idea. It proves the bound $n^2-2n$ . Idk if you can optimise it in the current form.\n\nOrder the numbers in increasing order. Fix $k \\geqslant 2$ . If $a_k \\leqslant 2(k-2)$ then by the Piegon-Hole Principle some two among $a_1, a_2, \\cdots, a_{k-... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 72,
"boxed": false,
"end_of_proof": false,
"n_reply": 16,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2811457.json"
} |
Given a positive integer $n$ , find all $n$ -tuples of real number $(x_1,x_2,\ldots,x_n)$ such that
\[ f(x_1,x_2,\cdots,x_n)=\sum_{k_1=0}^{2} \sum_{k_2=0}^{2} \cdots \sum_{k_n=0}^{2} \big| k_1x_1+k_2x_2+\cdots+k_nx_n-1 \big| \]
attains its minimum. | Good problem.
We claim that the minimum happens only when $x_k=\frac 1{n+1}$ for all $1\leq k\leq n$ . For the rest of the proof, assume $(x_1,x_2,\ldots,x_n)$ is a triple achieving the minimum.
We get rid of the case $n=1$ , where $f(x)=1+|x-1|+|2x-1|$ . Writing it out as a piece wise function, we easily see ... | [
"<details><summary>Sketch</summary>The answer is $x_1=\\cdots=x_n=\\frac{1}{n+1}$ only. \n\nClaim: minimum holds when and only when $x_i$ 's are equal.\n\nProof: Let $y=\\frac{x_1+x_2}{2}$ . I will prove that $f(x_1,x_2,x_3,\\cdots,x_n)\\ge f(y,y,x_3,\\cdots,x_n)$ Note for any $k,m,c\\in \\mathbb{R}$ , $|kx_... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 212,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2812216.json"
} |
Given a positive integer $n$ , let $D$ is the set of positive divisors of $n$ , and let $f: D \to \mathbb{Z}$ be a function. Prove that the following are equivalent:
(a) For any positive divisor $m$ of $n$ ,
\[ n ~\Big|~ \sum_{d|m} f(d) \binom{n/d}{m/d}. \]
(b) For any positive divisor $k$ of $n$ ,
\[ k ~\... | Nice problem.
We first claim the following:**[color=#ff0000]Claim.** For any positive integers $m$ and $k$ , we have
\[m\left|~\sum_{d\mid m}\mu\left(\frac md\right)\binom{kd-1}{d-1}\right.\]
where $\mu(j)$ is the classic mobs function.**Proof.** Consider any prime $p\mid m$ , and suppose $\nu_p(m)=\alpha\geq 1... | [
"<details><summary>Solution</summary>For convenience, write $g(d)=\\sum_{e \\mid d} f(e)$ so that $f(d)=\\sum_{e \\mid d} \\mu\\left(\\frac{d}{e}\\right) g(e)$ by Möbius Inversion. Then we need to prove that $k \\mid g(k)$ for all $k$ iff\n\\[n \\mid \\sum_{d \\mid m} \\binom{n/d}{m/d} \\sum_{e \\mid d} \\m... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 136,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2812219.json"
} |
Let $m,n$ be two positive integers with $m \ge n \ge 2022$ . Let $a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n$ be $2n$ real numbers. Prove that the numbers of ordered pairs $(i,j) ~(1 \le i,j \le n)$ such that
\[ |a_i+b_j-ij| \le m \]
does not exceed $3n\sqrt{m \log n}$ . | <blockquote>Let $m,n$ be two positive integers with $m \ge n \ge 2022$ . Let $a_1,a_2,\ldots,a_n,b_1,b_2,\ldots,b_n$ be $2n$ real numbers. Prove that the numbers of ordered pairs $(i,j) ~(1 \le i,j \le n)$ such that
\[ |a_i+b_j-ij| \le m \]does not exceed $3n\sqrt{m \log n}$ .</blockquote>
Consider the bipart... | [
"does anyone have any ideas?",
"Consider a bipartite graph $(U,V)$ with vertices $( \\{u_1,\\cdots,u_n\\}, \\{v_1,\\cdots,v_n\\})$ such that there is an edge between $u_i$ and $v_j$ if $|a_i+b_j-ij|\\le m$ .\n\nLet $E$ be the total number of edges, and $X$ be the number of 2-step walks $v_i\\leftrig... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 174,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2812220.json"
} |
Given two circles $\omega_1$ and $\omega_2$ where $\omega_2$ is inside $\omega_1$ . Show that there exists a point $P$ such that for any line $\ell$ not passing through $P$ , if $\ell$ intersects circle $\omega_1$ at $A,B$ and $\ell$ intersects circle $\omega_2$ at $C,D$ , where $A,C,D,B$ lie o... | Let $P$ be a point such that $\omega_1,\omega_2$ , and the point circle $P$ are coaxial. Let the radical axis of $\omega_1,\omega_2$ meet $\ell$ at $Q$ . Then the circle centered at $Q$ with radius $\sqrt{QA\cdot QB}=\sqrt{QC\cdot QD}=QP$ is an Appolonian circle with respect to $A,B$ and $C,D$ , so the... | [
"Let $P$ be a point such that after inversion wrt $P$ $\\omega_1^*$ and $\\omega_2^*$ are concentric, which is well known to exist (let $O$ be the new common center).\nNow, the angle condition gets simply translated to $\\angle A^*PC^*=\\angle B^*PD^*$ , where $(A^*B^*C^*D^*P)$ is cyclic. But this cycl... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 26,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2835360.json"
} |
Two positive real numbers $\alpha, \beta$ satisfies that for any positive integers $k_1,k_2$ , it holds that $\lfloor k_1 \alpha \rfloor \neq \lfloor k_2 \beta \rfloor$ , where $\lfloor x \rfloor$ denotes the largest integer less than or equal to $x$ . Prove that there exist positive integers $m_1,m_2$ such th... | Okay here is a complete solution. Thanks to Justanaccount for helping out!
Assume $\alpha,\beta \notin \mathbb{Q}$ Note the number of $k_1$ satisfying $\lfloor k_1 \alpha \rfloor < X$ is precisely $\lfloor \frac{X}{\alpha} \rfloor$ .
Let $a=\frac{1}{\alpha}$ and $b=\frac{1}{\beta}$ . From $\lfloor k_1\alpha... | [
"UPD: See solns at PM #4, #7\n\nSome ideas: assume $\\alpha,\\beta \\notin \\mathbb{Q}$ Note the number of $k_1$ satisfying $\\lfloor k_1 \\alpha \\rfloor < X$ is precisely $\\lfloor \\frac{X}{\\alpha} \\rfloor$ .\n\nLet $a=\\frac{1}{\\alpha}$ and $b=\\frac{1}{\\beta}$ . From $\\lfloor k_1\\alpha \\rfloor... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 172,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2835366.json"
} |
Given a positive integer $n \ge 2$ . Find all $n$ -tuples of positive integers $(a_1,a_2,\ldots,a_n)$ , such that $1<a_1 \le a_2 \le a_3 \le \cdots \le a_n$ , $a_1$ is odd, and
(1) $M=\frac{1}{2^n}(a_1-1)a_2 a_3 \cdots a_n$ is a positive integer;
(2) One can pick $n$ -tuples of integers $(k_{i,1},k_{i,2},\ldo... | <details><summary>Easy for P3, easiest problem on T3 for me</summary>Let $a$ be the number of odd terms in $a_i$ . The answer is $2^a|a_1-1$ only.
Call points $(x_1, \cdots, x_n)$ .
Claim: The maximum number of points that can be placed in $\times_{j=1}^n \mathbb{Z}_{a_j}$ is at most $\frac{1}{2^n} (a_1-1) \... | [] | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 118,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2835370.json"
} |
Find all positive integer $k$ such that one can find a number of triangles in the Cartesian plane, the centroid of each triangle is a lattice point, the union of these triangles is a square of side length $k$ (the sides of the square are not necessarily parallel to the axis, the vertices of the square are not neces... | Claim: the only such numbers are the multiples of $3$ .
If $k=3t$ pick an axis and grid aligned $k\times k$ square divided in $t\times t$ square each of side $3\times 3$ , and divide these in two triangles. In this way it is straight forward enough to see that the barycenters must have integer coordinates.
Lem... | [
"easy for China TST\n\n<details><summary>solution</summary>The answer is all $3|k$ .\n\nConstruction: \nIf $3|k$ then take two triangles. One with $(0, 0), (0, k), (k, 0)$ and one with $(k, k), (0, k), (k, 0)$ .\nTheir centroids are both lattice points.\n\nProof that others won't work:\nWe would prove that in... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 50,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2835373.json"
} |
Show that there exist constants $c$ and $\alpha > \frac{1}{2}$ , such that for any positive integer $n$ , there is a subset $A$ of $\{1,2,\ldots,n\}$ with cardinality $|A| \ge c \cdot n^\alpha$ , and for any $x,y \in A$ with $x \neq y$ , the difference $x-y$ is not a perfect square. | <blockquote>The main idea is **Trivial solutions --> Generalize**.
I'll present two similar approaches.
<details><summary>Solution 1)</summary>Denote $A$ as the set of numbers which contain only digits of $1,4,7,9,12$ under hexadecimal expression.
Obviously, we have chosen at least $0.0001n^{\log_{16}5}$ numbe... | [
"<blockquote>Show that there exist constants $c$ and $\\alpha > \\frac{1}{2}$ , such that for any positive integer $n$ , there is a subset $A$ of $\\{1,2,\\ldots,n\\}$ with cardinality $|A| \\ge c \\cdot n^\\alpha$ , and for any $x,y \\in A$ with $x \\neq y$ , the difference $x-y$ is not a perfect squ... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 152,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2835377.json"
} |
(1) Prove that, on the complex plane, the area of the convex hull of all complex roots of $z^{20}+63z+22=0$ is greater than $\pi$ .
(2) Let $a_1,a_2,\ldots,a_n$ be complex numbers with sum $1$ , and $k_1<k_2<\cdots<k_n$ be odd positive integers. Let $\omega$ be a complex number with norm at least $1$ . Prove... | I'll give a proof for Gauss Lucas Thm Here.
<blockquote>For any complex polynomial $P(z)$ , the roots of the derivative $P'(z)$ lies in the convex hull of the roots of $P(z).$ </blockquote>
Proof. let $P(z)=\lambda \prod_{k=1}^n(z-z_k)^{\alpha_k},$ then if $P'(Z)=0,$ $$ \frac{P'(Z)}{P(Z)}=\sum_{k=1}^n\frac{\al... | [
"<details><summary>hint</summary>Use the fact that the roots of $P'(x)$ lies in the convex hall of P(x).\nFor (2),consider $Q_0(z)=P(\\frac{1}{z})$ and recursively define $Q_{i+1}(z)$ be $Q'_{i}(z)$ with all the zero roots deleted.\nEstimate each $a_i$ by the constant factor of $Q_{n-i}(z)$ .</details>",... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 72,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2835381.json"
} |
Initially, each unit square of an $n \times n$ grid is colored red, yellow or blue. In each round, perform the following operation for every unit square simultaneously:
- For a red square, if there is a yellow square that has a common edge with it, then color it yellow.
- For a yellow square, if there is a blue squ... | The crux of this problem is the following claim.**<span style="color:#07f">Claim:</span>** There exists a cell that never changes color.
*Proof:* By the condition, the color of every cell must eventually become constant. Consider a cell $\mathcal{A}$ that first becomes a constant color, and assume WLOG that the colo... | [
"Assume wlog the blue \"army\" \"wins\" in the end, and consider one of the initial contiguous blue region which survives until the end. Assume that at any point of the \"expansion\" of this region, it meets a red \"soldier\". At each turn, the \"division\" (i.e the evolution of this initial red soldier/connected r... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 146,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2835389.json"
} |
Let $ABCD$ be a convex quadrilateral, the incenters of $\triangle ABC$ and $\triangle ADC$ are $I,J$ , respectively. It is known that $AC,BD,IJ$ concurrent at a point $P$ . The line perpendicular to $BD$ through $P$ intersects with the outer angle bisector of $\angle BAD$ and the outer angle bisector $... | <blockquote>Solved with **BarisKoyuncu**, **SerdarBozdag** and **sevket12**.
Let $AB \cap CD = X , AD \cap BC = Y , BI \cap DJ = M$ . Applying Desargues on triangles $\triangle AIB$ and $\triangle CJD$ , we obtain that the points $X , M , G = AI \cap CJ$ are collinear. Now note that since $AI$ is the exterior ... | [
"Is the problem right?\nCould you please check the original problem?\nI think there are some typos somewhere.\n\nHere is a proof that states : $AC,BD,IJ$ can never concurrent.\n\nLet $AC,BD$ intersect at $K$ Let incenter of triangle $ABK, CDK$ be $S,T$ .\nThen it is trivial that $S$ lies on segment $BI$ ... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 266,
"boxed": false,
"end_of_proof": false,
"n_reply": 14,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2835390.json"
} |
Find all functions $f: \mathbb R \to \mathbb R$ such that for any $x,y \in \mathbb R$ , the multiset $\{(f(xf(y)+1),f(yf(x)-1)\}$ is identical to the multiset $\{xf(f(y))+1,yf(f(x))-1\}$ .
*Note:* The multiset $\{a,b\}$ is identical to the multiset $\{c,d\}$ if and only if $a=c,b=d$ or $a=d,b=c$ . | I might have written a bit complicated :blush:
\[\{f(xf(y)+1),f(yf(x)-1)\}=\{xf(f(y))+1,yf(f(x))-1\}\]
Answers are $f(x)=x$ and $f(x)=-x$ which clearly fit. Let $P(x,y)$ be the assertion. Note that $P(0,0)$ gives $\{f(1),f(-1)\}=\{1,-1\}$ . If $f(1)=1$ , then $P(1,1)$ and $P(-1,1)$ imply $f(0)=0$ . If ... | [
"NOTE: THE PREVIOUS SOLUTION I POSTED HAS A SIGN ERROR, hopefully this one is correct.\n\nNote 2: This solves for f(1)=1 only \n\n<details><summary>Solution to f(1)=1 only</summary>The answer is $f(x)\\equiv x, f(x)\\equiv -x$ only. They both work. \n\nLet $P(x,y)$ denote the assertion that the multiset $\\{(f... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 434,
"boxed": false,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2835392.json"
} |
Find all positive integers $a,b,c$ and prime $p$ satisfying that
\[ 2^a p^b=(p+2)^c+1.\] | My 300th post!!!**Solution**
Notice that $2\mid 2^ap^b,$ we have $p\equiv 1\pmod 2.$ Let $c=2^{\theta}r$ where $2\nmid r,$ then $(p+2)^{2^{\theta}}+1\mid 2^ap^b.$ Lemma $:$ $(p+2)^{2^{\theta}}+1$ is divisible by $p.$ If it is not true $,$ $(p+2)^{2^{\theta}}+1=2^{\beta},$ then $\beta >2.$ Therefore ... | [
"If $c=0$ then obviously $a=1$ and $b=0$ Let $c>0$ $mod2$ gives $a>1$ If $p=2$ then $RHS\\equiv 1\\not\\equiv 0\\equiv LHS (mod 2)$ contradiction.So $p>=3$ If $b=0$ then we have:\nIf $c=1$ we have the solutionw on the form $(a,b,c,p):(a,0,1,2^a-3)$ with $p=prime$ If $c=odd>1$ then $2^a=(p+3)... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 1212,
"boxed": true,
"end_of_proof": true,
"n_reply": 9,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2835394.json"
} |
Let $n$ be a positive integer, $x_1,x_2,\ldots,x_{2n}$ be non-negative real numbers with sum $4$ . Prove that there exist integer $p$ and $q$ , with $0 \le q \le n-1$ , such that
\[ \sum_{i=1}^q x_{p+2i-1} \le 1 \mbox{ and } \sum_{i=q+1}^{n-1} x_{p+2i} \le 1, \]
where the indices are take modulo $2n$ .
*Not... | Here is another version of my proof above, which is simpler in my opinion. Moreover the gas station lemma is proved for completeness.
<details><summary>Solution</summary>Let us write $[n]\doteqdot\{1,2,\ldots,n\}$ for brevity.
*Lemma.* (gas station lemma) Let $n$ be a positive integer and let $a_1,\ldots,a_n$ b... | [
"Sketch and unLaTeXed since I am on mobile:\nLet S be the sum of x_i where i is even. Similarly let T be the sum of x_i where i is odd. WLOG S>=T. Consider a function f:1,...,n->R such that f(i)=S*x_(2i+1)-T*x_(2i). Then sum f(i)=0 so by gas station lemma, there exists an index a such that f(a),f(a+1),...,f(a+n-1) ... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 134,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2835400.json"
} |
Given a positive integer $n$ , let $D$ be the set of all positive divisors of $n$ . The subsets $A,B$ of $D$ satisfies that for any $a \in A$ and $b \in B$ , it holds that $a \nmid b$ and $b \nmid a$ . Show that
\[ \sqrt{|A|}+\sqrt{|B|} \le \sqrt{|D|}. \] | Solved with Luke Robitaille, Justin Lee, and Espen Slettnes.
Let \[V=\bigcup_{a\in A}\{\text{divisors of }a\} \quad\text{and}\quad W=\bigcup_{a\in A}\{\text{multiples of }a\}\cap D\]
<span style="color:red">**Claim:**</span> \(|V|\cdot|W|\ge|A|\cdot|D|\).
*Proof.* We induct on the number of prime divi... | [
"Oops this is pretty much the same as above\n\nLet $A_{down} = \\{ x \\text{ st } x|a \\text{ for some } a\\in A\\}$ and $A_{up} = \\{ \\text{ st } a|x|n \\text{ for some } a\\in A\\}$ . Define $B_{up}, B_{down}$ similarly.\n\nMain Claim: $|A| |D| \\le |A_{up}||A_{down}|$ Proof: We induct on $\\omega(n)$ . L... | [
"origin:aops",
"2022 China Team Selection Test",
"2022 Contests"
] | {
"answer_score": 100,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2022 Contests/2022 China Team Selection Test/2835403.json"
} |
Let $A_1A_2A_3A_4$ and $B_1B_2B_3B_4$ be two squares such that the boundaries of $A_1A_2A_3A_4$ and $B_1B_2B_3B_4$ does not contain any line segment. Construct 16 line segments $A_iB_j$ for each possible $i,j \in \{1,2,3,4\}$ . What is the maximum number of line segments that don't intersect the edges of $A... | There are two feasible interpretations of this problem, so I'll provide a solution for both. The construction for both of them is two concentric squares with sides parallel.
<details><summary>Solution with one interpretation</summary>If the problem means that an intersection is not counted if and only if the intersect... | [
"<details><summary>My Protest with 16</summary>[asy]\nlabel(\" $A_1$ \",(0,0),W);\nlabel(\" $A_2$ \",(4,0),E);\nlabel(\" $A_3$ \",(4,4),E);\nlabel(\" $A_4$ \",(0,4),W);\ndraw((0,0)--(4,0));\ndraw((4,0)--(4,4));\ndraw((4,4)--(0,4));\ndraw((0,0)--(0,4));\ndraw((0,0)--(4,4));\ndraw((0,4)--(4,0));\nlabel(\" $B_1$ \",(1... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1118,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791418.json"
} |
Find the smallest positive integer $n$ for which $315^2-n^2$ evenly divides $315^3-n^3$ .
*Proposed by Kyle Lee* | <details><summary>Solution</summary>First, we have
\begin{align*}
\left(315-n^2\right) \mid \left(315^3-n^3\right) &\implies (315-n)(315+n) \mid (315-n)(315^2+315n+n^2)
&\implies (315+n) \mid (315^2+315n+n^2)
&\implies (315+n) \mid (315^2+315n+n^2-n(315+n))
&\implies (315+n) \mid(315^2+315n+n^2-315n-n^2)
&\implies ... | [
"<details><summary>Euclidean Algorithm</summary>$$ 315^2-n^2=(315+n)(315-n) $$ $$ (315-n)(315^2+315n+n^2) $$ $$ (315+n)|(315^2+315n+n^2) $$ $$ (315+n)|(315^2) $$ $$ n=\\boxed{90} $$</details>"
] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1112,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791419.json"
} |
Let $ABCD$ be a rectangle with $AB=10$ and $AD=5.$ Suppose points $P$ and $Q$ are on segments $CD$ and $BC,$ respectively, such that the following conditions hold:
- $BD \parallel PQ$
- $\angle APQ=90^{\circ}.$
What is the area of $\triangle CPQ?$ *Proposed by Kyle Lee* | <details><summary>Solution</summary>Since $BD \parallel PQ$ and $AP \perp PQ,$ it follows that $AP \perp BD.$ Now, let us place rectangle $ABCD$ into the Cartesian Plane such that $$ A=(0,5), B=(10,5), C=(10,0), D=(0,0). $$ Since the slope of line $BD$ is
\begin{align*}
\frac{5-0}{10-0} &= \frac{5}{10}
&=... | [
"<details><summary>Solution</summary>Slope of $PQ$ is $\\frac{1}{2}$ thus, the slope of $AP$ is $-2.$ Therefore $PC=\\frac{15}{2}$ and $CQ=\\frac{15}{4}$ so our answer is $\\boxed{\\frac{225}{16}}$</details>"
] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1144,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791420.json"
} |
Let $\triangle ABC$ be equilateral with integer side length. Point $X$ lies on $\overline{BC}$ strictly between $B$ and $C$ such that $BX<CX$ . Let $C'$ denote the reflection of $C$ over the midpoint of $\overline{AX}$ . If $BC'=30$ , find the sum of all possible side lengths of $\triangle ABC$ .
*Pr... | <details><summary>Solution</summary>[asy]
pair C = dir(60), A = dir(180), B = dir(300);
draw(A--B--C--cycle);
pair X = (2B+C)/3;
draw(A--X);
pair Cp = A+X-C;
draw(A--Cp--X,red);
draw(Cp--B,blue);
pair P = extension(Cp,X,A,B);
dot(A^^B^^C^^X^^Cp^^P);
label(" $A$ ",A,dir(180));
label(" $B$ ",B,dir(300));
label(" $C$ ",... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1054,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791422.json"
} |
For any integer $a$ , let $f(a) = |a^4 - 36a^2 + 96a - 64|$ . What is the sum of all values of $f(a)$ that are prime?
*Proposed by Alexander Wang* | <details><summary>Solution</summary>First, we can factor the given function as $$ f(a) = |(a-2)(a-4)(a^2+6a-8)|. $$ For this to be prime, exactly two of $(a-2), (a-4),$ and $(a^2+6a-8)$ must be $\pm 1$ and the third must be $\pm p$ for a prime $p.$ We now split into cases:**a-2 is 1:** Then, we have $a=3$... | [
"With some playing around we have $$ f(a)=|(a-2)g(a)| $$ for some cubic $g(a)$ . In order for this to be prime we must either have $a=1$ or $a=3$ . The answer is $\\boxed{22}$ @below inspection shows that if $g(a) = \\pm 1$ then $a$ is not an integer",
"technically $g(a)=\\pm 1$ could be possible @... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1150,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791425.json"
} |
There are $9$ points arranged in a $3\times 3$ square grid. Let two points be adjacent if the distance between them is half the side length of the grid. (There should be $12$ pairs of adjacent points). Suppose that we wanted to connect $8$ pairs of adjacent points, such that all points are connected to each oth... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791426.json"
} | |
A $3\times2\times2$ right rectangular prism has one of its edges with length $3$ replaced with an edge of length $5$ parallel to the original edge. The other $11$ edges remain the same length, and the $6$ vertices that are not endpoints of the replaced edge remain in place. The resulting convex solid has $8$... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791429.json"
} | |
There are 36 contestants in the CMU Puyo-Puyo Tournament, each with distinct skill levels.
The tournament works as follows:
First, all $\binom{36}{2}$ pairings of players are written down on slips of paper and are placed in a hat.
Next, a slip of paper is drawn from the hat, and those two players pl... | pretty nice, solved this during my bio class
Let $1, 2, 3, \ldots, 36$ be the players, where a higher number represents a higher skill level. I'll use the notation $i-j$ to denote the match where $i$ plays $j$ . The key is to notice that only the matches of the form $k-(k+1)$ matter (ie, $1-2$ , $2-3$ , etc... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1166,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791430.json"
} |
For natural numbers $n$ , let $r(n)$ be the number formed by reversing the digits of $n$ , and take $f(n)$ to be the maximum value of $\frac{r(k)}k$ across all $n$ -digit positive integers $k$ .
If we define $g(n)=\left\lfloor\frac1{10-f(n)}\right\rfloor$ , what is the value of $g(20)$ ?
*Prop... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791432.json"
} | |
Adam places down cards one at a time from a standard 52 card deck (without replacement) in a pile. Each time he places a card, he gets points equal to the number of cards in a row immediately before his current card that are all the same suit as the current card. For instance, if there are currently two hearts on the t... | really fun problem to solve
Number the cards as follows for convenience: $\{A1, A2, \ldots, A13\}, \{B1, B2, \ldots, B13\}, \{C1, C2, \ldots, C13\}, \{D1, D2, \ldots, D13\}$ . The letter corresponds to the suit of the card. Suppose we place each card in a row, where cards to the left are placed earlier (ie, the leftm... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1166,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791433.json"
} |
Let $\{\varepsilon_i\}_{i\ge 1}, \{\theta_i\}_{i\ge 0}$ be two infinite sequences of real numbers, such that $\varepsilon_i \in \{-1,1\}$ for all $i$ , and the numbers $\theta_i$ obey $$ \tan \theta_{n+1} = \tan \theta_{n}+\varepsilon_n \sec(\theta_{n})-\tan \theta_{n-1} , \qquad n \ge 1 $$ and $\theta_0 = \f... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791434.json"
} | |
Let $ABCD$ be a cyclic quadrilateral with $AB=3, BC=2, CD=6, DA=8,$ and circumcircle $\Gamma.$ The tangents to $\Gamma$ at $A$ and $C$ intersect at $P$ and the tangents to $\Gamma$ at $B$ and $D$ intersect at $Q.$ Suppose lines $PB$ and $PD$ intersect $\Gamma$ at points $W \neq B$ and $X ... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791435.json"
} | |
Let $F_n$ denote the $n$ th Fibonacci number, with $F_0=0, F_1=1$ and $F_{n}=F_{n-1}+F_{n-2}$ for $n \geq 2$ . There exists a unique two digit prime $p$ such that for all $n$ , $p | F_{n+100} + F_n$ . Find $p$ .
*Proposed by Sam Rosenstrauch* | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791436.json"
} | |
Let a tree on $mn + 1$ vertices be $(m,n)$ -nice if the following conditions hold:
- $m + 1$ colors are assigned to the nodes of the tree
- for the first $m$ colors, there will be exactly $n$ nodes of color $i$ $(1\le i \le m)$
- the root node of the tree will be the unique node of color $m+1$ . \ite... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791437.json"
} | |
Let $ABC$ be a triangle with $AB = 5, BC = 13,$ and $AC = 12$ . Let $D$ be a point on minor arc $AC$ of the circumcircle of $ABC$ (endpoints excluded) and $P$ on $\overline{BC}$ . Let $B_1, C_1$ be the feet of perpendiculars from $P$ onto $CD, AB$ respectively and let $BB_1, CC_1$ hit $(ABC)$ ag... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791438.json"
} | |
Alice and Bob live on the same road. At time $t$ , they both decide to walk to each other's houses at constant speed. However, they were busy thinking about math so that they didn't realize passing each other. Alice arrived at Bob's house at $3:19\text{pm}$ , and Bob arrived at Alice's house at $3:29\text{pm}$ . Cha... | <details><summary>Solution</summary>First of all, clearly, Alice is faster than Bob. Now, let the ratio of Alice's speed to Bob's speed be $r.$ Note that, after passing each other, Alice has $r$ times less distance to get to Bob's house than Bob has to get to Alice's house and is also $r$ times faster. Therefore, ... | [
"ok this one was harder than #5\n",
"<details><summary>Solution</summary>Since they past each other at $3:11$ we know that it took 8 minutes for Alice to reach Bob's house after they met and 18 minutes for Bob to reach Alice's house after they met. Now let us say that they drove $x$ minutes before meeting eac... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1124,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791439.json"
} |
Arthur, Bob, and Carla each choose a three-digit number. They each multiply the digits of their own numbers. Arthur gets 64, Bob gets 35, and Carla gets 81. Then, they add corresponding digits of their numbers together. The total of the hundreds place is 24, that of the tens place is 12, and that of the ones place is... | <details><summary>Solution</summary>Let the numbers chosen by Arthur, Bob, and Carla be $\overline{a_1a_2a_3}, \overline{b_1b_2b_3},$ and $\overline{c_1c_2c_3}$ respectively, where $$ x_i \in \{1,2,\ldots,9\} \ \forall \ x \in \{a,b,c\}, i \in \{1,2,3\}. $$ It is not difficult to see that $$ \max(a_1) = 8, \ma... | [
"the first numbers i tested without reading the third sentence worked haha\n\n<details><summary>Solution</summary>Let the hundreds place of the three numbers be $a$ , $b$ , and $c$ , respectively. Note that $\\operatorname{max}(a)=8$ , $\\operatorname{max}(b)=7$ , and $\\operatorname{max}(c)=9$ . We have that... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1168,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791440.json"
} |
How many 4-digit numbers have exactly $9$ divisors from the set $\{1,2,3,4,5,6,7,8,9,10\}$ ?
*Proposed by Ethan Gu* | <details><summary>Solution</summary>It is not difficult to see that the excluded number must be $7,8,$ or $9.$ We now consider each case separately:**7:** Note that the number must be divisible by
\begin{align*}
\frac{\text{lcm} (1,2,3,\ldots,10)}{7} &= \frac{2520}{7}
&= 360.
\end{align*}
If we let the number be $... | [
"<details><summary>Solution</summary>Note that $\\operatorname{lcm}(1,2,3,\\cdots,10) = 2^3 \\cdot 3^2 \\cdot 5 \\cdot 7$ . **Case 1:** The divisor that is not included from the set is $7$ .\n\nThen the 4-digit number is a multiple of $\\tfrac{2520}{7} = 360$ . The smallest 4-digit multiple of $360$ is $360 \... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1168,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791441.json"
} |
A shipping company charges $.30l+.40w+.50h$ dollars to process a right rectangular prism-shaped box with dimensions $l,w,h$ in inches. The customers themselves are allowed to label the three dimensions of their box with $l,w,h$ for the purpose of calculating the processing fee. A customer finds that there are t... | <details><summary>Solution</summary>Let the dimensions of $B$ be $a \times b \times c.$ Then, suppose, without loss of generality, that we have $$ 0.3a+0.4b+0.5c=0.3b+0.4c+0.5a=8.1 $$ This implies that $2a=b+c.$ Note that if we have $2b=a+c$ or $2c=a+b,$ then it would follow that $a=b=c,$ contradiction. ... | [
"<details><summary>Systems of Equations Galore</summary>$$ 5l+3w+4h=81 $$ $$ 4l+5w+3h=81 $$ $$ 3l+5w+4h=87 $$ $$ 4l+3w+5h=87 $$ $$ l+h=2w $$ $$ h-l=6 $$ $$ 2w-2l=6 $$ $$ h=l+6 $$ $$ w=l+3 $$ $$ 5l+3(l+3)+4(l+6)=81 $$ $$ l=4 $$ $$ h=10 $$ $$ w=7 $$ $$ 2(4*7+4*10+7*10)=\\boxed{2... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1132,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791443.json"
} |
Alan is assigning values to lattice points on the 3d coordinate plane. First, Alan computes the roots of the cubic $20x^3-22x^2+2x+1$ and finds that they are $\alpha$ , $\beta$ , and $\gamma$ . He finds out that each of these roots satisfy $|\alpha|,|\beta|,|\gamma|\leq 1$ On each point $(x,y,z)$ where $x,y,$... | <blockquote><details><summary>Solution</summary>Let the given polynomial be $f(x).$
We have
\begin{align*}
\sum_{x,y,z \geq 0} \alpha^x \beta^y \gamma^z &= \left(\sum_{x \geq 0} \alpha^x \right)\left(\sum_{y \geq 0} \beta^y \right)\left(\sum_{z \geq 0} \gamma^z\right)
&= \frac{1}{1-\alpha} \cdot \frac{1}{1-\beta} ... | [
"<details><summary>Solution</summary>Note that $$ \\sum_{x=0}^{\\infty}\\sum_{y=0}^{\\infty}\\sum_{z=0}^{\\infty} \\alpha^x\\beta^y\\gamma^z = (\\alpha^0+\\alpha^1+\\cdots)(\\beta^0+\\beta^1+\\cdots)(\\gamma^0+\\gamma^1+\\cdots). $$ Since $\\alpha^0=\\beta^0=\\gamma^0=1$ and $|\\alpha|,|\\beta|,|\\gamma| \\l... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1108,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791444.json"
} |
Find the smallest positive integer $N$ such that each of the $101$ intervals $$ [N^2, (N+1)^2), [(N+1)^2, (N+2)^2), \cdots, [(N+100)^2, (N+101)^2) $$ contains at least one multiple of $1001.$ *Proposed by Kyle Lee* | <details><summary>Solution</summary>Intuitively, we see that the smallest desired $N$ is the smallest $N$ that guarantees that the desired result holds true minus a little bit, because the difference between consecutive squares getting smaller and smaller makes it harder and harder to find $101$ consecutive inter... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1152,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791445.json"
} |
For polynomials $P(x) = a_nx^n + \cdots + a_0$ , let $f(P) = a_n\cdots a_0$ be the product of the coefficients of $P$ . The polynomials $P_1,P_2,P_3,Q$ satisfy $P_1(x) = (x-a)(x-b)$ , $P_2(x) = (x-a)(x-c)$ , $P_3(x) = (x-b)(x-c)$ , $Q(x) = (x-a)(x-b)(x-c)$ for some complex numbers $a,b,c$ . Given $f(Q) =... | <details><summary>Storage</summary>$P_1 = (x - a)(x - b) = x^2 - (a + b)x + ab$ $P_2 = (x - a)(x - c) = x^2 - (c + a)x + ca$ $P_3 = (x - b)(x - c) = x^2 - (b + c)x + bc$ $Q(x) = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + bc + ca)x - abc$ $f(Q) = 8 \implies abc(a + b + c)(ab + bc + ca) = 8$ $\implies (a +... | [
"<details><summary>solution</summary>We have $f(Q) = (a+b+c)(ab + ac + bc)(abc) = 8$ and $f(P_1)+f_(P_2) + f(P_3) = -(a^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2) = 10$ , thus $a^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2 = -10$ .\n\nNow we see that $(a+b+c)(ab + ac + bc) = a^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2 + 3ab... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1028,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791446.json"
} |
Let $z$ be a complex number that satisfies the equation \[\frac{z-4}{z^2-5z+1} + \frac{2z-4}{2z^2-5z+1} + \frac{z-2}{z^2-3z+1} = \frac{3}{z}.\] Over all possible values of $z$ , find the sum of the values of \[\left| \frac{1}{z^2-5z+1} + \frac{1}{2z^2-5z+1} + \frac{1}{z^2-3z+1} \right|.\]
*Proposed by Justin Hsieh... | <details><summary>Solution</summary>First, multiplying both sides of the given equation by $z$ yields $$ \frac{z^2-4z}{z^2-5z+1} + \frac{2z^2-4z}{2z^2-5z+1} + \frac{z^2-2z}{z^2-3z+1}=3. $$ Then, subtracting $3$ from both sides of the equation yields
\begin{align*}
\frac{z^2-4z}{z^2-5z+1} + \frac{2z^2-4z}{2z^2-5... | [
"I fakesolved this by assuming all the terms were equal, for which the only solution is $z=1$ . Therefore the answer is $\\boxed{\\tfrac{11}{6}}$ ",
"oops I did same as above lol",
"<details><summary>DeToasty3's Solution</summary>Multiply both sides of the equation by $z$ . We get $$ \\frac{z^2-4z}{z^2-5z+... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1120,
"boxed": true,
"end_of_proof": true,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791447.json"
} |
Grant is standing at the beginning of a hallway with infinitely many lockers, numbered, $1, 2, 3, \ldots$ All of the lockers are initially closed. Initially, he has some set $S = \{1, 2, 3, \ldots\}$ .
Every step, for each element $s$ of $S$ , Grant goes through the hallway and opens each locker divisible by... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791448.json"
} | |
Find the probability such that when a polynomial in $\mathbb Z_{2027}[x]$ having degree at most $2026$ is chosen uniformly at random,
$$ x^{2027}-x | P^{k}(x) - x \iff 2021 | k $$ (note that $2027$ is prime).
Here $P^k(x)$ denotes $P$ composed with itself $k$ times.
*Proposed by Grant ... | This is a nice "cycles"/chains problem!
Let $p=2027$ . Note $P(x)$ can be any map from $\mathbb{Z}_p$ to (a subset of) $\mathbb{Z}_p$ .
Let $ord(z)$ be the minimal $k$ such that $P^k(z)=z$ . The condition is implying $ord(z)$ exists and $lcm_{z\in \mathbb{Z}_p} ord(z)=43\cdot 47$ .
This means all cycle... | [
"2021 was last year y'all gotta know it's not prime :( i lost so much time"
] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 58,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791450.json"
} |
Let $f(n)$ count the number of values $0\le k\le n^2$ such that $43\nmid\binom{n^2}{k}$ . Find the least positive value of $n$ such that $$ 43^{43}\mid f\left(\frac{43^{n}-1}{42}\right) $$ *Proposed by Adam Bertelli* | [
"this is too hard :("
] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791452.json"
} | |
Find the largest $c > 0$ such that for all $n \ge 1$ and $a_1,\dots,a_n, b_1,\dots, b_n > 0$ we have
$$ \sum_{j=1}^n a_j^4 \ge c\sum_{k = 1}^n \frac{\left(\sum_{j=1}^k a_jb_{k+1-j}\right)^4}{\left(\sum_{j=1}^k b_j^2j!\right)^2} $$ *Proposed by Grant Yu* | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791453.json"
} | |
An equilateral $12$ -gon has side length $10$ and interior angle measures that alternate between $90^\circ$ , $90^\circ$ , and $270^\circ$ . Compute the area of this $12$ -gon.
*Proposed by Connor Gordon* | <details><summary>Solution</summary>It is easy to see that the $12$ -gon is in the shape of a cross which is made up of $5$ congruent squares, each of side length $10.$ Hence, the area of the $12$ -gon is
\begin{align*}
5 \cdot 10^2 &= 5 \cdot 100
&= \boxed{500}.
\end{align*} $\square$</details> | [
"<details><summary>Solution</summary>Note you can split the $12$ -gon into five squares of side length $10$ , so the total area is $5(10^2)=\\boxed{500}$ [asy]\ndraw((0,0)--(1,0)--(1,1)--(2,1)--(2,0)--(3,0)--(3,-1)--(2,-1)--(2,-2)--(1,-2)--(1,-1)--(0,-1)--cycle);\n[/asy]</details>",
"This is quite literally th... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1110,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791457.json"
} |
A circle has radius $52$ and center $O$ . Points $A$ is on the circle, and point $P$ on $\overline{OA}$ satisfies $OP = 28$ . Point $Q$ is constructed such that $QA = QP = 15$ , and point $B$ is constructed on the circle so that $Q$ is on $\overline{OB}$ . Find $QB$ .
*Proposed by Justin Hsieh* | <details><summary>Solution</summary>First, note that
\begin{align*}
AP &= AO - PO
&= 52 - 28
&= 24.
\end{align*}
Let the foot of the altitude from $Q$ to $\overline{AO}$ be $X.$ Then, since $QA=QP,$ it follows that
\begin{align*}
AX = XP &= \frac{AP}{2}
&= \frac{24}{2}
&= 12.
\end{align*}
Therefore,
\begin{... | [
"<details><summary>Quick Projection</summary>Let the projection of $Q$ onto $AO$ be $M.$ Then $OM=40$ and $MQ=9$ thus $OQ=41$ and $QB=\\boxed{11}$</details>"
] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1110,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791458.json"
} |
Let $ABC$ be an acute triangle with $\angle ABC=60^{\circ}.$ Suppose points $D$ and $E$ are on lines $AB$ and $CB,$ respectively, such that $CDB$ and $AEB$ are equilateral triangles. Given that the positive difference between the perimeters of $CDB$ and $AEB$ is $60$ and $DE=45,$ what is the val... | Neat problem!
<details><summary>Solution</summary>Without loss of generality, suppose that $AB>BC.$ Then, it follows that $D$ is inside the triangle and $E$ is outside the triangle.
Now, since the perimeters of $CDB$ and $AEB$ differ by $60,$ it follows that their side lengths differ by $20,$ i.e. $AB-... | [
"<details><summary>Slick Solution</summary>WLOG let $D$ be outside triangle $ABC$ and $AB=x.$ Therefore we are looking for $x(x+20).$ We know that $AD=20.$ Law of Cosines on $\\triangle BED$ gives us $$ x^2+(x+20)^2-(x)(x+20)=45^2 $$ $$ x^2+20x=\\boxed{1625} $$</details>"
] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1124,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791461.json"
} |
Circle $\Gamma$ has diameter $\overline{AB}$ with $AB = 6$ . Point $C$ is constructed on line $AB$ so that $AB = BC$ and $A \neq C$ . Let $D$ be on $\Gamma$ so that $\overleftrightarrow{CD}$ is tangent to $\Gamma$ . Compute the distance from line $\overleftrightarrow{AD}$ to the circumcenter of $... | <details><summary>Solution</summary>Let $X$ be the center of circle $\Gamma,$ let $Y$ be the midpoint of $\overline{AD},$ and let $Z$ be the center of $(ADC).$ We wish to find $YZ.$ First, note that $AX=BX=DX=3,$ and we also have
\begin{align*}
CX &= BC+BX
&= 6+3
&= 9.
\end{align*}
Therefore, by Pythag... | [
"Let the center of circle $\\Gamma$ be $O$ , it is easy to get $OD=3,OC=9,\\cos\\angle{DOC}=\\frac{1}{3}$ Let the midpoint of $AD$ be $F$ , the intersection of perpendicular bisectors of $AD,AC$ , assuming $J$ , the length of $FJ$ is the desired value\n\nSo since $\\cos\\angle{AOD}=-\\cos\\angle{DOC};\\... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1124,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791463.json"
} |
Let $ABC$ be an equilateral triangle of unit side length and suppose $D$ is a point on segment $\overline{BC}$ such that $DB<DC.$ Let $M$ and $N$ denote the midpoints of $\overline{AB}$ and $\overline{AC},$ respectively. Suppose $X$ and $Y$ are the intersections of lines $AB$ and $ND,$ and lines... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791465.json"
} | |
A triangle $\triangle ABC$ satisfies $AB = 13$ , $BC = 14$ , and $AC = 15$ . Inside $\triangle ABC$ are three points $X$ , $Y$ , and $Z$ such that:
- $Y$ is the centroid of $\triangle ABX$
- $Z$ is the centroid of $\triangle BCY$
- $X$ is the centroid of $\triangle CAZ$
What is the area of $\... | <details><summary>Solution</summary>We proceed with barycentric coordinates. Let $\triangle{ABC}$ be the reference triangle with $$ A=(1,0,0), B=(0,1,0), C=(0,0,1). $$ Also, let $X=(m, n, 1-m-n)$ for some $0 < m,n < 1.$ Then, since $X, Y,$ and $Z$ are cyclical, it follows that $$ Y=(1-m-n, m, n), Z=(n, ... | [
"Define points $B(0,0), C(14,0), A(5,12).$ Then let $X(a,b)$ . Using the fact that the centroid in the coordinate plane is equal to the average of the x-coordinates and y-coordinates, we get $Y(\\frac{a+5}{3}, \\frac{b+12}3)$ and then $Z(\\frac{a+47}{9}, \\frac{b+12}{9})$ . We can solve again for $X$ , getti... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1136,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791468.json"
} |
Let $\Gamma_1, \Gamma_2, \Gamma_3$ be three pairwise externally tangent circles with radii $1,2,3,$ respectively. A circle passes through the centers of $\Gamma_2$ and $\Gamma_3$ and is externally tangent to $\Gamma_1$ at a point $P.$ Suppose $A$ and $B$ are the centers of $\Gamma_2$ and $\Gamma_3,$ ... | Solved with **Sleepy_Head**.
<details><summary>Solution</summary>[asy]
import geometry;
import olympiad;
size(5cm,0);
pair C = (0,0);
pair A = (0,3);
pair B = (4,0);
pair O = (3.5,3.5);
line CO = line(C,O);
path w1 = circle(C,1);
path w2 = circle(A,2);
path w3 = circle(B,3);
... | [
"this one was nice, just set random variables and things cancel out very nicely. ",
"Lol two applications of Stewart’s Theorem kills this problem\n\nThe answer is $\\frac{8}{15}$ ",
"Denote the circle that passes through the centers of $\\Gamma_2$ and $\\Gamma_3$ by $\\Omega$ , denote its center by $O$ ,... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1150,
"boxed": true,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791470.json"
} |
Let $A$ and $B$ be points on circle $\Gamma$ such that $AB=\sqrt{10}.$ Point $C$ is outside $\Gamma$ such that $\triangle ABC$ is equilateral. Let $D$ be a point on $\Gamma$ and suppose the line through $C$ and $D$ intersects $AB$ and $\Gamma$ again at points $E$ and $F \neq D.$ It is give... | Let altitude from $C$ to $AB$ meets $AB$ at $J$ , it is obvious to see that $CJ=\frac{\sqrt{30}}{2}$ , denote the distance $OJ=m,CD=DC=EF=x$ Since similarity between $\triangle{CJE},\triangle{EJO}, JE^2=CJ*OJ$ , we can now express Pythagorean theorem in $Rt\triangle{COE}, 4x^2+\frac{\sqrt{30}m}{2}+m^2=(\frac... | [
"<details><summary>kinda bashy</summary>Let the center of the circle be $O$ . Note that $OE$ perpendicularly bisects DF. Now AOBC is a kite, so $CO$ perpendicularly bisects $AB$ too. Let the midpoint of $AB$ be $M$ . Let the radius of the circle be $R$ , and let $CD=DE=EF=x$ . $\\Rightarrow OE^2 = R^... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1036,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791472.json"
} |
In triangle $ABC$ , let $I, O, H$ be the incenter, circumcenter and orthocenter, respectively. Suppose that $AI = 11$ and $AO = AH = 13$ . Find $OH$ .
*Proposed by Kevin You* | <blockquote>Nice problem. My solution is the same as **dchenmathcounts**'s but it has a different finish.
<details><summary>Sol</summary>$AH=AO$ means $2R\cos A=R$ so $\angle A=60^\circ$ . It's well-known that when $\angle A=60^\circ$ , $BHIOC$ is cyclic with center $M$ , where $M$ is the midpoint of minor ar... | [
"<details><summary>Good stuff</summary>We use complex coordinates. Let $O=0$ and note $H=a+b+c$ . Then $|O-a|=|H-a|$ or $|a|=|b+c|=13$ . Trig here gives $\\angle A = 60^{\\circ}$ . Now let $D$ be the arc midpoint of $BC$ (note this is equal to $|b+c|$ ) and note $AODH$ is a parallelogram. As $AD=AI+I... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 30,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791475.json"
} |
Let $\Gamma_1$ and $\Gamma_2$ be two circles with radii $r_1$ and $r_2,$ respectively, where $r_1>r_2.$ Suppose $\Gamma_1$ and $\Gamma_2$ intersect at two distinct points $A$ and $B.$ A point $C$ is selected on ray $\overrightarrow{AB},$ past $B,$ and the tangents to $\Gamma_1$ and $\Gamma_2$... | Note that since $C$ lies on the radical axis, we have $CP=CQ$ . Furthermore, due to the tangencies, $$ \angle PAB=\angle BPC,\angle QAB=\angle BQC. $$ Since $\angle BPC=\angle BQC$ as well since $\triangle CPQ$ is isosceles, we have that $AB$ is an angle bisector in $\triangle PAQ$ .
Thus, we remove all ... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 44,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791476.json"
} |
In acute $\triangle ABC,$ let $I$ denote the incenter and suppose that line $AI$ intersects segment $BC$ at a point $D.$ Given that $AI=3, ID=2,$ and $BI^2+CI^2=64,$ compute $BC^2.$ *Proposed by Kyle Lee* | One of my favorite computational geometry problems of all time.
<span style="color:blue">**First Solution, Trigonometry**</span> By the Law of Sines on $\triangle AIB$ , $$ 3 = AI = \frac{AB\sin \frac B2}{\sin\left(\frac A2+\frac B2\right)} = 4R\sin \frac B2 \sin \frac C2 $$ via the double angle formula. Similarl... | [
"<details><summary>Solution</summary>Let $AB = 3y$ and $AC = 3z.$ Then it follows that $BD = 2y$ and $CD = 2z$ . By Stewart's on $\\triangle{ABD}$ we have\n\\[ IB^2 = \\frac{(2y)^2\\cdot 3 + (3y)^2 \\cdot 2 - 5\\cdot 3 \\cdot 2}{5} = 6y^2 - 6\\]\nSimilarly $IC^2 = 6z^2-6,$ so\n\\[IB^2+IC^2 = 6y^2-6+6z^2 ... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1060,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791478.json"
} |
Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$ . Rays $\displaystyle \overrightarrow{OB}$ and $\displaystyle \overrightarrow{DC}$ intersect at $E$ , and rays $\displaystyle \overrightarrow{OC}$ and $\displaystyle \overrightarrow{AB}$ intersect at $F$ . Suppose that $AE = EC = CF = 4$ , and the... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791480.json"
} | |
A particle starts at $(0,0,0)$ in three-dimensional space. Each second, it randomly selects one of the eight lattice points a distance of $\sqrt{3}$ from its current location and moves to that point. What is the probability that, after two seconds, the particle is a distance of $2\sqrt{2}$ from its original loca... | <details><summary>Solution</summary>First, note that there are $$ 8^2=64 $$ possible paths that the particle takes in $2$ seconds.
The given condition of the particle traveling to a lattice point a distance of $\sqrt{3}$ from its current location is equivalent to the particle traveling along the vector $$ \l... | [
"<details><summary>Solution</summary>Note that the particle moves by $\\pm1$ unit in each of the $x,y,z$ coordinates each second. To be a distance of $2\\sqrt{2}$ from its original location after two seconds, it must be at $(\\pm2,\\pm2,0)$ or permutations. This requires it to travel the same direction twic... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1124,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791481.json"
} |
Starting with a $5 \times 5$ grid, choose a $4 \times 4$ square in it. Then, choose a $3 \times 3$ square in the $4 \times 4$ square, and a $2 \times 2$ square in the $3 \times 3$ square, and a $1 \times 1$ square in the $2 \times 2$ square. Assuming all squares chosen are made of unit squares inside th... | <details><summary>Solution</summary>Let us assume that the $1 \times 1$ square is in the bottom right corner of the $2 \times 2$ square and, by symmetry, we just need to multiply by $4$ in the end.
Now, if the $2 \times 2$ square is in the bottom right, bottom left, top left, and top right of the $3 \times 3$... | [
"<details><summary>Casework</summary>It doesn't matter which $4\\times4$ square you choose. \nWLOG let's assume you chose the bottom left $4\\times4$ square. Then you have $4$ cases.\n\n\n- Top right $3\\times3$ square\n- Top left $3\\times3$ square\n- Bottom right $3\\times3$ square\n- Bottom left $3\... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1118,
"boxed": true,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791483.json"
} |
We say that a set $S$ of $3$ unit squares is \textit{commutable} if $S = \{s_1,s_2,s_3\}$ for some $s_1,s_2,s_3$ where $s_2$ shares a side with each of $s_1,s_3$ . How many ways are there to partition a $3\times 3$ grid of unit squares into $3$ pairwise disjoint commutable sets?
*Proposed by Srinivasan ... | <details><summary>Solution</summary>Let us define an L as a commutable set in which $s_1$ and $s_3$ share sides with $s_2$ that are *adjacent* and let an I be a commutable set in which $s_1$ and $s_3$ share sides with $s_2$ that are [/i]opposite[/i]. (Note that the letters L and I resemble the shapes of the... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1142,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791485.json"
} |
Dilhan is running around a track for $12$ laps. If halfway through a lap, Dilhan has his phone on him, he has a $\frac{1}{3}$ chance to drop it there. If Dilhan runs past his phone on the ground, he will attempt to pick it up with a $\frac{2}{3}$ chance of success, and won't drop it for the rest of the lap. He st... | <details><summary>Solution</summary>If, before running the $12$ th lap, Dilhan has his phone, then there is a $\tfrac{2}{3}$ chance that he will still have it by the end of the $12$ th lap.
On the other hand, if Dilhan *doesn't* have his phone before running the $12$ th lap, then there is still a $\tfrac{2}{3}$ ... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1116,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791486.json"
} |
Daniel, Ethan, and Zack are playing a multi-round game of Tetris. Whoever wins $11$ rounds first is crowned the champion. However Zack is trying to pull off a "reverse-sweep", where (at-least) one of the other two players first hits $10$ wins while Zack is still at $0$ , but Zack still ends up being the first to r... | <details><summary>Solution</summary>There are $2$ ways to choose which person hits $10$ while Zack is still at $0$ (both other players can do so but we will correct for overcount later), call them Player $A.$ Then, we can view the problem as arranging $21$ $X$ 's and $10$ $Y$ 's if we use an $X$ to repr... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1154,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791487.json"
} |
A sequence of pairwise distinct positive integers is called averaging if each term after the first two is the average of the previous two terms. Let $M$ be the maximum possible number of terms in an averaging sequence in which every term is less than or equal to $2022$ and let $N$ be the number of such distinct s... | <details><summary>Solution</summary>Note that the length of an averaging sequence with first two terms $a$ and $b$ for $a, b \in \mathbb{Z}_{\ge1}$ is $2+\nu_2|a-b|$ because after the first two terms, the positive difference between consecutive terms of the sequence is cut in half every time, which can happen a... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1150,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791489.json"
} |
For a family gathering, $8$ people order one dish each. The family sits around a circular table. Find the number of ways to place the dishes so that each person’s dish is either to the left, right, or directly in front of them.
*Proposed by Nicole Sim* | <details><summary>Solution</summary>We proceed with casework on the number of people who have their own dish. This number clearly must be even (because the remaining people, if any, have to split into pairs of adjacent people who have each other's dishes).**0 people have their own dish:** Then, either the family's dish... | [
"oops casework",
"<details><summary>Standard Casework</summary>There are many cases so let's break it down. We first have $2$ easy cases where everyone rotates right or everyone rotates left. \nThen, we need to count the number of *swaps* that occur which is when $2$ adjacent people will swap their dishes. A ... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1134,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791490.json"
} |
The CMU Kiltie Band is attempting to crash a helicopter via grappling hook. The helicopter starts parallel (angle $0$ degrees) to the ground. Each time the band members pull the hook, they tilt the helicopter forward by either $x$ or $x+1$ degrees, with equal probability, if the helicopter is currently at an an... | <details><summary>Solution</summary>First, note that the expected number of times the band must pull the hook in order to tilt it to a $1$ degree angle is $2$ because $0$ can go either to itself or to $1,$ each with $\tfrac{1}{2}$ probability.
Then, it is not difficult to see that each of the numbers $32, 3... | [
"<details><summary>Solution</summary>Let $E_n$ be the expected number of pulls if it is currently $n$ degrees. We have $$ E_n = \\tfrac{1}{2}E_{2n} + \\tfrac{1}{2}E_{2n+1}+1. $$ Now, we have $0=E_{90} = E_{91}=\\cdots$ so $1= E_{45} = E_{46} = \\cdots = E_{89}$ . Now, we can compute $E_{44} = 3/2$ . Con... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1126,
"boxed": true,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791492.json"
} |
At CMIMC headquarters, there is a row of $n$ lightbulbs, each of which is connected to a light switch. Daniel the electrician knows that exactly one of the switches doesn't work, and needs to find out which one. Every second, he can do exactly one of 3 things:
- Flip a switch, changing the lightbulb from of... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791493.json"
} | |
Barry has a standard die containing the numbers 1-6 on its faces.
He rolls the die continuously, keeping track of the sum of the numbers he has rolled so far, starting from 0. Let $E_n$ be the expected number of time he needs to until his recorded sum is at least $n$ .
It turns out that there exist positiv... | [
"picture this u are a professor in descriptive set theory and giving a talk at your local math cl.ub. You rae talking about unfriendly colorings of graphs and then all of a sudden, you see him there. Barack Obama. You say: holy cow, i never expected barack obama to come to my talk! And then barack obama says given ... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791496.json"
} | |
In a class of $12$ students, no two people are the same height. Compute the total number of ways for the students to arrange themselves in a line such that:
- for all $1 < i < 12$ , the person in the $i$ -th position (with the leftmost position being $1$ ) is taller than exactly $i\pmod 3$ of their adj... | <details><summary>Solution</summary>First, note that from the first condition, the heights of the people in line have to follow this pattern: $$ \underline{\phantom{n}} < \underline{\phantom{n}} > \underline{\phantom{n}} < \underline{\phantom{n}} < \underline{\phantom{n}} > \underline{\phantom{n}} < \underline{\phant... | [] | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 1072,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791497.json"
} |
Daniel has a (mostly) standard deck of 54 cards, consisting of 4 suits each containing the ranks 1 to 13 as well as 2 jokers. Daniel plays the following game: He shuffles the deck uniformly randomly and then takes all of the cards that end up strictly between the two jokers. He then sums up the ranks of all the cards h... | <details><summary>solution</summary>The key is to build a generating function with two formal variables: $x$ . whose exponent represents the number of cards strictly between the two jokers, and $y$ . whose exponent represents the sum of the ranks of the cards strictly between the jokers.
To begin, note that if there... | [
"Solved this as a 1st grader on the CMIMC.\n\n<details><summary>Combinatorial Approach</summary>Let the number of cards between the jokers be $0\\leq d\\leq 52$ . Call a configuration good if the score is a multiple of 13.\n\nFor $d \\neq 0,13,26,39,52$ , we will use the cycle approach. Consider one arrangement o... | [
"origin:aops",
"2022 CMIMC",
"2022 Contests"
] | {
"answer_score": 98,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 CMIMC/2791498.json"
} |
Find all integer values of $x$ for which the value of the expression
\[x^2+6x+33\]
is a perfect square. | $(x+3)^2+24 =m^2$ $(m-x-3)(m+x+3) = 24$ $x = (-8,-4,-2,2)$ | [
"StarLex1, the question asks for all <u>integer</u> solutions. You missed the negative solutions. (Of course your method also yields those as well.)",
"<blockquote>Find all integer values of $x$ for which the value of the expression \n\\[x^2+6x+33\\] \nis a perfect square.</blockquote>\nToo easy for HSO\nShould... | [
"origin:aops",
"2022 Contests",
"2022 Cyprus JBMO TST"
] | {
"answer_score": 6,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2785510.json"
} |
Let $ABCD$ be a square. Let $E, Z$ be points on the sides $AB, CD$ of the square respectively, such that $DE\parallel BZ$ . Assume that the triangles $\triangle EAD, \triangle ZCB$ and the parallelogram $BEDZ$ have the same area.
If the distance between the parallel lines $DE$ and $BZ$ is equal to $1$... | Area of paralelogram $BEDZ$ is equal to $DE*1=DE$ . On the other hand same area is equal to $EB*AB$ . Area of whole square is $AB^2$ and Is also equal to sum of area of triangles $EAD, ZCB$ and parallelogram $BEDZ$ so we have $AB^2=3AB*EB$ . Since $DE=EB*AB$ we have $DE=\frac{AB^2}{3}$ . From triangle $E... | [] | [
"origin:aops",
"2022 Contests",
"2022 Cyprus JBMO TST"
] | {
"answer_score": 26,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2785511.json"
} |
If $x,y$ are real numbers with $x+y\geqslant 0$ , determine the minimum value of the expression
\[K=x^5+y^5-x^4y-xy^4+x^2+4x+7\]
For which values of $x,y$ does $K$ take its minimum value?
| $K=(x+y)((x-y)^2(x^2+y^2))+(x+2)^2+3\ge 0+0+3=3$ | [
"[url]https://artofproblemsolving.com/community/c6h2776804p24363062"
] | [
"origin:aops",
"2022 Contests",
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] | {
"answer_score": 2,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2785512.json"
} |
Consider the digits $1, 2, 3, 4, 5, 6, 7$ .
(a) Determine the number of seven-digit numbers with distinct digits that can be constructed using the digits above.
(b) If we place all of these seven-digit numbers in increasing order, find the seven-digit number which appears in the $2022^{\text{th}}$ position.
| [
"a) 7!\nb) 1_ _ _ _ _ _ = 720\n 2 _ _ _ _ _ _ = 720 = 1440\n 3 1 _ _ _ _ _ = 120 \n 3 2 _ _ _ _ _ = 120\n 3 4 _ _ _ _ _ = 120\n 3 5 _ _ _ _ _ = 120 = 480\n 3 6 1 _ _ _ _ = 24\n 3 6 2 _ _ _ _ = 24\n 3 6 4 _ _ _ _ = 24\n 3 6 5 _ _ _ _ = 24\n 3 6 7 1 _ _ _ = 6\n3671542\n\n \n\... | [
"origin:aops",
"2022 Contests",
"2022 Cyprus JBMO TST"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2785513.json"
} | |
Determine all real numbers $x\in\mathbb{R}$ for which
\[
\left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x}{3} \right\rfloor=x-2022.
\]
The notation $\lfloor z \rfloor$ , for $z\in\mathbb{R}$ , denotes the largest integer which is less than or equal to $z$ . For example:
\[\lfloor 3.98 \rfloor =3 \qua... | <blockquote>Determine all real numbers $x\in\mathbb{R}$ for which
\[
\left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{x}{3} \right\rfloor=x-2022
\]
</blockquote> $LHS\in\mathbb Z$ and so $RHS\in\mathbb Z$ and so $x\in\mathbb Z$ If $x=6n$ , equation is $3n+2n=6n-2022$ and so $n=2022$ and $x=12132$... | [] | [
"origin:aops",
"2022 Contests",
"2022 Cyprus JBMO TST"
] | {
"answer_score": 1058,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2785525.json"
} |
Determine all pairs of prime numbers $(p, q)$ which satisfy the equation
\[
p^3+q^3+1=p^2q^2
\]
| We claim that $\boxed {(p,q)=(3,2)}$ is the only solution.
WLOG $p \ge q$ ,
If $p=q$ $2p^3+1=p^4 \Rightarrow p^3(p-2)=1$ which is not possible.
Hence, $p >q$ Now, $q^3+1 \equiv 0 \pmod{p} \Rightarrow q^6 \equiv 1 \pmod{p}$ So, $\operatorname{ord}_p(q)=1,2,3,6$ **<span style="color:#f00">Case 1:-</span>** $\ope... | [
"Without loss of generality assume that $p \\geq q$ . If $p=q$ then $2p^3+1=p^4$ , but $1=p^3(p-2) \\geq p^3>1,$ absurd.\n\nNow let $p>q$ . If $q=3$ , then $p^3+28=9p^2$ , i.e. $(p-2)(p^2-7p-14)=0$ , so $p=2$ , absurd. If now $q \\neq 3$ , then taking $\\pmod p^2$ we obtain $p^2 \\mid (q^3+1)=(q+1)(q^2-... | [
"origin:aops",
"2022 Contests",
"2022 Cyprus JBMO TST"
] | {
"answer_score": 1062,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2785526.json"
} |
Let $ABC$ be an acute-angled triangle, and let $D, E$ and $K$ be the midpoints of its sides $AB, AC$ and $BC$ respectively. Let $O$ be the circumcentre of triangle $ABC$ , and let $M$ be the foot of the perpendicular from $A$ on the line $BC$ . From the midpoint $P$ of $OM$ we draw a line parallel... | Statement is much scarier than the problem itself.
a) Just note that $DMEK$ is isosceles trapezoid, since $D,M,E,K$ are concyclic and $\overline{MK}\parallel \overline{DE}$ . Thus, perpendicular bisectors of $\overline{DE},\overline{MK}$ coincide. Therefore, $Z$ is the center of $(ADEO)$ as $D$ lies on $... | [
"Since $ZP \\parallel AM$ and $OP=PM$ $\\implies AZ=ZO$ . And $AD=DB$ , $AE=EC$ $\\to DZ\\parallel OB$ and $ZE\\parallel OC$ . From Thales, we have $2DZ=OB , 2ZE=OC$ $\\to DZ=ZE$ .\n And $$ E_{DZE}=\\frac{DE\\cdot TZ}{2}=\\frac{\\tfrac{BC}{2}\\cdot \\tfrac{OK}{2}}{2}=\\frac{BC\\cdot OK}{2}. $$ and... | [
"origin:aops",
"2022 Contests",
"2022 Cyprus JBMO TST"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2785531.json"
} |
Let $A$ be a subset of $\{1, 2, 3, \ldots, 50\}$ with the property: for every $x,y\in A$ with $x\neq y$ , it holds that
\[\left| \frac{1}{x}- \frac{1}{y}\right|>\frac{1}{1000}.\]
Determine the largest possible number of elements that the set $A$ can have. | We begin with the following Claim:**Claim:** If $x,x+1 \in A$ , then $x \leq 31$ . Furthermore, if $x,x+2 \in A$ , then $x \leq 43$ .
*Proof:* Easy to see from the given hypothesis $\blacksquare$ Therefore, by partitioning the elements of the set $X=\{1, 2, 3, \ldots, 50\}$ in 10 groups as follows,**Group 1:** ... | [] | [
"origin:aops",
"2022 Contests",
"2022 Cyprus JBMO TST"
] | {
"answer_score": 150,
"boxed": false,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2785535.json"
} |
Prove that for every natural number $k$ , at least one of the integers
\[ 2k-1, \quad 5k-1 \quad \text{and} \quad 13k-1\]
is not a perfect square.
| <blockquote>With process of elimination, if $k$ is $1$ , then $2k-1$ would be a perfect square.
If $k$ is $13$ , then $5k-1$ can't be a perfect square.
Looking closely at the wording *at least one* of the integers can't be a perfect square for any natural number $k$ .
So, our answer is $13k-1$ </blockquot... | [
"If $k$ is even, then $2k-1$ cannot be a perfect square. If $k\\equiv 3\\pmod{4}$ , $k\\equiv 5\\pmod{16}$ , or $k\\equiv 9\\pmod{16}$ , then $5k-1$ cannot be a perfect square. If $k\\equiv 1\\pmod{16}$ or $k\\equiv 13\\pmod{16}$ , then $13k-1$ cannot be a perfect square. As we have exhausted all resi... | [
"origin:aops",
"2022 Contests",
"2022 Cyprus JBMO TST"
] | {
"answer_score": 46,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2810356.json"
} |
In a triangle $ABC$ with $\widehat{A}=80^{\circ}$ and $\widehat{B}=60^{\circ}$ , the internal angle bisector of $\widehat{C}$ meets the side $AB$ at the point $D$ . The parallel from $D$ to the side $AC$ , meets the side $BC$ at the point $E$ .
Find the measure of the angle $\angle EAB$ . | [
"https://artofproblemsolving.com/community/c6h2828368p25017716 Romania JBMO tst"
] | [
"origin:aops",
"2022 Contests",
"2022 Cyprus JBMO TST"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2810357.json"
} | |
If $a,b,c$ are positive real numbers with $abc=1$ , prove that
(a) \[2\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\right) \geqslant \frac{9}{ab+bc+ca}\]
(b)\[2\left(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\right) \geqslant \frac{9}{a^2 b+b^2 c+c^2 a}\] | (a) Using $abc=1$ , and AM-HM inequality,
\[
\sum \frac{ab}{a+b} = \sum \frac{1}{c(a+b)} \ge \frac{9}{2(ab+bc+ca)}.
\]
(b) It suffices to show $a^2b + b^2c + c^2 a\ge ab+bc+ca$ . To that end, let $a=x/y$ , $b=y/z$ and $c=z/x$ for some $x,y,z>0$ . After some algebra, it boils down verifying $x^3+y^3+z^3 \ge x^2... | [
"As in Bulgaria JBMO 2022 TST. In a), write $ab = 1/c$ etc. and finish by Titu's lemma. In b) we wish to show $a^2b + b^2c + c^2a \\geq ab + bc + ca$ which follows by the AM-GMs $a^2b + c^2a + c^2a \\geq 3\\sqrt[3]{a^4bc^4} = 3ac$ ."
] | [
"origin:aops",
"2022 Contests",
"2022 Cyprus JBMO TST"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2810360.json"
} |
The numbers $1, 2, 3, \ldots , 10$ are written on the blackboard. In each step, Andrew chooses two numbers $a, b$ which are written on the blackboard such that $a\geqslant 2b$ , he erases them, and in their place writes the number $a-2b$ .
Find all numbers $n$ , such that after a sequence of steps as above, at ... | Can the remaining number be present multiple times? If not, then note that the sum of numbers is preserved mod $3$ and so the remaining
number can be at most $1$ , $4$ , $7$ , $10$ (and there are examples for all of these, I guess). | [] | [
"origin:aops",
"2022 Contests",
"2022 Cyprus JBMO TST"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Cyprus JBMO TST/2810362.json"
} |
Find all pairs of integers $(m, n)$ which satisfy the equation
\[(2n^2+5m-5n-mn)^2=m^3n\] | My try $(2n^2+5m-5n-mn)^2=m^3n\Longrightarrow$ either $m,n\le0$ , or $m,n\ge0$ 1) Case with $m=n$ Replacing $m$ with $n\Longrightarrow n^4=n^4\Rightarrow$ infinite solutions $(n,m)=(t,t), (-t,-t)$ 2) Case with $m\neq n$ The LHS is a perfect square, $m^3n$ should also be a square. This occurs if $mn$ i... | [
"<blockquote>(WLOG $m>n),\\Longrightarrow m=nk^2\\Rightarrow mn=n^2k^2$ Subst. $m=nk^2$ in the title equation</blockquote>\n1) you can not say \"WLOG $m>n$ \" since there is no symetrie\n2) $mn$ perfect square does not imply $m=nk^2$ (look for example at $(m,n)=(18,8)$ )\n\n",
" $(2n^2+5m-5n-mn)^2=m^3n$... | [
"origin:aops",
"2022 Contests",
"2022 Cyprus TST"
] | {
"answer_score": 64,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Cyprus TST/2785624.json"
} |
Determine for how many positive integers $n\in\{1, 2, \ldots, 2022\}$ it holds that $402$ divides at least one of
\[n^2-1, n^3-1, n^4-1\] | I claim the answer is $31$ Note that $402 = 2\cdot 3\cdot 67$ . From here, we immediately obtain that $n\equiv 1\pmod{2}$ . Next, assume $402\mid n^2-1$ . Then, $n\in\{1,2\}\pmod{3}$ and $n\in\{\pm 1\}\pmod{67}$ . There are four such residue classes in modulo $402$ . Likewise, if $402\mid n^4-1=(n^2-1)(n^2+1)$... | [
"@above you can't have $n \\equiv 2 \\pmod{3}$ and $n \\equiv 29, 37 \\pmod{67}$ , so there are $6$ residue classes.",
"No, you certainly can: if $n\\equiv 2\\pmod{3}$ , then $3\\mid n^2-1$ . Likewise, if $n\\in\\{29,37\\}$ then $67\\mid n^2+n+1\\implies 67\\mid n^3-1$ .**Edit.** I was wrong, thx to pos... | [
"origin:aops",
"2022 Contests",
"2022 Cyprus TST"
] | {
"answer_score": 52,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Cyprus TST/2785626.json"
} |
Let $ABC$ be an obtuse-angled triangle with $ \angle ABC>90^{\circ}$ , and let $(c)$ be its circumcircle. The internal angle bisector of $\angle BAC$ meets again the circle $(c)$ at the point $E$ , and the line $BC$ at the point $D$ . The circle of diameter $DE$ meets the circle $(c)$ at the point $H$... | Another solution**a)** Let $\angle DAC = a , \angle DCA= b$ then $\angle ADB = a+b = \angle ADK$ also $\angle DCE=a \implies
\angle ACE=a+b \implies \angle AHK=a+b$ Hence $\angle AHK= \angle ADK \implies ADHK$ cyclic.**b)**Let $M$ be midpoint $BC$ .Obviously $EB=EC \implies EM \bot BC \implies M$ lies on ... | [
"[https://mathematica.gr/forum/viewtopic.php?f=58&t=71185#p345811](https://mathematica.gr/forum/viewtopic.php?f=58&t=71185#p345811)\n",
"For (a), just $\\angle KHA = \\angle ACE = \\angle ACB + \\angle EAC = \\angle KDA$ . Let $O$ be the circumcenter of this circle.\n\nThen $O$ is the midpoint of $KD$ (and... | [
"origin:aops",
"2022 Contests",
"2022 Cyprus TST"
] | {
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Cyprus TST/2785629.json"
} |
Let
\[M=\{1, 2, 3, \ldots, 2022\}\]
Determine the least positive integer $k$ , such that for every $k$ subsets of $M$ with the cardinality of each subset equal to $3$ , there are two of these subsets with exactly one common element. | Say that a set of triples, $V$ , is "good" iff each pair of triples $u,v\in V$ is disjoint or has two elements in common. Construct a graph $\mathcal{G}$ on good $V$ by drawing an edge $u\sim v$ between each $u,v\in V$ with $|u\cap v|=2$ . The relation $\sim$ is obviously reflexive and, in a good graph,... | [
"Hello @above,\nThe subsets {1,2,3},{1,2,4},…,{1,2,2022} shows that $k \\geq 2021$ .\nAlso could you please explain what you mean by efficiency?\nBest regards ",
"Call the \"repeated\" number set of a number $n \\in M$ the set of duplicates of n in the k elements (Causing n to be counted more than once).\nFor ... | [
"origin:aops",
"2022 Contests",
"2022 Cyprus TST"
] | {
"answer_score": 1072,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Cyprus TST/2785630.json"
} |
Find all pairs of real numbers $(x,y)$ for which
\[
\begin{aligned}
x^2+y^2+xy&=133
x+y+\sqrt{xy}&=19
\end{aligned}
\] | Squaring and simplifying, we get $(x+y+ \sqrt{xy})^2=133+38 \sqrt{xy}$ . So we obtain $\sqrt{xy}=6 \Rightarrow xy=36$ . Thus, $x+y=19- \sqrt{xy}=13$ . So $x,y$ are roots of $u^2-13u+36=(u-4)(u=9)=0$ by Vieta. So $\boxed{(x,y)=(4,9),(9,4)}$ . | [
" $x^2+y^2+xy=(x+y+\\sqrt{xy})(x+y-\\sqrt{xy})\\implies \\frac{133}{19}=7=x+y-\\sqrt{xy}$ $(x+y+\\sqrt{xy})+(x+y-\\sqrt{xy})=19+7\\implies x+y=13\\implies \\sqrt{xy}=6\\implies xy=36$ By Vieta’s formula, $x,y$ are roots of $$ a^2-13a+36=0\\implies (a-4)(a-9)=0\\implies a=4,9\\implies \\boxed{(x,y)=(4,9),(9,4)}... | [
"origin:aops",
"2022 Contests",
"2022 Cyprus TST"
] | {
"answer_score": 1012,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Cyprus TST/2787231.json"
} |
Let $n, m$ be positive integers such that
\[n(4n+1)=m(5m+1)\]
(a) Show that the difference $n-m$ is a perfect square of a positive integer.
(b) Find a pair of positive integers $(n, m)$ which satisfies the above relation.
Additional part (not asked in the TST): Find all such pairs $(n,m)$ . | a) Clearly, we have $n>m$ . We get $4n^2+n=5m^2+m$ .
Factorising, we obtain $(n-m)(4n+4m+1)=m^2$ and $(n-m)(5m+5n+1)=n^2$ .
Letting $gcd(m,n)=d \Rightarrow gcd(m^2,n^2)=d^2$ we obtain $\gcd(m^2,n^2)=(n-m)*\gcd(4n+4m+1,5m+5n+1)=(n-m)*\gcd(4m+4n+1,m+n)=(n-m)=d^2$ . Hence proved.
b) Let $n=dx, m=dy$ . Thus $n-... | [
" $(a)\\quad n(4n+1)=m(5m+1)\\implies (n-m)(4m+4n+1)=m^2, (n-m)(5m+5n+1)=n^2$ $\\gcd(m^2,n^2)=(n-m)\\gcd(4m+4n+1,5m+5n+1)=n-m\\implies \\gcd(m,n)^2=n-m$ Thus, $n-m$ is perfect square.",
"(b) $(n,m)=(38,34)$ ",
"Clearly $m>n$ . Let $d={\\rm gcd}(m,n)$ with $m=dm_1$ and $n=dn_1$ . Then,\n\\[\n4dn_1^2 + ... | [
"origin:aops",
"2022 Contests",
"2022 Cyprus TST"
] | {
"answer_score": 1134,
"boxed": true,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Cyprus TST/2787233.json"
} |
Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$ and let $(c)$ be its circumcircle with center $O$ . Let $M$ be the midpoint of $BC$ . The line $AM$ meets the circle $(c)$ again at the point $D$ . The circumcircle $(c_1)$ of triangle $\triangle MDC$ intersects the line $AC$ at the point... | <blockquote>Could you tell how to verify that BZ is perpendicular to AC，thanks so much</blockquote>
Since $\angle CZM=\angle AIM=\angle BDM=\angle BCA$ we have $ZM=MC$ but $BN \perp NC, BM=MC$ , so we have $MN=BM=MC=MZ$ $\therefore B,N,Z,C$ concyclic.
btw: There is something different between the posts by @Deme... | [
"Easy to verify $BZ\\bot AC$ , which implies $N,Z,C,B$ in a circle.",
"\nCould you tell how to verify that BZ is perpendicular to AC，thanks so much\n",
"No problem ,I can learn it ,thanks"
] | [
"origin:aops",
"2022 Contests",
"2022 Cyprus TST"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Cyprus TST/2787241.json"
} |
Let $m, n$ be positive integers with $m<n$ and consider an $n\times n$ board from which its upper left $ m\times m$ part has been removed. An example of such board for $n=5$ and $m=2$ is shown below.
Determine for which pairs $(m, n)$ this board can be tiled with $3\times 1$ tiles. Each tile can be pos... | <details><summary>Hint: Necessary condition:</summary>$3 \mid n^2-m^2$ then $3 \mid (n-m)(n+m)$</details>
<details><summary>Hint: Extension</summary>If $(n,m)$ is possible so is $(n+3,m)$ and so is $(n+3,m+3)$</details> | [] | [
"origin:aops",
"2022 Contests",
"2022 Cyprus TST"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Cyprus TST/2787243.json"
} |
Charles has some marbles. Their colors are either red, green, or blue. The total number of red and green marbles is $38\%$ more than that of blue marbles. The total number of green and blue marbles is $150\%$ more than that of red marbles. If the total number of blue and red marbles is more than that of green marbl... | <details><summary>Solution</summary>Let $r$ denote red marbles, $g$ denote green marbles, and $b$ denote blue marbles. So, $$ r+g=\frac{69}{50}b $$ $$ g+b=\frac{5}{2}r $$ Subtracting, we get $$ r-b=\frac{69}{50}b-\frac{5}{2}r \implies r=\frac{17}{25}b $$ Plugging into the first equation, $$ \frac{34}{50}... | [
"First! $\\;$ ",
"lol\n\nthis problem is rlly good for a #1 btw",
"I agree! Good job, **stayhomedomath**!",
"<blockquote>I agree! Good job, **stayhomedomath**!</blockquote>\n\nwhy did you bold the name like five times",
"i don't know **pog**",
"<blockquote>i don't know **pog**</blockquote>\n\nVery cool *... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1034,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2022 Contests/2022 DIME/2673235.json"
} |
An up-right path from lattice points $P$ and $Q$ on the $xy$ -plane is a path in which every move is either one unit right or one unit up. The probability that a randomly chosen up-right path from $(0,0)$ to $(10,3)$ does not intersect the graph of $y=x^2+0.5$ can be written as $\tfrac mn$ , where $m$ and... | Although the probability that some path is chosen is the same, the number of ways to get to that path is different, thus altering the final probability.
Trivially there are $\binom{13}{3}$ ways to get to $(10,3)$ . Now, the final “correct” grid is an $8 \times 4$ grid connected by $3$ points on the lower left c... | [
"i'm getting 210/286 = 105/143 => 248 from block walking and also its 2*10c2+10c3=210??",
"…I think the author messed up. Yeah the answer should be 248 I'm pretty sure. I will correct it when I get back home.",
"k thanks :)",
"Actually, each path is not equally likely. Try testing on a 3x3 grid :)\n\nThe answ... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1026,
"boxed": false,
"end_of_proof": false,
"n_reply": 30,
"path": "Contest Collections/2022 Contests/2022 DIME/2673236.json"
} |
Given a regular hexagon $ABCDEF$ , let point $P$ be the intersection of lines $BC$ and $DE$ , and let point $Q$ be the intersection of lines $AP$ and $CD$ . If the area of $\triangle QEP$ is equal to $72$ , find the area of regular hexagon $ABCDEF$ .
*Proposed by **DeToasty3*** | Suppose $s$ is the side length of the hexagon. Assign coordinates to $P$ as the origin: we get two equations: $$ y=x \sqrt{3} + s \sqrt{3} $$ $$ y=-\sqrt{3}{2}s $$ Solving yields the coordinates of $Q$ is $(\tfrac{-2}{3} s, \tfrac{s \sqrt{3}}{3})$ , so the area of $\triangle QEP$ is $\frac{\sqrt{3}}{3}... | [
"how about u catch this ratio",
"coordbash gives 72*9/2=324",
"<blockquote>coordbash gives 72*9/2=324</blockquote>\n\nsimilar triangles + triangle median $\\Longrightarrow 72\\cdot\\frac{1}{2}\\cdot\\frac{3}{2}\\cdot6=324$ ",
":|\n\ni thought that $72\\cdot\\frac{1}{2}\\cdot\\frac{3}{2}\\cdot6 = 432$ ",
... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1024,
"boxed": true,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2022 Contests/2022 DIME/2673237.json"
} |
The four-digit base ten number $\underline{a}\;\underline{b}\;\underline{c}\;\underline{d}$ has all nonzero digits and is a multiple of $99$ . Additionally, the two-digit base ten number $\underline{a}\;\underline{b}$ is a divisor of $150$ , and the two-digit base ten number $\underline{c}\;\underline{d}$ is a ... | I'm gonna use $ab$ and $cd$ cus im too lazy. We can represent $abcd \equiv 100ab + cd \equiv ab + cd \pmod{99}$ . Then we have that the two digit factors of $150$ are: $\{10, 15, 30, 50, 75\}$ and the factors of $168$ are: $\{12, 14, 24, 28, 42, 56, 84\}$ We want these two added up to be divisible by $99$ , a... | [
"7524, 1584",
"just bash to get that the two numbers are 7524, 1584",
"wait i just realized the title is misleading this problem isn't about bases it's about digits\n\nok changed :P",
"Bash for two-digit divisors of $168$ and $150$ , bash the condition for multiple of $11$ , and then the multiple of $9$ ... | [
"origin:aops",
"2022 Contests",
"2022 DIME"
] | {
"answer_score": 1032,
"boxed": true,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 DIME/2673238.json"
} |
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