problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
Given an acute triangle $ABC$ . with $H$ as its orthocenter, lines $\ell_1$ and $\ell_2$ go through $H$ and are perpendicular to each other. Line $\ell_1$ cuts $BC$ and the extension of $AB$ on $D$ and $Z$ respectively. Whereas line $\ell_2$ cuts $BC$ and the extension of $AC$ on $E$ and $X$... | [
"Same problem as [Iranian Geometry Olympiad 2015 Advanced P3](https://artofproblemsolving.com/community/c6h1137975p5325937)"
] | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2735507.json"
} | |
Prove that there exists a set $X \subseteq \mathbb{N}$ which contains exactly 2022 elements such that for every distinct $a, b, c \in X$ the following equality:
\[ \gcd(a^n+b^n, c) = 1 \] is satisfied for every positive integer $n$ . | Let $a_1 = 21$ and inductively add new elements by the following procedure, suppose we have until $a_k$ . We're going to always have $a_i$ to have two new $3 \pmod 4$ primes dividing it. So let the primes so far be $p_1, p_2, \cdots, p_{2k}$ . Construct two new primes $p,q$ such that both satisfy $\left(\fra... | [
"<details><summary>Solution</summary><details><summary>FLT On Steroids</summary>Let $p_1, p_2, \\cdots, p_{2022}$ be the smallest odd primes and let \\[ N = \\prod_{i=1}^{2022} (p_i - 1). \\] Then, the set $X = \\{ p_1^N, p_2^N, \\ldots, p_{2022}^N\\}$ works. Since $p_i - 1 \\vert N$ for every such $i$ , by ... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 138,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2735508.json"
} |
Let $a, b, c$ be positive real numbers such that $abc = 1$ . Prove that $$ (a + b + c)(ab + bc + ca) + 3\ge 4(a + b + c). $$ | <blockquote>Let $a, b, c$ be positive real numbers such that $abc = 1$ . Prove that $$ (a + b + c)(ab + bc + ca) + 3\ge 4(a + b + c). $$ </blockquote>
pqr method: $p = a + b + c, q = ab + bc + ca, r = abc = 1$ .
We need to prove that $pq + 3 \ge 4p$ .
Using $q^2\ge 3pr$ , we have $q\ge \sqrt{3pr} = \sqrt{3p... | [
"deleted ....",
"<blockquote>**Rewrite as** $(a+b+c)(ab+bc+ca-4)>=-3$ AM-GM $(a+b+c)>=3$ AM-HM $\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}>=3$ $(ab+bc+ca)>=3$ </blockquote>\nPlease, stop writing nonsense. You can not use $a+b+c\\ge 3$ when $ab+bc+ca-4$ is negative. \n",
"<blockquote><blockquote>**Rewrite as** ... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2742895.json"
} |
Distinct pebbles are placed on a $1001 \times 1001$ board consisting of $1001^2$ unit tiles, such that every unit tile consists of at most one pebble. The *pebble set* of a unit tile is the set of all pebbles situated in the same row or column with said unit tile. Determine the minimum amount of pebbles that must b... | [
"silimar with [2021Benelux](https://artofproblemsolving.com/community/c6h2549153p21775820)"
] | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2743542.json"
} | |
Given that $ABC$ is a triangle, points $A_i, B_i, C_i \hspace{0.15cm} (i \in \{1,2,3\})$ and $O_A, O_B, O_C$ satisfy the following criteria:
a) $ABB_1A_2, BCC_1B_2, CAA_1C_2$ are rectangles not containing any interior points of the triangle $ABC$ ,
b) $\displaystyle \frac{AB}{BB_1} = \frac{BC... | The number of participants that got this one correct during the test is lower than expected...
I think this is because of the format of the test, which consists of all the four topics ACGN, and is not ordered by difficulty.
Anyway here are some solutions.
<details><summary>Solution 1 (Finding the concurrency point ex... | [
"Let $M_A,M_B,M_C$ be the midpoints of $BC,CA,AB$ respectively, let $G$ be the centroid of $ABC$ , let $K$ be the symmedian point of $ABC$ , and $K_A,K_B,K_C$ its projections on $BC,CA,AB$ .\nSince the barycentric coordinates of $K$ are $(a^2:b^2:c^2)$ it follows that $|\\vec{KK_A}|:BC=|\\vec{KK_B... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 292,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2743549.json"
} |
Let $n$ be a natural number, with the prime factorisation
\[ n = p_1^{e_1} p_2^{e_2} \cdots p_r^{e_r} \] where $p_1, \ldots, p_r$ are distinct primes, and $e_i$ is a natural number. Define
\[ rad(n) = p_1p_2 \cdots p_r \] to be the product of all distinct prime factors of $n$ . Determine all polynomials ... | Since $0<rad (n) \leq n$ for all $n\in \mathbb{Z}$ , $0<P(n)\leq n$ for infinitely many $n$ . This leads to $\lim_{n\to \infty}\frac{P(n)}{n} \leq 1$ and hence $\deg P(x)\leq 1$ .**Case 1:** $P(x)\equiv c \in \mathbb{Q}$ , it's easy to see that $c$ is square-free is all we need.**Case 2:** $\deg P(x)=1$ , ... | [
"When you have zero effort in making test problems. \n<blockquote>\nLet $n$ be a positive integer, with prime factorization $$ n = p_1^{e_1}p_2^{e_2} \\cdots p_r^{e_r} $$ for distinct primes $p_1, \\ldots, p_r$ and $e_i$ positive integers. Define $$ rad(n) = p_1p_2\\cdots p_r, $$ the product of all distin... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 70,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2743550.json"
} |
Let $a$ and $b$ be two positive reals such that the following inequality
\[ ax^3 + by^2 \geq xy - 1 \] is satisfied for any positive reals $x, y \geq 1$ . Determine the smallest possible value of $a^2 + b$ .
*Proposed by Fajar Yuliawan* | <blockquote> Let $a$ and $b$ be two positive reals such that the inequality
\[ ax^3 + by^2 \geq xy - 1 \] holds for any positive reals $x, y \geq 1$ . Determine the smallest possible value of $a^2 + b$ .
</blockquote>
---------------------
The answer is $\boxed{\frac{2}{3\sqrt{3}}}$ , which is achieved by $(a,b... | [
"Here's an elementary solution using calculus.\n\nDefine $f(x,y)=ax^3+by^2-xy+1$ . Taking partial derivatives, this is stationary at $3ax^2-y=0$ and $2by-x=0$ , that is at $f(0,0)=1$ and $f\\left(\\tfrac{1}{6ab},\\tfrac{1}{12ab^2}\\right)=1-\\tfrac{1}{432a^2b^3}$ . The latter is clearly the minimum, and si... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 1140,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2744232.json"
} |
A $3 \times 3 \times 4$ cuboid is constructed out of 36 white-coloured unit cubes. Then, all six of the cuboid's sides are coloured red. After that, the cuboid is dismantled into its constituent unit cubes. Then, randomly, all said unit cubes are constructed into the cuboid of its original size (and position).
... | a) I think the answer is $3^88!$ although I'm not so sure
b)The corner cubes must permute, and so also the edge cubes and the face cubes. So we have a probability of $$ \frac{3^8\cdot 8!2^{16}16!4^{10}10!24^{2}2!}{24^{36}\cdot 36!} $$ by considering the possible permutations of the four classes (corner, edge, face... | [] | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 6,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2744233.json"
} |
Let $AB$ be the diameter of circle $\Gamma$ centred at $O$ . Point $C$ lies on ray $\overrightarrow{AB}$ . The line through $C$ cuts circle $\Gamma$ at $D$ and $E$ , with point $D$ being closer to $C$ than $E$ is. $OF$ is the diameter of the circumcircle of triangle $BOD$ . Next, construct $CF$... | [
"Chinese Western Mathematical Olympiad 2006, Problem 6\nhttps://artofproblemsolving.com/community/c6h118089p670265"
] | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2744235.json"
} | |
Given positive odd integers $m$ and $n$ where the set of all prime factors of $m$ is the same as the set of all prime factors $n$ , and $n \vert m$ . Let $a$ be an arbitrary integer which is relatively prime to $m$ and $n$ . Prove that:
\[ o_m(a) = o_n(a) \times \frac{m}{\gcd(m, a^{o_n(a)}-1)} \] where... | <blockquote>Given positive odd integers $m$ and $n$ where $\text{rad}(m) = \text{rad}(n)$ , and $n \vert m$ . Let $a$ be an arbitrary integer which is relatively prime to $m$ and $n$ . Prove that:
\[ o_m(a) = o_n(a) \times \frac{m}{\gcd(m, a^{o_n(a)}-1)} \] where $o_k(a)$ denotes the smallest positive ... | [
"Let $p$ be a prime dividing $n$ . Note that $o_n(a)|o_m(a)$ and $v_p(a^{ko_n(a)}-1)=v_p(a^{o_n(a)}-1)+v_p(k)$ .\nTherefore if $v_p(a^{o_n(a)}-1)\\geq v_p(m)$ we don't need any factors of $p$ in $k$ but otherwise we must have $p^{v_p(m)-v_p(a^{o_n(m)}-1)}=p^{v_p(\\frac{m}{\\gcd(m,a^{o_n(a)}-1)})}$ The... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 56,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2744236.json"
} |
For each natural number $n$ , let $f(n)$ denote the number of ordered integer pairs $(x,y)$ satisfying the following equation:
\[ x^2 - xy + y^2 = n. \]
a) Determine $f(2022)$ .
b) Determine the largest natural number $m$ such that $m$ divides $f(n)$ for every natural number $n$ . | [
"Gosh, this is just Turkey 2000 TST just change into 2022:\nhttps://artofproblemsolving.com/community/q2h396239p2203313"
] | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2744240.json"
} | |
In a nonisosceles triangle $ABC$ , point $I$ is its incentre and $\Gamma$ is its circumcircle. Points $E$ and $D$ lie on $\Gamma$ and the circumcircle of triangle $BIC$ respectively such that $AE$ and $ID$ are both perpendicular to $BC$ . Let $M$ be the midpoint of $BC$ , $N$ be the midpoint of a... | <details><summary>Solution</summary>[asy]
size(10cm);
defaultpen(fontsize(9pt));
pair A = dir(130), B = dir(215), C = dir(-35), O = circumcenter(A, B, C);
pair A1 = -A;
pair H = orthocenter(A, B, C);
pair E = foot(A, B, C) * 2 - H;
pair I = incenter(A, B, C), D1 = foot(I, B, C), N = dir(90), P = dir(270), E1 = foot(I, ... | [
"I like this problem :D**<span style=\"color:#f00\">Claim 1:</span>** $G$ is the A-sharkydevil point.**<span style=\"color:#f00\">Proof:</span>** Let $F'$ the touch point of the incircle of $\\triangle ABC$ with $BC$ and let $N'$ the midpoint of the minor arc $BC$ of $\\Gamma$ using I-E lemma we do an... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 232,
"boxed": false,
"end_of_proof": true,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2744243.json"
} |
Let $A$ be a subset of $\{1,2,\ldots,2020\}$ such that the difference of any two distinct elements in $A$ is not prime. Determine the maximum number of elements in set $A$ . | I think the answer is $\boxed{505}$ . We can form $A$ as: $A=\{1,5,9,...,2017\}$ , $A=\{2,6,10,...,2018\}$ $A=\{3,7,11,...,2019\}$ and $A=\{4,8,12,...,2020\}$ Assume that $|A|>505$ . We will seperate $\{1,2,...,2020\}$ into 505 subsets which form as $\{4k+1,4k+2,4k+3,4k+4\}, k=0,1,2,...,504$ . Because $|A|>50... | [
"any solution?\n",
"How about $|a - b| = 1$ ? @above",
"If I'm not mistake at most $5$ of $20$ consecutive numbers can be element of $A$ , which $\\frac{1}{4}$ of $20$ . So the result follows immediately. (Answer is $505$ )",
"The answer is $505$ , first order the set and let $n$ be the size of ... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 1026,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2744244.json"
} |
Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying
\[ f(a^2) - f(b^2) \leq (f(a)+b)(a-f(b)) \] for all $a,b \in \mathbb{R}$ . | Let $P(a,b)$ be the given assertion. $P(0,0)\Rightarrow f(0)^2\le0\Rightarrow f(0)=0$ $P(x,0)\Rightarrow f(x^2)\le xf(x)$ $P(0,x)\Rightarrow-f(x^2)\le-xf(x)\Rightarrow f(x^2)\ge xf(x)$ So $f(x^2)=xf(x)$ , hence $f$ is odd. Now $P(x,y)$ gives the new assertion $Q(x,y):f(x)f(y)\le xy$ . $Q(x,-y)\Rightarrow-f(x)... | [
"[MEMO 2021 T-1](https://artofproblemsolving.com/community/c6h2665236p23091906)",
"Let $P(a,b)$ denote the assertion for this functional inequality. $P(0,0)\\implies (f(0))^2\\leq 0$ and hence, $f(0)=0$ .\nNow, $P(a,a)\\implies (f(a))^2\\leq a^2$ . $P(0,a)$ and $P(a,0)$ gives $a\\cdot f(a)=f(a^2)$ which... | [
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] | {
"answer_score": 1030,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2744245.json"
} |
Non-zero polynomials $P(x)$ , $Q(x)$ , and $R(x)$ with real coefficients satisfy the identities $$ P(x) + Q(x) + R(x) = P(Q(x)) + Q(R(x)) + R(P(x)) = 0. $$ Prove that the degrees of the three polynomials are all even. | <blockquote>Non-zero polynomials $P(x)$ , $Q(x)$ , and $R(x)$ with real coefficients satisfy the identities $$ P(x) + Q(x) + R(x) = P(Q(x)) + Q(R(x)) + R(P(x)) = 0. $$ Prove that the degrees of the three polynomials are all even.</blockquote> $$ P(x)= a_n \cdot x^n + ... + a_2 \cdot x^2 + a_1 \cdot x + a_0 $$ ... | [
"This one was a really good free problem. \nWLOG $deg(P)>deg(Q),deg(R)$ , then from $$ P+Q+R=0 $$ we'll get that $P=-(Q+R)$ which is a contradiction since they're non-zero. Then WLOG $deg(Q)=deg(P)=n$ , $deg(R)=m$ . But it's easy to see that from $$ P(Q)+Q(R)+R(P)=0 $$ we get: $$ P(Q)=-(Q(R)+R(P)) \\... | [
"origin:aops",
"2022 Contests",
"2022 International Zhautykov Olympiad"
] | {
"answer_score": 1216,
"boxed": false,
"end_of_proof": true,
"n_reply": 12,
"path": "Contest Collections/2022 Contests/2022 International Zhautykov Olympiad/2783537.json"
} |
A ten-level $2$ -tree is drawn in the plane: a vertex $A_1$ is marked, it is connected by segments with two vertices $B_1$ and $B_2$ , each of $B_1$ and $B_2$ is connected by segments with two of the four vertices $C_1, C_2, C_3, C_4$ (each $C_i$ is connected with one $B_j$ exactly); and so on, up to $... | Make the question general such that we are working on $n$ -level $2$ -tree.Notice that we have $3$ kinds of points in $2-tree$ .
$(1)$ Vertices connected with $2$ segments (Starting vertex)
$(2)$ Vertices connected with $1$ segments (Last level vertices of tree)
$(3)$ vertices connected with $3$ ... | [
"if I'm not mistaken, the answer is $M=g(n)=2^{2^{n-3}}$ for $n$ -level trees ( $n\\geq 3$ ).\nfirst of all it is clear that all such \"permutations\" are in fact uniquely achievable from the main tree by the way of \"swapping\", where we define a \"swap\" as taking the two subtrees descending from a vertex of a... | [
"origin:aops",
"2022 Contests",
"2022 International Zhautykov Olympiad"
] | {
"answer_score": 170,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 International Zhautykov Olympiad/2783538.json"
} |
In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$ , and a point $M$ on the segment $CN$ so that $AB = BM = CM$ . Point $K$ is the reflection of $N$ in line $MD$ . The line $MK$ meets the segment $AD$ at point $L$ . Let $P$ be the common point of the circumcir... | <blockquote>In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$ , and a point $M$ on the segment $CN$ so that $AB = BM = CM$ . Point $K$ is the reflection of $N$ in line $MD$ . The line $MK$ meets the segment $AD$ at point $L$ . Let $P$ be the common point of t... | [
"Let $\\angle MBC=\\angle MCB=\\alpha$ . Then, $\\angle MKD=\\angle MND=\\angle MCB=\\alpha$ .\nLet $\\angle NDM=\\angle KDM=\\beta$ . Since $DM\\perp NK$ , we find that $\\angle DKN=90^\\circ-\\beta$ . Hence, $\\angle MNK=\\angle MKN=90^\\circ-\\alpha-\\beta$ . Since $|CM|=|AB|=|CD|$ , we find that $\\angl... | [
"origin:aops",
"2022 Contests",
"2022 International Zhautykov Olympiad"
] | {
"answer_score": 1238,
"boxed": false,
"end_of_proof": true,
"n_reply": 17,
"path": "Contest Collections/2022 Contests/2022 International Zhautykov Olympiad/2783539.json"
} |
A polynomial $f(x)$ with real coefficients of degree greater than $1$ is given. Prove that there are infinitely many positive integers which cannot be represented in the form \[f(n+1)+f(n+2)+\cdots+f(n+k)\]
where $n$ and $k$ are positive integers. | Assume the leading coefficient of $f$ is positive. Let $f_k(n)=f(n)+\cdots+f(n+k-1)$ , so we want to show that there are infinitely many positive integers not representable as $f_k(n)$ for some $(k,n)$ .
Consider some large interval $[1,N]$ and suppose $f_k(n) \in [1,N]$ . This clearly implies that $n$ is ... | [
"Reposting my solution from yesterday:\nSuppose that $f$ has degree $d \\ge 2$ . Clearly, we may assume that the leading coefficient of $f$ is positive. In that case $f(n)+C \\gg n^d$ for all $n$ where $C$ is a suitable constant depending only on $f$ and hence\n\\[f(n+1)+\\dots+f(n+k) +Ck \\gg (n+1)^d+... | [
"origin:aops",
"2022 Contests",
"2022 International Zhautykov Olympiad"
] | {
"answer_score": 130,
"boxed": false,
"end_of_proof": false,
"n_reply": 16,
"path": "Contest Collections/2022 Contests/2022 International Zhautykov Olympiad/2783540.json"
} |
In triangle $ABC$ , a point $M$ is the midpoint of $AB$ , and a point $I$ is the incentre. Point $A_1$ is the reflection of $A$ in $BI$ , and $B_1$ is the reflection of $B$ in $AI$ . Let $N$ be the midpoint of $A_1B_1$ . Prove that $IN > IM$ . | $\angle ABI = \angle CBI = \angle A_1BI \implies $ $B,C,A_1$ collinear.Similarity $A,C,B_1$ collinear. $$ AI= A_1I = x , BI = B_1I = y $$ $$ IN = \frac{1}{2} \cdot \sqrt{2(x^2+y^2)- A_1B_1^2} $$ $$ IM = \frac{1}{2} \cdot \sqrt{2(x^2+y^2)- AB^2} $$ it suffices to prove $AB > A_1B_1$ $\angle ACB = \al... | [
"First note that $A_1,B_1$ lie on $BC,AC$ and $AI=A_1I, BI=B_1I$ . calculating the medians in two triangles $AIB, A_1IB_1$ , we see that $$ 4IN^2=2AI^2+2BI^2-A_1B_1^2, 4IM^2=2AI^2+2BI^2-AB^2 $$ so its enough to prove $AB>A_1B_1$ , Set the middles of arcs $ACB, AC$ on the circumcircle as $T,R$ . by a spi... | [
"origin:aops",
"2022 Contests",
"2022 International Zhautykov Olympiad"
] | {
"answer_score": 140,
"boxed": false,
"end_of_proof": false,
"n_reply": 14,
"path": "Contest Collections/2022 Contests/2022 International Zhautykov Olympiad/2783541.json"
} |
Do there exist two bounded sequences $a_1, a_2,\ldots$ and $b_1, b_2,\ldots$ such that for each positive integers $n$ and $m>n$ at least one of the two inequalities $|a_m-a_n|>1/\sqrt{n},$ and $|b_m-b_n|>1/\sqrt{n}$ holds? | For each $i$ , draw a square of side length $1/\sqrt{i}$ with sides parallel to the coordinate axes centered at $(a_i,b_i)$ . Then no squares overlap, since for $m>n$ we have $\max\{|a_m-a_n|,|b_m-b_n|\}>1/\sqrt{n}>1/(2\sqrt{n})+1/(2\sqrt{m})$ . On the other hand, the sum of the areas of the squares diverges, he... | [
"This is a really nice one. \n\nPlot all $(a_i, b_i)$ on the coordinate plane. Center a square with side length $1/\\sqrt{i}$ and sides parallel to the coordinate axes at each $(a_i,b_i).$ For any $m > n$ we have $|a_m-a_n|>1/\\sqrt{n} > 1/(2\\sqrt{n}) + 1/(2\\sqrt{m})$ and/or $|b_m-b_n|>1/\\sqrt{n} > ... | [
"origin:aops",
"2022 Contests",
"2022 International Zhautykov Olympiad"
] | {
"answer_score": 124,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2022 Contests/2022 International Zhautykov Olympiad/2783542.json"
} |
Three parallel lines $L_1, L_2, L_2$ are drawn in the plane such that the perpendicular distance between $L_1$ and $L_2$ is $3$ and the perpendicular distance between lines $L_2$ and $L_3$ is also $3$ . A square $ABCD$ is constructed such that $A$ lies on $L_1$ , $B$ lies on $L_3$ and $C$ lies o... | Uh oh... That $90^\circ$ angle between $A,B,C$ screams for complex come on! For sure unnecessary, but if first post can be trig, then this is fine as well.
*Solution:* Toss this onto the complex number with $B$ as origin. By the problem condition, assume $c = m + 3i$ for some real $m$ . Then $A$ is precisel... | [
"Say the side length is $x$ , now say $\\angle (L_3, AB)=\\theta$ , this means $\\angle (BC, L_3)=90^{\\circ}- \\theta$ , now you have $x\\cos{\\theta}=3$ and $x\\sin{\\theta}=6$ , squaring and adding them you get $x^2=45$ .",
"18 gang :( ",
" $\\frac{x^2}{2}=3\\times\\frac{15}{2}$ but i somehow calcul... | [
"origin:aops",
"2022 Contests",
"2022 IOQM India"
] | {
"answer_score": 1130,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796887.json"
} |
Ria writes down the numbers $1,2,\cdots, 101$ in red and blue pens. The largest blue number is equal to the number of numbers written in blue and the smallest red number is equal to half the number of numbers in red. How many numbers did Ria write with red pen? | $If$ the largest blue number is $x$ and the number of numbers written in blue is also $x$ then $x+1$ is the smallest number written in red(since every number before that was written in blue) $101-x$ is the number of numbers written in red
then by the question,
... | [
"I lost 2 marks in this just because I got 101-33=78 :wallbash_red: ",
"its $68$ , the idea is to keep the min of one thing to be as close to the max of the other thing.",
"Let $x$ be the number of blue pens. Then the number of red pens will be $101-x$ . **Since the largest blue pen marking will be the num... | [
"origin:aops",
"2022 Contests",
"2022 IOQM India"
] | {
"answer_score": 24,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796888.json"
} |
Consider the set $\mathcal{T}$ of all triangles whose sides are distinct prime numbers which are also in arithmetic progression. Let $\triangle \in \mathcal{T}$ be the triangle with least perimeter. If $a^{\circ}$ is the largest angle of $\triangle$ and $L$ is its perimeter, determine the value of $\frac{a}{... | A lengthier approach.
Ofcourse the triangle is a $3,5,7$ triangle. Thus using heron's formula, the area of the traingle is $$ \frac{\sqrt{(15)(9)(1)(5)}}{4} = \frac{1}{2} \times 5 \times 3 \times \sin (\angle \text{opposite to 7}). $$ Ofcourse the angle opposite to $7$ is the largest. $$ \Longrightarrow \si... | [
"Notice that triangle with sides $3$ , $5$ and $7$ is required triangle $\\triangle$ belonging to the set $\\mathcal{T}$ with least perimeter $L$ that is 15. So, we basically need to find angle opposite to side with length $7$ . By LoC, we get $\\cos(a)=-1/2$ <span style=\"font-size:50%\">and you mess... | [
"origin:aops",
"2022 Contests",
"2022 IOQM India"
] | {
"answer_score": 1028,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796889.json"
} |
Consider the set of all 6-digit numbers consisting of only three digits, $a,b,c$ where $a,b,c$ are distinct. Suppose the sum of all these numbers is $593999406$ . What is the largest remainder when the three digit number $abc$ is divided by $100$ ? | $a+b+c = 22$ so yesh | [
"...\\[98\\]",
"Let the set denoted by S\nThen we have 3^6=729 numbers belong to S.\nNow if we look at the central tendency we get the mean (593999406÷729)=814814\nClearly a,b,c are 8,4,1 \nThen the largest reminder is 84",
"We get the expression:\n(a+b+c)(1+10+100+1000+10000)(3.3.3.3.3)= $593999406$ \nSimpli... | [
"origin:aops",
"2022 Contests",
"2022 IOQM India"
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"answer_score": 2,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796890.json"
} |
In parallelogram $ABCD$ , the longer side is twice the shorter side. Let $XYZW$ be the quadrilateral formed by the internal bisectors of the angles of $ABCD$ . If the area of $XYZW$ is $10$ , find the area of $ABCD$ | I will just post a solution because I am mastermind who blundered it in exam even after seeing the bash on exam.
Let $\overline{AB}$ be the longer side of the parallelogram. Let $\angle DAB = 2\theta$ . One can angle chase to find that $XYZW$ is a rectangle. Let $B$ bisector meet $C$ bisector at $X$ and $Y$... | [
"The key is that the angle bisectors meet on the midpoint of the opposite sides.\nand also form a rectangle.\nfrom here it is easy",
"i just assumed that the parallelogram is a rectangle and from there it is easy",
"@numbertheorydog @mathlearner2357\neven if you dont, its still a simple question, you just have ... | [
"origin:aops",
"2022 Contests",
"2022 IOQM India"
] | {
"answer_score": 144,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796891.json"
} |
Let $x,y,z$ be positive real numbers such that $x^2 + y^2 = 49, y^2 + yz + z^2 = 36$ and $x^2 + \sqrt{3}xz + z^2 = 25$ . If the value of $2xy + \sqrt{3}yz + zx$ can be written as $p \sqrt{q}$ where $p,q \in \mathbb{Z}$ and $q$ is squarefree, find $p+q$ . | create a triangle $ABC$ , let $O$ be a point inside $ABC$ such that $\angle{AOB}=\frac{\pi}{2}$ , $\angle{AOC}=\frac{2\pi}{3}$ and $\angle{BOC}=\frac{5\pi}{6}$ , now denote $AO=y, BO=x, OC=z$ , and $AB=7,AC=6,BC=5$ . now $[AOC]+[BOC]+[AOC]=[ABC]$ , so $\frac{xy}{2}+\frac{yz\sin{\frac{2\pi}{3}}}{2}+\frac{... | [
"Cosine identities go brrrrrrrr.... and I obviously didn't notice them in the test. $30$ is the answer.",
"More like [CMI 2019/5](https://artofproblemsolving.com/community/c6h1944124p13395044)"
] | [
"origin:aops",
"2022 Contests",
"2022 IOQM India"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796892.json"
} |
Find the number of maps $f: \{1,2,3\} \rightarrow \{1,2,3,4,5\}$ such that $f(i) \le f(j)$ whenever $i < j$ . | When $f(1)=1$ , you have 15 possible cases and similarly, number of such maps for a fixed $f(1)$ is a triangular number, and summing all of them up you get $15+10+6+3+1=35$ . | [
"\\[\\sum_{i=1}^{5} x_{i}=3\\] which gives \\[{5+3-1}\\choose{3}\\]",
"@bove That's C++ right? And [c] documenting?",
"Sol:- $1 \\leq f(1)<f(2)+1<f(3)+2\\leq 7$ . So we just need to select 3 numbers from $1,2..,7$ . So ans is $C(7,3)=35$ ",
"By stars and bars there are 35 ways to pick a set of 3 values fr... | [
"origin:aops",
"2022 Contests",
"2022 IOQM India"
] | {
"answer_score": 6,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796893.json"
} |
For any real number $t$ , let $\lfloor t \rfloor$ denote the largest integer $\le t$ . Suppose that $N$ is the greatest integer such that $$ \left \lfloor \sqrt{\left \lfloor \sqrt{\left \lfloor \sqrt{N} \right \rfloor}\right \rfloor}\right \rfloor = 4 $$ Find the sum of digits of $N$ . | The first nest of the floor and square root implies $$ 16 \leq \left \lfloor \sqrt{\left \lfloor \sqrt{N} \right \rfloor} \right \rfloor < 25. $$ The next nest of floor and square root implies $$ 256 \leq \left \lfloor \sqrt{N} \right \rfloor < 625 $$ Thus, $$ 256^2 \leq N < 625^2. $$ The maximum value is $... | [
"hahaha just take $5^8-1$ and obviously its sum of digits is $25$ . ",
"<blockquote>hahaha just take $5^8-1$ and obviously its sum of digits is $25$ .</blockquote>\n\nLol, same mistake done here"
] | [
"origin:aops",
"2022 Contests",
"2022 IOQM India"
] | {
"answer_score": 1016,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796894.json"
} |
A $12 \times 12$ board is divided into $144$ unit squares by drawing lines parallel to the sides. Two rooks placed on two unit squares are said to be non-attacking if they are not in the same column or same row. Find the least number $N$ such that if $N$ rooks are placed on the unit squares, one rook per square... | fill $6$ rows/columns with $6*12$ rooks and place another in an empty row/col. | [
"easiest problem on the test\n",
"how to interpret this? i did not get the wording",
"You have to find the least number of rooks which when placed on the board in any manner will always result in 7 non attacking rooks. \nSo we fill any six columns/rows completely with 12*6=72 rooks(this is basicaly the worst wa... | [
"origin:aops",
"2022 Contests",
"2022 IOQM India"
] | {
"answer_score": 4,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796895.json"
} |
Let $P_0 = (3,1)$ and define $P_{n+1} = (x_n, y_n)$ for $n \ge 0$ by $$ x_{n+1} = - \frac{3x_n - y_n}{2}, y_{n+1} = - \frac{x_n + y_n}{2} $$ Find the area of the quadrilateral formed by the points $P_{96}, P_{97}, P_{98}, P_{99}$ . | We claim that the area of the convex quadrilateral formed by $P_{i}, P_{i +1}, P_{i + 2}, P_{i+3}$ is the same as that formed by $P_{i + 1}, P_{i + 2}, P_{i +3}, P_{i +4}$ .
Let $\vec{v}_i = \begin{pmatrix}x_i y_i\end{pmatrix}$ for all $i\ge 0$ and $M =\begin{pmatrix}-3/2 & 1/2 -1/2 & -1/2\end{pmatrix}$ and... | [
"Scam, apparently the coordinates of those points are just those of the first four and you get $8$ as the answer.",
"Nope they are not those of the first four. However the area remains the same",
"Why this problem has been declared bonus by HBCSE?",
"Can anyone please clarify why this problem has been decla... | [
"origin:aops",
"2022 Contests",
"2022 IOQM India"
] | {
"answer_score": 1040,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796896.json"
} |
Suppose that $P$ is the polynomial of least degree with integer coefficients such that $$ P(\sqrt{7} + \sqrt{5}) = 2(\sqrt{7} - \sqrt{5}) $$ Find $P(2)$ . | Say $t_1=\sqrt{7}+\sqrt{5}$ Since , $P(\sqrt{7}+\sqrt{5})=2(\sqrt{7}-\sqrt{5})$ . We try to write the RHS in terms of $t_1$ . It is easy to see through rationalisation that $t_1=\frac{2}{\sqrt{7}-\sqrt{5}}\implies \sqrt{7}-\sqrt{5}=\frac{2}{t_1}$ . Now , we define : $Q(t)=tP(t)-4$ such that $Q(t_1)=0$ from... | [
"yes, you try $deg(P)=2$ , it doesn't work, you try $deg(P)=3$ and you get $24x-x^3$ so $P(2)=40$ .",
" $P(\\sqrt7 + \\sqrt5) = 2(\\sqrt7 - \\sqrt5)$ \nwe have : $(\\sqrt7 + \\sqrt5)(\\sqrt7 - \\sqrt5) = 2 $ now we have : $P(\\sqrt7 + \\sqrt5) = (\\sqrt7 + \\sqrt5)(\\sqrt7 - \\sqrt5)^2 $ \nWe also have... | [
"origin:aops",
"2022 Contests",
"2022 IOQM India"
] | {
"answer_score": 1036,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796897.json"
} |
In how many ways can four married couples sit in a merry-go-round with identical seats such that men and women occupy alternate seats and no husband seats next to his wife? | just seat 4 ppl of the same gender in $(4-1)!$ ways and u get $2$ choices to seat another (as seating one of em fixes the whole arrangement).
So, $3!*2$ . | [
"I had already solve this problem from isi tomato book and in exam I did, mistake , very disappointing "
] | [
"origin:aops",
"2022 Contests",
"2022 IOQM India"
] | {
"answer_score": 6,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 IOQM India/2796898.json"
} |
For positive integers $a$ and $b$ , if the expression $\frac{a^2+b^2}{(a-b)^2}$ is an integer, prove that the expression $\frac{a^3+b^3}{(a-b)^3}$ is an integer as well. | Let $\gcd(a,b)=d$ . Then, $\frac{(da)^2+(db)^2}{(da-db)^2} = \frac{a^2+b^2}{(a-b)^2}$ and $\frac{(da)^3+(db)^3}{(da-db)^3} = \frac{a^3+b^3}{(a-b)^3}$ so it suffices to solve the problem for $a,b$ such that $\gcd(a,b)=1$ . Assume WLOG, $a>b$ . Now, note that if $a-b\geq 2$ there exists a prime $p\mid a-b$ . ... | [
"WLOG $(a,b)=1$ . We have $$ (a-b)^2|a^2+b^2-(a-b)^2=2ab $$ Also, $gcd((a-b)^2, a)= gcd((a-b)^2, b)=1$ . Hence, $(a-b)^2|2$ . Therefore, $a-b=\\pm 1$ . The rest follows.",
" $$ (a-b)^2|a^2+b^2-(a-b)^2=2ab \\Rightarrow \\frac{2ab}{(a-b)^2}\\in \\mathbb{Z}\\Rightarrow \\frac{2a^2b-2ab^2}{(a-b)^3}\\in \\mathb... | [
"origin:aops",
"2022 Contests",
"2022 JBMO TST - Turkey"
] | {
"answer_score": 46,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802471.json"
} |
For a real number $a$ , $[a]$ denotes the largest integer not exceeding $a$ .
Find all positive real numbers $x$ satisfying the equation $$ x\cdot [x]+2022=[x^2] $$ | Let $[x]=a$ and $x=a+b$ where $0\leq b<1$ . We have $$ (a+b)a+2022=[(a+b)^2]\Rightarrow a^2+ab+2022=a^2+[2ab+b^2]\Rightarrow ab+2022=[2ab+b^2] $$ Clearly, RHS is an integer. Hence, $ab$ should be an integer too. Then, $[2ab+b^2]=2ab+[b^2]=2ab$ . So the original equation becomes $$ ab+2022=2ab\Rightarrow 202... | [
"we see that if 2022 is whole and [x^2] is whole too then x*[x] is whole too. meanwhile , if x is natural we get x^2+2022=x^2 which is imossible. Then let for satisfying x let's take x=b/a and x is not included to Z or b is not divisible by a ; a,b are natural and [x]=a. then we get that a+1>b/a>a<==>a^2<b<a^2+a so... | [
"origin:aops",
"2022 Contests",
"2022 JBMO TST - Turkey"
] | {
"answer_score": 26,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802473.json"
} |
Each of the $29$ people attending a party wears one of three different types of hats. Call a person *lucky* if at least two of his friends wear different types of hats. Show that it is always possible to replace the hat of a person at this party with a hat of one of the other two types, in a way that the total number... | <details><summary>Solution</summary>Consider the obvious graph interpretation, where vertices represent people, edges represent friendships and we write $0,1$ or $2$ on each of the vertices. Call the graph $G$ and for a vertex $v \in V(G)$ , we denote $d(v)$ and $N(v)$ as the degree of $v$ and the set of n... | [
"@hakN How do you obtain that $d_j(v) \\geq 2$ for every $v$ ? We do indeed have $d_i(v) \\leq 2$ and $d_i(v) + d_j(v) \\geq 2$ but it is quite unclear how to show $d_j(v) \\geq 2$ . Also, the claim \" $d_i(v) = 2$ only if $d_i(v) + d_j(v) = 2$ \" is correct but does not really fit with $d_j(v) \\geq 2$... | [
"origin:aops",
"2022 Contests",
"2022 JBMO TST - Turkey"
] | {
"answer_score": 168,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802483.json"
} |
Given a convex quadrilateral $ABCD$ such that $m(\widehat{ABC})=m(\widehat{BCD})$ . The lines $AD$ and $BC$ intersect at a point $P$ and the line passing through $P$ which is parallel to $AB$ , intersects $BD$ at $T$ . Prove that $$ m(\widehat{ACB})=m(\widehat{PCT}) $$ | $AB\cap CD=Q$ We will use method of moving points.
Take $QBCD$ fixed. Animate $A$ over $QB$ . Denote $R(XY,ZT)$ as reflection of $XY$ to $ZT$ .
\[f:A\rightarrow AC\rightarrow R(AC,BC)\rightarrow R(AC,BC)\cap BQ\]
\[g: A\rightarrow AD\rightarrow AD\cap BC=T\rightarrow T(QB)_{\infty}\cap BD=P\rightarrow PC\cap... | [
"Let the line passing through $D$ and parallel to $AB$ intersects $BP$ and $AT$ at $K$ and $L$ , respectively. We have $$ \\frac{|DK|}{|AB|}=\\frac{|PD|}{|PA|}=\\frac{|PD|}{|PA|}\\cdot\\frac{|PT|}{|DL|}\\cdot\\frac{|DL|}{|PT|}=\\frac{|PD|}{|PA|}\\cdot\\frac{|AP|}{|AD|}\\cdot\\frac{|DL|}{|PT|}= $$ $$ ... | [
"origin:aops",
"2022 Contests",
"2022 JBMO TST - Turkey"
] | {
"answer_score": 170,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802490.json"
} |
Each of the $n$ students writes one of the numbers $1,2$ or $3$ on each of the $29$ boards. If any two students wrote different numbers on at least one of the boards and any three students wrote the same number on at least one of the boards, what is the maximum possible value of $n$ ? | <details><summary>Solution</summary>The answer is $3^{28}$ .
Instead of writing a number in each board, assume that each student chooses a $29$ digit number where each digit is either $1,2$ or $3$ .
Constructing an example is not hard. Assume that the students wrote every $29$ digit number starting with $1$ ... | [] | [
"origin:aops",
"2022 Contests",
"2022 JBMO TST - Turkey"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802498.json"
} |
Let $c$ be a real number. If the inequality $$ f(c)\cdot f(-c)\ge f(a) $$ holds for all $f(x)=x^2-2ax+b$ where $a$ and $b$ are arbitrary real numbers, find all possible values of $c$ . | I claim the answer is $c=\pm 1/2$ . This is easily seen to work. Next, note that $f(c)f(-c)\ge f(a)$ iff $g(b)\ge 0$ where
\[
g(b) = b^2 + (2c^2-1)b +c^4 -4a^2c^2 + a^2.
\]
We want the quadratic $g(b)\ge 0$ for all choices of $a,b$ . This holds iff its discriminant is non-positive: $(2c^2-1)^2 - 4(c^4+a^2-4a^... | [] | [
"origin:aops",
"2022 Contests",
"2022 JBMO TST - Turkey"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802500.json"
} |
In a triangle $\triangle ABC$ with $\angle ABC < \angle BCA$ , we define $K$ as the excenter with respect to $A$ . The lines $AK$ and $BC$ intersect in a point $D$ . Let $E$ be the circumcenter of $\triangle BKC$ . Prove that
\[\frac{1}{|KA|} = \frac{1}{|KD|} + \frac{1}{|KE|}.\] | We need to prove that $\frac{|KA|}{|KD|}+\frac{|KA|}{|KE|}=1$ .
Simple Angle-Chasing gives us $\angle ADC=\frac{C-B}2$ . Hence, $$ \frac{|KA|}{|KD|}=\frac{|AC|}{|CD|}=\frac{\sin{ADC}}{\sin{CAD}}=\frac{\sin{(\frac{C-B}2)}}{\cos{(\frac A2)}} $$ Also, $$ \frac{|KA|}{|KE|}=\frac{2|KA|}{2R_{BKC}}=\frac{2|KA|\cdot\sin... | [
" \n we know that if $ (A,B;C,D)=-1 $ then $ \\frac{2}{AB}=\\frac{1}{CB}+\\frac{1}{DB}$ ( with signed distance)\n just consider $T$ the symmetric of $K$ in $E$ which is on $(E)$ ...\n\nMy regards**RH HAS**"
] | [
"origin:aops",
"2022 Contests",
"2022 JBMO TST - Turkey"
] | {
"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802503.json"
} |
Find all prime numbers $p$ such that the number $$ 3^p+4^p+5^p+9^p-98 $$ has at most $6$ positive divisors. | <blockquote>See that $p=2,3,5$ works. Let $p\ge 7$ .
See that $3^p+4^p\equiv 5^p+9^p\equiv -98\equiv 0\pmod{7}$ as $p$ is odd.
Also, $11|3^p+4^p+5^p+9^p-98$ as $p$ is odd and not divisible by $5$ .
Since this number has at most $6$ positive divisors, we find that $3^p+4^p+5^p+9^p-98\leq 11^2\cdot 7=847$ ... | [
"See that $p=2,3,5$ works. Let $p\\ge 7$ .\nSee that $3^p+4^p\\equiv 5^p+9^p\\equiv -98\\equiv 0\\pmod{7}$ as $p$ is odd.\nAlso, $11|3^p+4^p+5^p+9^p-98$ as $p$ is odd and not divisible by $5$ .\nSince this number has at most $6$ positive divisors, we find that $3^p+4^p+5^p+9^p-98\\leq 11^2\\cdot 7=84... | [
"origin:aops",
"2022 Contests",
"2022 JBMO TST - Turkey"
] | {
"answer_score": 22,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 JBMO TST - Turkey/2802505.json"
} |
Represent $\frac{1}{2021}$ as a difference of two irreducible fractions with smaller denominators.
*(Proposed by Bogdan Rublov)* | <blockquote>Represent $\frac{1}{2021}$ as a difference of two irreducible fractions with smaller denominators.</blockquote>
From $\frac ab-\frac cd=\frac 1{2021}$ , we get $ad-bc=\frac{bd}{2021}$ And so $(b,d)\in\{(2021u,v),(u,2021v),(43u,47v),(47u,43v)\}$ I dont know what is the order relation you use when you sp... | [] | [
"origin:aops",
"2022 Contests",
"2022 Kyiv City MO Round 1"
] | {
"answer_score": 1018,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 1/2764246.json"
} |
There are $n$ sticks which have distinct integer length. Suppose that it's possible to form a non-degenerate triangle from any $3$ distinct sticks among them. It's also known that there are sticks of lengths $5$ and $12$ among them. What's the largest possible value of $n$ under such conditions?
*(Proposed b... | <details><summary>Solution</summary>The answer is $6$ . Clearly, all further lengths must be at least $8$ and at most $16$ to form a non-degenerate triangle with $5$ and $12$ .
In particular, $5$ is the smallest length. Now, if $n$ is the next smallest length, then the largest length can be at most $n+4$ ... | [
"<details><summary>Answer</summary>Answer --> 6</details>"
] | [
"origin:aops",
"2022 Contests",
"2022 Kyiv City MO Round 1"
] | {
"answer_score": 36,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 1/2764253.json"
} |
You are given $n$ not necessarily distinct real numbers $a_1, a_2, \ldots, a_n$ . Let's consider all $2^n-1$ ways to select some nonempty subset of these numbers, and for each such subset calculate the sum of the selected numbers. What largest possible number of them could have been equal to $1$ ?
For example, i... | <blockquote>You are given $n$ not necessarily distinct real numbers $a_1, a_2, \ldots, a_n$ . Let's consider all $2^n-1$ ways to select some nonempty subset of these numbers, and for each such subset calculate the sum of the selected numbers. What largest possible number of them could have been equal to $1$ ?
Fo... | [] | [
"origin:aops",
"2022 Contests",
"2022 Kyiv City MO Round 1"
] | {
"answer_score": 1058,
"boxed": true,
"end_of_proof": false,
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In some magic country, there are banknotes only of values $3$ , $25$ , $80$ hryvnyas. Businessman Victor ate in one restaurant of this country for $2024$ days in a row, and each day (except the first) he spent exactly $1$ hryvnya more than the day before (without any change). Could he have spent exactly $10000... | [] | [
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Consider $5$ distinct positive integers. Can their mean be
a)Exactly $3$ times larger than their largest common divisor?
b)Exactly $2$ times larger than their largest common divisor?
| <blockquote>Consider $5$ distinct positive integers. Can their mean be
a)Exactly $3$ times larger than their largest common divisor?</blockquote>
Yes, choose for example $1,2,3,10,14$ <blockquote>b)Exactly $2$ times larger than their largest common divisor?</blockquote>
Yes, choose for example $1,2,3,4,10$ | [
"<blockquote><blockquote>Consider $5$ distinct positive integers. Can their mean be \n\na)Exactly $3$ times larger than their largest common divisor?</blockquote>\nYes, choose for example $1,2,3,10,14$ </blockquote>\nI'm not sure, I understand. It seems to me that the largest common divisor is $1$ while the ... | [
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In triangle $ABC$ $\angle B > 90^\circ$ . Tangents to this circle in points $A$ and $B$ meet at point $P$ , and the line passing through $B$ perpendicular to $BC$ meets the line $AC$ at point $K$ . Prove that $PA = PK$ .
*(Proposed by Danylo Khilko)* | [] | [
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What's the largest number of integers from $1$ to $2022$ that you can choose so that no sum of any two different chosen integers is divisible by any difference of two different chosen integers?
*(Proposed by Oleksii Masalitin)* | [] | [
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$2022$ teams participated in an underwater polo tournament, each two teams played exactly once against each other. Team receives $2, 1, 0$ points for win, draw, and loss correspondingly. It turned out that all teams got distinct numbers of points. In the final standings the teams were ordered by the total number of... | [] | [
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What's the smallest possible value of $$ \frac{(x+y+|x-y|)^2}{xy} $$ over positive real numbers $x, y$ ? | Let $x, y$ be positive real numbers. Prove that $$ \frac{(x+y+|x-y|)^2}{xy}\geq 4 $$ $$ \frac{(3x+y+|x-2y|)^2}{xy} \geq 24 $$ <details><summary>*</summary><blockquote>What's the smallest possible value of $$ \frac{(x+y+|x-y|)^2}{xy} $$ over positive real numbers $x, y$ ?</blockquote></details> | [
"wlog cheese guys??? \nby symmetry $ x \\geq y$ Then inequality just $\\frac{4x^2}{xy}$ Then $\\frac{4x^2}{xy} \\geq \\frac{4x^2}{x^2} = 4$ "
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For any reals $x, y$ , show the following inequality: $$ \sqrt{(x+4)^2 + (y+2)^2} + \sqrt{(x-5)^2 + (y+4)^2} \le \sqrt{(x-2)^2 + (y-6)^2} + \sqrt{(x-5)^2 + (y-6)^2} + 20 $$ *(Proposed by Bogdan Rublov)* | In the cartesian system of coordinates $xOy$ consider the variable point $M(x,y)$ , ( $x,y\in\mathbb{R}$ ) and the fixed points: $A(-4,-2);B(5,-4);C(2,6);D(5,6)$ . $\sqrt{(x+4)^2 + (y+2)^2}=MA;\;\sqrt{(x-5)^2 + (y+4)^2}=MB$ ; $\sqrt{(x-2)^2 + (y-6)^2}=MC;\;\sqrt{(x-5)^2 + (y-6)^2}=MD$ .
We observe: $AC=BD=10$ .
Th... | [
"<blockquote>For any reals $x, y$ , show the following inequality: $$ \\sqrt{(x+4)^2 + (y+2)^2} + \\sqrt{(x-5)^2 + (y+4)^2} \\le \\sqrt{(x-2)^2 + (y-6)^2} + \\sqrt{(x-5)^2 + (y-6)^2} + 20 $$ *(Proposed by Bogdan Rublov)*</blockquote>\nHardest algebra ever?\nis it?\n",
"@above there's something called sarcasm a... | [
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Let $AL$ be the inner bisector of triangle $ABC$ . The circle centered at $B$ with radius $BL$ meets the ray $AL$ at points $L$ and $E$ , and the circle centered at $C$ with radius $CL$ meets the ray $AL$ at points $L$ and $D$ . Show that $AL^2 = AE\times AD$ .
*(Proposed by Mykola Moroz)* | [
"Dear Mathlinkers,\n\nhttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=530917\n\ninspire a new approach...\n\nSincerely\nJean-Louis"
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Let's call integer square-free if it's not divisible by $p^2$ for any prime $p$ . You are given a square-free integer $n>1$ , which has exactly $d$ positive divisors. Find the largest number of its divisors that you can choose, such that $a^2 + ab - n$ isn't a square of an integer for any $a, b$ among chosen ... | Let $n = p_1 \cdot p_2 \dots p_k$ where $p_i$ are primes and $k \geq 1$ . Then $d = 2^k$ and we claim that the answer is $2^{k-1}$ .
It is clear that we can't choose two divisors $a,b$ of $n$ so that $ab = n$ , so this means we can choose at most $\frac{d}{2} = 2^{k-1}$ divisors.
Now if we choose the d... | [] | [
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$n\ge 2$ teams participated in an underwater polo tournament, each two teams played exactly once against each other. A team receives $2, 1, 0$ points for a win, draw, and loss correspondingly. It turned out that all teams got distinct numbers of points. In the final standings, the teams were ordered by the total nu... | [] | [
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Does there exist a quadratic trinomial $ax^2 + bx + c$ such that $a, b, c$ are odd integers, and $\frac{1}{2022}$ is one of its roots? | <blockquote>Does there exist a quadratic trinomial $ax^2 + bx + c$ such that $a, b, c$ are odd integers, and $\frac{1}{2022}$ is one of its roots?</blockquote>
No, since $2022^2c+2022b+a=0$ implies $a$ is even
| [
" $2022^2c+2022b+a=0 \\implies a$ is evan $\\implies$ contradiction",
" this problem is very easy"
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You are given $2n$ distinct integers. What's the largest integer $C$ such that you can always form at least $C$ pairs from them, so that no integer is in more than one pair, and the sum of integers in each pair is a composite number?
*(Proposed by Anton Trygub)* | Bump! I think answer is $C=n-1$ (look at parity) | [
"<blockquote>Bump! I think answer is $C=n-1$ (look at parity)</blockquote>\n\nBump!"
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Diagonals of a cyclic quadrilateral $ABCD$ intersect at point $P$ . The circumscribed circles of triangles $APD$ and $BPC$ intersect the line $AB$ at points $E, F$ correspondingly. $Q$ and $R$ are the projections of $P$ onto the lines $FC, DE$ correspondingly. Show that $AB \parallel QR$ .
*(Propos... | Cute problem.
By angle chase, $\measuredangle PCD = \measuredangle FCP$ and $\measuredangle PDC = \measuredangle EDP$ . If $W$ is the projection of $P$ onto $CD$ , it is easy to see that $PR=PW=PQ$ . Another angle chase shows that $\measuredangle DEF = \measuredangle EFC$ . Now, if $CF || DE$ , the desired ... | [
"I assume you mean the projections of $E,F$ .\nIf so, since $EFQR$ is cyclic of diameter $EF$ , it suffices to probe that the two angles at the base of the quadrilateral are equal. Indeed $\\measuredangle QFE=\\measuredangle CFB=\\measuredangle CPB=\\measuredangle DPA=\\measuredangle DEA=\\measuredangle REF$ ,... | [
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For any nonnegative reals $x, y$ show the inequality $$ x^2y^2 + x^2y + xy^2 \le x^4y + x + y^4 $$ . | For any nonnegative reals $x, y$ show the inequality $$ x^2y^2 + x^2y + xy^2 \le x^3y^2 + x^2 + y^3 $$ | [
" $(6x^4y+2x+5y^4)+(5x^4y+6x+2y^4)+(2x^4y+5x+6y^4)\\ge 13(x^2y^2+x^2y+xy^2)$ ",
"Adding a dummy third variable makes the inequality look less weird: $x^4y+y^4z+z^4x \\ge x^2y^2z + x^2yz^2 + xy^2z^2$ . The rest of the proof is a simple application of Cauchy with weights $(\\frac{6}{13}, \\frac{5}{13}, \\frac{2}{... | [
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There is a black token in the lower-left corner of a board $m \times n$ ( $m, n \ge 3$ ), and there are white tokens in the lower-right and upper-left corners of this board. Petryk and Vasyl are playing a game, with Petryk playing with a black token and Vasyl with white tokens. Petryk moves first.
In his move, a pl... | [] | [
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The teacher wrote $5$ distinct real numbers on the board. After this, Petryk calculated the sums of each pair of these numbers and wrote them on the left part of the board, and Vasyl calculated the sums of each triple of these numbers and wrote them on the left part of the board (each of them wrote $10$ numbers). C... | [] | [
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Let $H$ and $O$ be the orthocenter and the circumcenter of the triangle $ABC$ . Line $OH$ intersects the sides $AB, AC$ at points $X, Y$ correspondingly, so that $H$ belongs to the segment $OX$ . It turned out that $XH = HO = OY$ . Find $\angle BAC$ .
*(Proposed by Oleksii Masalitin)* | By the way, the property $XH=OY$ is enough for the result. The argumentation could be the following.
Triangles $AXH$ and $AOY$ are equal because $XH=OY$ , $\angle XAH=\angle YAO$ and these triangles have the same length of the altitude from $A$ . Thus, $R=AH$ . At the same time $R=\frac{a}{2\sin\alpha}$ a... | [] | [
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You are given $n\ge 4$ positive real numbers. It turned out that all $\frac{n(n-1)}{2}$ of their pairwise products form an arithmetic progression in some order. Show that all given numbers are equal.
*(Proposed by Anton Trygub)* | [] | [
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Find the smallest integer $n$ for which it's possible to cut a square into $2n$ squares of two sizes: $n$ squares of one size, and $n$ squares of another size.
*(Proposed by Bogdan Rublov)* | Bump! Is answer $n=20$ ? | [
"Could you tell me where you get the problems from?",
"<blockquote>Could you tell me where you get the problems from?</blockquote>\n\nThey are published in Ukrainian on http://matholymp.com.ua.",
"<blockquote>Bump! Is answer $n=20$ ?</blockquote>\n\nBump! Has anyone got the official solutions? I can't find the... | [
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a) Do there exist positive integers $a$ and $d$ such that $[a, a+d] = [a, a+2d]$ ?
b) Do there exist positive integers $a$ and $d$ such that $[a, a+d] = [a, a+4d]$ ?
Here $[a, b]$ denotes the least common multiple of integers $a, b$ . | a) No. Note that otherwise one would have
\[
\frac{a(a+d)}{(a,d)} = \frac{a(a+2d)}{(a,2d)},
\]
where we used $(a,a+d)=(a,d)$ and $(a,a+2d)=(a,2d)$ via Euclidean algorithm. Now, note that $(a,2d)=k(a,d)$ with $k\in\{1,2\}$ . The case $k=1$ is clearly contradictory, whereas for $k=2$ , we obtain $2a+2d=a+2d$ ,... | [
"deleted ....\n@below pointed out already",
"@above\nb) is wrong for $a=2d$ we have $[a, a+d] =[2d, 3d]=6d, [a,a+4d]=[2d,6d]=6d$ Solution in a) has error because $\\gcd(a,2b) = \\gcd(a,b) $ only for odd $a$ "
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There is a central train station in point $O$ , which is connected to other train stations $A_1, A_2, \ldots, A_8$ with tracks. There is also a track between stations $A_i$ and $A_{i+1}$ for each $i$ from $1$ to $8$ (here $A_9 = A_1$ ). The length of each track $A_iA_{i+1}$ is equal to $1$ , and the le... | [] | [
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In triangle $ABC$ the median $BM$ is equal to half of the side $BC$ . Show that $\angle ABM = \angle BCA + \angle BAC$ .
*(Proposed by Anton Trygub)* | Let $N$ be the midpoint of $BC$ and let $D$ be the reflection of $N$ wrt $B$ . Then $BD=BM=BN$ , so $DM\perp MN\parallel AB$ , so $\angle ABM=\angle ABD=\angle BCA+\angle BAC$ , as desired. | [
"Double the median, i.e let $N \\in BM$ be such that $MB = MN$ . Then $ABCN$ is a parallelogram and $BN = BC$ , from which the result follows easily."
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Fedir and Mykhailo have three piles of stones: the first contains $100$ stones, the second $101$ , the third $102$ . They are playing a game, going in turns, Fedir makes the first move. In one move player can select any two piles of stones, let's say they have $a$ and $b$ stones left correspondently, and remove... | [] | [
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Find all triples $(a, b, c)$ of positive integers for which $a + [a, b] = b + [b, c] = c + [c, a]$ .
Here $[a, b]$ denotes the least common multiple of integers $a, b$ .
*(Proposed by Mykhailo Shtandenko)* | <details><summary>Solution</summary>W.l.o.g. $\text{gcd}(a,b,c)=1$ since otherwise we can just divide it out.
If $p \mid a,b$ , then $p \mid a+[a,b]-[c,a]=c$ , contradiction!
Similarly, we can deduce that the three numbers are pairwise coprime. But then $[a,b]=ab$ etc. so the equations just become $a+ab=b+bc=c+c... | [
"Note that $a\\mid a+[a,b]=c+[c,a]\\implies a\\mid c$ . Therefore $a \\mid c \\mid b \\mid a$ , or $a=b=c$ , which works."
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Monica and Bogdan are playing a game, depending on given integers $n, k$ . First, Monica writes some $k$ positive numbers. Bogdan wins, if he is able to find $n$ points on the plane with the following property: for any number $m$ written by Monica, there are some two points chosen by Bogdan with distance exactly... | Claim: Monica wins if and only if $n\leq k$ .
If $n\geq k+1$ , then Bogdan picks the origin and the points $(x_i,0)$ ( $1\leq i\leq k$ ) where the $x_i$ are the specified distances.
If $n\leq k$ then it suffices for Monica to pick the distances $1,2,2^2,...,2^{k-1}$ . Assume Bogdan could win.
Consider the grap... | [] | [
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Nonzero real numbers $x_1, x_2, \ldots, x_n$ satisfy the following condition: $$ x_1 - \frac{1}{x_2} = x_2 - \frac{1}{x_3} = \ldots = x_{n-1} - \frac{1}{x_n} = x_n - \frac{1}{x_1} $$ Determine all $n$ for which $x_1, x_2, \ldots, x_n$ have to be equal.
*(Proposed by Oleksii Masalitin, Anton Trygub)* | <blockquote>Nonzero real numbers $x_1, x_2, \ldots, x_n$ satisfy the following condition: $$ x_1 - \frac{1}{x_2} = x_2 - \frac{1}{x_3} = \ldots = x_{n-1} - \frac{1}{x_n} = x_n - \frac{1}{x_1} $$ Determine all $n$ for which $x_1, x_2, \ldots, x_n$ have to be equal.</blockquote>
Let us look for cases where all $... | [] | [
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Points $D, E, F$ are selected on sides $BC, CA, AB$ correspondingly of triangle $ABC$ with $\angle C = 90^\circ$ such that $\angle DAB = \angle CBE$ and $\angle BEC = \angle AEF$ . Show that $DB = DF$ .
*(Proposed by Mykhailo Shtandenko)* | [] | [
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Find all triples $(a, b, c)$ of positive integers for which $a + (a, b) = b + (b, c) = c + (c, a)$ .
Here $(a, b)$ denotes the greatest common divisor of integers $a, b$ .
*(Proposed by Mykhailo Shtandenko)* | Solved it ORALLY..
WLOG $gcd(a, b, c) =1$ , now if a prime $p \mid gcd(a, b) $ , then $p \mid c$ , which is not possible, so $a, b, c$ are pairwise coprime, hence condition is $a+1 = b+1 =c+1$ , implying $a=b=c$ , so only solutions that work are $(k, k, k) $ for all natural $k$ ... | [] | [
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$2022$ points are arranged in a circle, one of which is colored in black, and others in white. In one operation, The Hedgehog can do one of the following actions:
1) Choose two adjacent points of the same color and flip the color of both of them (white becomes black, black becomes white)
2) Choose two points of opp... | No. Label the positions of the points consecutively $1,-1,1,-1,\dots$ and let $\ell_i,i=1,\dots,2022$ be the values od these labels. Set $c_i:=1$ if the color of the $i$ -th point is white, otherwise $c_i=-1$ . Consider $S:=\sum_{i=1}^{2022}\ell_i c_i.$ This value is invariant under the allowed recolorings. ... | [] | [
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"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 2/2769226.json"
} |
Let $\omega$ denote the circumscribed circle of triangle $ABC$ , $I$ be its incenter, and $K$ be any point on arc $AC$ of $\omega$ not containing $B$ . Point $P$ is symmetric to $I$ with respect to point $K$ . Point $T$ on arc $AC$ of $\omega$ containing point $B$ is such that $\angle KCT = \a... | <blockquote>Let $\omega$ denote the circumscribed circle of triangle $ABC$ , $I$ be its incenter, and $K$ be any point on arc $AC$ of $\omega$ not containing $B$ . Point $P$ is symmetric to $I$ with respect to point $K$ . Point $T$ on arc $AC$ of $\omega$ containing point $B$ is such that $\an... | [
"<blockquote>Am I tired or something but this problem also does not make any sense for me, could you check your problem.</blockquote>\n\nSorry, I checked and everything seems right, what exactly is your problem?",
"<blockquote><blockquote>Let $\\omega$ denote the circumscribed circle of triangle $ABC$ , $I$ ... | [
"origin:aops",
"2022 Contests",
"2022 Kyiv City MO Round 2"
] | {
"answer_score": 176,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 2/2769227.json"
} |
Positive reals $x, y, z$ satisfy $$ \frac{xy+1}{x+1} = \frac{yz+1}{y+1} = \frac{zx+1}{z+1} $$ Do they all have to be equal?
*(Proposed by Oleksii Masalitin)* | $$ xyz + yz + x + 1 = xy^2 + xy + y + 1,
xyz + xy + y + 1 = y z^2 + yz + z + 1,xyz + xy + z + 1 =z x^2 + zx + x + 1 $$ Summing all equations and after canceling we get $$ xy^2 + yz^2+zx^2=3xyz $$ By AM - GM , we have $$ xy^2 + yz^2+zx^2\geq 3xyz $$
So $ x = y = z.$ <details><summary>Very nice.</summary><... | [
" $\\frac{xy+1}{yz+1}=\\frac{x+1}{y+1} \\to \\frac{y(x-z)}{yz+1}=\\frac{x-y}{y+1}$ same way $\\frac{z(y-x)}{zx+1}=\\frac{y-z}{z+1},\\frac{x(z-y)}{xy+1}=\\frac{z-x}{x+1}$ If we multiply it, then we get $ \\frac{xyz(x-z)(y-x)(z-y)}{(xy+1)(yz+1)(zx+1)}= \\frac{(x-y)(y-z)(z-x)}{(x+1)(y+1)(z+1)}$ or $(x-y)(y-z)(z-x)... | [
"origin:aops",
"2022 Contests",
"2022 Kyiv City MO Round 2"
] | {
"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 2/2769228.json"
} |
Let $AH_A, BH_B, CH_C$ be the altitudes of triangle $ABC$ . Prove that if $\frac{H_BC}{AC} = \frac{H_CA}{AB}$ , then the line symmetric to $BC$ with respect to line $H_BH_C$ is tangent to the circumscribed circle of triangle $H_BH_CA$ .
*(Proposed by Mykhailo Bondarenko)* | [asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(12cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /*... | [
"Let $D$ lie on $BC$ s.t. $AC//DH_C$ .\nBy the condition, we have $DH_B//AB$ .\nThis implies $(AH_BH_C)$ is symmetric to $(DH_BH_C)$ wrt $H_BH_C$ .\nSince $AH\\bot BC$ , we have the tangent at $D$ wrt $(DH_BH_C)$ is parallel to $BC$ , which implies $BC$ is tangent to $(DH_BH_C)$ .\nSo we're done... | [
"origin:aops",
"2022 Contests",
"2022 Kyiv City MO Round 2"
] | {
"answer_score": 86,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 2/2769231.json"
} |
Prime $p>2$ and a polynomial $Q$ with integer coefficients are such that there are no integers $1 \le i < j \le p-1$ for which $(Q(j)-Q(i))(jQ(j)-iQ(i))$ is divisible by $p$ . What is the smallest possible degree of $Q$ ?
*(Proposed by Anton Trygub)* | <blockquote>Prime $p>2$ and a polynomial $Q$ with integer coefficients are such that there are no integers $1 \le i < j \le p-1$ for which $(Q(j)-Q(i))(jQ(j)-iQ(i))$ is divisible by $p$ . What is the smallest possible degree of $Q$ ?
*(Proposed by Anton Trygub)*</blockquote>
<details><summary>Theorem</summa... | [] | [
"origin:aops",
"2022 Contests",
"2022 Kyiv City MO Round 2"
] | {
"answer_score": 66,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 2/2769232.json"
} |
Initially memory of computer contained a single polynomial $x^2-1$ . Every minute computer chooses any polynomial $f(x)$ from its memory and writes $f(x^2-1)$ and $f(x)^2-1$ to it, or chooses any two distinct polynomials $g(x), h(x)$ from its memory and writes polynomial $\frac{g(x) + h(x)}{2}$ to it (no pol... | <blockquote>Initially memory of computer contained a single polynomial $x^2-1$ . Every minute computer chooses any polynomial $f(x)$ from its memory and writes $f(x^2-1)$ and $f(x)^2-1$ to it, or chooses any two distinct polynomials $g(x), h(x)$ from its memory and writes polynomial $\frac{g(x) + h(x)}{2}$ t... | [] | [
"origin:aops",
"2022 Contests",
"2022 Kyiv City MO Round 2"
] | {
"answer_score": 26,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 2/2769233.json"
} |
Find the largest $k$ for which there exists a permutation $(a_1, a_2, \ldots, a_{2022})$ of integers from $1$ to $2022$ such that for at least $k$ distinct $i$ with $1 \le i \le 2022$ the number $\frac{a_1 + a_2 + \ldots + a_i}{1 + 2 + \ldots + i}$ is an integer larger than $1$ .
*(Proposed by Oleksii... | <details><summary>Answer</summary>$k = 1011$</details>
<details><summary>Solution</summary>Let $b_i = \frac{a_1 + a_2 + \ldots + a_i}{1 + 2 + \ldots + i}$ .
Now, note that $k = 1011$ is achievable by letting $a_i = 2i~~\forall i \leq 1011$ and completing the permutation in any way. This will have $b_i = 2$ for ... | [] | [
"origin:aops",
"2022 Contests",
"2022 Kyiv City MO Round 2"
] | {
"answer_score": 54,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 2/2769234.json"
} |
Let $ABCD$ be the cyclic quadrilateral. Suppose that there exists some line $l$ parallel to $BD$ which is tangent to the inscribed circles of triangles $ABC, CDA$ . Show that $l$ passes through the incenter of $BCD$ or through the incenter of $DAB$ .
*(Proposed by Fedir Yudin)* | Do I miss something or the problem is not true. Take any triangle $ABC$ with $BA\neq BC$ , let $D$ be reflection of $B$ over $AC$ . For this quadrilateral, the statement does not hold even though $ABCD$ has incircle and there is line $l$ parallel to $BD$ tangent to incircles of triangles $ABC,CDA$ by sy... | [
"<blockquote>Do I miss something or the problem is not true. Take any triangle $ABC$ with $BA\\neq BC$ , let $D$ be reflection of $B$ over $AC$ . For this quadrilateral, the statement does not hold even though $ABCD$ has incircle and there is line $l$ parallel to $BD$ tangent to incircles of triangles... | [
"origin:aops",
"2022 Contests",
"2022 Kyiv City MO Round 2"
] | {
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Kyiv City MO Round 2/2769235.json"
} |
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that \[ f(a-b)f(c-d) + f(a-d)f(b-c) \leq (a-c)f(b-d) \] for all real numbers $a, b, c,$ and $d$ . | We claim the only solution is given by $f(x)=x$ or $f(x)=0$ . Clearly these work.
Assume $f$ is not identically $0$ . If we plug in $a=c$ , we get
\[f(a-d)\left[f(a-b)+f(b-a)\right]\leq 0\]
Letting $y=a-d$ and $x=b-a$ (and noting these are independent), we get
\[f(y)(f(x)+f(-x))\leq 0\]
Assuming that $f(x)... | [
" $P(a,a,a,a)$ gives $f(0)=0$ .\n\nIf there's a $x$ such that $f(x)<0$ . $P(a,a,c,d)$ gives $f(a-d)f(a-c)<=(a-c)f(a-d)$ \nNow if we take $d$ such that $a-d=x$ we have $f(a-c)<=a-c$ Of we take $d$ such that $f(a-d)>0$ we have $f(a-c)>=a-c$ So $f(x)=x$ If there is no $x$ such that $f(x)<0$ the... | [
"origin:aops",
"2022 Contests",
"2022 Philippine MO"
] | {
"answer_score": 44,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Philippine MO/2804292.json"
} |
The PMO Magician has a special party game. There are $n$ chairs, labelled $1$ to $n$ . There are $n$ sheets of paper, labelled $1$ to $n$ .
- On each chair, she attaches exactly one sheet whose number does not match the number on the chair.
- She then asks $n$ party guests to sit on the chairs so that eac... | Really, the problem is asking for us to create a bunch of cycles (of length $\ge 2$ ) on $n$ vertices such that every vertex is part of exactly one cycle and such that the lcm of cycle lengths is $m$ . If $m$ is not a prime power, say its $m = st$ where $s,t$ are coprime and $> 1$ , since $m > st - s - t$ ,... | [
"@above I think you made a typo when you wrote $m=sx+ty$ . It should be $n=sx+ty$ . Nice solution btw.",
"oh wait pmo is done already?"
] | [
"origin:aops",
"2022 Contests",
"2022 Philippine MO"
] | {
"answer_score": 52,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Philippine MO/2804299.json"
} |
Call a lattice point *visible* if the line segment connecting the point and the origin does not pass through another lattice point. Given a positive integer $k$ , denote by $S_k$ the set of all visible lattice points $(x, y)$ such that $x^2 + y^2 = k^2$ . Let $D$ denote the set of all positive divisors of $202... | If $(x, y)$ is a visible lattice point, then $x$ and $y$ are integers such that $gcd(x, y) = 1$ .
If $gcd(x, y) = 1$ , then $gcd(x, k^2) = gcd(y, k^2) = 1$ . So there exists an integer $m$ such that $y \equiv mx \pmod{k^2}$ . Then $m^2 \equiv -1 \pmod{k^2} \implies m^2 \equiv -1 \pmod{k}$ .
If a prime num... | [
"We can note that the visible condition rewrites to $(x,y)$ is *visible* if and only if $\\gcd(x,y)=1$ .\n\nThus, we're asked to find the number of primitive solutions to $x^2+y^2=k^2$ as $k$ ranges over the divisors of $2021\\cdot 2025$ .\n\nWe first assume $x=0$ (where the case $y=0$ is identical, and... | [
"origin:aops",
"2022 Contests",
"2022 Philippine MO"
] | {
"answer_score": 1074,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Philippine MO/2804304.json"
} |
Let $\triangle ABC$ have incenter $I$ and centroid $G$ . Suppose that $P_A$ is the foot of the perpendicular from $C$ to the exterior angle bisector of $B$ , and $Q_A$ is the foot of the perpendicular from $B$ to the exterior angle bisector of $C$ . Define $P_B$ , $P_C$ , $Q_B$ , and $Q_C$ similarly... | Let $M_A, M_B, M_C$ be the midpoints of $BC, CA, AB$ respectively and let $I_A, I_B, I_C$ be the $A, B, C$ -excenters of $\triangle ABC$ respectively.
Note that $\angle BP_AC = 90^{\circ}$ , so $M_AB = M_AC = M_AP_A$ . So $\angle BM_AP_A = 180^{\circ} - 2\angle CBI_A = 180^{\circ} - 2\left(90^{\circ} - \fr... | [
"Let $I_A,I_B,I_C$ be the excenter of $ABC$ and $X,Y,Z$ be the midpoint of $BC,AC,AB$ .\n\n :arrow: Taylor circle of $I_AI_BI_C$ gives $P_CQ_CP_AQ_AP_BQ_B$ are on a circle.\n\n :arrow: $P_CQ_C//I_AI_B$ :arrow: $XQ_B//AB\\Rightarrow X,Y,Q_B \\text{ collinear}$ :arrow: $XP_C=XQ_B$ :arrow: If $J$ ... | [
"origin:aops",
"2022 Contests",
"2022 Philippine MO"
] | {
"answer_score": 72,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Philippine MO/2804308.json"
} |
Find all positive integers $n$ for which there exists a set of exactly $n$ distinct positive integers, none of which exceed $n^2$ , whose reciprocals add up to $1$ . | Let $S_n = \{a_1, a_2, \dots, a_n\}$ be a set of $n$ distinct positive integers where $a_1 < a_2 <\dots< a_n \le n^2$ and $\sum_{i=1}^n \frac{1}{a_i} = 1$ .
For $n = 1$ , let $S_1 = \{1\}$ .
For $n = 2$ , $\frac{1}{a_1} + \frac{1}{a_2} = 1 \iff (a_1 - 1)(a_2 - 1) = 1 \implies (a_1, a_2) = (2, 2)$ . But $a_... | [
"<details><summary>'Solution'</summary>We claim that all $n \\neq 2$ work. \nIt's not hard to see that $\\frac{1}{a_1} + \\frac{1}{a_2} = 1$ has no solution in positive integers by multiplying and factoring the resulting expression.\nTo prove that all other $n$ work, we will construct the numbers $a_1$ to ... | [
"origin:aops",
"2022 Contests",
"2022 Philippine MO"
] | {
"answer_score": 1056,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Philippine MO/2805000.json"
} |
In $\triangle ABC$ , let $D$ be the point on side $BC$ such that $AB+BD=DC+CA.$ The line $AD$ intersects the circumcircle of $\triangle ABC$ again at point $X \neq A$ . Prove that one of the common tangents of the circumcircles of $\triangle BDX$ and $\triangle CDX$ is parallel to $BC$ . | Let $E$ and $F$ be the intersections of the angle bisectors of $\angle DXB$ and $\angle DXC$ with $(DXB)$ and $(DXC)$ respectively ( $E, F \ne D$ ). Let $Y$ and $Z$ be the projections from $E$ and $F$ to $BC$ respectively. Let $I$ be the incenter of $\triangle ABC$ and $P$ be the foot from ... | [
"Notice that $D$ is the $\\text{A - excenter touch point}$ in $\\triangle ABC$ , and perform a $\\mathcal{I} \\left (D, -\\sqrt{DB\\cdot DC} \\right) $ inversion to finish.\n"
] | [
"origin:aops",
"2022 Contests",
"2022 Philippine MO"
] | {
"answer_score": 90,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Philippine MO/2805001.json"
} |
Let $a, b,$ and $c$ be positive real numbers such that $ab + bc + ca = 3$ . Show that \[ \dfrac{bc}{1 + a^4} + \dfrac{ca}{1 + b^4} + \dfrac{ab}{1 + c^4} \geq \dfrac{3}{2}. \] | $$ bc=x,ab=z,ac=y $$ $\frac{bc}{1+a^4} = \frac{x^3}{x^2+y^2z^2}= x - \frac{xy^2z^2}{x^2+y^2z^2}$ $$ \sum \frac{xy^2z^2}{x^2+y^2z^2} \leq \sum \frac{xy^2z^2}{2xyz} \leq \frac{3}{2} $$ $\blacksquare$ | [
"I can't believe I had to do this again.**Claim:** $\\tfrac{1}{1 + c^2} \\geq 1 - \\tfrac c2$ for $c \\geq 0$ . **Proof:** Just expand, and get $2 \\geq (2-c)(1 + c^2) \\iff c(c-1)^2 \\geq 0$ . Equality when $c = 0, 1$ . \n\nHence, $\\tfrac{1}{1 + x^4} \\geq \\tfrac{2 - x^2}{2}$ with equality at $x = 1$ fo... | [
"origin:aops",
"2022 Contests",
"2022 Philippine MO"
] | {
"answer_score": 112,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2022 Contests/2022 Philippine MO/2805004.json"
} |
The set $S = \{1, 2, \dots, 2022\}$ is to be partitioned into $n$ disjoint subsets $S_1, S_2, \dots, S_n$ such that for each $i \in \{1, 2, \dots, n\}$ , exactly one of the following statements is true:
(a) For all $x, y \in S_i$ , with $x \neq y, \gcd(x, y) > 1.$ (b) For all $x, y \in S_i$ , with $x \neq y... | Since there is no solution still, I will post sketch of my solution. I hope I didn't make any mistake.
The answer is $14$ . To construct $S_i$ s, let's first construct $S'_i$ s. Let $p_i$ be $i$ th prime number. $S'_i$ be set of numbers that are less than $2023$ and divisible by $p_i$ , for $i=1,2,...,12$ .... | [
" $S_{14}'$ seems to have $1, 43$ and $43^2$ , which are not in any of the lower $S_i'$ , so $S_{14}$ fails the condition: ie $gcd(1,43)=1$ and $gcd(43,43^2)=43$ .\n\nThe official solutions doesn't seem to have a complete proof either, I can sketch one at some point if there is enough interest for me to b... | [
"origin:aops",
"2022 Contests",
"2022 Philippine MO"
] | {
"answer_score": 58,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Philippine MO/2805005.json"
} |
Given quadrilateral $ABCD$ inscribed into a circle with diagonal $AC$ as diameter. Let $E$ be a point on segment $BC$ s.t. $\sphericalangle DAC=\sphericalangle EAB$ . Point $M$ is midpoint of $CE$ . Prove that $BM=DM$ . | <details><summary>Solution</summary>Let $P=AE\cap (ABCD)$ $\textbf{Claim:}$ $M$ on $DP$ $\emph{Proof.}$ Let $M'=DP\cap EC$ . Thus $\angle CEP=\angle AEB=\frac \pi 2 -\angle EAB=\frac \pi 2 -\angle DAC=\angle ACD=\angle APD$ . So $\triangle M'PE$ is isosceles. Because $\angle APC=\angle ADC$ , $\triangle ... | [
"Let $AE\\cap (ABCD)=D'$ . Then, because $M$ is the midpoint of $CE$ , $\\angle MD'C = \\angle DAC$ , so $D,M,D'$ collinear. $\\triangle DCM \\cong BD'M$ which finishes the proof.",
" $AE$ and $(ABCD)$ intersection $X$ . $XD$ and $BC$ intersection $Y$ .Very easy $Y$ is midpoint of $CE$ ",
"... | [
"origin:aops",
"2022 Contests",
"2022 Poland - Second Round"
] | {
"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Poland - Second Round/2779163.json"
} |
Let $n$ be an positive integer. We call $n$ $\textit{good}$ when there exists positive integer $k$ s.t. $n=k(k+1)$ . Does there exist 2022 pairwise distinct $\textit{good}$ numbers s.t. their sum is also $\textit{good}$ number?
| <blockquote>Let $n$ be an positive integer. We call $n$ $\textit{good}$ when there exists positive integer $k$ s.t. $n=k(k+1)$ . Does there exist 2022 pairwise distinct $\textit{good}$ numbers s.t. their sum is also $\textit{good}$ number?</blockquote>
Yes,
Choose $2021$ any parwise distinct good numbers... | [
"an example can be constructed for this problem or for any natural number in general",
"This problem was proposed by **[Burii](https://artofproblemsolving.com/community/user/100466)**.",
"Let $G$ be the set of all good numbers.\n\nObviously we must have that if $n\\in G$ then $n$ must be even.\n<span styl... | [
"origin:aops",
"2022 Contests",
"2022 Poland - Second Round"
] | {
"answer_score": 128,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Poland - Second Round/2779164.json"
} |
$n$ players took part in badminton tournament, where $n$ is positive and odd integer. Each two players played two matches with each other. There were no draws. Each player has won as many matches as he has lost. Prove that you can cancel half of the matches s.t. each player still has won as many matches as he has... | <details><summary>Click to expand</summary>Define a graph, where there is a directed edge $A \rightarrow B$ , when $A$ won 2 times with $B$ , and undirected edge $A - B$ , when $A$ won one time and $B$ also won one time. Note that for every pair of vertices there exist exactly one (un)directed edge. Now let's ... | [
"I think its solvable easly with Ore's theorem [https://en.wikipedia.org/wiki/Ore%27s_theorem](https://en.wikipedia.org/wiki/Ore%27s_theorem). Credits to Franek W."
] | [
"origin:aops",
"2022 Contests",
"2022 Poland - Second Round"
] | {
"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Poland - Second Round/2779165.json"
} |
Find all real quadruples $(a,b,c,d)$ satisfying the system of equations $$ \left\{ \begin{array}{ll}
ab+cd = 6
ac + bd = 3
ad + bc = 2
a + b + c + d = 6.
\end{array} \right. $$ | Adding the first two equations yields $(a+d)(b+c)=9.$ Since $a+d+b+c=6,$ \begin{align*}(a+d)(6-(a+d))=9&\implies (a+d)^2-6(a+d)+9=0&\implies (a+d-3)^2=0\end{align*} and $a+d=b+c=3.$ Similarly, $a+c=\{2,4\}$ and $a+b=\{1,5\}.$ Let $a+c=k_1,b+d=6-k_1,a+b=k_2,c+d=6-k_2$ by Vieta. We see $k_1=a+c=a+3-b$ and ... | [
"<blockquote>Find all real quadruples $(a,b,c,d)$ satisfying the system of equations $$ \\left\\{ \\begin{array}{ll}\nab+cd = 6 \nac + bd = 3 \nad + bc = 2 \na + b + c + d = 6.\n\\end{array} \\right.\t $$ </blockquote>\nCute problem. \nWe have $(a + d)(b + c) = (ab + cd) + (ac + bd) = 9$ and $(a + d) + (b + ... | [
"origin:aops",
"2022 Contests",
"2022 Poland - Second Round"
] | {
"answer_score": 130,
"boxed": false,
"end_of_proof": true,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Poland - Second Round/2779436.json"
} |
Given a cyclic quadriteral $ABCD$ . The circumcenter lies in the quadriteral $ABCD$ . Diagonals $AC$ and $BD$ intersects at $S$ . Points $P$ and $Q$ are the midpoints of $AD$ and $BC$ . Let $p$ be a line perpendicular to $AC$ through $P$ , $q$ perpendicular line to $BD$ through $Q$ and $s$ pe... | Let $p,q$ cut $BD,AC$ again at $E,F$ . It is enough to prove $EF \parallel CD$ by orthocenter $\Leftrightarrow \frac{SC}{SD} = \frac{CF}{DE}$ but $\frac{CF}{DE} = \frac{CQ}{DP} = \frac{BC}{AD} = \frac{SC}{SD} \ \blacksquare.$ | [
"This problem was proposed by **[Burii](https://artofproblemsolving.com/community/user/100466)**.",
"Let $p$ meet $AC$ at $K$ and $q$ meet $BD$ at $T$ and $p,q$ meet at $X$ and $XS$ meet $DC$ at $H$ . we want to prove $\\angle XHD = \\angle 90$ . we will prove $XTHD$ is cyclic. $\\angle T... | [
"origin:aops",
"2022 Contests",
"2022 Poland - Second Round"
] | {
"answer_score": 112,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Poland - Second Round/2779445.json"
} |
Positive integers $a,b,c$ satisfying the equation $$ a^3+4b+c = abc, $$ where $a \geq c$ and the number $p = a^2+2a+2$ is a prime. Prove that $p$ divides $a+2b+2$ . | Solved with **[proxima1681](https://artofproblemsolving.com/community/user/849747)**.
We first of all start by finding the parity of $a,b,c$ . Clearly $a$ must be odd because $p$ is prime. Now one can check that $c$ being odd is forced. The parity of $c$ and $a$ combined gives $b$ even.
Rearranging the e... | [
"Interesting. \n\nWe have $$ a\\geq c=\\frac{a^3+4b}{ab-1}. $$ The inequality actually gives us that $a^2b\\geq a^3+4b+a>a^3\\implies b>a$ , which is useful later on. The divisibility gives us that $$ ab-1 \\mid (a^3+4b)a-4(ab-1)=a^4+4. $$ Observe that $p=a^2+2a+2\\mid a^4+4$ since $$ a^2+2a+2\\mid (a^2+... | [
"origin:aops",
"2022 Contests",
"2022 Poland - Second Round"
] | {
"answer_score": 148,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Poland - Second Round/2779448.json"
} |
Let be given a parallelogram $ ABCD$ and two points $ A_1$ , $ C_1$ on its sides $ AB$ , $ BC$ , respectively. Lines $ AC_1$ and $ CA_1$ meet at $ P$ . Assume that the circumcircles of triangles $ AA_1P$ and $ CC_1P$ intersect at the second point $ Q$ inside triangle $ ACD$ . Prove that $ \angle PDA... | Nice angle chasing exercise! :)
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(14.182cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default ... | [
"<span style=\"color:darkblue\">From [**Virgil Nicula**'s extension](http://www.mathlinks.ro/viewtopic.php?p=1486021#1486021) we have $ \\frac {\\sin \\angle ABQ}{\\sin \\angle CBQ} \\equal{} \\frac {AA_1}{CC_1}$ Denote by $ E$ the intersection of the lines $ AD$ and $ CP$ . Then $ \\dfrac{\\sin \\angle ADP}{... | [
"origin:aops",
"2022 Contests",
"2022 Romania EGMO TST"
] | {
"answer_score": 148,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2022 Contests/2022 Romania EGMO TST/276011.json"
} |
A finite set $M$ of real numbers has the following properties: $M$ has at least $4$ elements, and there exists a bijective function $f:M\to M$ , different from the identity, such that $ab\leq f(a)f(b)$ for all $a\neq b\in M.$ Prove that the sum of the elements of $M$ is $0.$ | <blockquote>A finite set $M$ of real numbers has the following properties: $M$ has at least $4$ elements, and there exists a bijective function $f:M\to M$ , different from the identity, such that $ab\leq f(a)f(b)$ for all $a\neq b\in M.$ Prove that the sum of the elements of $M$ is $0.$ </blockquote>
(Not... | [
"Let $M=\\{a_1,a_2,...,a_n\\}$ and $f(a_i)=b_i$ for all $1\\le i\\le n$ . Then $\\sum_{1\\le i<j\\le n}a_ia_j\\le \\sum_{1\\le i<j\\le n}b_ib_j= \\sum_{1\\le i<j\\le n }a_ia_j$ .So $a_ia_j=b_ib_j$ , for all $i\\ne j$ . And it means that $c=\\frac{b_i}{a_i}$ for all $i$ , where $c$ is constant. So $\\s... | [
"origin:aops",
"2022 Contests",
"2022 Romania EGMO TST"
] | {
"answer_score": 60,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Romania EGMO TST/2781220.json"
} |
At first, on a board, the number $1$ is written $100$ times. Every minute, we pick a number $a$ from the board, erase it, and write $a/3$ thrice instead. We say that a positive integer $n$ is *persistent* if after any amount of time, regardless of the numbers we pick, we can find at least $n$ equal numbers ... | We claim that the greatest $n$ is $67$ .
First, we show that we can always find at least $67$ equal numbers. Suppose that at some point that there are at most $66$ of each number, and that there are $m$ numbers written on the board. Then, note that each number written on the board is of the form $\frac1{3^i}$... | [
"gg nice problem\n<details><summary>Solution</summary>Clearly any number that ever appears on the board is of the form $1/3^{i}$ for some $i$ . Further the sum of the numbers on the board, is fixed, it is always $100$ . Letting $a_i$ denote the number of copies of $1/3^{i}$ at a certain fixed moment we have... | [
"origin:aops",
"2022 Contests",
"2022 Romania EGMO TST"
] | {
"answer_score": 68,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Romania EGMO TST/2781223.json"
} |
Let $ABCD$ be a convex quadrilateral and let $O$ be the intersection of its diagonals. Let $P,Q,R,$ and $S$ be the projections of $O$ on $AB,BC,CD,$ and $DA$ respectively. Prove that \[2(OP+OQ+OR+OS)\leq AB+BC+CD+DA.\] | By symmetry, we only need to prove $2(OP+OR) \le AD+BC$ . It follows from these facts:
- Let the angle bisectors of $\angle AOB$ and $\angle COD$ be $OK$ and $OL$ ( $K,L$ lie on $AB,CD$ , respectively). Then $OP \le OK$ , $OQ \le OL$ .
- Let $M,N$ be the midpoint of $AB,CD$ , then $MN=|\overrightarro... | [] | [
"origin:aops",
"2022 Contests",
"2022 Romania EGMO TST"
] | {
"answer_score": 36,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Romania EGMO TST/2781225.json"
} |
For every positive integer $N\geq 2$ with prime factorisation $N=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ we define \[f(N):=1+p_1a_1+p_2a_2+\cdots+p_ka_k.\] Let $x_0\geq 2$ be a positive integer. We define the sequence $x_{n+1}=f(x_n)$ for all $n\geq 0.$ Prove that this sequence is eventually periodic and determin... | <blockquote><blockquote>Let $N$ be a composite number such that $N=p_1\cdots p_m$ , with $p_i$ being prime. Then the function is basically $1+\sum_{i=1}^m p_i$ .
Since $m$ is $\geq 2$ and $p_i\geq 2$ for all $i$ , this is bounded by $f(N)\leq 1+2(m-1)+\frac{N}{2^{m-1}}$ (note that $\frac{N}{2^m}\geq 2$ ... | [
"interesting",
"Let $N$ be a composite number such that $N=p_1\\cdots p_m$ , with $p_i$ being prime. Then the function is basically $1+\\sum_{i=1}^m p_i$ .\nSince $m$ is $\\geq 2$ and $p_i\\geq 2$ for all $i$ , this is bounded by $f(N)\\leq 1+2(m-1)+\\frac{N}{2^{m-1}}$ (note that $\\frac{N}{2^m}\\... | [
"origin:aops",
"2022 Contests",
"2022 Romania EGMO TST"
] | {
"answer_score": 90,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Romania EGMO TST/2781227.json"
} |
Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that all real numbers $x$ and $y$ satisfy \[f(f(x)+y)=f(x^2-y)+4f(x)y.\] | Let $P(x,y)$ be the assertion.
<span style="color:#f00">Claim 1</span>: $f(0)=0$ Proof: $P(0,0)$ gives $f(f(0))=f(0)$ $P(0,-f(0))$ and using above fact gives that $f(0)=0$ . $\blacksquare$ <span style="color:#f00">Claim 2</span> : $\forall x$ , $f(x)=0$ or $x^2$
Proof: $P\left(x,\frac{x^2-f(x)}{2}\right)$... | [
"See [here](https://artofproblemsolving.com/community/q1h1141874p19124638).",
"<blockquote>Determine all functions $f:\\mathbb{R}\\to\\mathbb{R}$ such that all real numbers $x$ and $y$ satisfy \\[f(f(x)+y)=f(x^2-y)+4f(x)y.\\]</blockquote>\n\nCool!\n\nLet \\(P(x,y)\\) denote the assertion of the given functi... | [
"origin:aops",
"2022 Contests",
"2022 Romania EGMO TST"
] | {
"answer_score": 1200,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2022 Contests/2022 Romania EGMO TST/2781228.json"
} |
On a board there is a regular polygon $A_1A_2\ldots A_{99}.$ Ana and Barbu alternatively occupy empty vertices of the polygon and write down triangles on a list: Ana only writes obtuse triangles, while Barbu only writes acute ones.
At the first turn, Ana chooses three vertices $X,Y$ and $Z$ and writes down $\tr... | <blockquote><blockquote>Ana always wins.
Let's say $A$ and $B$ to the Ana and Barbu,respectively. Since $99$ is odd, $A$ will play the last move, so after any move of $B$ , there is at least $1$ unchosen vertex of polygon. Let's show that if $B$ chooses triangle $XYZ$ and $L$ is unchosen point, then at... | [
"Ana wins. At first turn Ana takes A1, A2, A3. Now pair up the other vertices (A4, A5), (A6, A7)…. (A98, A99). I claim that every time Barbu takes some vertice, Ana can take its pair. Wlog assume Barbu makes a triangle XYZ with Y being the new vertice. Let G be the pair of Y. And say Y and G are on different sides ... | [
"origin:aops",
"2022 Contests",
"2022 Romania EGMO TST"
] | {
"answer_score": 90,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Romania EGMO TST/2781229.json"
} |
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