problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
In a triangle, the edges are extended past both vertices by the length of the edge opposite to the respective vertex.
Show that the area of the resulting hexagon is at least $13$ times the area of the original triangle. | if we represent the 3 sides of the triangle as $a,b,c$ then calculate the area of every individual region of the hexagon then we have to prove: $$ \frac{(a+b)^2}{ab}+\frac{(b+c)^2}{bc}+\frac{(c+a)^2}{ca}+\frac{a}c+\frac{c}b+\frac{b}a\ge 15 $$ which is always true. | [
"The only 'tricky' part in this problem was to get rid of the sin's after calculating the area, which can be done by the extended law of sin's (using the circum radius of $ABC$ . Then Am-Gm suffices.",
"<blockquote>The only 'tricky' part in this problem was to get rid of the sin's after calculating the area, whi... | [
"origin:aops",
"2023 German National Olympiad",
"2023 Contests"
] | {
"answer_score": 6,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 German National Olympiad/3092051.json"
} |
For a competition a school wants to nominate a team of $k$ students, where $k$ is a given positive integer. Each member of the team has to compete in the three disciplines juggling, singing and mental arithmetic. To qualify for the team, the $n \ge 2$ students of the school compete in qualifying competitions, det... | I tried to translate Chunji Wang's solution. If there are any mistakes in the translation, let me know. (Question: Why should we take the student with $(a_{i_0},b_{n+1-i_0}$ )? His juggling skill of $a_{i_0}$ is at least as strong as $a_{i_0+1}$ . So shouldn't we take $(a_{i_0+1},b_{n-i_0})$ or am I missing somet... | [
"The answer to $a)$ is <details><summary>Spoiler a)</summary>$n=2,\\ldots, 6$</details> and for $b)$ <details><summary>Spoiler b)</summary>all $n$</details>. Note that $a)$ was the $11$ th grade version, $b)$ the $12$ th grade version, which was significantly harder. Out of all $25$ students in grade ... | [
"origin:aops",
"2023 German National Olympiad",
"2023 Contests"
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"answer_score": 68,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 German National Olympiad/3092053.json"
} |
Determine all triples $(a,b,c)$ of real numbers with
\[a+\frac{4}{b}=b+\frac{4}{c}=c+\frac{4}{a}.\] | I claim that the answers are the permutations of the following: $\boxed{\left(\pm2, \mp4, \mp1\right)}$ , $\boxed{\left(x, x, x\right)}$ , where $x\in\mathbb{R}-\{0\}$ , and $\boxed{\left(\frac{-2(x+2)}{x}, x, \frac{-4}{x+2}\right)}$ , $\boxed{\left(\frac{2(x-2)}{x}, x, \frac{-4}{x-2}\right)}$ , where $x\in\mathb... | [
"<details><summary>Wrong</summary>WLOG $a\\ge b,c$ Case 1: $b \\ge c$ $c\\le a, \\frac{4}{a}\\le \\frac{4}{b}$ so $a=b=c$ Case 2: $ c \\ge b$ $b \\le a, \\frac{4}{c} \\le \\frac{4}{b}$ so $a=b=c$</details>",
"<blockquote>WLOG $a\\ge b,c$ Case 1: $b \\ge c$ $c\\le a, \\frac{4}{a}\\le \\frac{4}{b}$ </bl... | [
"origin:aops",
"2023 German National Olympiad",
"2023 Contests"
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"answer_score": 1186,
"boxed": true,
"end_of_proof": true,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 German National Olympiad/3092054.json"
} |
Let $ABC$ be an acute triangle with altitudes $AA'$ and $BB'$ and orthocenter $H$ . Let $C_0$ be the midpoint of the segment $AB$ . Let $g$ be the line symmetric to the line $CC_0$ with respect to the angular bisector of $\angle ACB$ . Let $h$ be the line symmetric to the line $HC_0$ with respect to ... | $\color{blue}\boxed{\textbf{SOLUTION}}$ $g$ is the symmedian of $\triangle ABC$ and $A'B'$ is antiparallel to $AB$
So, $g$ pass through the midpoint of $A'B'$ call $K$ We need to show $h$ pass through the midpoint of $A'B',K$ that is $\angle AHC_0=\angle BHK$ We have, $\triangle ABH$ is similar to... | [
"<span style=\"color:#00f\">**Solution.**</span> Naturally, $g$ and $h$ are the $A$ and $H$ -symmedians of $\\bigtriangleup ABC$ and $\\bigtriangleup AHB$ , respectively. Since $A'B'$ is antiparallel to $AB$ , it follows that both $g$ and $h$ pass through the midpoint of $A'B'$ . $\\square$ ",
... | [
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"answer_score": 1136,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 German National Olympiad/3092056.json"
} |
The equation $x^3-3x^2+1=0$ has three real solutions $x_1<x_2<x_3$ . Show that for any positive integer $n$ , the number $\left\lceil x_3^n\right\rceil$ is a multiple of $3$ . | Solved with **L567, starchan, Siddharth03**.
The idea is quite clear so this may be a little easy for a 3/6 but oh well. The underlying intuition is that $F_n = \lceil \frac{\varphi^n}{\sqrt{5}} \rceil$ where $F_n$ is the Fibonacci sequence.
<details><summary>Investigation of the three roots </summary>Before begin... | [
"Denote $P(x)=x^3-3x^2+1$ . $P(-1)<0,\\,P(0)>0,\\;P(1)<0\\Longrightarrow x_1\\in(-1,0);\\;x_2\\in(0,1)$ . $P(2)<0,\\,P(3)>0\\Longrightarrow x_3\\in(2,3)$ .\nA more precise calculus gives us: $x_1\\in\\left(-\\dfrac{6}{10},-\\dfrac{5}{10}\\right);\\;x_2\\in\\left(\\dfrac{6}{10},\\dfrac{7}{10}\\right)$ .\nResults: $... | [
"origin:aops",
"2023 German National Olympiad",
"2023 Contests"
] | {
"answer_score": 280,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 German National Olympiad/3092057.json"
} |
Let $ABCD$ be a cyclic quadrilateral. Assume that the points $Q, A, B, P$ are collinear in this order, in such a way that the line $AC$ is tangent to the circle $ADQ$ , and the line $BD$ is tangent to the circle $BCP$ . Let $M$ and $N$ be the midpoints of segments $BC$ and $AD$ , respectively. Prove th... | <details><summary>complex bash</summary>Let $ABCD$ be inscribed in the unit circle, so that
\begin{align*}
|a|=|b|=|c|=|d|&=1
m &= \frac{b+c}2
n &= \frac{a+d}2
\end{align*}
Let $X, Y$ be the circumcenters of $\triangle ADQ, \triangle BCP$ , respectively. Then $X$ lies on the perpendicular bisector of $\overli... | [
"Trig let’s goo ",
"Angle chasing let's goo\n[asy]\nsize(9cm);\nimport olympiad;\nimport geometry;\ndefaultpen(fontsize(10pt));\npair A = dir(230);\npair B = dir(310);\npair C = dir(30);\npair D = dir(170);\npair M = (B+C)/2;\npair N = (A+D)/2;\npair T = (C+D)/2;\npair X = 2 * foot(circumcenter(A,B,T), C,D) - T;\... | [
"origin:aops",
"2023 Contests",
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] | {
"answer_score": 188,
"boxed": false,
"end_of_proof": false,
"n_reply": 44,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3107322.json"
} |
Let $(a_n)_{n\geq 1}$ be a sequence of positive real numbers with the property that $$ (a_{n+1})^2 + a_na_{n+2} \leq a_n + a_{n+2} $$ for all positive integers $n$ . Show that $a_{2022}\leq 1$ . | Notice that the inequality from the problem statement is equivalent to $$ a_{n+1}^2-1\leqslant(a_n-1)(1-a_{n+2})\text{.} $$ Let's consider sequences of symbols $<$ and $>$ , where we assign the symbol $>$ to each $a_i$ iff $a_i > 1$ , and we assign the symbol $<$ to each $a_i$ iff $a_i \leqslant 1$ . The... | [
"<details><summary>Solution</summary>It will be helpful to rewrite the inequality as $$ (1-a_n)(1-a_{n+2}) \\leq 1-a_{n+1}^2 $$ Assume $a_{2022}>1$ . We first claim that\n\n<span style=\"color:red\">**Claim:**</span> We have either\n\n\n- $a_{2020} < 1$ , $a_{2021} > 1$ , $a_{2022} > 1$ , $a_{2023} < 1$ , or\... | [
"origin:aops",
"2023 Contests",
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"answer_score": 1156,
"boxed": false,
"end_of_proof": true,
"n_reply": 39,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3107323.json"
} |
In each square of a garden shaped like a $2022 \times 2022$ board, there is initially a tree of height $0$ . A gardener and a lumberjack alternate turns playing the following game, with the gardener taking the first turn:
- The gardener chooses a square in the garden. Each tree on that square and all the surroundi... | "this is a very skull strat" - mathboy100
Solved with **shreyasharma**.
The gardener can ensure $\boxed{K = \frac{5}{9} \cdot 2022^2}$ majestic trees.**<span style="color:#f00">Gardener's Strategy</span>**
Consider partitioning the $202 2 \times 2022$ grid into $674^2$ individual $3 \times 3$ squares. Also d... | [
"This felt rather straightforward. <details><summary>Solution</summary>Let us prove that the answer is $5 \\cdot 674^2$ . In general, let us replace $2022$ by $3k$ and prove that the answer is $5k^2$ .\n\nLet us first prove that the lumberjack can prevent more thant $5k^2$ trees becoming majestic. To this e... | [
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"answer_score": 1194,
"boxed": false,
"end_of_proof": false,
"n_reply": 29,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3107325.json"
} |
Let $\mathbb Z_{\ge 0}$ be the set of non-negative integers, and let $f:\mathbb Z_{\ge 0}\times \mathbb Z_{\ge 0} \to \mathbb Z_{\ge 0}$ be a bijection such that whenever $f(x_1,y_1) > f(x_2, y_2)$ , we have $f(x_1+1, y_1) > f(x_2 + 1, y_2)$ and $f(x_1, y_1+1) > f(x_2, y_2+1)$ .
Let $N$ be the number of pair... | C9 is not so intimidating after all :D
Call $(a,b)$ an *increasing vector* if $a, b \neq 0$ are of different signs and there exists some $(x_0,y_0)$ such that $f(x_0+a,y_0+b)>f(x_0,y_0)$ . then the problem condition implies that $f(x+a,y+b)>f(x,y)$ for all $x \geq x_0, y \geq y_0$ . Also, say that an incre... | [
"<blockquote>Clearly $f(x,y)<f(x,y+1)$ and $f(x,y)<f(x+1,y)$ ,otherwise $f(x,y)>f(x,y+1)>f(x,y+2)\\cdots$ ,contradiction.\nNow notice that for each pair of $(a,b)$ , the sign of $f(x+a,y+b)-f(x,y)$ (if well defined) are all equal(they are equal to the smallest such vector). We call $(a,b)$ good if that sign... | [
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"answer_score": 232,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3107328.json"
} |
In the acute-angled triangle $ABC$ , the point $F$ is the foot of the altitude from $A$ , and $P$ is a point on the segment $AF$ . The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$ , respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$ , respect... | Oh man... Such a cutie!
*Solution:* Define $T \coloneqq AF \cap XB$ and $Q$ to be the second intersection of $\odot(ABD)$ and $\odot(AEC)$ . Observe that $\triangle PED \sim \triangle ABC$ .
[asy]
import geometry;
import olympiad;
defaultpen(fontsize(11pt));
size(250);
pair A = dir(105);
pair B = dir(210);
p... | [
"This is ISL 2022 G2.\n\n<details><summary>Solution</summary>The solution is done in just two steps:\n\n\n- $\\tfrac{FD}{FC} = \\tfrac{FP}{FA} = \\tfrac{FE}{FB}$ , so $FB\\cdot FD=FC\\cdot FE$ , implying that $F$ lies on the radical axis of $\\odot(ABD)$ and $\\odot(ACE)$ .\n- $BC$ externally bisects both $... | [
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"boxed": false,
"end_of_proof": false,
"n_reply": 58,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3107329.json"
} |
A number is called *Norwegian* if it has three distinct positive divisors whose sum is equal to $2022$ . Determine the smallest Norwegian number.
(Note: The total number of positive divisors of a Norwegian number is allowed to be larger than $3$ .) | <details><summary>Solution</summary>The answer is $\boxed{1344}$ , which works since $1344, 672$ , and $6$ are positive divisors of $1344$ and $$ 1344+ 672 + 6 = 2022. $$ Now, assume there existed a smaller $a < 1344$ which was Norwegian. Then, there exist distinct divisors $x_1, y_1, z_1$ of $a$ for wh... | [
"<blockquote>\nLet $a,b,c$ be distinct naturals so that $a+b+c=2022$ . Minimize $\\text{lcm}(a,b,c)$ .\n</blockquote>\nThe answer is $1344$ , achieved when $a=6, b=672, c=1344$ . Assume $c>\\max(a,b)$ . Because $c>674$ we must have $a,b|c$ if we have $\\text{lcm}(a,b,c)<1344$ .\n\nHence two divisors of ... | [
"origin:aops",
"2023 Contests",
"2023 Germany Team Selection Test"
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"answer_score": 1262,
"boxed": false,
"end_of_proof": false,
"n_reply": 37,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3107332.json"
} |
Prove that $5^n-3^n$ is not divisible by $2^n+65$ for any positive integer $n$ . | <blockquote>Prove that $5^n-3^n$ is not divisible by $2^n+65$ for any positive integer $n$ .</blockquote> $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Here $\left(\frac{a}{b}\right)$ is the Jacobi symbol
Suppose there exists $n$ such that $$ 2^n+65|5^n-3^n $$ If $n=1\Rightarrow 67... | [
"Note that if $n$ is even then $3\\mid2^n+65$ but $3\\not | 5^n-3^n$ , impossible, so $n$ is odd, and clearly $n=1$ fails. Now we claim all $n>1$ fail. To see why, note that\n\n\\[-1 = \\left( \\frac{2^n+65}{5} \\right) = \\left( \\frac{5}{2^n+65} \\right) = \\left( \\frac{5^n}{2^n+65} \\right),\\]\n\nan... | [
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"boxed": false,
"end_of_proof": false,
"n_reply": 37,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3107338.json"
} |
Let $k\ge2$ be an integer. Find the smallest integer $n \ge k+1$ with the property that there exists a set of $n$ distinct real numbers such that each of its elements can be written as a sum of $k$ other distinct elements of the set. | Solved with **GrantStar, OronSH,** and **pikapika007**.
The answer is $\boxed{k + 4}$ .
Construction is \[\begin{cases}
\{-(t+2), -(t+1), \ldots, -1, 1, 2, \ldots, t+2 \}& \text{if } k = 2t
\{-(t+2), -(t+1), \ldots, -1,0,1,2,\ldots, t + 2\} & \text{if } k = 2t + 1
\end{cases}\] for each positive integer $t$ (... | [
"<details><summary>Answer</summary>$n=k+4$</details>\n<details><summary>Solution</summary>Let us first prove that $n \\ge k+4$ . Let the numbers be $a_1<a_2<\\dots<a_n$ so that\n\\[a_1 \\ge a_2+a_3+\\dots+a_{k+1}>a_1+a_3+a_4+\\dots+a_{k+1}\\]\nand hence $a_3+a_4+\\dots+a_{k+1}<0$ . Similarly,\n\\[a_n \\le a_{n-... | [
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"answer_score": 1194,
"boxed": false,
"end_of_proof": false,
"n_reply": 26,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3107342.json"
} |
Let $ABC$ be a triangle and $\ell_1,\ell_2$ be two parallel lines. Let $\ell_i$ intersects line $BC,CA,AB$ at $X_i,Y_i,Z_i$ , respectively. Let $\Delta_i$ be the triangle formed by the line passed through $X_i$ and perpendicular to $BC$ , the line passed through $Y_i$ and perpendicular to $CA$ , and th... | <details><summary>MOHS Estimate</summary>15M, easy side of JMO 3/6 (most JMO 2/5 geometry is 5M, so I can't really say that this would be good for a JMO2/5)
The writeup took almost as much time as the solve</details>
First, we slightly modify the statement that we want to prove. Let $\theta$ be a fixed direction, an... | [
"I remembered this question in my TST and was very disappointed.\n(PS: It was placed as the 3/6 slot)\n\nSketch:\n1) $\\Delta_1$ and $\\Delta_2$ are homothetic\n2) Show that the centre of homothety is indeed the point of tangency by directed angle chase and Simson line. (The centre of homothety lies on $(ABC)$... | [
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"boxed": false,
"end_of_proof": false,
"n_reply": 35,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3107344.json"
} |
Two triangles $ABC, A’B’C’$ have the same orthocenter $H$ and the same circumcircle with center $O$ . Letting $PQR$ be the triangle formed by $AA’, BB’, CC’$ , prove that the circumcenter of $PQR$ lies on $OH$ . | Let $\omega$ be the ellipse which is the locus of points $X$ such that $OX+HX$ is equal to the radius of $(ABCA'B'C')$ , which is tangent to $\overline{BC},\overline{CA},\overline{AB},\overline{B'C'},\overline{C'A'},$ and $\overline{A'B'}$ since the reflection of $H$ over these lines is on $(ABCA'B'C')$ .... | [
"[https://artofproblemsolving.com/community/c6h85652p498701?fbclid=IwAR39bJvyCLUNBID0--tNmKb0qTKKZDT304AjDznnmeHDq43eyvN7BLOf9PY](https://artofproblemsolving.com/community/c6h85652p498701?fbclid=IwAR39bJvyCLUNBID0--tNmKb0qTKKZDT304AjDznnmeHDq43eyvN7BLOf9PY)\n\nIt is an old problem.",
"what if this used as tst que... | [
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"answer_score": 356,
"boxed": false,
"end_of_proof": true,
"n_reply": 19,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3107352.json"
} |
Find all positive integers $n>2$ such that $$ n! \mid \prod_{ p<q\le n, p,q \, \text{primes}} (p+q) $$ | <details><summary>Solution</summary>I assume that an empty product has value $1$ .
The answer is $n = \boxed{1, 7}$ . $n = 1$ works because $\prod_{p < q\leq 1, p, q, \text{ primes }}(p+q)$ is an empty product with value $1$ , and $1! = 1$ , so $1$ divides $1$ . $n = 7$ works because $$ \prod_{p < q\leq... | [
"I claim that only $n=7$ works and it's easy to verify.\n\nLet $p$ be the largest prime atmost $n$ . Since $p\\mid n!$ , there exists two primes $p_1,p_2$ , such that $p\\mid p_1+p_2<2p$ . Hence $p_1+p_2=p$ . Since $p$ is odd, $p_1=2$ . Call $p_2=q$ from now on.\n\nNote similarly, $q\\mid n!$ , so th... | [
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"answer_score": 1348,
"boxed": false,
"end_of_proof": false,
"n_reply": 53,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3107356.json"
} |
Let $m,n \geqslant 2$ be integers, let $X$ be a set with $n$ elements, and let $X_1,X_2,\ldots,X_m$ be pairwise distinct non-empty, not necessary disjoint subset of $X$ . A function $f \colon X \to \{1,2,\ldots,n+1\}$ is called *nice* if there exists an index $k$ such that \[\sum_{x \in X_k} f(x)>\sum_{x \... | It was also South Korea TST D1 P2. I will post my solution on the test :D
Let $A=\{f\mid f:X\rightarrow \{1,2,\cdots,n\}\}$ , and $B$ be a set of *nice* function.
We will prove that there exists an injective map $\phi:A\rightarrow B$ .
If $f_0\in A$ is *nice*, then let $\phi(f_0)=f_0$ .
If $f_0\in A$ is not ... | [
"<details><summary>Solution</summary>The basic idea is that $n^n$ is the number of functions $g:X \\to \\{1,2,\\dots,n\\}$ so that we should try to find an injective map from such functions $g$ to nice functions $f$ . Indeed, this works. For given such $g$ , let us define\n\\[f(x)=\\begin{cases} g(x)+1 & x ... | [
"origin:aops",
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"answer_score": 242,
"boxed": false,
"end_of_proof": false,
"n_reply": 20,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3107370.json"
} |
Let $ABC$ be an acute triangle and let $\omega$ be its circumcircle. Let the tangents to $\omega$ through $B,C$ meet each other at point $P$ . Prove that the perpendicular bisector of $AB$ and the parallel to $AB$ through $P$ meet at line $AC$ . | Let the perpendicular bisector of $\overline{AB}$ intersect $\overline{AC}$ at $K$ , so $\triangle ABK$ is $K$ -isosceles, and $\measuredangle PBC=\measuredangle BAK$ , hence $\triangle ABK \sim \triangle CBP$ , and therefore $\measuredangle BPC=\measuredangle BKA=\measuredangle BKC$ , so $BCPK$ is cyclic.... | [
"The trick is to note that the point of concurrence is actually a point of intersection of circumcircle of $BCP$ and line $AC$ .",
"Trivial by complex numbers, let $A=a^2,B = b^2,C=c^2,P = \\frac{2b^2c^2}{b^2 + c^2}$ . Now consider both arc midpoints of $BA$ , being $ab$ , $-ab$ . We then calculate the int... | [
"origin:aops",
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"answer_score": 124,
"boxed": false,
"end_of_proof": true,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3111755.json"
} |
Let $f(x)$ be a monic polynomial of degree $2023$ with positive integer coefficients.
Show that for any sufficiently large integer $N$ and any prime number $p>2023N$ , the product
\[f(1)f(2)\dots f(N)\]
is at most $\binom{2023}{2}$ times divisible by $p$ .
*Proposed by Ashwin Sah* | This problem is beautiful. I tried it last year, struggled a lot, and came up with something sketchy and handwavy. Now I'm back with a polished solution.
Let $f(x)=a_{2023}x^{2023}+\cdots+a_0$ . I am going to prove that for each $0 \leq m \leq 2022$ there are at most $m$ roots of $f(x)$ modulo $p^{2023-m}$ ... | [
"What if P(1) is a root?",
"Sorry, I missed the condition that the coefficients are **positive**integers.",
"Here's progress that gives an upper bound of $\\tbinom{2024}{2}$ . I'm not sure if this can be improved to the desired bound.\n\nWe claim that for $k=1,2,\\dots,2023$ , at most $k$ of $f(1)$ , $f(2... | [
"origin:aops",
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"answer_score": 274,
"boxed": false,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3111820.json"
} |
Let $P$ be a polynomial with integer coefficients. Assume that there exists a positive integer $n$ with $P(n^2)=2022$ . Prove that there cannot be a positive rational number $r$ with $P(r^2)=2024$ . | <details><summary>Solution</summary>Let $$ P(x)=(x-n^2)Q(x)+2022=(x-n^2)(a_0+a_1x+\ldots+a_dx^d)+2022 $$ for integers $a_0, a_1, \ldots, a_d$ and let $r=\frac{p} {q}$ for $(p, q)=1$ . The second equation $P(r^2)=2024$ rewrites as $$ (p^2-q^2n^2)(a_0q^{2d}+a_1p^2q^{2d-2}+\ldots+a_dp^{2d})=2q^{2d+2}. $$ Si... | [] | [
"origin:aops",
"2023 Contests",
"2023 Germany Team Selection Test"
] | {
"answer_score": 22,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3189760.json"
} |
Does there exist a positive odd integer $n$ so that there are primes $p_1$ , $p_2$ dividing $2^n-1$ with $p_1-p_2=2$ ? | <details><summary>Solution</summary>It is sufficient to find a pair of twin primes $p, q$ such that the orders of $2$ modulo $p$ and $q$ are odd, as then we can take $n$ to be the product of the two orders. Notice that it is a necessary condition that $2^{\frac{p-1}{2}}\equiv 1 \pmod p$ and $2^{\frac{q-1}{... | [
" $n=315$ , $p_1=73$ , $p_2=71$ "
] | [
"origin:aops",
"2023 Contests",
"2023 Germany Team Selection Test"
] | {
"answer_score": 36,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3189765.json"
} |
Let $A$ be a non-empty set of integers with the following property: For each $a \in A$ , there exist not necessarily distinct integers $b,c \in A$ so that $a=b+c$ .
(a) Proof that there are examples of sets $A$ fulfilling above property that do not contain $0$ as element.
(b) Proof that there exist $a_1,\... | [
"Part a) is easy, just take $\\mathbb Z\\setminus\\{0\\}$ ."
] | [
"origin:aops",
"2023 Contests",
"2023 Germany Team Selection Test"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3189770.json"
} | |
Let $ABC$ be an acute angled triangle with orthocenter $H$ and $AB<AC$ . The point $T$ lies on line $BC$ so that $AT$ is a tangent to the circumcircle of $ABC$ . Let lines $AH$ and $BC$ meet at point $D$ and let $M$ be the midpoint of $HC$ . Let the circumcircle of $AHT$ meets $CH$ in $P \not=... | Oh wow, what a problem, its just angle chase but unless u have a fixed plan to solve this its hard to find motivation for the following solution:
Let $AC \cap (AHT)=J$ , first note that $M$ is center of $(CDH)$ so we have $\angle CPQ=\angle MDC=\angle MCD$ so $PQ=QC$ , now notice that $\angle AJP=180-\angle AH... | [] | [
"origin:aops",
"2023 Contests",
"2023 Germany Team Selection Test"
] | {
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3189775.json"
} |
Let $ABC$ be a acute angled triangle and let $AD,BE,CF$ be its altitudes. $X \not=A,B$ and $Y \not=A,C$ lie on sides $AB$ and $AC$ , respectively, so that $ADXY$ is a cyclic quadrilateral. Let $H$ be the orthocenter of triangle $AXY$ .
Prove that $H$ lies on line $EF$ . | I'm so sorry
By phantom points it suffices to prove that for every point $H' \in \overline{EF}$ , if we let $X',Y'$ lie on $\overline{AB},\overline{AC}$ respectively such that $\overline{X'H'} \perp \overline{AC}$ and $\overline{Y'H'} \perp \overline{AB}$ then $AX'Y'D$ is cyclic. Place the problem in the co... | [
"Let $(ADXY)$ intersect $BC$ again at $Z$ . As $Z$ moves linearly, $ZX$ and $ZY$ are parallel to fixed lines, so $X$ and $Y$ also move linearly. The $X$ and $Y$ altitudes of $AXY$ move linearly since they are parallel to $CF$ and $BE$ , which means that $H$ also moves linearly. When $Z=B$... | [
"origin:aops",
"2023 Contests",
"2023 Germany Team Selection Test"
] | {
"answer_score": 182,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/3189778.json"
} |
In a triangle $\triangle ABC$ with orthocenter $H$ , let $BH$ and $CH$ intersect $AC$ and $AB$ at $E$ and $F$ , respectively. If the tangent line to the circumcircle of $\triangle ABC$ passing through $A$ intersects $BC$ at $P$ , $M$ is the midpoint of $AH$ , and $EF$ intersects $BC$ at $G$... | Fortunately, trig and radical axes are both not required. :D
<details><summary>Solution</summary>Define $D$ to be the intersection of $AH$ and $BC$ and $Q$ to be the intersection of $DF$ and $AP$ .
Since $F$ is on $AB$ , $\angle FAP=\angle BAP$ . Since $AP$ is tangent to the circumcircle of $\triang... | [
"This problem can be killed by just a few lines of trigonometry,but since this is my friend's problem,I shall solve it in a nice way. :) \n\nLet $M_a$ be the midpoint of $BC$ . $FE$ is the radical axis of the nine-point circle and that of $BCEF$ ,while $BC$ is the radical axis of the nine-point circle and $... | [
"origin:aops",
"2023 Contests",
"2023 Germany Team Selection Test"
] | {
"answer_score": 168,
"boxed": false,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Germany Team Selection Test/617846.json"
} |
Find all quadruplets (x, y, z, w) of positive real numbers that satisfy the following system: $\begin{cases}
\frac{xyz+1}{x+1}= \frac{yzw+1}{y+1}= \frac{zwx+1}{z+1}= \frac{wxy+1}{w+1}
x+y+z+w= 48
\end{cases}$ | <blockquote>Find all quadruplets (x, y, z, w) of positive real numbers that satisfy the following system: $\begin{cases}
\frac{xyz+1}{x+1}= \frac{yzw+1}{y+1}= \frac{zwx+1}{z+1}= \frac{wxy+1}{w+1}
x+y+z+w= 48
\end{cases}$ </blockquote>
Let $u>0$ be the common value of first line. So $xyz+1=u(x+1)$ , $yzw+1=u(y+1)$ , ... | [
" $\\frac{xyz+1}{x+1}= \\frac{yzw+1}{y+1}$ $\\implies$ $xy^2z+xyz+y+1=xyzw+yzw+x+1$ $(1)$ $\\frac{yzw+1}{y+1}= \\frac{zwx+1}{z+1}$ $\\implies$ $yz^2w+yzw+z+1=xyzw+zwx+y+1$ $(2)$ $\\frac{zwx+1}{z+1}= \\frac{wxy+1}{w+1}$ $\\implies$ $zw^2x+zwx+w+1=xyzw+wxy+z+1 $ $(3)$ $\\frac{wxy+1}{w+1}=\\frac{xy... | [
"origin:aops",
"2023 Contests",
"2023 Greece National Olympiad"
] | {
"answer_score": 1064,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Greece National Olympiad/3017817.json"
} |
Find all positive integers $N$ that are perfect squares and their decimal representation consists of $n$ digits equal to 2 and one digit equal to 5, where $n$ takes positive integer values. | <blockquote>Find all positive integers $N$ that are perfect squares and their decimal representation consists of $n$ digits equal to 2 and one digit equal to 5, where $n$ takes positive integer values.</blockquote>
Let $N=22\cdots 25$ , the number of ${}2$ is ${}n+1$ , where $n\geq 0$ .
Then $N=\frac{2(10^... | [
" $(x+5)(x-5) = 2^{n+1} \\cdot 5^{n}$ easily solvable as $gcd(x+5,x-5) = gcd(x+5,10) = 2,5,10$ and then easy bounding.",
"<blockquote> $(x+5)(x-5) = 2^{n+1} \\cdot 5^{n}$ easily solvable as $gcd(x+5,x-5) = gcd(x+5,10) = 2,5$ and then easy bounding.</blockquote>\n It turns out that $gcd(x - 5 , x + 5) = 10$... | [
"origin:aops",
"2023 Contests",
"2023 Greece National Olympiad"
] | {
"answer_score": 140,
"boxed": false,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Greece National Olympiad/3017820.json"
} |
A triangle $ABC$ with $AB>AC$ is given, $AD$ is the A-angle bisector with point $D$ on $BC$ and point $I$ is the incenter of triangle $ABC$ . Point M is the midpoint of segment $AD$ and point $F$ is the second intersection of $MB$ with the circumcirle of triangle $BIC$ . Prove that $AF\bot FC$ . | Note that the condition that $AF \perp FC$ simply says that $F$ lies on the circle with diameter $AC$ . Thus, we wish to show that points $A$ , $F$ , $H$ and $C$ are concyclic, where $H$ is the foot of the $A-$ altitude. Let $N_A$ denote the $BC$ arc midpoint which is well known to be the center of ... | [
"Obviously $AD \\cap (BIC)=I_A$ is the $A$ -excenter. We want $\\angle AFC=90^{o}$ and we have that $\\angle MFC=90^{o}-\\frac{\\alpha}{2}$ , so we want $\\angle AFM =\\angle BAM \\iff MD^2=MA^2=MF \\cdot MB=MI \\cdot MI_A$ , but the last one easily follows since $(AD; II_A)=-1$ .",
"Let $L$ be the midp... | [
"origin:aops",
"2023 Contests",
"2023 Greece National Olympiad"
] | {
"answer_score": 96,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 Greece National Olympiad/3017822.json"
} |
A class consists of 26 students with two students sitting on each desk. Suddenly, the students decide to change seats, such that every two students that were previously sitting together are now apart. Find the maximum value of positive integer $N$ such that, regardless of the students' sitting positions, at the end t... | We have a $2$ -regular graph $G$ with $26$ vertices, such that there is one blue and one red edge incident from every vertex (and the two sets of $13$ monochromatic edges form perfect matchings). We claim $G$ is bipartite. Indeed, otherwise there is an odd cycle and thus there are two adjacent edges with the s... | [
"See [here](https://artofproblemsolving.com/community/c6h520389p2931044). Quite an unoriginal problem."
] | [
"origin:aops",
"2023 Contests",
"2023 Greece National Olympiad"
] | {
"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Greece National Olympiad/3017823.json"
} |
Let $\mathbb{Q}^{+}$ denote the set of positive rational numbers. Find, with proof, all functions $f:\mathbb{Q}^+ \to \mathbb{Q}^+$ such that, for all positive rational numbers $x$ and $y,$ we have \[f(x)=f(x+y)+f(x+x^2f(y)).\] | <details><summary>Solution</summary>The answer is $\boxed{f(x) = \frac{1}{x}}$ only, which works since it maps $\mathbb{Q}^{+}$ to $\mathbb{Q}^{+}$ , and for all $x,y \in \mathbb{Q}^{+}$ ,
\begin{align*}
f(x+y) + f(x+x^2f(y)) &= \frac{1}{x+y} + \frac{1}{x+\frac{x^2}{y}}
&= \frac{x}{x^2+xy}... | [
"See [2017 Swiss TST P6](https://artofproblemsolving.com/community/c6h1828555p12238401)",
"bruh are you kidding me rn",
"For the record I had no faith that this problem is original. Found the thread in under 15 seconds with [Approach0](https://approach0.xyz/search/).",
"Despite the fact that the exact same FE... | [
"origin:aops",
"2023 Contests",
"2023 HMIC"
] | {
"answer_score": 1264,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 HMIC/3059756.json"
} |
A prime number $p$ is mundane if there exist positive integers $a$ and $b$ less than $\tfrac{p}{2}$ such that $\tfrac{ab-1}{p}$ is a positive integer. Find, with proof, all prime numbers that are not mundane. | <details><summary>Solution</summary>The answer is $p = \boxed{2, 3, 5, 7, 13}$ .
Note that if $a$ and $b$ are positive integers, then $\frac{ab-1}{p} = 0$ if $a=b=1$ and $\frac{ab-1}{p} > 0$ otherwise. Thus, a prime $p$ is mundane if and only if there exists positive integers $1 < a,b < \frac{p}{2}$ for... | [
"2,3,5,7,13",
"...............",
"<blockquote>2,3,5,7,13</blockquote>\n\nThe proof is left as an exercise to the reader.\n",
"When $p<20$ check by hand that only $2,3,5,7,$ and $13$ work. When $p>20$ , take \\[(a,b)=\\begin{cases}\n\\left(\\frac{2p-2}{5},\\frac{p-5}{2}\\right) & p \\equiv 1 \\pmod{5} \... | [
"origin:aops",
"2023 Contests",
"2023 HMIC"
] | {
"answer_score": 1314,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 HMIC/3059757.json"
} |
Triangle $ABC$ has incircle $\omega$ and $A$ -excircle $\omega_A.$ Circle $\gamma_B$ passes through $B$ and is externally tangent to $\omega$ and $\omega_A.$ Circle $\gamma_C$ passes through $C$ and is externally tangent to $\omega$ and $\omega_A.$ If $\gamma_B$ intersects line $BC$ again at ... | My problem! Here is the original statement:
<blockquote>
A circle through $B$ and a circle through $C$ are both externally tangent to the incircle and $A$ -excircle of $\triangle ABC$ . Prove $\overline{BC}$ cuts the circles into congruent chords.
</blockquote>
<details><summary>Diagram</summary>[asy]
size(300... | [
"Hope this is not a fakesolve.\nset BP=x CQ=y and using casey on incircle excircle and point circles B and P (C ans Q) we can get two functions.\nSince the tangency points of incircle and excircle are symmetric wrt midpoint of BC and the way the ciecles $\\gamma_B$ and $\\gamma_C$ get tangent are the same the t... | [
"origin:aops",
"2023 Contests",
"2023 HMIC"
] | {
"answer_score": 98,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 HMIC/3059758.json"
} |
Let $n>1$ be a positive integer. Claire writes $n$ distinct positive real numbers $x_1, x_2, \dots, x_n$ in a row on a blackboard. In a $\textit{move},$ William can erase a number $x$ and replace it with either $\tfrac{1}{x}$ or $x+1$ at the same location. His goal is to perform a sequence of moves such t... | Reindex to $0$ . We note that one main roadblock is that the $+1$ operation quickly becomes expensive as numbers get big, so we begin with a crucial claim:**<span style="color:#f00">Claim 1:</span>**
It is possible to transform any sequence $x_0, x_1, \dots , x_{n-1}$ into $y_0, y_1, \dots , y_{n-1}$ such that ... | [
"::::::::::::::::",
"(a) Divide and conquer. Put the first half of the numbers in decreasing order, add $1$ to all of them, then take reciprocals so that they form an increasing sequence of numbers below $1$ . Put the second half of the numbers in increasing order, and add $1$ to all of them so that they for... | [
"origin:aops",
"2023 Contests",
"2023 HMIC"
] | {
"answer_score": 80,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 HMIC/3059764.json"
} |
Let $a_1, a_2, \dots$ be an infinite sequence of positive integers such that, for all positive integers $m$ and $n,$ we have that $a_{m+n}$ divides $a_ma_n-1.$ Prove that there exists an integer $C$ such that, for all positive integers $k>C,$ we have $a_k=1.$ | Solved <details><summary>with</summary>**GuvercinciHoca**, **mhmdredstone** and **SerdarBozdag**</details>
<details><summary>Solution</summary>Observe that all terms of the sequence are pairwise coprime, so if the sequence is bounded we are done. Assume otherwise that the sequence is unbounded.
1) There exists $j$ s... | [
"This is a nice one. If I didn't fakesolve it, I think the hardest part is to focus on the idea of bound the function above.\n\n*Solution:*\n\nLet $a_n=f(n)$ . The equation is the same as $$ f(m+n) | f(m)f(n)-1 $$ <span style=\"color:#f00\">**Claim 1:**</span> $f(m) \\le f(1)^m$ for all $m$ .\n\n*Proof:* Indu... | [
"origin:aops",
"2023 Contests",
"2023 HMIC"
] | {
"answer_score": 240,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 HMIC/3059765.json"
} |
Mandy needs to wake up early for attending a mathematics contest. She has set an alarm in her smartphone every 15 minutes since 5:30 am. If an alarm is not pressed off by her or her mother (or anything else), it will ring for a while, stop for a while, then will ring again 9 minutes later as the first ring, and so on (... | [
"was it 2023 tst or 2022's\n",
"Mandy had set her alarm at 5:30 a.m.\nMandy had wake up at 6:30 a.m.\nEvery 15 minutes it will ring.\nIt will ring 9 minutes gap.\nTherefore,15-9=6\nx=6"
] | [
"origin:aops",
"2023 Contests",
"2023 Hong Kong Team Selection Test"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Hong Kong Team Selection Test/2926852.json"
} | |
Let $n$ be a positive integer. Show that if p is prime dividing $5^{4n}-5^{3n}+5^{2n}-5^{n}+1$ , then $p\equiv 1 \;(\bmod\; 4)$ . | Let $n$ be a positive integer. Show that if p is prime dividing $5^{4n}-5^{3n}+5^{2n}-5^{n}+1$ , then $p\equiv 1 \;(\bmod\; 4)$ .
Note that $p>2$ and
\[p\mid\frac{5^{5n}+1}{5^n+1}.\]
We split into cases.**Case 1.** $p\mid 5^n+1$ . Then $\nu_p$ of the RHS thing is exactly $\nu_p(5)$ by LTE, so we must have ... | [
"Clearly $p\\equiv 1\\pmod{10}$ , as $(5^{n})^5\\equiv -1\\pmod p$ and is not $-1$ . If $p\\equiv 3\\pmod 4$ , note that any primitive 10th root of unity, one of which is $5^n$ , is not a quadratic residue (as if $g$ is a generator, and $g^k=5^n$ , then $10k=p-1$ so $k$ is odd). However, by quadratic r... | [
"origin:aops",
"2023 Contests",
"2023 Hong Kong Team Selection Test"
] | {
"answer_score": 158,
"boxed": false,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Hong Kong Team Selection Test/2926857.json"
} |
A point $P$ lies inside an equilateral triangle $ABC$ such that $AP=15$ and $BP=8$ . Find the maximum possible value of the sum of areas of triangles $ABP$ and $BCP$ . | $[ABP]+[BCP] = \frac{8*s*\sin(60-x)+8s*\sin(x)}{2} \leq^{Jensen} \frac{16s*\sin(30)}{2} =4s$
achieved when $x=30$ $15^2 =8^2+s^2-2*8*s*\cos(30)$ $s=4\sqrt{3}+\sqrt{209}$ $[ABP]+[BCP] = \frac{8*s*\sin(60-x)+8s*\sin(x)}{2} \leq^{Jensen} \frac{16s*\sin(30)}{2} =4s=4(4\sqrt{3}+\sqrt{209})$
is there something wrong... | [] | [
"origin:aops",
"2023 Contests",
"2023 Hong Kong Team Selection Test"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Hong Kong Team Selection Test/2926860.json"
} |
Let $x$ , $y$ , $z$ be real numbers such that $x+y+z \ne 0$ . Find the minimum value of $\frac{|x|+|x+4y|+|y+7z|+2|z|}{|x+y+z|}$ | WLOG, let $x+y+z=1$ so $\frac{|x|+|x+4y|+|y+7z|+2|z|}{|x+y+z|}=|x|+2|z|+|4-3x-4z|+|1-x+6z|$ suppose $f(z)=|x|+2|z|+|4-3x-4z|+|1-x+6z|$ it's trival that $f(z) \geq f(\dfrac {x-1} 6)$ so $|x|+2|z|+|4-3x-4z|+|1-x+6z| \geq |x|+\dfrac {|x-1|} 3 +|\dfrac {11x} 3-\dfrac {14} 3|$ suppose $g(x)=|x|+\dfrac {|x-1|} 3 +|\dfr... | [
"Posted previously by me: [https://artofproblemsolving.com/community/u488422h2923667p26140565](https://artofproblemsolving.com/community/u488422h2923667p26140565)"
] | [
"origin:aops",
"2023 Contests",
"2023 Hong Kong Team Selection Test"
] | {
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Hong Kong Team Selection Test/2926896.json"
} |
Let $ABCD$ be a convex quadrilateral with $AB=5$ , $AD=17$ , and $CD=6$ . If the angle bisector of $\angle{BAD}$ and $\angle{ADC}$ intersect at the midpoint of $BC$ , find the area of $ABCD$ . | [
"This is AIME 2022 II/11 but with different numbers.\n\nI believe Dr. Ching does not encourage posting problems that are reused, but given that this is for the collection, that should be fine I guess."
] | [
"origin:aops",
"2023 Contests",
"2023 Hong Kong Team Selection Test"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Hong Kong Team Selection Test/2926925.json"
} | |
(a) Find the smallest number of lines drawn on the plane so that they produce exactly 2022 points of intersection. (Note: For 1 point of intersection, the minimum is 2; for 2 points, minimum is 3; for 3 points, minimum is 3; for 4 points, minimum is 4; for 5 points, the minimum is 4, etc.)
(b) What happens if the lines... | It is obvious that using $k$ lines, one can produce at most $\frac{k(k-1)}{2}$ points, so we claim that for all numbers $n$ such that $\frac{(k-1)k}{2} < n \leq \frac{k(k+1)}{2}$ the answer is $k+1$ .
We will prove it by induction on $k$ . Consider all numbers from $\frac{(k-2)(k-1)}{2} + 1$ to $\frac{(k-... | [] | [
"origin:aops",
"2023 Contests",
"2023 Hong Kong Team Selection Test"
] | {
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Hong Kong Team Selection Test/2926927.json"
} |
Given a $24 \times 24$ square grid, initially all its unit squares are coloured white. A move consists of choosing a row, or a column, and changing the colours of all its unit squares, from white to black, and from black to white. Is it possible that after finitely many moves, the square grid contains exactly $574$ ... | **Answer:** No, it is impossible to make the square grid contains exactly 574 black units.
Denote the move that changing the colours of all unit squares in row $n$ is $A_n$ , the move that changing the colours of all unit squares in column $m$ is $B_m$ .
For any moves $C_1,C_2,\cdots,C_j\in\{A_1,A_2,\cdots,A_{2... | [
"We have 24×24=576 squares in the 24×24 square grid.\nWe have to colour all squares black or white.\nTherefore,exactly 574 is not possible.",
"<blockquote>We have 24×24=576 squares in the 24×24 square grid.\nWe have to colour all squares black or white.\nTherefore,exactly 576 is not possible.</blockquote>\n\nwhat... | [
"origin:aops",
"2023 Contests",
"2023 Hong Kong Team Selection Test"
] | {
"answer_score": 50,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Hong Kong Team Selection Test/2946152.json"
} |
Giiven $\Delta ABC$ , $\angle CAB=75^{\circ}$ and $\angle ACB=45^{\circ}$ . $BC$ is extended to $T$ so that $BC=CT$ . Let $M$ be the midpoint of the segment $AT$ . Find $\angle BMC$ . | Let us draw altitude from vertex $A$ to $BC$ and call intersection point $H$ .Then $\angle BAH = 30^\circ$ and $\angle HAC = 45^\circ$ , So if $BH = x$ then $AB = 2x , AH = x\sqrt3 , HC = x\sqrt3 , AC = x\sqrt6$ .Since $BC=CT , AM=MT\implies$ $MC$ is midsegment of $\triangle ABT(MC =x)$ .Thus $BM$ an... | [
" $C$ is the midpoint of $BT$ , so $CM$ and $AB$ are parallel.\nLet $H$ be the orthogonal projection $A$ to $BC$ , and $\\angle ABC = 180^\\circ - 45^\\circ - 75^\\circ = 60^\\circ$ , so if $AB = 2a$ , then $BH = a$ , $AH = a\\sqrt3 = HC$ , $AC = a\\sqrt6$ . \nLet $D$ be the intersection of $AC$ ... | [
"origin:aops",
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"path": "Contest Collections/2023 Contests/2023 Hong Kong Team Selection Test/2946154.json"
} |
Let $n\ge 4$ be a positive integer. Consider any set $A$ formed by $n$ distinct real numbers such that the following condition holds: for every $a\in A$ , there exist distinct elements $x, y, z \in A$ such that $\left| x-a \right|, \left| y-a \right|, \left| z-a \right| \ge 1$ . For each $n$ , find the great... | [] | [
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} | |
A two digit number $s$ is special if $s$ is the common two leading digits of the decimal expansion of $4^n$ and $5^n$ , where $n$ is a certain positive integer. Given that there are two special number, find these two special numbers. | Firstly, note that $4^n\cdot (5^n)^2=10^{2n}$ , so if we let $4^n=a_n\cdot 10^{\alpha_n}$ and $5^n=b_n\cdot 10^{\beta_n}$ , with $1\leq a_n,b_n<10$ , we must have $a_nb_n^2=10^t$ where $0\leq t<3$ . Using the fact that $s=\lfloor 10a_n\rfloor=\lfloor 10b_n\rfloor$ , we must have $10^3\leq s^3\leq 10^{t+3}<(s+... | [] | [
"origin:aops",
"2023 Contests",
"2023 Hong Kong Team Selection Test"
] | {
"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Hong Kong Team Selection Test/2946156.json"
} |
Suppose $a$ , $b$ and $c$ are nonzero real numberss satisfying $abc=2$ . Prove that among the three numbers $2a-\frac{1}{b}$ , $2b-\frac{1}{c}$ and $2c-\frac{1}{a}$ , at most two of them are greater than $2$ . | <details><summary>solution</summary>We assume that $2a - \frac{1}{b}, 2b - \frac{1}{c}, 2c - \frac{1}{a} > 2.$ Multiplying the inequalities together, we have
\[(2a - \frac{1}{b})(2b - \frac{1}{c})(2c - \frac{1}{a}) > 8.\]
Expanding, we get
\[8abc - 4(a+b+c) + 2(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) - \frac{1}{abc} ... | [
"Suppose $2b-\\frac{1}{c}$ , $2a-\\frac{1}{b}$ , $2b-\\frac{1}{c}$ and $2c-\\frac{1}{a} >2$ Multiply and expanding: $8abc-\\frac{1}{abc}-4(a+b+c)+\\frac{2}{a}+\\frac{2}{b}+\\frac{2}{c}>8$ Summing up and expanding: $2a+2b+2c-\\frac{1}{a}-\\frac{1}{b}-\\frac{1}{c}>6$ Denote $A=2a+2b+2c-\\frac{1}{a}-\\frac{1}{... | [
"origin:aops",
"2023 Contests",
"2023 Hong Kong Team Selection Test"
] | {
"answer_score": 114,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2023 Contests/2023 Hong Kong Team Selection Test/2971450.json"
} |
Find the period of the repetend of the fraction $\frac{39}{1428}$ by using *binary* numbers, i.e. its binary decimal representation.
(Note: When a proper fraction is expressed as a decimal number (of any base), either the decimal number terminates after finite steps, or it is of the form $0.b_1b_2\cdots b_sa_1a_2\c... | <details><summary>Outline of solution</summary>First, reduce this fraction into $\frac{13}{476}$ . Note that $476 = 119\cdot 4$ and hence the binary expansion of $\frac{13}{476}$ should be exactly the same as $\frac{13}{119}$ except it is shifted two digits rightwards. Then consider the form given in the questio... | [
"btw some dude calculated $\\frac{39}{1428}$ :agent: ",
"The period is order of $2$ modulo $119$ ."
] | [
"origin:aops",
"2023 Contests",
"2023 Hong Kong Team Selection Test"
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"answer_score": 24,
"boxed": false,
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"n_reply": 3,
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} |
Given a $2023 \times 2023$ square grid, there are beetles on some of the unit squares, with at most one beetle on each unit square. In the first minute, every beetle will move one step to its right or left adjacent square. In the second minute, every beetle will move again, only this time, in case the beetle moved ri... | In my opinion very high quality. Color the grid in a checkerboard pattern. Let a beetle be happy if it is in a black cell whose row has $1012$ black cells, sad if in a black cell whose row has $1011$ black cells. A beetle in a black cell must alternate between happy and sad every two moves, so we can have at most ... | [
"This is also [Dutch TST 2024 3.1](https://artofproblemsolving.com/community/c6h3347970).",
"<blockquote>Given a $2023 \\times 2023$ square grid, there are beetles on some of the unit squares, with at most one beetle on each unit square. In the first minute, every beetle will move one step to its right or left ... | [
"origin:aops",
"2023 Contests",
"2023 Hong Kong Team Selection Test"
] | {
"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Hong Kong Team Selection Test/2971455.json"
} |
Let $ABCD$ be a quadrilateral inscribed in a circle $\Gamma$ such that $AB=BC=CD$ . Let $M$ and $N$ be the midpoints of $AD$ and $AB$ respectively. The line $CM$ meets $\Gamma$ again at $E$ . Prove that the tangent at $E$ to $\Gamma$ , the line $AD$ and the line $CN$ are concurrent. | <details><summary>Lemma</summary>If $(D, A; E, B) = -1$ , and it forms a cyclic quadrilateral that is harmonic, then the tangents at $E$ and $B$ are concurrent with line segment $AD$ .</details>
We just have to show that $CN$ intersects point $P$ , where $P$ is the intersection of the tangent lines to $AD$ ... | [
"I found this familiar during the test. Nevertheless, I couldn't find the synthetic solution, and I length chased it (compute $XE^2$ by LoC twice on $XEM$ and $DMC,$ where $X=AD\\cap CN.$ ). In fact, this is equivalent to JMO 2011/5.",
"First, note that $(D,A;E,B)=C(D,A;M,\\infty)=-1$ , so $ABDE$ is har... | [
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"2023 Contests",
"2023 Hong Kong Team Selection Test"
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"answer_score": 132,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Hong Kong Team Selection Test/2971456.json"
} |
Let $r > 0$ be a real number. All the interior points of the disc $D(r)$ of radius $r$ are colored with one of two colors, red or blue.
- If $r > \frac{\pi}{\sqrt{3}}$ , show that we can find two points $A$ and $B$ in the interior of the disc such that $AB = \pi$ and $A,B$ have the same color
- Does t... | In both cases, we can find such $A,B$ . Let $k$ be sufficiently large integer. Then observe that for a regular $2k+1$ gon, with vertices $A_1A_2 \cdots A_{2k+1}$ inscribed in a circle of radius $R$ , the ratio $\frac{A_1A_{k+1}}{R}$ can be made $> 2- \epsilon$ for any positive $\epsilon$ . Since we know $... | [
"I will skip the details.\n\nAssume otherwise. If $O$ is the center of the disk, construct a circle $\\omega$ with center $O$ radius $\\frac{\\pi}{2}<R<r$ . Then for any close enough two points on $\\omega$ we can find another point in disk which has the same distance $\\pi$ to these points. Thus these t... | [
"origin:aops",
"2023 Contests",
"2023 India EGMO TST"
] | {
"answer_score": 132,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 India EGMO TST/2975763.json"
} |
Alice has an integer $N > 1$ on the blackboard. Each minute, she deletes the current number $x$ on the blackboard and writes $2x+1$ if $x$ is not the cube of an integer, or the cube root of $x$ otherwise. Prove that at some point of time, she writes a number larger than $10^{100}$ .
*Proposed by Anant Mudga... | FTSoC suppose that there exists a maximum number $M^3$ we can reach(note that this maximum number has to be a perfect cube otherwise the next number on the board will be $2M+1>M$ , which is a contradiction).
Call **type 1 operation** the change $x\rightarrow 2x+1$ and **type 2 operation** the change $x^3\rightar... | [
"Note that $2x+1+1 = 2(x+1)$ and $x^3 + 1 = (x+1)(x^2 - x + 1)$ with $x^2 - x + 1$ odd always. So the $\\nu_2(n+1)$ goes up by $1$ when $2x+1$ is done and stays the same when you take the cube root. So the only way the numbers Alice writes are bounded is if $\\nu_2(n+1)$ remains constant eventually, b... | [
"origin:aops",
"2023 Contests",
"2023 India EGMO TST"
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"answer_score": 172,
"boxed": false,
"end_of_proof": true,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 India EGMO TST/2975768.json"
} |
Let $N \geqslant 3$ be an integer. In the country of Sibyl, there are $N^2$ towns arranged as the vertices of an $N \times N$ grid, with each pair of towns corresponding to an adjacent pair of vertices on the grid connected by a road. Several automated drones are given the instruction to traverse a rectangular pa... | The answer is $N$ if $N$ is odd and $N-1$ if $N$ is even. Clearly it suffices to show the bound for when $N$ is odd.
Consider all roads connecting a town on the border to a town on the inner $(N-2) \times (N-2)$ grid. There are $4(N-2)$ such roads. Call these *good*. Some of these are vertical, and the ... | [
"Consider the leftmost $n-1$ vertical edges of the bottom row and topmost $n-1$ horizontal edges of the last column.Any drone removed,removes at most two edges ,so directly proves the bound for even.It also characterises the potential equality cases for odd $n$ ,as any two of the edges uniquely define a drone.... | [
"origin:aops",
"2023 Contests",
"2023 India EGMO TST"
] | {
"answer_score": 154,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 India EGMO TST/2975769.json"
} |
Let $f, g$ be functions $\mathbb{R} \rightarrow \mathbb{R}$ such that for all reals $x,y$ , $$ f(g(x) + y) = g(x + y) $$ Prove that either $f$ is the identity function or $g$ is periodic.
*Proposed by Pranjal Srivastava* | <details><summary>Solution</summary>We proceed with casework on whether or not $f$ is injective.
Firstly, if $g$ is not injective, then there exist two distinct real numbers $a$ and $b$ such that $g(a)=g(b).$ Then, we get
\begin{align*}
g(a+x) &= f(g(a)+x)
&= f(g(b)+x)
&= g(b+x),
\end{align*}
... | [
"If $g(x) = x+c$ for some constant $c$ , then $f(x+y+c) = x+y+c$ so it is the identity, so ignore this case.\n\nThen, there exists some $k$ such that $p = g(k)-k-g(0) \\neq 0$ . Then we have that $f(g(k) + y) = f(k+y) = f(g(0) + k+y)$ so $f$ is periodic with period $p$ .\n\nThen $g(y+p) = f(g(0)+y+p) ... | [
"origin:aops",
"2023 Contests",
"2023 India EGMO TST"
] | {
"answer_score": 152,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2023 Contests/2023 India EGMO TST/2975770.json"
} |
Let $k$ be a positive integer. A sequence of integers $a_1, a_2, \cdots$ is called $k$ -pop if the following holds: for every $n \in \mathbb{N}$ , $a_n$ is equal to the number of distinct elements in the set $\{a_1, \cdots , a_{n+k} \}$ . Determine, as a function of $k$ , how many $k$ -pop sequences there ar... | I loved this problem!
The answer is $\boxed{2^k}$ .
Define $g_i$ to be the $i^{\text{th}}$ 'gap', i.e., $g_i = a_{i + 1} - a_i$ .
Note that $0 \leq g_i \leq 1$ for every $i$ .
The key observation is the following:**<span style="color:#f00">Claim:</span>** In any $k$ -pop sequence, $g_{i + k} = g_i$ for all... | [
"First, note that the sequence is non-decreasing. Also, $a_{i+1} - a_i$ must be either $0$ or $1$ , and it depends on whether $a_{i+1+k} - a_{i+k} = 0$ or $1$ .\n\nDefine $b_i = a_{i+1} - a_i$ , then the condition means that $b_{i+k} = b_i$ , and so each residue class mod $k$ has a constant value of $b... | [
"origin:aops",
"2023 Contests",
"2023 India EGMO TST"
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"answer_score": 1042,
"boxed": true,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 India EGMO TST/2975772.json"
} |
Let $ABC$ be an isosceles triangle with $AB = AC$ . Suppose $P,Q,R$ are points on segments $AC, AB, BC$ respectively such that $AP = QB$ , $\angle PBC = 90^\circ - \angle BAC$ and $RP = RQ$ . Let $O_1, O_2$ be the circumcenters of $\triangle APQ$ and $\triangle CRP$ . Prove that $BR = O_1O_2$ .
*Propo... | **Amazing angle chasing Problem!!**
Let $O$ be center of $\triangle ABC$ and for sake of niceness, let $\measuredangle BAE = 2\theta$ . Now $$ \measuredangle PBA = 90 - \theta - (90 - 2\theta) = \theta = 90 - (90 -\theta) = 90 - \measuredangle ACB $$ Which implies $O$ lie on $BP$ .
<span style="color:#00f... | [
"Let $O$ be the circumcenter of $ABC$ . Then triangles $OBQ$ and $OAP$ are congruent, so there is a rotation at $O$ that sends one of these triangles to the other. As a corollary, $\\angle POQ = \\angle AOB$ , so $O$ is on $(APQ)$ . $OP = OQ$ means $OR$ is the perpendicular bisector of $PQ$ . So ... | [
"origin:aops",
"2023 Contests",
"2023 India EGMO TST"
] | {
"answer_score": 218,
"boxed": false,
"end_of_proof": true,
"n_reply": 11,
"path": "Contest Collections/2023 Contests/2023 India EGMO TST/2975773.json"
} |
For a positive integer $n$ we denote by $s(n)$ the sum of the digits of $n$ . Let $P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ be a polynomial, where $n \geqslant 2$ and $a_i$ is a positive integer for all $0 \leqslant i \leqslant n-1$ . Could it be the case that, for all positive integers $k$ , $s(k)$ and ... | huge revenge arc. every polynomial digits problem is the same?
The answer is no. In fact, we only need that $P$ has positive leading coefficient (rephrasing the problem to ask about sufficiently large $k$ ). The key claim is the following:**<span style="color:#00f">Claim:</span>** There exist positive integers $C... | [
"That was a nice one! Here is my solution which is different from the official solution and does not need that $P$ is monic. Indeed, it is quite natural if one starts by considering the case of small degree. <details><summary>Solution</summary>The key is to note that\n\\[P(x+10^t)=P(x)+10^tP'(x)+10^{2t} \\frac{P'... | [
"origin:aops",
"2023 Contests",
"2023 India IMO Training Camp"
] | {
"answer_score": 254,
"boxed": false,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107335.json"
} |
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $f(x) + y$ and $f(y) + x$ have the same number of $1$ 's in their binary representations, for any $x,y \in \mathbb{N}$ . | A story about this problem: It was originally meant to be D3 P3 (basically the hardest problem in the TSTs), but a few days before the test, some of us found an easier solution while trying. Hence the problem had to be demoted to D3 P1. D4 P3 at that time, which was a very hard geo, was shifted to D3 P3 (I think it was... | [
" $P(2^n-f(x),x)$ means $x=2^k-f(2^n-f(x))$ for some $k$ , for all $x,n$ . $P(2^a+2^k-f(x),x)$ means that $f(2^a+2^k-f(x))+x$ is of the form $2^c+2^d$ , where $c\\neq d$ when $a\\neq k$ and $c=d$ when $a=k$ . Replacing $x$ in the previous equation with $2^n-f(x)$ , $f(2^a+x)+2^n-f(x)$ either h... | [
"origin:aops",
"2023 Contests",
"2023 India IMO Training Camp"
] | {
"answer_score": 1270,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107382.json"
} |
In triangle $ABC$ , with orthocenter $H$ and circumcircle $\Gamma$ , the bisector of angle $BAC$ meets $\overline{BC}$ at $K$ . Point $Q$ lies on $\Gamma$ such that $\overline{AQ} \perp \overline{QK}$ . Circumcircle of $\triangle AQH$ meets $\overline{AC}$ at $Y$ and $\overline{AB}$ at $Z$ . Let ... | **<span style="color:#00f">Lemma: </span>** $ABC$ is a triangle with orthocenter $H$ , $AH\cap BC=D$ . Let a circle passing through $A,H$ meet $AB,AC$ at $E,F$ . If $M,N$ are the midpoints of $BE,CF$ then $A,D,E,F$ are concyclic.**<span style="color:#f00">Proof: </span>**Invert at $A$ with radius $\sqr... | [
"[bary bash solution to be added oop] \n\nComments - I bary bashed this problem in contest LoL (wasn't too bad! OnLy 15 PaGeS)",
"Problem was proposed by Anant Mudgal. Here are the two official solutions:**Official Solution 1.** (by Anant Mudgal) \n\nWe shall only deal with the cases when $\\triangle ABC$ is ac... | [
"origin:aops",
"2023 Contests",
"2023 India IMO Training Camp"
] | {
"answer_score": 234,
"boxed": false,
"end_of_proof": false,
"n_reply": 16,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107385.json"
} |
In a school, every pair of students are either friends or strangers. Friendship is mutual, and no student is friends with themselves. A sequence of (not necessarily distinct) students $A_1, A_2, \dots, A_{2023}$ is called *mischievous* if
$\bullet$ Total number of friends of $A_1$ is odd.
$\bullet$ $A_i$... | My problem! The above is easily the cleanest solution (plus it was found during the exam). Here is my original solution:
We first put the problem in graph theoretic terms:
Consider a finite simple graph $G$ . A walk of length $2022$ is called *mischievous* if the degree of its starting vertex is odd, and the degree... | [
"my solution from test\n<details><summary>solution</summary>Rephrase to graph theory with friendships denoting edges in the corresponding statement. Let $f(u, v, k)$ denote the number of walks of length $k$ starting from vertex $u$ and ending in vertex $v$ . Let $d_u$ denote the degree of vertex $u$ . We ... | [
"origin:aops",
"2023 Contests",
"2023 India IMO Training Camp"
] | {
"answer_score": 276,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107388.json"
} |
Suppose an acute scalene triangle $ABC$ has incentre $I$ and incircle touching $BC$ at $D$ . Let $Z$ be the antipode of $A$ in the circumcircle of $ABC$ . Point $L$ is chosen on the internal angle bisector of $\angle BZC$ such that $AL = LI$ . Let $M$ be the midpoint of arc $BZC$ , and let $V$ be ... | Let the angle bisector of $\angle BZC$ meet $\odot ABC$ at $M'$ . Let $X$ denote the midpoint of $AI$ . Let $O$ denote the centre of $\odot ABC$ . Thus, $AI \perp XL$ , and $MOM'$ are collinear.**<span style="color:#f00">Claim:</span>** $AMZM'$ is a parallelogram**<span style="color:#0ff">Proof:</span>*... | [
"<details><summary>Trig</summary>Let $N$ be the midpoint of arc $ABC$ containing $A$ , I claim that $\\triangle MIV \\stackrel{-}{\\sim} \\triangle MNL$ which finishes because $$ \\angle DVM = \\angle VMN = \\angle IMN + \\angle VMI = \\angle IMN + \\angle NML = \\angle IML. $$ Let $S \\equiv MD \\cap Z... | [
"origin:aops",
"2023 Contests",
"2023 India IMO Training Camp"
] | {
"answer_score": 152,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107472.json"
} |
In triangle $ABC$ , let $D$ be the foot of the perpendicular from $A$ to line $BC$ . Point $K$ lies inside triangle $ABC$ such that $\angle KAB = \angle KCA$ and $\angle KAC = \angle KBA$ . The line through $K$ perpendicular to like $DK$ meets the circle with diameter $BC$ at points $X,Y$ . Prove th... | Solution I submitted in the test : Humpty Dumpty Points :)
[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(15cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(12); default... | [
"Nice config problem! Here is my solution.\n\nFirst, we set up the stage by defining a few points and making basic observations.\n\n\n- Let $L$ be the reflection of $A$ across $K$ . Since $K$ is the Dumpty point, it follows that $L\\in\\odot(ABC)$ and $ABLC$ is harmonic quadrilateral.\n- Let $A_1$ be t... | [
"origin:aops",
"2023 Contests",
"2023 India IMO Training Camp"
] | {
"answer_score": 284,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107478.json"
} |
In the fictional country of Mahishmati, there are $50$ cities, including a capital city. Some pairs of cities are connected by two-way flights. Given a city $A$ , an ordered list of cities $C_1,\ldots, C_{50}$ is called an *antitour* from $A$ if
- every city (including $A$ ) appears in the list exactly once, ... | Look at the city as a graph $\mathcal{G}$ with $50$ vertices, with each vertex representing a city, and each edge representing a two-way flight between two cities. We first make some observations about the nature of Mahishmati.
<span style="color:#f00">**Claim :**The graph $\mathcal{G}$ is a bipartite graph with... | [
"Consider a graph with $50$ vertices, the cities representing the vertices, and edges representing the flights.\n\n Call a graph *good* if there are *antitours* from each city.\n <span style=\"color:#0ff\">**Claim:**</span> A *good* graph cannot have any odd cycle.\n <span style=\"color:#f00\">**Proof:**... | [
"origin:aops",
"2023 Contests",
"2023 India IMO Training Camp"
] | {
"answer_score": 124,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107622.json"
} |
Let $g:\mathbb{N}\to \mathbb{N}$ be a bijective function and suppose that $f:\mathbb{N}\to \mathbb{N}$ is a function such that:
- For all naturals $x$ , $$ \underbrace{f(\cdots (f}_{x^{2023}\;f\text{'s}}(x)))=x. $$
- For all naturals $x,y$ such that $x|y$ , we have $f(x)|g(y)$ .
Prove that $f(x)=x$ .
... | Claim: $f$ is bijective.
Proof: Suppose $f(a) = f(b)$ . Then note that $f^{a^{2023}}(a) = a$ and $f^{b^{2023}}(b) = f^{b^{2023}}(a) = b $ . Now we see that $(ab)^{2023}$ is a multiple of both $a^{2023}$ and $b^{2023}$ , so $f^{(ab)^{2023}}(a)$ equals both $a$ and $b$ , which means $a = b$ , so $f$ ... | [
"I present a slightly long but detailed solution!\n\nFor $x=1$ , the first condition give us $f(1)=1$ . Furthermore, by the second condition, $x | x\\Rightarrow f(x) | g(x)$ , for all $x\\in\\mathbb{N}$ .\n\nDenote $f^n(x)=\\underbrace{f(f(\\dots f}_{n\\,\\,f'\\text{s}} (x)\\dots))$ .\n\nNote that, if $f^n(x)... | [
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"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107625.json"
} |
Let $\mathbb{Z}_{\ge 0}$ be the set of non-negative integers and $\mathbb{R}^+$ be the set of positive real numbers. Let $f: \mathbb{Z}_{\ge 0}^2 \rightarrow \mathbb{R}^+$ be a function such that $f(0, k) = 2^k$ and $f(k, 0) = 1$ for all integers $k \ge 0$ , and $$ f(m, n) = \frac{2f(m-1, n) \cdot f(m, n-1... | Let $g(m,n) = \frac{1}{f(m,n)}$ . Then $g(0,k) = \frac{1}{2^k}$ , $g(k,0) = 1$ for all integers $k \ge 0$ , and $$ 2 \cdot g(m,n) = g(m-1,n) + g(m, n-1) $$ for all integers $m,n \ge 1$ .
Let $G(x,y) = \sum_{m,n \ge 0} g(m,n) x^m y^n$ be the generating function of $g(m,n)$ .
<span style="color:#00f">Clai... | [
"Let $g(m,n)=1/f(m,n)$ . Consider a random walk starting from $(m+1,n)$ where at each step we go left or down 1 unit with equal probability. Then, $g(m,n)$ is the probability that we touch the line $y=0$ before we touch the line $x=0$ .\n\nSo we want to estimate the following quantity: start a random walk f... | [
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"2023 Contests",
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"answer_score": 1058,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107628.json"
} |
Let $ABC$ be a triangle, and let $D$ be the foot of the $A-$ altitude. Points $P, Q$ are chosen on $BC$ such that $DP = DQ = DA$ . Suppose $AP$ and $AQ$ intersect the circumcircle of $ABC$ again at $X$ and $Y$ . Prove that the perpendicular bisectors of the lines $PX$ , $QY$ , and $BC$ are concur... | **<span style="color:#f00">Coordinates FTW!</span>**
We employ cartesian coordinates with $B$ as the origin $(0,0)$ and the $x$ -axis along the line $BC$ . Let $A \equiv (b,c)$ , $B \equiv (0,0)$ and $C \equiv (a,0)$ .The coordinates of various points are as follows:
$D \equiv (b,0)$
$P \equiv (b+c,0)$... | [
"Let $K$ be the intersection of the perpendicular bisectors of $PX,QY$ . Let $E,F$ be the midpoints of $PX,QY$ . Let $EK,FK$ meet $BC$ at $M,N$ . Then angle chasing gives $\\angle PME = \\angle KMD = \\frac{\\pi}{4}$ , and symmetric angle chasing gives the same for $\\angle KND$ . Hence to finish, we j... | [
"origin:aops",
"2023 Contests",
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"answer_score": 1050,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107632.json"
} |
Prove that for all integers $k>2$ , there exists $k$ distinct positive integers $a_1, \dots, a_k$ such that $$ \sum_{1 \le i<j \le k} \frac{1}{a_ia_j} =1. $$ *Proposed by Anant Mudgal* | Extremely cursed problem. Here's what I came up with — if someone knows a better construction, please post it and restore my faith in symmetric polynomials.
<details><summary>Solution</summary>We'll replace $k$ with $n$ so that I don't have to fix my original write-up. Let us first introduce some notation: given p... | [
"@Ankoganit, Very Beautiful. @Aops, I want to know who proposed this, because I want to know how he/she came up with the motivation of making this problem. "
] | [
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"answer_score": 52,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107633.json"
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The numbers $1,2,3,4,\ldots , 39$ are written on a blackboard. In one step we are allowed to choose two numbers $a$ and $b$ on the blackboard such that $a$ divides $b$ , and replace $a$ and $b$ by the single number $\tfrac{b}{a}$ . This process is continued till no number on the board divides any other n... | [
"Take the primes that appear an odd number of times in this totally.\n2, 7, 11, 13 23, 29, 31, 37. 8 primes. 2*7=14 which is less than 39, so our 7 final numbers are 14 11 13 23 29 31 37. everyone solved this there...",
"2 is the least possible number\nDivide the numbers such that the least we can get\n38/19,18/3... | [
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"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107634.json"
} | |
Let $\mathbb R^+$ be the set of all positive real numbers. Find all functions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ satisfying \[f(x+y^2f(x^2))=f(xy)^2+f(x)\] for all $x,y \in \mathbb{R}^+$ .
*Proposed by Shantanu Nene* | The answer is $f\equiv Vx$ forall $x\in\mathbb{R^+}$ for some constant positive $V$ . Let $P(x,y)$ be the assertion above
Let any $t>0$ then $P\left(x,\sqrt{\frac{t}{f(x^2)}}\right)$ implies $f$ is strictly increasing, hence $f$ injective. $\textbf{Claim 1:}$ $f$ is not bounded on $\mathbb{R}^+$ Let... | [
"The answer is $f \\equiv cx$ for all $x \\in \\mathbb{R^+}$ and some constant $c$ . Verifying is easy.\nLet $P(x, y)$ be the assertion above. $\\textbf{Claim: }$ $f$ is strictly increasing, hence injective. $\\textit{Proof}$ Trivial as $(x, x+y^2f(x^2))$ can attain any two pairs of values in $\\mat... | [
"origin:aops",
"2023 Contests",
"2023 India IMO Training Camp"
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"answer_score": 1298,
"boxed": false,
"end_of_proof": true,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107637.json"
} |
Let $n$ be any positive integer, and let $S(n)$ denote the number of permutations $\tau$ of $\{1,\dots,n\}$ such that $k^4+(\tau(k))^4$ is prime for all $k=1,\dots,n$ . Show that $S(n)$ is always a square. | <details><summary>Solution</summary>For all $1\leq i\leq n$ , let $i^4 + \tau(i)^4 = p_i$ , for primes $p_i$ . Then, note that $$ p_1 + p_2 + \cdots + p_n = 1^4 + 2^4 + \cdots + n^4 + (\tau(1)^4 + \tau(2)^4 + \cdots + \tau(n^4)) = 2(1^4+2^4+\cdots + n^4). $$
Now, suppose that $n$ was even. Then, $p_1+p_2+\c... | [
"Can anyone observe that $S(4)=2$ ? $(\\text{Odd})^4+(\\text{Even})^4=\\text{Odd}$ expect the case $1^4+1^4=2$ but then parity implies this fact that $S(4)=2$ . \n\nI might be wrong can someone check?",
"No, $S(4)=4$ . There are four such permutations: $(2, 1, 4, 3), (2, 3, 4, 1),\n(4, 1, 2, 3),\n(4, 3... | [
"origin:aops",
"2023 Contests",
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"answer_score": 286,
"boxed": false,
"end_of_proof": true,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/3107638.json"
} |
For a positive integer $k$ , let $s(k)$ denote the sum of the digits of $k$ . Show that there are infinitely many natural numbers $n$ such that $s(2^n) > s(2^{n+1})$ . | Suppose not and there exists $N$ where for all $n \ge N$ we had $s(2^n) \le s(2^{n+1})$ .
Claim: For all $n\ge N$ , $s(2^{n+6}) \ge s(2^n) + 27$ .
Proof: Just spam the fact that $s(a) - s(b)$ is at least the remainder when $a-b$ is divided by $9$ for $a > b$ (so for example, $s(2^{n+1}) - s(2^n)$ is... | [
"For a positive integer $k$ written with $d$ digits, thus $10^{d-1} \\leq k < 10^d$ , we have $s(k) \\leq 9d \\leq 9(\\log_{10} k + 1)$ . \nTherefore we must have $s(2^n) \\leq 9(n\\log_{10} 2 + 1)$ .\n\nOn the other hand, we have $s(k) \\equiv k \\pmod{9}$ . The residues modulo $9$ of the numbers $2^n$ ... | [
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"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 India IMO Training Camp/621440.json"
} |
Let $S$ be a finite set of positive integers. Assume that there are precisely 2023 ordered pairs $(x,y)$ in $S\times S$ so that the product $xy$ is a perfect square. Prove that one can find at least four distinct elements in $S$ so that none of their pairwise products is a perfect square.
*Note:* As an exa... | My solution but essentially same as everybody else's
We say $x\sim y$ if $xy$ is a square. Now observe that $x\sim y$ , $y\sim z$ $\implies x\sim z$ and $x\sim y\implies y\sim x$ . Also $x\sim x$ for all $x$ . Thus, $\sim$ is an equivalence relation
and we need to just show that $S$ has atleast $4$ ... | [
"Needlessly pretentious write-up :D\n\nLet $p_1<p_2<\\dots$ be the sequence of prime numbers. Denote by $\\mathbb{F}_2^{\\omega}$ the $\\mathbb{F}_2$ vector space of all binary sequences $(a_1, a_2, \\dots)$ with entries in $\\mathbb{F}_2$ . Consider the set map $\\Phi: \\mathbb{N} \\rightarrow \\mathbb{F... | [
"origin:aops",
"2023 India National Olympiad",
"2023 Contests"
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"answer_score": 150,
"boxed": false,
"end_of_proof": true,
"n_reply": 23,
"path": "Contest Collections/2023 Contests/2023 India National Olympiad/2995076.json"
} |
Suppose $a_0,\ldots, a_{100}$ are positive reals. Consider the following polynomial for each $k$ in $\{0,1,\ldots, 100\}$ : $$ a_{100+k}x^{100}+100a_{99+k}x^{99}+a_{98+k}x^{98}+a_{97+k}x^{97}+\dots+a_{2+k}x^2+a_{1+k}x+a_k, $$ where indices are taken modulo $101$ , *i.e.*, $a_{100+i}=a_{i-1}$ for any $i$ in ... | Another solution with a slightly different finish I had in contest
Let $P_k$ denote the polynomial $a_{100+k}x^{100}+100a_{99+k}x^{99}+a_{98+k}x^{98}+a_{97+k}x^{97}+\dots+a_{2+k}x^2+a_{1+k}x+a_k$ . Notice that all roots of $P_k$ are negative. **Claim:** $\left( \frac{a_{i}}{a_{i+1}} \right)^{100} = \left(\frac{a_... | [
"Assume to the contrary that all of the polynomials have all real roots. Clearly, all roots must be negative real numbers. For a polynomial $P$ , denote by $\\sigma_i(P)$ the sum of products of roots of $P$ with $i$ of them taken at a time, for $i=1, 2, \\dots, \\text{deg} \\, P$ . Denote by $P_k$ , the $... | [
"origin:aops",
"2023 India National Olympiad",
"2023 Contests"
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"answer_score": 184,
"boxed": false,
"end_of_proof": false,
"n_reply": 25,
"path": "Contest Collections/2023 Contests/2023 India National Olympiad/2995077.json"
} |
Let $\mathbb N$ denote the set of all positive integers. Find all real numbers $c$ for which there exists a function $f:\mathbb N\to \mathbb N$ satisfying:
- for any $x,a\in\mathbb N$ , the quantity $\frac{f(x+a)-f(x)}{a}$ is an integer if and only if $a=1$ ;
- for all $x\in \mathbb N$ , we have $|f(x... | Solution with Pranjal Srivastava, Anant Mudgal and Sutanay Bhattacharya.
We call a function $f$ *good* if it satisfies the given conditions. If a function $f$ is *good* such that $f(x)-cx$ is bounded then $g(x)=f(x)-\lfloor{c\rfloor}x$ is also *good* with $g(x)-{c}x$ bounded. We allow functions to be from $... | [
"redacted.",
"The following is my solution, though the title is entirely by Sutanay, so go get him. I'll let him or someone else post the official solution. \n\nReplace condition (b) with \"bounded\" (the example we show for $c$ which work will work for the $2023$ bound). Clearly, $c>0$ else $f$ will even... | [
"origin:aops",
"2023 India National Olympiad",
"2023 Contests"
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"answer_score": 190,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2023 Contests/2023 India National Olympiad/2995078.json"
} |
Let $k \geq 1$ and $N>1$ be two integers. On a circle are placed $2N+1$ coins all showing heads. Calvin and Hobbes play the following game. Calvin starts and on his move can turn any coin from heads to tails. Hobbes on his move can turn at most one coin that is next to the coin that Calvin turned just now from ta... | My solution (also the official one).
Calvin wins if $k \in \{1, 2, \dots, N+1\}$ and Hobbes wins otherwise.
Label the coins $1, 2, \dots, 2N+1$ . Note that if $k \geq N+2$ then Hobbes wins as follows: he pairs the coins $2i-1$ and $2i$ for $1 \leq i \leq N$ . If Calvin in a move makes both coins in a pair... | [
"I am not sure what is done above but here's what I did: Pair $\\{1,2,\\ldots,2N\\}$ (so they form $N$ pairs) and box $2N+1$ (home alone). Easy to see Calvin can flip $N$ coins no two adjacent. Now show that he can manage to flip the unpaired square by cleverly forming $TT$ . For Hobbes, he can ensure each... | [
"origin:aops",
"2023 India National Olympiad",
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"answer_score": 146,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 India National Olympiad/2995079.json"
} |
Euler marks $n$ different points in the Euclidean plane. For each pair of marked points, Gauss writes down the number $\lfloor \log_2 d \rfloor$ where $d$ is the distance between the two points. Prove that Gauss writes down less than $2n$ distinct values.
*Note:* For any $d>0$ , $\lfloor \log_2 d\rfloor$ is... | Tejaswi points out the following. It was too late to consider the one-dimensional variant though.
<blockquote>
Let $n$ be a natural number with $n \ge 2$ and let $S$ be a set of $n$ distinct real numbers. For elements $s, t$ in $S$ we define $d(s, t)$ to be the integer such that $2^{d(s,t)} \le |s-t| <2... | [
"Solution. (Pranjal Srivastava) \n\nWe change Euler and Gauss to Barista and coffee cups to the original phrasing of the proposal.\n\nWe first prove that the barista writes down at most $n$ even numbers.\n\nFor each even number $2k$ that the barista writes down, choose a single pair of cups whose distance lies ... | [
"origin:aops",
"2023 India National Olympiad",
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"answer_score": 162,
"boxed": false,
"end_of_proof": true,
"n_reply": 18,
"path": "Contest Collections/2023 Contests/2023 India National Olympiad/2995080.json"
} |
Euclid has a tool called *cyclos* which allows him to do the following:
- Given three non-collinear marked points, draw the circle passing through them.
- Given two marked points, draw the circle with them as endpoints of a diameter.
- Mark any intersection points of two drawn circles or mark a new point o... | Solution by Rohan Goyal, Anant Mudgal and Daniel Hu
We begin by proving a series of lemmas.**Lemma 1** Given a non-right angled triangle $ABC$ , we can draw the nine-point circle and mark the orthocentre $H$ using only a cyclos.**Proof** Draw circles $(BC), (CA), (AB)$ and mark their intersections to get the thre... | [
"Wow this was a very nice problem. Took me a lot of time during the exam and still couldn't solve it. can someone post a solution?",
"We have following properties**Property 1** If we have any $3$ marked points $A,B,C$ then we can mark its orthocenter $H$ , nine point circle of $(ABC)$ ,midpoints $M,N,P$ o... | [
"origin:aops",
"2023 India National Olympiad",
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"answer_score": 424,
"boxed": false,
"end_of_proof": false,
"n_reply": 26,
"path": "Contest Collections/2023 Contests/2023 India National Olympiad/2995081.json"
} |
Peter has a deck of $1001$ cards, and with a blue pen he has written the numbers $1,2,\ldots,1001$ on the cards (one number on each card). He replaced cards in a circle so that blue numbers were on the bottom side of the card. Then, for each card $C$ , he took $500$ consecutive cards following $C$ (clockwise o... | Very cute!
Let $S$ be the set of all cards which have $500$ written in red. There must exist a card that has all other elements of $S$ among the next $500$ from it, and this must be the minimal number among all the cards (Initially this number is $1$ ). Now turn over this card $c$ with and write $c+1001$ ... | [
"[REDACTED]\n\nWell I guess I need to practice writing better.",
"I think yes, since the competition finished...\n\n\nAnyway, AoPS deleted P2 and P3. I am not sure now... day1 was held yesterday.",
"Yes, allowed",
"<blockquote>Yes, allowed</blockquote>\n\nМедеубек Кунгожин сказал что пока нельзя",
"В<blockq... | [
"origin:aops",
"2023 Contests",
"2023 International Zhautykov Olympiad"
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"answer_score": 126,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2023 Contests/2023 International Zhautykov Olympiad/3007982.json"
} |
We call a positive integer $n$ is $good$ , if there exist integers $a,b,c,x,y$ such that $n=ax^2+bxy+cy^2$ and $b^2-4ac=-20$ . Prove that the product of any two good numbers is also a good number. | <span style="color:#f00">**Claim 1:**</span>Given a prime $p$ with $-5$ as a $QR$ , there exists $a,b$ such that $\frac{a^2+5b^2}{p}=1$ or $2$ .
*Proof (Fermat's Christmas Theorem, modified).* This can be easily verified for $p\le5$ . Let $p\mid r^2+5$ . Consider all $(\lfloor \sqrt{p} \rfloor+1)^2$ pairs... | [
"The main idea is to prove that $n$ is good if and only if for every prime p, that $n \\vdots p$ , $x^2 \\equiv -5(mod p)$ ",
"<details><summary>s</summary>**Lemma:** For any odd $p\\geq5,$ if $\\exists \\, t: p \\mid t^2+5,$ then $\\exists \\, a,b \\in Z: a^2+5b^2=p$ or $a^2+5b^2=2p.$ *Proof:* $\\, \... | [
"origin:aops",
"2023 Contests",
"2023 International Zhautykov Olympiad"
] | {
"answer_score": 236,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 International Zhautykov Olympiad/3008863.json"
} |
Let $a_1, a_2, \cdots, a_k$ be natural numbers. Let $S(n)$ be the number of solutions in nonnegative integers to $a_1x_1 + a_2x_2 + \cdots + a_kx_k = n$ . Suppose $S(n) \neq 0$ for all big enough $n$ . Show that for all sufficiently large $n$ , we have $S(n+1) < 2S(n)$ . | Here's a less explicit solution. Say two functions $P,Q$ are approximately equal if $\frac{P(x)}{Q(x)} \rightarrow 1$ for big enough $x$ .
<span style="color:#f00">**Claim:**</span> For a fixed $k$ , there exists a polynomial $P_k(x)$ such that $S(n)$ and $P_k(n)$ are approximately equal.
<span style="col... | [
"I have two solutions to Problem 3, one via recursion and one via generating functions.\n\n<details><summary>2 Solutions</summary><blockquote>Given positive integers $a_1,\\cdots,a_k$ such that $\\gcd(x_1,\\cdots,x_k)=d$ . Let $S(n)$ denote the number of solution $(x_1,\\cdots,x_k)\\in \\mathbb{N}_0^k$ of th... | [
"origin:aops",
"2023 Contests",
"2023 International Zhautykov Olympiad"
] | {
"answer_score": 174,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 International Zhautykov Olympiad/3009623.json"
} |
The sum of $n > 2$ nonzero real numbers (not necessarily distinct) equals zero. For each of the $2^n - 1$ ways to choose one or more of these numbers, their sums are written in non-increasing order in a row. The first number in the row is $S$ . Find the smallest possible value of the second number. | <blockquote>The sum of $n > 2$ nonzero real numbers (not necessarily distinct) equals zero. For each of the $2^n - 1$ ways to choose one or more of these numbers, their sums are written in non-increasing order in a row. The first number in the row is $S$ . Find the smallest possible value of the second number.</bl... | [
"Answer is $S(1-\\frac{1}{m})$ , where $m=[\\frac{(n + 1)}{2}]$ . ( $[m]$ is the floor function). WLOG assume $a_1 \\geq a_2 \\geq ... \\geq a_{n}$ and let's say $S=a_1+..+a_k=|a_{k+1}+...+a_n|$ . It is not hard to notice that the second number in the list is $T = max(s-a_k,s+a_{k+1})$ . Let's try and find m... | [
"origin:aops",
"2023 Contests",
"2023 International Zhautykov Olympiad"
] | {
"answer_score": 1044,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 International Zhautykov Olympiad/3009626.json"
} |
The tangent at $C$ to $\Omega$ , the circumcircle of scalene triangle $ABC$ intersects $AB$ at $D$ . Through point $D$ , a line is drawn that intersects segments $AC$ and $BC$ at $K$ and $L$ respectively. On the segment $AB$ points $M$ and $N$ are marked such that $AC \parallel NL$ and $BC \pa... | Note that $DM = \frac{DB \cdot DK}{DL}$ and $DN = \frac{DA \cdot DL}{DK}$ so $DM \cdot DN = DB \cdot DA = DC^2$ . So $D$ lies on the radical axis of $(AQB)$ and $(MNPQ)$ , meaning it suffices to show that these two circles are tangent.
We have $\angle A + \angle BCM = \angle DCM = \angle CNM = \angle A + \... | [
"We will use the following well-known fact: For a $\\triangle ABC$ and a point $D$ on $AB$ , $CD$ is tangent to $(ABC)$ if and only if $\\frac{DA}{DB}=\\frac{CA^2}{CB^2}$ .\n\nWe claim that $CM$ and $CN$ are isogonal with respect to $\\angle ACB$ . By Menelaus theorem, $\\frac{DA}{DB}\\frac{KB}{KC}\... | [
"origin:aops",
"2023 Contests",
"2023 International Zhautykov Olympiad"
] | {
"answer_score": 138,
"boxed": false,
"end_of_proof": true,
"n_reply": 13,
"path": "Contest Collections/2023 Contests/2023 International Zhautykov Olympiad/3009627.json"
} |
Several blue and green rectangular napkins (perhaps of different sizes) with vertical and horizontal sides were placed on the plane. It turned out that any two napkins of different colors can be crossed by a vertical or horizontal line (perhaps along the border). Prove that it is possible to choose a color, two horizon... | Call a set of napkins *good* if it can be cut with a horizontal line.
Take the rightmost vertical line $l$ so that the green napkins strictly to the left of $l$ are good, and the blue napkins strictly to the left of $l$ are good too.
This means there are two same color napkins (not strictly) to the left of $... | [
"I think this is different than the one above: \n<details><summary>Click to expand</summary>Call a set of napkins \"good\" if it can be cut with a horizontal line, and \"bad\" otherwise. Note that \"bad\" sets are precisely the ones that contain two napkins, which can be divided by a horizontal line.\n\nEach vertic... | [
"origin:aops",
"2023 Contests",
"2023 International Zhautykov Olympiad"
] | {
"answer_score": 154,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 International Zhautykov Olympiad/3009628.json"
} |
1. In right triangle $ABC$ with $\angle{A}= \textdegree{90}$ , point $P$ is chosen. $D \in BC$ such that $PD \perp BC$ .
Let the intersection of $PD$ with $AB$ and $AC$ be $E$ and $F$ respectively.
Denote by $X$ and $Y$ as the intersection of $(APE)$ and $(APF)$ with $BP$ and $CP$ respecti... | We first establish the following key claim.
<span style="color:#f00">**Claim :**Points $X$ and $Y$ lie on the circle $(ABC)$ .</span>
*Proof :* Observe that $(AFPY)$ and $(AFDB)$ are clearly cyclic (one is given the other is due to right angles). Note that
\[CP \cdot CY = CF \cdot CA = CD\cdot CB\]
Thus, $(... | [
"By angle chasing $\\angle{AXP} = \\angle{ACB}$ and $\\angle{AYC} = \\angle{ABC} \\implies X, Y \\in (BC) \\implies P$ is the orthocenter of the triangle formed by $BC, BY, CX$ , which finishes. ",
"Concurrency of Altitudes in any triangle will solve the problem…",
"Let $BY\\cap AC=T$ Just to prove $P$ ... | [
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"path": "Contest Collections/2023 Contests/2023 Iran MO (2nd Round)/3074017.json"
} |
2. Prove that for any $2\le n \in \mathbb{N}$ there exists positive integers $a_1,a_2,\cdots,a_n$ such that $\forall i\neq j: \text{gcd}(a_i,a_j) = 1$ and $\forall i: a_i \ge 1402$ and the given relation holds. $$ [\frac{a_1}{a_2}]+[\frac{a_2}{a_3}]+\cdots+[\frac{a_n}{a_1}] = [\frac{a_2}{a_1}]+[\frac{a_3}{a_2}... | <blockquote>Just use $n!-1,2(n!)-1,\cdots, n(n!)-1$ .</blockquote>
Following this hint we let $$ a_i = i\cdot k! - 1 $$ and choose $k$ sufficiently large such that $k > n$ and $a_i \geq 1402$ for all $1 \leq i \leq n$ . Check that for all $i > j$ $$ (a_i, a_j) = (i\cdot k! - 1, j\cdot k! - 1) = ((i - j)\c... | [
"Just use $n!-1,2(n!)-1,\\cdots, n(n!)-1$ .",
"See [here](https://artofproblemsolving.com/community/c6h3065930p27662599).....",
"<blockquote>Just use $n!-1,2(n!)-1,\\cdots, n(n!)-1$ .</blockquote> $\\forall i: a_i \\ge 1402$ ",
"Also it could be solved by using the fact that for each $2\\leq N $ there ex... | [
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"path": "Contest Collections/2023 Contests/2023 Iran MO (2nd Round)/3074019.json"
} |
4. A positive integer n is given.Find the smallest $k$ such that we can fill a $3*k$ gird with non-negative integers
such that: $\newline$ $i$ ) Sum of the numbers in each column is $n$ .
$ii$ ) Each of the numbers $0,1,\dots,n$ appears at least once in each row. | The minimum $k$ is $k=\lceil \frac{3(n+1)}{2} \rceil $ First, sum of numbers in every column in $n$ As there are $k$ columns such so sum of all numbers would be $kn$ .
Now every row contains each number from $0$ to $n$ atleast once.
So sum of every number $\geq \frac{3n(n+1)}{2}$ [as there are 3 rows]
So no... | [
"The sum of all the numbers in the grid is $n(k)$ and it is at least $0+1+2+...+n=\\frac{n(n+1)}{2}$ , so we have $k \\geq \\frac{(n+1)}{2}$ So we have $2$ cases depending on the parity of $n$ .\nFor each column we will denote the three numbers on it as $(a, b, c)$ Case 1) if $n$ is even, let $n=2p$ $\\... | [
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"2023 Contests",
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"answer_score": 210,
"boxed": false,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Iran MO (2nd Round)/3074022.json"
} |
3. We have a $n \times n$ board. We color the unit square $(i,j)$ black if $i=j$ , red if $i<j$ and green if $i>j$ . Let $a_{i,j}$ be the color of the unit square $(i,j)$ . In each move we switch two rows and write down the $n$ -tuple $(a_{1,1},a_{2,2},\cdots,a_{n,n})$ . How many $n$ -tuples can we obtain... | Non-black houses on the main diagonal are called colored houses. We claim that we can achieve a coloring of the main diameter with the allowed movement of the problem if and only if, in this coloring, the first colored house on the main diameter is red and the last colored house on the main diameter is green. To prove ... | [
"<details><summary>Answer</summary>$\\frac{3^n+3-2n}{4}$ . And the main idea is, all $n-$ tuples need to have\n the first color from the left, other than black, must be green and\n the first color from the right, other than black, must be red</details>"
] | [
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"path": "Contest Collections/2023 Contests/2023 Iran MO (2nd Round)/3074024.json"
} |
6. Circles $W_{1}$ and $W_{2}$ with equal radii are given. Let $P$ , $Q$ be the intersection of the circles.
points $B$ and $C$ are on $W_{1}$ and $W_{2}$ such that they are inside $W_{2}$ and $W_{1}$ respectively.
Points $X$ , $Y$ $\neq$ $P$ are on $W_{1}$ and $W_{2}$ respectively, such th... | **<span style="color:#00f">Claim 1:</span>** $B$ is the orthocenter of triangle $PQY$ . **<span style="color:#f00">Proof:</span>** First notice that since the radius of $w_{1}$ is equal to the radius of $w_{2}$ the orthocenter of $PQY$ lies on $(BPQ)$ . Also $\frac{BQ}{\sin\angle QPB}=R_{(PQB)} \implies \fra... | [
"Note that $X, Y$ are the intersections of $\\omega_1, \\omega_2$ with the lines through $B, C$ perpendicular to $PQ$ . Thus, $(XPC), (YPB)$ are $\\omega_1, \\omega_2$ translated by $QC, QB$ in the direction of $QC, QB$ respectively, which means that they both pass through the point which forms a par... | [
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"answer_score": 166,
"boxed": false,
"end_of_proof": true,
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"path": "Contest Collections/2023 Contests/2023 Iran MO (2nd Round)/3074025.json"
} |
5. We call $(P_n)_{n\in \mathbb{N}}$ an arithmetic sequence with common difference $Q(x)$ if $\forall n: P_{n+1} = P_n + Q$ $\newline$ We have an arithmetic sequence with a common difference $Q(x)$ and the first term $P(x)$ such that $P,Q$ are monic polynomials with integer coefficients and don't share an ... | <blockquote>5. We call $(P_n)_{n\in \mathbb{N}}$ an arithmetic sequence with common difference $Q(x)$ if $\forall n: P_{n+1} = P_n + Q$ $\newline$ We have an arithmetic sequence with a common difference $Q(x)$ and the first term $P(x)$ such that $P,Q$ are monic polynomials with integer coefficients and don... | [
"The proof of $a)$ is same as above. And for the proof of $b)$ ,\nLet $\\frac{P}{Q}=R$ and $deg(R)=d$ . Since $P_{n+1}=Q.(R+n)$ , $R$ is surjective on negative integers which also means $d$ is odd as $R$ is monic. On the other hand, there exists an integer $N$ such that $(x+N)^d\\leq R(x) <(x+N+1)^d... | [
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"answer_score": 1114,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Iran MO (2nd Round)/3074028.json"
} |
Find all function $ f: \mathbb{R}^{+} \to \mathbb{R}^{+}$ such that for every three real positive number $x,y,z$ : $$ x+f(y) , f(f(y)) + z , f(f(z))+f(x) $$ are length of three sides of a triangle and for every postive number $p$ , there is a triangle with these sides and perimeter $p$ .
*Proposed by Amirho... | Let $p_n$ be any sequence of positive numbers such that $p_n \to 0^+$ as $n \to \infty$ . Let $x_n, y_n, z_n$ be such that $$ x_n+f(y_n) + f(f(y_n)) + z_n + f(f(z_n))+f(x_n) = p_n $$ Since $f$ gets only positive values, we see that $x_n$ , $f(x_n)$ , $f(y_n)$ , $f(f(y_n))$ , $z_n$ and $ f(f(z_n))$ ... | [
"Sorry for the unrelated comment, but could you please post the rest of the problems from the first two days of Iranian TSTs (unless they are from ISL, of course)? ",
"<blockquote>Sorry for the unrelated comment, but could you please post the rest of the problems from the first two days of Iranian TSTs (unless th... | [
"origin:aops",
"2023 Iran Team Selection Test",
"2023 Contests"
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"answer_score": 50,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Iran Team Selection Test/3032794.json"
} |
Suppose that $d(n)$ is the number of positive divisors of natural number $n$ . Prove that there is a natural number $n$ such that $$ \forall i\in \mathbb{N} , i \le 1402: \frac{d(n)}{d(n \pm i)} >1401 $$ *Proposed by Navid Safaei and Mohammadamin Sharifi* | For both (at least this is the stronger version and also true as my solution shows).
<details><summary>Solution</summary>Let $p_1<p_2<\dots<p_m$ be the primes less than $1402$ and $p_{m+1}<p_{m+2}<\dots$ the sequence of other primes.
Consider $n=AB$ where $A=(p_1p_2\dots p_m)^{1402}$ and $b=p_{m+1}\dots p_k$... | [
"This means that the inequality should be true for one appropriate choice of the sign, or for both $ n \\pm i$ ?",
"It seems like another NT construction problem by Mr. Safaei had proposed, I'll try to solve it tomorrow.",
"Alright here is a solution that is similar to $Post \\ 3.$ Let $p_i$ the $i-th$ p... | [
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"2023 Iran Team Selection Test",
"2023 Contests"
] | {
"answer_score": 54,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Iran Team Selection Test/3032955.json"
} |
Suppose that $n\ge3$ is a natural number. Find the maximum value $k$ such that there are real numbers $a_1,a_2,...,a_n \in [0,1)$ (not necessarily distinct) that for every natural number like $j \le k$ , sum of some $a_i$ -s is $j$ .
*Proposed by Navid Safaei* | <details><summary>Solution</summary>Solved with **Cookierookie** and **sevket12**.
The answer is $n-2$ .
First, assume that $k=n-1$ works, we must have $a_1+\dots+a_n=n-1$ , and $\sum_{i\neq j} a_i = n-2$ for some $j\in \{1,2,\dots,n\}$ , but these two would imply $a_j=1$ , contradiction.
Now we show $k=n-... | [
"We'll show that the answer is $n-2$ for each number $n \\ge 3$ . \nFirst note that if we can produce all numbers $1 , 2 , ... , n-1$ with $0 \\le a_1 , ... , a_n < 1$ , then since the sum of each $n-1$ numbers between $a_1 , ... , a_n$ is less than $n-1$ , we must have $a_1+a_2+...+a_n=n-1$ and simi... | [
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"2023 Iran Team Selection Test",
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"answer_score": 168,
"boxed": false,
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"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Iran Team Selection Test/3032959.json"
} |
$ABCD$ is cyclic quadrilateral and $O$ is the center of its circumcircle. Suppose that $AD \cap BC = E$ and $AC \cap BD = F$ . Circle $\omega$ is tanget to line $AC$ and $BD$ . $PQ$ is a diameter of $\omega$ that $F$ is orthocenter of $EPQ$ . Prove that line $OE$ is passing through center of $\ome... | **Lemma 1 (Well known) :** Let $\omega$ and $\omega'$ be two circles which intersect each other at points $A , B$ .Then for two arbitrary points $X,Y$ in the plane the quadrilateral $XABY$ is cyclic , if and only if we have : $$ \frac{P_{\omega}{(X)}}{P_{\omega}{(Y)}}=\frac{P_{\omega'}{(X)}}{P_{\omega'}{(Y)}}... | [
"Let $AB\\cap CD=G$ and denote the mid-point of $PQ$ as $M$ . We know $GF\\perp EO$ from brocard thus it's enough to show $EM\\perp GF$ . Let the intersections of $AC,BD$ with $\\omega$ be $Y,X$ respectively and denote $K=XY\\cap PQ$ . $FE$ is the polar of $K$ hence $(K,E;X,Y)=-1$ which implies... | [
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"answer_score": 186,
"boxed": false,
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"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Iran Team Selection Test/3032970.json"
} |
Arman, starting from a number, calculates the sum of the cubes of the digits of that number, and again calculates the sum of the cubes of the digits of the resulting number and continues the same process. Arman calls a number $Good$ if it reaches $1$ after performing a number of steps. Prove that there is an arithm... | <details><summary>ignore this</summary>Here's an sketch(forgive me if it's wrong it's around 1 a.m here and I'm posting this after like a minute since I saw the problem :))
Clearly the final number would be sth between $1$ , $9$ for any positive integer .so distribute them into 9 different sets. Note that it's enough... | [
"This seems hard\n\nOnly useful thing I can think of is that $370$ or $371$ never reaches one so you can't just say that every number is good",
"@above\nIt’s not necessary that the final number is between 1 and 9. \nFor example 133 -> 55 -> 250 -> 133",
"First we prove that there exists an arithmetic progre... | [
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} |
Suppose that $n\ge2$ and $a_1,a_2,...,a_n$ are natural numbers that $ (a_1,a_2,...,a_n)=1$ . Find all strictly increasing function $f: \mathbb{Z} \to \mathbb{R} $ that: $$ \forall x_1,x_2,...,x_n \in \mathbb{Z} : f(\sum_{i=1}^{n} {x_ia_i}) = \sum_{i=1}^{n} {f(x_ia_i}) $$ *Proposed by Navid Safaei and... | Is this the only solution?
Write $(a_2, ..., a_n)=d, a_1=d'$ where $(d, d')=1$ . Take positive reals $0=c_0<c_1<c_2<...<c_{dd'}$ . $$ f(x)= \left\lfloor \frac{x}{dd'} \right\rfloor c_{dd'} +c_{ \left\{ \frac{x}{dd'} \right\} dd'} $$ The hard part is to prove for $n=2$ .
It is obvious that $f(0)=0$ , thus pu... | [
"<blockquote>Is this the only solution?\n\nWrite $(a_2, ..., a_n)=d, a_1=d'$ where $(d, d')=1$ . Take positive reals $0=c_0<c_1<c_2<...<c_{dd'}$ . $$ f(x)= \\left\\lfloor \\frac{x}{dd'} \\right\\rfloor c_{dd'} +c_{ \\left\\{ \\frac{x}{dd'} \\right\\} dd'} $$ </blockquote>\n\nI Agree with the rest but does th... | [
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"2023 Iran Team Selection Test",
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"answer_score": 66,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Iran Team Selection Test/3032984.json"
} |
$ABC$ is an acute triangle with orthocenter $H$ . Point $P$ is in triangle $BHC$ that $\angle HPC = 3 \angle HBC $ and $\angle HPB =3 \angle HCB $ . Reflection of point $P$ through $BH,CH$ is $X,Y$ . if $S$ is the center of circumcircle of $AXY$ , Prove that:
$$ \angle BAS = \angle CAP $$ *Propos... | Let $\beta = \angle HBC$ and $\gamma = \angle HCB$ Let $\phi$ be the $\sqrt{bc}$ inversion of triangle $BHC$ (with the center $H$ .) Let $P' = \phi(P)$ . Thus, $\angle CBP' = 2\beta$ and $\angle BCP' = 2 \gamma .$ [rule]**<span style="color:#00f">Claim 1:</span>** $A$ is the Miquel point of quadrilatera... | [
"Love this one !\nLet $O$ be the center of circumcircle of triangle $\\triangle ABC$ and suppose that $T$ is the reflection point of $O$ respect to $BC$ ( The center of circumcircle of triangle $\\triangle BHC$ ). We define points $X,Y$ such that $X,C$ and $Y,B$ are located on same side of $AH$ ... | [
"origin:aops",
"2023 Iran Team Selection Test",
"2023 Contests"
] | {
"answer_score": 206,
"boxed": false,
"end_of_proof": true,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 Iran Team Selection Test/3033473.json"
} |
Suppose that we have $2n$ non-empty subset of $ \big\{0,1,2,...,2n-1\big\} $ that sum of the elements of these subsets is $ \binom{2n+1}{2}$ . Prove that we can choose one element from every subset that some of them is $ \binom{2n}{2}$ *Proposed by Morteza Saghafian and Afrouz Jabalameli* | :) Very Beautiful Problem:
First we recall the following result, <blockquote>Consider the sets of integers $X_1, X_2, \dots, X_m$ . Then let $$ X=\{x_1+x_2+\dots+x_m |x_1\in X_1,x_2\in X_2,\dots ,x_m\in X_m\} $$ The following inequality holds $$ |X|\geq |X_1|+|X_2|+\dots+|X_m|-(m-1) $$ </blockquote>
Now lab... | [
"Are you sure the statement is correct?",
"This statement is not correct. Consider the following counterexample for $n=3$ :\nSubsets: {0}, {1}, {2}, {3}, {1,2,3}, {2,3,4}",
"<blockquote>This statement is not correct. Consider the following counterexample for $n=3$ :\nSubsets: {0}, {1}, {2}, {3}, {1,2,3}, {2,3... | [
"origin:aops",
"2023 Iran Team Selection Test",
"2023 Contests"
] | {
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 Iran Team Selection Test/3033477.json"
} |
Find all injective $f:\mathbb{Z}\ge0 \to \mathbb{Z}\ge0 $ that for every natural number $n$ and real numbers $a_0,a_1,...,a_n$ (not everyone equal to $0$ ), polynomial $\sum_{i=0}^{n}{a_i x^i}$ have real root if and only if $\sum_{i=0}^{n}{a_i x^{f(i)}}$ have real root.
*Proposed by Hesam Rajabzadeh* | The only answer is $\boxed{f(n)=cn}$ for some odd integer $c$ . It is not hard to see that these functions work.**<span style="color:#6AA84F">Claim:</span>** $f(0)=0$ Consider the polynomial $1$ .**<span style="color:#6AA84F">
Claim:</span>** $f(n)$ and $n$ have the same parity
Consider the polynomial $x^n+1$... | [
"<details><summary>Solution</summary>All such functions are $f(n)=tn \\ \\forall n\\in \\mathbb Z_{\\geq 0}$ where $t\\in \\mathbb{N}$ is odd constant. These functions can easily be seen to work since $x\\mapsto x^t$ is bijective.\n\nFirst, putting $n$ even and $a_n>0,a_{n-1}=\\dots=a_1=0,a_0>0$ , we get ... | [
"origin:aops",
"2023 Iran Team Selection Test",
"2023 Contests"
] | {
"answer_score": 1050,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Iran Team Selection Test/3033514.json"
} |
line $l$ through the point $A$ from triangle $ABC$ . Point $X$ is on line $l$ . ${\omega}_b$ and ${\omega}_c$ are circles that through points $X,A$ and respectively tanget to $AB$ adn $AC$ . tangets from $B,C$ respectively to ${\omega}_b$ and ${\omega}_c$ meet them in $Y,Z$ . Prove that by chang... | Let $O_b,O_c$ are centers of $\omega_b$ and $\omega_c$ respectively.
Obviously $X,Y,Z$ are reflections of $A$ according to $O_bO_c,BO_b,CO_c$ .
Take an inversion centered $A$ with radius $1$ .We want to show that circles $X^* Y^* Z^*$ are coaxial.
We can easily see that $X^*$ is circiumcenter of $AO^... | [
"Is this exam 2 P5 or the polynomial problem (there are two P5's now)?",
"<blockquote>Is this exam 2 P5 or the polynomial problem (there are two P5's now)?</blockquote>\n\nSorry. Geometry problem was problem 4 in the second exam. I edited.",
"<details><summary>Solution</summary>The first fixed point is the refl... | [
"origin:aops",
"2023 Iran Team Selection Test",
"2023 Contests"
] | {
"answer_score": 164,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Iran Team Selection Test/3033523.json"
} |
Suppose $\frac{1}{2} < s < 1$ . An insect flying on $[0,1]$ . If it is on point $a$ , it jump into point $ a\times s$ or $(a-1) \times s +1$ . For every real number $0 \le c \le 1$ , Prove that insect can jump that after some jumps , it has a distance less than $\frac {1}{1402}$ from point $c$ .
*Propos... | I hope this is right.
The key thing is problem is symmetric to points $0$ and $1$ since moves are $|0-a|\rightarrow |0-as|$ or $|1-a|\rightarrow |1-as|$ Lets restate the problem. $\frac{1}{2} < s < 1$ is a real number. There is a point in $[0,1]$ . We apply homothety with ratio s on the points and the center i... | [
"<span style=\"color:#f00\">Lemma</span>: $\\forall(l, r)\\subset[0, 1],\\ \\exists m\\geq0$ and $a_0, a_1, \\dots, a_m\\in\\{0, 1\\}$ s.t. $(1-s)\\sum_{i=0}^ma_is^i\\in(l, r)$ .\n\nProof:\nLet's recursively construct $a_k$ as follow: $a_k:=\\begin{cases}\n1\\text{, if }(1-s)(s^k+\\sum_{i=0}^{k-1}a_is^i)<r\n... | [
"origin:aops",
"2023 Iran Team Selection Test",
"2023 Contests"
] | {
"answer_score": 170,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Iran Team Selection Test/3033579.json"
} |
The game of **Hive**is played on a regular hexagonal grid (as shown in the figure) by 3 players. The grid consists of $k$ layers (where $k$ is a natural number) surrounding a regular hexagon, with each layer constructed around the previous layer. The figure below shows a grid with 2 layers.
The players, *Ali*, *S... | With just seven hexagons, clearly team P1 + P2 wins. P3 never gets a chance to move.
On bigger grids, P1 + P2 claim the six-hexagon ring, then for each move P3 makes, P1 and P2 follow that with $120^\circ$ and $240^\circ$ rotations of that around the central hexagon. Since P1 + P2 will always be able to move, th... | [] | [
"origin:aops",
"2023 Iran Team Selection Test",
"2023 Contests"
] | {
"answer_score": 4,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Iran Team Selection Test/3156939.json"
} |
A positive integer is *totally square* is the sum of its digits (written in base $10$ ) is a square number. For example, $13$ is totally square because $1 + 3 = 2^2$ , but $16$ is not totally square.
Show that there are infinitely many positive integers that are not the sum of two totally square integers. | Any positive integer congruent to $3$ or $6$ mod $9$ will do; the main idea is that $S(n) \equiv n \pmod{9}$ can't be a perfect square if $n$ is a quadratic nonresidue mod $9$ ; i.e. if $n \in \{2, 3, 5, 6, 8\}$ . But $3$ and $6$ cannot be written as the sum of two quadratic residues, so any number in t... | [
"Consider the number $10^k - 1$ , any two numbers that sum to it must have no carries in addition. So if $x + y = 10^k - 1$ with $s(x) = a^2$ and $s(y) = b^2$ , then $a^2 + b^2 = s(x)+s(y) = s(10^k - 1) = 9k$ . But if $k$ has a $3 \\pmod 4$ prime divisor with odd $\\nu_p$ , then clearly this is impossib... | [
"origin:aops",
"2023 Irish Math Olympiad",
"2023 Contests"
] | {
"answer_score": 120,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Irish Math Olympiad/3072024.json"
} |
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