problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
Aisling and Brendan take alternate moves in the following game. Before the game starts, the number $x = 2023$ is written on a piece of paper. Aisling makes the first move. A move from a positive integer $x$ consists of replacing $x$ either with $x + 1$ or with $x/p$ where $p$ is a prime factor of $x$ .
Th... | Aisling wins
First she divides $2023$ by $17$ . Brendan makes a move on $119$ . He can't divide it by $7$ or $17$ or he loses the next move, so he makes it $120$ . Now, Aisling makes it $121$ . Again, Brendan has to make it $122$ otherwise he loses immediately. Now, Aisling makes it $123$ . Again, Brendan... | [
"<blockquote>Aisling and Brendan take alternate moves in the following game. Before the game starts, the number $x = 2023$ is written on a piece of paper. Aisling makes the first move. A move from a positive integer $x$ consists of replacing $x$ either with $x + 1$ or with $x/p$ where $p$ is a prime fac... | [
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"2023 Contests"
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"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Irish Math Olympiad/3072057.json"
} |
Suppose that $a, b, c$ are positive real numbers and $a + b + c = 3$ . Prove that $$ \frac{a+b}{c+2} + \frac{b+c}{a+2} + \frac{c+a}{b+2} \geq 2 $$ and determine when equality holds. | [quote name="sqing" url="/community/p27743329"]
Suppose that $a, b, c$ are positive real numbers and $a + b + c = 3$ . Prove that $$ \frac{a+b}{c+k} + \frac{b+c}{a+k} + \frac{c+a}{b+k} \geq \frac{6}{k+1} $$ Where $k>0.$ </blockquote> $\frac{a+b}{c+k}=\frac{3-c}{c+k}$ so we can rewrite our inequality as $$ f(a)... | [
" $\\sum_{cyc} \\frac{a+b+c+2}{c+2}\\ge 5 \\iff \\sum_{cyc} \\frac{1}{c+2} \\ge 1$ . By C-S, $(a+2+b+2+c+2)(\\sum_{cyc} \\frac{1}{c+2})\\ge 9$ , so we are done. Equality holds iff $a=b=c=1$ . ",
" $$ \\frac{a+b}{c+2} + \\frac{b+c}{a+2} + \\frac{c+a}{b+2} \\geq 2 $$ $$ \\frac{a}{c+2} + \\frac{b}{a+2} + \\fra... | [
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"boxed": false,
"end_of_proof": true,
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"path": "Contest Collections/2023 Contests/2023 Irish Math Olympiad/3072113.json"
} |
The triangle $ABC$ has circumcentre $O$ and circumcircle $\Gamma$ . Let $AI$ be a diameter of $\Gamma$ . The ray $AI$ extends to intersect the circumcircle $\omega$ of $\triangle BOC$ for the second time at a point $P$ .
Let $AD$ and $IQ$ be perpendicular to $BC$ , with $D$ and $Q$ on $BC$ . L... | [
"(a) True by power of a point at D, since the reflection of I over the perpendicular bisector of BC is on AD.\n\n(b) Let the line through I parallel to BC intersect OM at J; then it suffices to show that MI || JP, which is equivalent to OM / OJ = OI / OP, which is obvious because if P* is the inverse of P with resp... | [
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"path": "Contest Collections/2023 Contests/2023 Irish Math Olympiad/3072236.json"
} | |
Caitlin and Donal play a game called *Basketball Shoot-Out*. The game consists of $10$ rounds. In each round, Caitlin and Donal both throw a ball simultaneously at each other's basket. If a player's ball falls into the basket, that player scores one point; otherwise, they score zero points. The scoreboard shows the c... | [
"Write the sequence like this:\n\nBegin with a C. Then, for each round, if Caitlin scored, write a C, otherwise, write a D; after this, if Donal scored, write a D, otherwise, write a C. If n rounds are played, the sequence has length 2n+1.\n\nThe sequence is valid iff when read from LEFT to RIGHT, D's never outnumb... | [
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"path": "Contest Collections/2023 Contests/2023 Irish Math Olympiad/3072239.json"
} | |
We are given a triangle $ABC$ such that $\angle BAC < 90^{\circ}$ . The point $D$ is on the opposite side of the line $AB$ to $C$ such that $|AD| = |BD|$ and $\angle ADB = 90^{\circ}$ . Similarly, the point $E$ is on the opposite side of $AC$ to $B$ such that $|AE| = |CE|$ and $\angle AEC = 90^{\ci... | $|EX| $ $=$ $ |DB|$ , $|EC|=|XD|$ and $\angle XEC = \angle XDB$ then we get $\triangle XEC$ and $\triangle BDX$ congruent from there we get $|CX|=|XB|$ | [] | [
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"path": "Contest Collections/2023 Contests/2023 Irish Math Olympiad/3072500.json"
} |
For $n \geq 3$ , a *special n-triangle* is a triangle with $n$ distinct numbers on each side such that the sum of the numbers on a side is the same for all sides. For instance, because $41 + 23 + 43 = 43 + 17 + 47 = 47 + 19 + 41$ , the following is a special $3$ -triangle: $$ 41 $$ $$ 23\text{ }\text{ }\text{ ... | [] | [
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} | |
Let $A, B, C, D, E$ be five points on a circle such that $|AB| = |CD|$ and $|BC| = |DE|$ . The segments $AD$ and $BE$ intersect at $F$ . Let $M$ denote the midpoint of segment $CD$ . Prove that the circle of center $M$ and radius $ME$ passes through the midpoint of segment $AF$ . | Let $N$ denote the midpoint of $AF$ , and reflect $F$ across $D$ to obtain a point $G$ . By power of a point we have \[BF \cdot FE = AF \cdot FD = NF \cdot FG,\] so $BNEG$ is cyclic. It remains to show that this circle has center $M$ , i.e. that $ME = MG$ (since we already have $MB = ME$ by symmetry). Bu... | [
"Let $S$ be the midpoint of $BF$ and let $T$ be the midpoint of segment $AF$ . We will prove that $\\triangle MST \\cong \\triangle EDM$ . $AB = CD \\Rightarrow AD\\|BC$ and $ \\angle BAD = \\angle CDA = \\alpha $ . $ED = BC \\Rightarrow DC \\| BE$ . Now $DFBC$ is a parallelogram $\\Rightarrow \\angle... | [
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"path": "Contest Collections/2023 Contests/2023 Irish Math Olympiad/3073154.json"
} |
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ with the property that $$ f(x)f(y) = (xy - 1)^2f\left(\frac{x + y - 1}{xy - 1}\right) $$ for all real numbers $x, y$ with $xy \neq 1$ . | <blockquote>Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ with the property that $$ P(x, y): f(x)f(y) = (xy - 1)^2f\left(\frac{x + y - 1}{xy - 1}\right) $$ for all real numbers $x, y$ with $xy \neq 1$ .</blockquote> $P(x,0) \land P(1-x, 0) \implies f(x)(f(0)^2-1)=0$ $\forall x \in \mathbb{R}$ .
If ... | [] | [
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"path": "Contest Collections/2023 Contests/2023 Irish Math Olympiad/3073156.json"
} |
The positive integers $a, b, c, d$ satisfy
(i) $a + b + c + d = 2023$ (ii) $2023 \text{ } | \text{ } ab - cd$ (iii) $2023 \text{ } | \text{ } a^2 + b^2 + c^2 + d^2.$ Assuming that each of the numbers $a, b, c, d$ is divisible by $7$ , prove that each of the numbers $a, b, c, d$ is divisible by $17$ . | <blockquote>:huh:
Let $a=17x,b=17y,c=17z,d=17w$ , and so we have $x+y+z+w=17^2, 17^2 \mid (xy-zw), 17^2 \mid (x^2+y^2+z^2+w^2)$ .
Note that if one of $x,y,z,w,$ is a multiple of $17$ , let it be $x$ , then $17 \mid zw$ and so if WLOG $17 \mid z$ we obtain $17 \mid (y+w)$ and $17 \mid (y^2+w^2),$ from wh... | [
"I am doing prealgebra 2 and this feels like I just entered another realm of maths",
":huh: \n\nLet $a=7x,b=7y,c=7z,d=7w$ , and so we have $x+y+z+w=17^2, 17^2 \\mid (xy-zw), 17^2 \\mid (x^2+y^2+z^2+w^2)$ .\n\nNote that if one of $x,y,z,w,$ is a multiple of $17$ , let it be $x$ , then $17 \\mid zw$ and so i... | [
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} |
2000 people are sitting around a round table. Each one of them is either a truth-sayer (who always tells the truth) or a liar (who always lies). Each person said: "At least two of the three people next to me to the right are liars". How many truth-sayers are there in the circle? | For a truth teller, at least two of the three people next to him to the right are liars. For a lier, no more than one of the three people next to him to the right are liars. Bases on these two conditions we find that the only two possibilities are:
[asy]
size(5cm);
label("T",dir(0));label("T",dir(40));label("T",dir(160... | [] | [
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} |
The non-negative integers $x,y$ satisfy $\sqrt{x}+\sqrt{x+60}=\sqrt{y}$ . Find the largest possible value for $x$ . | Square both sides to get $2x+60+2\sqrt{x^2+60x}=y$ . So $\sqrt{x^2+60}$ is an integer. Let $x^2+60=m^2$ to get $$ (x+30)^2-m^2=(x+30+m)(x+30-m)=900. $$ To make $x$ largest, we need to maximize $$ x+30=\frac12(x+30+m+x+30-m). $$ Since their product is fixed, their sum reaches maximum when they're the farthe... | [] | [
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"path": "Contest Collections/2023 Contests/2023 Israel National Olympiad/2979784.json"
} |
A triangle $ABC$ is given together with an arbitrary circle $\omega$ . Let $\alpha$ be the reflection of $\omega$ with respect to $A$ , $\beta$ the reflection of $\omega$ with respect to $B$ , and $\gamma$ the reflection of $\omega$ with respect to $C$ . It is known that the circles $\alpha, \beta, \... | [] | [
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For each positive integer $n$ , find all triples $a,b,c$ of real numbers for which
\[\begin{cases}a=b^n+c^n
b=c^n+a^n
c=a^n+b^n\end{cases}\] | First of all, $a=b=c=0$ and $a=b=c=2^{-1/(n-1)}$ are always solutions and, for odd $n$ , $a=b=c=-2^{-1/(n-1)}$ is a solution.
If any two variables are positive, then they all are.
If $n$ is even, then other solutions must be purely positive.
If $n$ is odd, the solutions other than $(0,0,0)$ come in opposi... | [] | [
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} |
Let $ABC$ be an equilateral triangle whose sides have length $1$ . The midpoints of $AB,BC$ are $M,N$ respectively. Points $K,L$ were chosen on $AC$ so that $KLMN$ is a rectangle. Inside this rectangle are three semi-circles with the same radius, as in the picture (the endpoints are on the edges of the rec... | [] | [
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Determine if there exists a set $S$ of $5783$ different real numbers with the following property:
For every $a,b\in S$ (not necessarily distinct) there are $c\neq d$ in $S$ so that $a\cdot b=c+d$ . | Solution: <details><summary>Click to expand</summary>Just take $S=\{0, \pm 1, \pm x, \cdots, \pm x^{2890}\}$ where $x^{2891}=x+1$ and $x\in (1,2)$ .</details> | [] | [
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A real number is written next to each vertex of a regular pentagon. All five numbers are different. A triple of vertices is called **successful** if they form an isosceles triangle for which the number written on the top vertex is either larger than both numbers written on the base vertices, or smaller than both. Find ... | Let be $A_0,A_1,A_2,A_3,A_4$ the vertices of the regular pentagon, in this order.
For each $k\in\{0,1,2,3,4\}$ , there are exactly $2$ isosceles triangles with the top vertex $A_k:\; A_kA_{k-1}A_{k+1}$ and $ A_kA_{k-2}A_{k+2}$ (we consider the indexes modulo $5$ ).
Denote $a_k$ the number associated to the ... | [] | [
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} |
Let $SABCDE$ be a pyramid whose base $ABCDE$ is a regular pentagon and whose other faces are acute triangles. The altitudes from $S$ to the base sides dissect them into ten triangles, colored red and blue alternatingly. Prove that the sum of the squared areas of the red triangles is equal to the sum of the square... | [] | [
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Find all functions $f:\mathbb{Z}\to \mathbb{Z}_{>0}$ for which
\[f(x+f(y))^2+f(y+f(x))^2=f(f(x)+f(y))^2+1\]
holds for any $x,y\in \mathbb{Z}$ . | Pretty nice, especially the "proof" of injectivity (of course at the end of the day $f$ will be seen not to be injective).
We claim that $f \equiv 1$ is the only solution. As usual, let $f^k(x) = f(f^{k-1}(x))$
The proof will proceed in the following steps, none of them too unnatural but not too straight forwa... | [
"Gorgeous!!**Case 1:** $f$ is not injective.\nSuppose $f(a)=f(b)$ , then we easily get $f(a+f(y))=f(b+f(y))$ . Repeating this, we have $f(a+\\sum f(y_i))=f(b+\\sum f(y_i))$ for any collection of $y_i$ s. Since $2f(x+f(x))^2=f(2f(x))^2+1$ we can find two coprime values in the range of $f$ , hence by chicken... | [
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} |
For positive integers $n$ , let $f_2(n)$ denote the number of divisors of $n$ which are perfect squares, and $f_3(n)$ denotes the number of positive divisors which are perfect cubes. Prove that for each positive integer $k$ there exists a positive integer $n$ for which $\frac{f_2(n)}{f_3(n)}=k$ . | Notice that it suffices to write $k$ as the product of some terms of the form $g(n)=\frac{[\frac{n}{2}]+1}{[\frac{n}{3}]+1}$ . For this, we use strong induction. For $n=1, g(0)=1$ . Now, suppose this holds for all $n<k$ , let's show that it holds for $n=k$ . First, notice that $g(2)=2$ , $g(4)=\frac32$ , $g(6k... | [] | [
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} |
In an $8 \times 8$ grid of squares, each square was colored black or white so that no $2\times 2$ square has all its squares in the same color. A sequence of distinct squares $x_1,\dots, x_m$ is called a **snake of length $m$** if for each $1\leq i <m$ the squares $x_i, x_{i+1}$ are adjacent and are of diffe... | [
"Bump Bump",
"<details><summary>Answer</summary>8</details>\n<details><summary>Main Claim</summary>We can always find a snake which either goes from the bottom to the top of the board or from the far right to the far left.</details>\n"
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} | |
In triangle $ABC$ the orthocenter is $H$ and the foot of the altitude from $A$ is $D$ . Point $P$ satisfies $AP=HP$ , and the line $PA$ is tangent to $(ABC)$ . Line $PD$ intersects lines $AB, AC$ at points $X,Y$ respectively.
Prove that $\angle YHX = \angle BAC$ or $\angle YHX+\angle BAC= 180^\c... |
<details><summary>My Solution</summary>Let the intersection of the line $AD$ and $(ABC)$ be $H'$ . $\mathcal{H}$ is the homothety which takes $H'$ to $H$ with center $A$ . $R$ and $S$ is the image of $B,C$ under $\mathcal{H}$ respectively. Let's define $X' = AR\cap HS$ , $Y'=AS \cap HR$ and $Z=... | [
"<details><summary>Solution</summary>Assume that $PD$ intersects rays $AB, AC$ . $D$ is the midpoint of $BB',CC'$ and $HH'$ . $M$ is the midpoint of $AH$ . $A'$ is the antipode of $A$ in $(ABC)$ . $PD$ intersects $(ABC)$ at $U,V$ . $O$ is the circumcenter. $\\frac{PA}{PM}=\\frac{AH'}{AA'}=\\... | [
"origin:aops",
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"path": "Contest Collections/2023 Contests/2023 Israel TST/3038106.json"
} |
Toph wants to tile a rectangular $m\times n$ square grid with the $6$ types of tiles in the picture (moving the tiles is allowed, but rotating and reflecting is not). For which pairs $(m,n)$ is this possible?
**Attachments:**
![tstp1.png](https://cdn.artofproblemsolving.com/attachments/2/0/a9a57d8f0768576417ced... | [] | [
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For each positive integer $n$ , define $A(n)$ to be the sum of its divisors, and $B(n)$ to be the sum of products of pairs of its divisors. For example,
\[A(10)=1+2+5+10=18\]
\[B(10)=1\cdot 2+1\cdot 5+1\cdot 10+2\cdot 5+2\cdot 10+5\cdot 10=97\]
Find all positive integers $n$ for which $A(n)$ divides $B(n)$ . | <details><summary>solution</summary>Let C(n) be the sum of the squares of the divisors. We have $2A(n)|A(n)^2-C(n)$ . Assume that $2 |A(n)$ . $2 \prod \frac{p^{a+1}-1}{p-1} | \prod \frac{p^{2(a+1)}-1}{p^2-1} \implies 2|\prod \frac{p^{a+1}+1}{p+1}$ . Note that $v_2(\frac{p^{a+1}+1}{p+1}) \le 0$ for any $p$ , contr... | [] | [
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"2023 Israel TST",
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Let $ABC$ be a fixed triangle. Three similar (by point order) isosceles trapezoids are built on its sides: $ABXY, BCZW, CAUV$ , such that the sides of the triangle are bases of the respective trapezoids. The circumcircles of triangles $XZU, YWV$ meet at two points $P, Q$ . Prove that the line $PQ$ passes throug... | <details><summary>Complex Bash</summary>WLOG $A, B, C$ on the unit circle, and denote a lowercase variable complex number as the coordinate of a point labelled with the corresponding uppercase letter. We claim that $G$ , the centroid of $\triangle ABC$ , is the fixed point.
Let $r = AU / AC$ and $\omega$ be th... | [] | [
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A regular polygon with $20$ vertices is given. Alice colors each vertex in one of two colors. Bob then draws a diagonal connecting two opposite vertices. Now Bob draws perpendicular segments to this diagonal, each segment having vertices of the same color as endpoints. He gets a fish from Alice for each such segment ... | The answer is $4$ . For an example, paint $2$ adjacent vertices with the same color alternately.
Let $a$ be the number of vertices painted with one of the colors, $a \geq 10$ wlog. Let's calc the average number of segments drawned by Bob over all the $10$ simmetry axis in an arbitrary config. If this number i... | [] | [
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} |
Let $ABC$ be an isosceles triangle, $AB=AC$ inscribed in a circle $\omega$ . The $B$ -symmedian intersects $\omega$ again at $D$ . The circle through $C,D$ and tangent to $BC$ and the circle through $A,D$ and tangent to $CD$ intersect at points $D,X$ . The incenter of $ABC$ is denoted $I$ . Prove t... | Symmedians, the worst of them all. Anyway,
Let $M$ be the midpoint of $AC$ . It is well-known that $X \in (MDC)$ . Now, $\angle CXB= 2\pi - \angle CXD - \angle DXA - \angle AXB
= (\pi-\angle CXD) + (\pi - \angle DXA) - \angle AXB
= (\angle AMB) + (\pi - \angle CBA) - \angle AXB = \frac{\pi}{2} + \frac{\angle A}... | [
"By design, $ABCD$ is harmonic, so there exists a (unique) Brokard point. To see that $X$ is this point, note that it satisfies the condition that three of the angles $XAD, XDC, XCB$ are equal, and only the Brokard point satisfies these.\nHaving concluded that $X$ is the Brokard point of harmonic quadrilate... | [
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"path": "Contest Collections/2023 Contests/2023 Israel TST/3038347.json"
} |
Given a polynomial $P$ and a positive integer $k$ , we denote the $k$ -fold composition of $P$ by $P^{\circ k}$ . A polynomial $P$ with real coefficients is called **perfect** if for each integer $n$ there is a positive integer $k$ so that $P^{\circ k}(n)$ is an integer. Is it true that for each perfect ... | <blockquote>The answer is negative.</blockquote>
You'r polynomial doesn't satisfy the condition as $P(-1)=\frac{3}{2}$ and $\frac{3}{2}$ is a fixed point.
The general idea is correct though. | [
"Bumpbump",
"The answer is negative.\n\n**Attachments:**\n\n[Israel Tst 2023 Day 5 P3.pdf](https://cdn.artofproblemsolving.com/attachments/d/1/75d8da2afc5d41ea1df5d3c97413764596c619.pdf)",
"<blockquote><blockquote>The answer is negative.</blockquote>\n\nYou'r polynomial doesn't satisfy the condition as $P(-1)=... | [
"origin:aops",
"2023 Israel TST",
"2023 Contests"
] | {
"answer_score": 4,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Israel TST/3038352.json"
} |
Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that for all $x, y\in \mathbb{R}$ the following holds:
\[f(x)+f(y)=f(xy)+f(f(x)+f(y))\]
| <blockquote>Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that for all $x, y\in \mathbb{R}$ the following holds:
\[f(x)+f(y)=f(xy)+f(f(x)+f(y))\]</blockquote>
Let $P(x,y)$ be the assertion $f(x)+f(y)=f(xy)+f(f(x)+f(y))$ .
Note that all functions of the form $\boxed{f(x)=c}$ , $c\in\mathbb R$ work, oth... | [
"Very nice!\n\nLet $P(x,y)$ be the given assertion,.\n\nClearly $f\\equiv c$ works, so let's now assume that $f$ is non-constant. $P(0,0): f(0)=f(2f(0))$ $P(x,0): f(x)=f(f(x)+f(0))$ $P(x,2f(0)): f(x)+f(2f(0))=f(2f(0)x)+f(f(x)+f(2f(0))\\Leftrightarrow f(2f(0)x)=f(0)$ so $f(0)=0$ otherwise $f$ must be co... | [
"origin:aops",
"2023 Israel TST",
"2023 Contests"
] | {
"answer_score": 1062,
"boxed": true,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 Israel TST/3068944.json"
} |
Let $n>3$ be an integer. Integers $a_1, \dots, a_n$ are given so that $a_k\in \{k, -k\}$ for all $1\leq k\leq n$ . Prove that there is a sequence of indices $1\leq k_1, k_2, \dots, k_n\leq n$ , not necessarily distinct, for which the sums
\[a_{k_1}\]
\[a_{k_1}+a_{k_2}\]
\[a_{k_1}+a_{k_2}+a_{k_3}\]
\[\vdots\]
\[... | problem good very, me likey much
<details><summary>solution</summary>Let $m = 2n+1$ and make a directed graph $G$ on $\mathbb{Z}/m\mathbb{Z}$ where $u \mapsto v$ if and only if there exists some $a_j$ congruent to $(v-u) \bmod m$ . Note that every vertex in $G$ has outdegree exactly $n$ now, so there are... | [
"bump this..."
] | [
"origin:aops",
"2023 Israel TST",
"2023 Contests"
] | {
"answer_score": 180,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Israel TST/3068951.json"
} |
Let $ABC$ be an acute-angled triangle with circumcenter $O$ and incenter $I$ . The midpoint of arc $BC$ of the circumcircle of $ABC$ not containing $A$ is denoted $S$ . Points $E, F$ were chosen on line $OI$ for which $BE$ and $CF$ are both perpendicular to $OI$ . Point $X$ was chosen so that $X... | <details><summary>solution</summary>We denote line $OI {}$ as $\ell$ .
Let $L$ denote the intersection $\ell \cap BC$ . Without losing generality, we assume that $L$ lies on ray $CB$ past $B$ .
[center][img width = "40"]https://i.imgur.com/oTaBNnu.png[/img][/center]
From the perpendicularities, we note th... | [
"There is a generalized problem: Let $ABCD$ be a cyclic quadrilateral inscribed in $(O)$ , $X$ is a point on $AC$ , then the orthopole of $XO$ wrt $ABD$ , $CBD$ , and the foot of $X$ on $BD$ are collinear.\n\nLet the line connecting the midpoint of $AB,AD$ be $\\ell_A$ , the line connecting the mid... | [
"origin:aops",
"2023 Israel TST",
"2023 Contests"
] | {
"answer_score": 230,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Israel TST/3068953.json"
} |
Let $a, b$ be positive integers such that $54^a=a^b$ . Prove that $a$ is a power of $54$ . | <details><summary>Solution</summary>Clearly both sides are positive integers, so we can take the $2$ -adic valuation of both sides to get $v_2(54^a) = v_2 (a^b).$ Using the basic properties of p-adic valuation we can rewrite this equation as $a = bv_2(a)$ which means that $v_2(a) = \frac{a}{b}.$ This means that ... | [
"It's easy to see that $v_2(a)=\\dfrac{a}{b}$ and $v_3(a)=\\dfrac{3a}{b}$ and $\\dfrac{a}{b}$ is integer\nSo $a$ is power of 54",
" $54^a=a^b \\Rightarrow \\frac{a}{b}=\\log_{54}{a}$ Intuitively both sides must be positive integers, so the result follows.",
"<blockquote>Intuitively</blockquote>\nAs an ... | [
"origin:aops",
"2023 ITAMO",
"2023 Contests"
] | {
"answer_score": 122,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 ITAMO/3067504.json"
} |
Let $s(n)$ denote the sum of the digits of $n$ .
a) Do there exist distinct positive integers $a, b$ , such that $2023a+s(a)=2023b+s(b)$ ?
b) Do there exist distinct positive integers $a, b$ , such that $a+2023s(a)=b+2023s(b)$ ? | Bruh wut
a) $10^{2024}-1$ and $10^{2024}+8$ .
b) $95555$ and $113762$
literally anything works | [
"Well, quite trivially, the answer is yes for both a) and b), as we can just choose $a=b$ .\nPossibly you missed the word \"distinct\"?"
] | [
"origin:aops",
"2023 ITAMO",
"2023 Contests"
] | {
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 ITAMO/3067505.json"
} |
Fix circle with center $O$ , diameter $AB$ and a point $C$ on it, different from $A, B$ . Let a point $D$ , different from $A, B$ , vary on the arc $AB$ not containing $C$ . Let $E$ lie on $CD$ such that $BE \perp CD$ . Prove that $CE \cdot ED$ is maximal exactly when $BOED$ is cyclic. | Note $B$ spirals $CE$ to $AD$ and $CA$ to $ED$ .
Hence $\frac{ED}{CA}=\frac{BD}{AB}$ and $\frac{EC}{AD}=\frac{BC}{AB}$ . This means $$ ED\times EC=\frac{BD\times BC\times CA\times AD}{AB^2}. $$ The only thing that matters is $BD\times AD$ , to maximise $D$ is the midpoint of arc. | [
"I'm assuming $O$ is the center of the circle?",
"<blockquote>I'm assuming $O$ is the center of the circle?</blockquote>\n\n<blockquote>Fix circle with center $O$ , diameter $AB$ </blockquote>\n\nit says O is the center",
"I edited it after his comment. ",
"Come on! This is an easy complex bash.",
"Le... | [
"origin:aops",
"2023 ITAMO",
"2023 Contests"
] | {
"answer_score": 22,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 ITAMO/3067506.json"
} |
Let $a, b, c$ be reals satisfying $a^2+b^2+c^2=6$ . Find the maximal values of the expressions
a) $(a-b)^2+(b-c)^2+(c-a)^2$ ;
b) $(a-b)^2 \cdot (b-c)^2 \cdot (c-a)^2$ .
In both cases, describe all triples for which equality holds. | notice for (a) it is equal to $12-2(ab+bc+ca)$ $(a^2+b^2+c^2) = 6 $ by cauchy $-3\sqrt{2} \leq(a+b+c) \leq 3\sqrt{2}$ $ab+bc+ca = \frac{(a+b+c)^2-a^2-b^2-c^2}{2}=\frac{(a+b+c)^2-6}{2}$ $2ab+2bc+2ca=(a+b+c)^2-6$ $12-2ab-2bc-2ca = 18-(a+b+c)^2 \leq 18$ Equality cases $a+b+c =0$ $a^2+b^2+c^2=6$ $ab+bc+ca=-3$ $b... | [
"Let $a, b, c$ be reals satisfying $a^2+b^2+c^2=6$ . Prove that $$ (a-b)^2+(b-c)^2+(c-a)^2\\leq 18 $$ $$ (a-b)^2 (b-c)^2 (c-a)^2\\leq 108 $$ ",
"Let's say A to this polynom. $ab+ac+bc=((a+b+c)^2-a^2-b^2-c^2)/2$ For $(a+b+c)^2$ , we can get this inequality; $3\\sqrt{2}>=(a+b+c)^2>=0$ If we make it bigg... | [
"origin:aops",
"2023 ITAMO",
"2023 Contests"
] | {
"answer_score": 24,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 ITAMO/3067507.json"
} |
Dedalo buys a finite number of binary strings, each of finite length and made up of the binary digits 0 and 1. For each string, he pays $(\frac{1}{2})^L$ drachmas, where $L$ is the length of the string. The Minotaur is able to escape the labyrinth if he can find an infinite sequence of binary digits that does not c... | Take a big $n$ and use the strings with $n+1$ $1$ s and the string with $n+1$ $0$ s. These cost $\frac{1}{2^n}$ total.
Now any possible valid string can only have at most $n$ consecutive bits that are equal. Replace the binary string with an integer array depicting the length of consecutive equal blocks. Fo... | [
"The only possible values for $c$ are $0, 25, 50, 75, 100, 125, 150, 175 \\dots $ because $\\frac{c}{100}$ must be a fraction of a power of 2. I also see that we can inflate the price at any time by buying random strings that don't actually affect the game. For example we can achieve 125 by buying $0, 00, 01... | [
"origin:aops",
"2023 ITAMO",
"2023 Contests"
] | {
"answer_score": 50,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 ITAMO/3067509.json"
} |
Let $n$ be a positive integer. On a blackboard, Bobo writes a list of $n$ non-negative integers. He then performs a sequence of moves, each of which is as follows:
-for each $i = 1, . . . , n$ , he computes the number $a_i$ of integers currently on the board that are at most $i$ ,
-he erases all integers on t... | We claim that the process eventually ends in $1,2,\ldots,n$ in $2n$ moves. Note that if the process reaches this point, it indeed terminates.
We induct on $n$ :**Base Case.** $n=1$ . In this case, if the number he starts with is at most $1$ , it becomes $1$ . Otherwise, it becomes a $0$ , which on the next mo... | [
" $a)$ Notice that after $2$ moves $n$ will be written on the board and there will be no number bigger than $n$ so we can kinda proceed by induction I'm to lazy to write up. Anyways we should get that max amount of moves is less than or equal $2n$ $b)$ Construction for $2n$ is all zeroes."
] | [
"origin:aops",
"2023 ITAMO",
"2023 Contests"
] | {
"answer_score": 54,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 ITAMO/3067511.json"
} |
Let $n,k$ are integers and $p$ is a prime number. Find all $(n,k,p)$ such that $|6n^2-17n-39|=p^k$ | we have $|2n+3|\cdot |3n-13|=p^{k}$ .
now clearly we must have $k \geqslant 0$ Consider $|2n+3|=p^{\alpha} , |3n-13|=p^{\beta}$ , where $\alpha , \beta \in \mathbb{Z}_{0}^{+}$ clearly we can't have $\alpha=\beta=0$ , so:
- ***Case 1:-*** $\alpha=0$ then we have $|2n+3|=1$ which gives $n=-1$ or $-2$ , p... | [
"\\[|(2n+3)(3n-13)| = p^k\\] $gcd(2n+3,3n-13) = gcd(2n+3, 6n-26) = gcd(2n+3,35) = 1,5,7,35$ so finitely many $n$ to check.\nif $gcd(2n+3,3n-13) = 1$ , then either $n=-1$ or $n=4$ (one of the factors has to be one). This gives $(n,k,p) = (4,1,11)$ .\n\nif $gcd(2n+3,3n-13) = 5$ , then one has to equal $\\p... | [
"origin:aops",
"2023 JBMO TST - Turkey",
"2023 Contests"
] | {
"answer_score": 1160,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 JBMO TST - Turkey/3063100.json"
} |
Let $ABC$ is acute angled triangle and $K,L$ is points on $AC,BC$ respectively such that $\angle{AKB}=\angle{ALB}$ . $P$ is intersection of $AL$ and $BK$ and $Q$ is the midpoint of segment $KL$ . Let $T,S$ are the intersection $AL,BK$ with $(ABC)$ respectively. Prove that $TK,SL,PQ$ are concurre... | $\angle{AKB} = \angle{ALB} \iff A,K,B,L$ are cyclic $\iff \measuredangle{ALK} = \measuredangle{ABK} \iff \measuredangle{ALK} = \measuredangle{ABS} \iff \measuredangle{TLK} = \measuredangle{ABS} \iff \measuredangle{TLK} = \measuredangle{ATS} \iff \measuredangle{TLK} = \measuredangle{LTS} \iff KL \parallel ST$
We ha... | [
"You sure the problem is correct?",
"corrected,thanks.",
"So easy, LK//ST",
"Let $ABC$ be not is you are creating a contest and K and L are points not is correct your grammatic mistakes",
"<blockquote>Let $ABC$ be not is you are creating a contest and K and L are points not is correct your grammatic mis... | [
"origin:aops",
"2023 JBMO TST - Turkey",
"2023 Contests"
] | {
"answer_score": 54,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2023 Contests/2023 JBMO TST - Turkey/3063101.json"
} |
Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+f(x))=f(-x)$ and for all $x \leq y$ it satisfies $f(x) \leq f(y)$ | Just consider the longest interval $(x_1,x_2)$ .Such that $f(x_1)=f(x_2)=c$ and $c\in (x_1,x_2)$ .
Then try to prove $-x_1+f(-x_1)\not \in (x_1,x_2)$ or $-x_2+f(-x_2)\not \in (x_1,x_2)$ .
(And you need to prove such interval is exists)
Then answer is $\boxed{f(x)\equiv c\quad c\in \mathbb{R}}$ | [
"don't know how to do\nfor all $x \\leq y$ it satisfies $f(x) \\leq f(y)$ means it increase in $R$ but i just found $f(x)=-2x$ which not fits.",
"<blockquote>don't know how to do\nfor all $x \\leq y$ it satisfies $f(x) \\leq f(y)$ means it increase in $R$ but i just found $f(x)=-2x$ which not fits.</blo... | [
"origin:aops",
"2023 JBMO TST - Turkey",
"2023 Contests"
] | {
"answer_score": 1012,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 JBMO TST - Turkey/3063103.json"
} |
Prove that for all $a,b,c$ positive real numbers $\dfrac{a^4+1}{b^3+b^2+b}+\dfrac{b^4+1}{c^3+c^2+c}+\dfrac{c^4+1}{a^3+a^2+a} \ge 2$ | $\frac{a^4+1}{2}+\frac{a^4+a^4+a^4+1}{4}+\frac{a^4+1+1+1}{4}=\frac{6*a^4+6}{4}=\frac{3*(a^4+1)}{2} \ge a^3+a^2+a$ $\frac{a^4+1}{a^3+a^2+a} \ge \frac{2}{3}$ Applying $AM-GM$ to $LHS$ ; $LHS \ge 3* \frac{2}{3}=2 \ge 2$ Done! | [
"Just note that $a^4+1 \\geq \\dfrac{2}{3} \\cdot (a^3+a^2+a)$ (it is equivalent to $(a-1)^2(3a^2+4a+3) \\geq 0)$ , and use AM-GM.",
"My solution from the exam. $LHS \\geq 3\\sqrt[3]{\\frac{(a^4+1)(b^4+1)(c^4+1)}{(a^3+a^2+a)(b^3+b^2+b)(c^3+c^2+c)}}$ Let's prove that $\\frac{a^4+1}{a^3+a^2+a} \\geq \\frac{2}{3... | [
"origin:aops",
"2023 JBMO TST - Turkey",
"2023 Contests"
] | {
"answer_score": 10,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 JBMO TST - Turkey/3063104.json"
} |
Let $ABC$ is triangle and $D \in AB$ , $E \in AC$ such that $DE//BC$ . Let $(ABC)$ meets with $(BDE)$ and $(CDE)$ at the second time $K,L$ respectively. $BK$ and $CL$ intersect at $T$ . Prove that $TA$ is tangent to the $(ABC)$ | [center]
[asy]
import graph; size(13cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -1.85536646698403, xmax = 35.12511020013301, ymin = -5.30602092... | [
"Note that $T,D,E$ is collinear, $D,L,C,E$ is cyclic\nSo $TD\\cdot TE=TL\\cdot TC$ Which meas that $Pow(T,\\odot(ABC))=Pow(T,\\odot(ADE))$ Hence $T$ is on the redical axis of $\\odot(ABC)$ and $\\odot(ADE)\\Rightarrow$ $TA$ is tangent to the $\\odot(ABC)$ ",
"My solution from the exam.\nLet the tangent... | [
"origin:aops",
"2023 JBMO TST - Turkey",
"2023 Contests"
] | {
"answer_score": 138,
"boxed": false,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 JBMO TST - Turkey/3063105.json"
} |
For a prime number $p$ . Can the number of n positive integers that make the expression \[\dfrac{n^3+np+1}{n+p+1}\] an integer be $777$ ? | The answer is no!
Assume the contrary, and observe that $p>2$ . We first notice $n\equiv -p-1\pmod{n+p+1}$ . So,
\[
0\equiv n^3+np+1\equiv -(p+1)^3 - p(p+1)+1\pmod{n+p+1},
\]
which implies
\[
n+p+1\mid n^3+np+1\Leftrightarrow n+p+1\mid p(p+2)^2.
\]
As $n\ge 1$ , the number of all such $n$ is clearly the number o... | [] | [
"origin:aops",
"2023 JBMO TST - Turkey",
"2023 Contests"
] | {
"answer_score": 60,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 JBMO TST - Turkey/3063107.json"
} |
Initially, Aslı distributes $1000$ balls to $30$ boxes as she wishes. After that, Aslı and Zehra make alternated moves which consists of taking a ball in any wanted box starting with Aslı. One who takes the last ball from any box takes that box to herself. What is the maximum number of boxes can Aslı guarantee to t... | We will say a box is "good" if it contains an odd number of balls. Let $a_n$ be the number of good boxes after $2n$ moves and let $b_n$ be the number of moves where Aslı took a ball from a good box in her first $n$ moves. Assume that Zehra always takes balls from good boxes. (obviously she can do that) In that ... | [
"Answer is 15 i will add solution later",
"Aslı'nın herhangi bir top diziliminde çift sayıda çift toplu kutu ve tek toplu kutu olacağı barizdir. Bu durumda tek toplu kutuya tek toplu kutu ve çift toplu kutuya çift toplu kutu eşleştirmesi yapılabilir.Zehra aşağıdaki algoritma ile Aslı'nın mümkün olduğunca az kutu ... | [
"origin:aops",
"2023 JBMO TST - Turkey",
"2023 Contests"
] | {
"answer_score": 1038,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 JBMO TST - Turkey/3063108.json"
} |
A marble is placed on each $33$ unit square of a $10*10$ chessboard. After that, the number of marbles in the same row or column with that square is written on each of the remaining empty unit squares. What is the maximum sum of the numbers written on the board?
| **Answer:** $438$ .**Example:** $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
X&X&X&&&&&& \hline
X&X&X&&&&&& \hline
X&X&X&&&&&& \hline
&&&X&X&X&&&\hline
&&&X&X&X&&&\hline
&&&X&X&X&&&\hline
&&&&&&X&X&X&X \hline
&&&&&&X&X&X&X\hline
&&&&&&X&X&X&X \hline
&&&&&&X&X&X& \hline
\end{array} $$ **Proof:**
Construct the b... | [
"Answer is 438 i will ad solution later",
"We will prove that the answer to the problem is $438$ . Let $x_i$ and $y_i$ be the number of marbles in the $i-$ row and column of the board, respectively. Note that $\\displaystyle \\sum_{i=1}^{10} x_i=\\sum_{i=1}^{10} y_i=10$ . The total sum of the numbers writ... | [
"origin:aops",
"2023 JBMO TST - Turkey",
"2023 Contests"
] | {
"answer_score": 164,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 JBMO TST - Turkey/3063113.json"
} |
Find all pairs $(a,b)$ of positive integers such that $a!+b$ and $b!+a$ are both powers of $5$ .
*Nikola Velov, North Macedonia* | We claim the only working ordered pairs are $(a,b)=\boxed{(1,4),(4,1),(5,5)}$ . It is easy to see that they work; we now prove they are the only ones.
It is not hard to verify that these are the only solutions when $a,b\le 5$ , so henceforth, FTSOC, assume $a,b>5$ . We must have $a!+b = 5^x$ and $a+b! = 5^y$ , f... | [
"WLOG assume that $a \\geq b$ . We split into two cases.**Case 1:** $a=b$ . Then, $a!+a=5^k$ with $k \\geq 1$ . If $k=1$ then obiously $(a,b)=(5,5)$ works. If $k \\geq 2$ , then $(a-1)!+1 \\equiv 1 \\pmod 5$ , and since $5^k=a!+a=a((a-1)!+1),$ we must have $(a-1)!+1=1,$ a contradiction.**Case 2:** $... | [
"origin:aops",
"2023 Junior Balkan Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 1156,
"boxed": false,
"end_of_proof": false,
"n_reply": 21,
"path": "Contest Collections/2023 Contests/2023 Junior Balkan Mathematical Olympiad/3098577.json"
} |
Prove that for all non-negative real numbers $x,y,z$ , not all equal to $0$ , the following inequality holds $\displaystyle \dfrac{2x^2-x+y+z}{x+y^2+z^2}+\dfrac{2y^2+x-y+z}{x^2+y+z^2}+\dfrac{2z^2+x+y-z}{x^2+y^2+z}\geq 3.$ Determine all the triples $(x,y,z)$ for which the equality holds.
*Milan Mitreski, Serbia* | <blockquote>Prove that for all non-negative real numbers $x,y,z$ , not all equal to $0$ , the following inequality holds $\displaystyle \dfrac{2x^2-x+y+z}{x+y^2+z^2}+\dfrac{2y^2+x-y+z}{x^2+y+z^2}+\dfrac{2z^2+x+y-z}{x^2+y^2+z}\geq 3.$ Determine all the triples $(x,y,z)$ for which the equality holds.
*Milan Mitreski... | [
"Notice that its equivalent to $\\frac{a}{b+c} + \\frac{b}{c+a} +\\frac{c}{a+b} \\geq \\frac{3}{2} $ \n\nfor $a=2x^2-x+y+z $ , $b=2y^2 +x-y+z$ , $c=2z^2+x+y-z$ This can be rewritten as $\\sum \\frac{(a-b)^2}{(c+a)(c+b)} \\geq 0$ So since $a+b , b+c , c+a >0$ it is true and equality holds when $a=b=c \\i... | [
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Alice and Bob play the following game on a $100\times 100$ grid, taking turns, with Alice starting first. Initially the grid is empty. At their turn, they choose an integer from $1$ to $100^2$ that is not written yet in any of the cells and choose an empty cell, and place it in the chosen cell. When there is no e... | Solved with **MathLuis** even though we both finished individually me ending up headsolving on the way to the garage. It turns that the problem for a general $n\times n$ grid is drastically easier for even $n$ . It should be interesting to see what happens when $n$ is odd.
We claim that Bob always wins, when the ... | [
"The second player have the wining strategy.He place horizondal domino in the row $1,2,3,...,99$ but not in $100$ and make the couples: $(100^2,1),(100^2-1,2),.....,(100^2/2+1,100^2/2)$ Τhen it follows if $A$ plays a number on someone domino then $B$ plays the number of his pair on the same domino.If $A$ ... | [
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Let $ABC$ be an acute triangle with circumcenter $O$ . Let $D$ be the foot of the altitude from $A$ to $BC$ and let $M$ be the midpoint of $OD$ . The points $O_b$ and $O_c$ are the circumcenters of triangles $AOC$ and $AOB$ , respectively. If $AO=AD$ , prove that points $A$ , $O_b$ , $M$ and $O... | **Lemma:** In isosceles triangle $ABC$ with $BA=BC$ , let $H$ be the orthocenter. $AH\cap BC=D$ . Let $\omega$ be the circle centered at $C$ with radius $CM$ where $M$ is the midpoint of $AC$ . The tangent at $D$ to $(M)$ intersects $\omega$ at $K,L$ . The reflections of $C$ to $ML,MK$ are $X... | [
"<details><summary>Solution</summary>Let $AM \\cap BC=P, OO_c \\cap BC=X, OO_b \\cap BC=Y$ . We have that $\\angle AO_bY=180^{o}-2\\gamma=\\angle AXY$ , so $AO_bXY$ is cyclic and similarly $AO_cXY$ is cyclic. Hence, we are left to prove that $AMXY$ is cyclic. We have that $AM$ is the perpendicular bisecto... | [
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"path": "Contest Collections/2023 Contests/2023 Junior Balkan Mathematical Olympiad/3098582.json"
} |
Find all prime $x,y$ and $z,$ such that $x^5 +y^3 -(x+y)^2=3z^3$ | Since $x^5$ has the same parity as $x$ , $y^3$ as $y$ , $(x+y)^2$ as $x+y$ , we deduce that the left-hand side is even, so $3z^3$ must also be even and hence $z=2$ . We are left to solve $x^5 + y^3 - 24 = (x+y)^2$ . If $x\geq y$ , then $x^5 + y^3 - 24 \leq 4x^2$ , so $x^2(x^3 - 4) - 24 \leq 0$ , which i... | [
"Sol\n\n<details><summary>Click to expand</summary>$x^5+y^3-(x+y)^2=3z^3$ Taking the $LHS \\mod{3}$ , the only \"fitting\" possibilities for $x,y>3$ are: \n\nCase $1: (3k+1)^5+(3p+2)^3-(3k+3p+3)^2=3z^3$ Case $2: (3k+2)^5+(3p+1)^3-(3k+3p+3)^2=3z^3$ As shown below, $2\\mid LHS\\Longrightarrow z=2$ is the only ... | [
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"answer_score": 48,
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} |
Let $ AD $ , $ BE $ and $ CF $ be the altitudes of $ \Delta ABC $ . The points $ P, \, \, Q, \, \, R $ and $ S $ are the feet of the perpendiculars drawn from the point $ D $ on the segments $ BA $ , $ BE $ , $ CF $ and $ CA $ , respectively. Prove that the points $ P, \, \, Q, \, \, R $ and $ S $ ... | <details><summary>bash through point</summary>Put $B (0,0)$ then $A (m,n)$ , $C(p,0)$ consequently just play with the gradient $D(m,0) $ , $E (\frac{n^2*p}{(m-p)^2+n^2},\frac{n*p(p-m)}{(p-m)^2+n^2})$ , $F=(\frac{m^2*p}{m^2+n^2},\frac{m*p*n}{m^2+n^2})$ $P=(\frac{m^3}{m^2+n^2} ,\frac{m^2n}{m^2+n^2})$ , $Q = ... | [
"PQS is the Simson line of D wrt ABE, while PQR is the Simpon line of D wrt AFC, so we are done"
] | [
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The positive integer $ n $ verifies $$ \frac{1}{1\cdot(\sqrt{2}+\sqrt{1})+\sqrt{1}}+\frac{1}{2\cdot(\sqrt{3}+\sqrt{2})+\sqrt{2}}+\cdots+\frac{1}{n\cdot(\sqrt{n+1}+\sqrt{n})+\sqrt{n}}=\frac{2022}{2023}. $$ Find the sum of digits of number $ n $ . | Note that all our terms are of the form $$ \dfrac{1}{a\sqrt{b}+b\sqrt{a}} $$ where $a=n$ and $b=n+1$ for integer $n.$ We rationalize the denominator to get $$ \dfrac{a\sqrt{b}-b\sqrt{a}}{a^2b-b^2a}=\dfrac{a\sqrt{b}-b\sqrt{a}}{ab}\cdot\dfrac{1}{a-b}=\left(\dfrac{1}{\sqrt{b}}-\dfrac{1}{\sqrt{a}}\right)\cdot -1=\... | [
" $$ n=2023^2-1 $$ "
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Prove that the number $A=2024^{n+1}-2023n-2024$ has at least $15$ different positive divisors for every nonnegative integer $ n $ . | [
"A=2024^(n+1)-2023n-2024=2024(2024^n-1)-2023n=2024*2023(2024^(n-1)+...+2024+1)-2023n=2023(2024(2024^(n-1)+...+2024+1)-n)=2023^2*T, where T - natural number. Thus σ(A)>=σ(7^2*17^4)=3*5=15"
] | [
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The positive integers $ a, b, c $ are the lengths of the sides of a right triangle. Prove that $abc$ is divisible by $60$ . | $x^2\equiv 0, 1\pmod{3}$ . $x^2\equiv 0, 1\pmod{4}$ . $x^2\equiv 0, 1, 4\pmod{5}$ . $a^2=b^2+c^2$ | [] | [
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Real numbers $a,b,c$ with $a\neq b$ verify $$ a^2(b+c)=b^2(c+a)=2023. $$ Find the numerical value of $E=c^2(a+b)$ . | We have $a^2 (b+c)-b^2 (c+a)=0$ or $(a-b)(ab+bc+ca)=0$ But $a \neq b$ , so $ab+bc+ca=0$ .
We have $a^2 (b+c)=b^2 (c+a)=c^2 (a+b)=-abc$ Finally, $E=2023$ | [] | [
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Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=3. $ Prove that $$ \frac{a^4+3ab^3}{a^3+2b^3}+\frac{b^4+3bc^3}{b^3+2c^3}+\frac{c^4+3ca^3}{c^3+2a^3}\leq4. $$ | $\frac{a^4+3ab^3}{a^3+2b^3}+\frac{b^4+3bc^3}{b^3+2c^3}+\frac{c^4+3ca^3}{c^3+2a^3}= \sum \frac{a^4+3ab^3}{a^3+2b^3}
=\sum \frac{a(a^3+3b^3)}{a^3+2b^3}=\sum \frac{a(a^3+2b^3+b^3)}{a^3+2b^3}=\sum \frac{a(a^3+2b^3)+ab^3}{a^3+2b^3}=$ $=\sum \frac{a(a^3+2b^3)}{a^3+2b^3}+\sum \frac{ab^3}{a^3+2b^3}=\sum a + \sum \frac{ab^3}{... | [
"[Solution](https://artofproblemsolving.com/community/c6h1750587p15786648)",
"Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=3. $ Prove that $$ \\frac{a^4+2ab^3}{a^3+ b^3}+\\frac{b^4+2 bc^3}{b^3+ c^3}+\\frac{c^4+2ca^3}{c^3+ a^3} \\leq\\frac{9}{2} $$ \n\n"
] | [
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Let $\Omega$ be the circumscribed circle of the acute triangle $ABC$ and $ D $ a point the small arc $BC$ of $\Omega$ . Points $E$ and $ F $ are on the sides $ AB$ and $AC$ , respectively, such that the quadrilateral $CDEF$ is a parallelogram. Point $G$ is on the small arc $AC$ such that lines $D... | Let $\angle BAC = \alpha$ and $\angle AFE = \beta$ , then $\angle FCD = \beta = \angle FED$ .
And $\angle AEF = 180^\circ - \alpha - \beta = \angle ABF = \angle FCG$ .
Let $GF \cap \odot ABC = K$ . We will prove that $K$ , $E$ , and $D$ are collinear.
We have $\angle AKD = 180^\circ - \angle ACD = 180^\ci... | [
"Since $BG \\parallel CD$ , it's easy to show that $BD=CG$ .\nWe have $\\angle AEF = \\angle ABG = \\angle GCF$ Since $\\angle ADE = \\angle DAC = \\angle DBC$ , $\\angle EAD = \\angle BCD$ , $\\triangle ADE \\sim \\triangle CBD$ Therefore, $\\frac{BD}{DC} = \\frac{DE}{AE}$ or $\\frac{CG}{CF}=\\frac{AE}{AF... | [
"origin:aops",
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"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Moldova/3066400.json"
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On the board there are three real numbers $(a,b,c)$ . During a $procedure$ the numbers are erased and in their place another three numbers a written, either $(c,b,a)$ or every time a nonzero real number $ d $ is chosen and the numbers $(a, 2ad+b, ad^2+bd+c)$ are written.
1) If we start with $(1,-2,-1)$ writ... | [] | [
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Let $ABCD$ be a trapezoid with bases $ AB$ and $CD$ $(AB>CD)$ . Diagonals $AC$ and $BD$ intersect in point $ N$ and lines $AD$ and $BC$ intersect in point $ M$ . The circumscribed circles of $ADN$ and $BCN$ intersect in point $ P$ , different from point $ N$ . Prove that the angles $AMP$ and ... | Quite hard for JBMO, if not mistaken. It seems the easiest way is to argue that $MN$ bisects $AB$ and $CD$ by Steiner's theorem for trapezoids and hence we need to show that $MP$ is a symmedian in $CMD$ , i.e. that $\frac{S_{DMP}}{S_{CMP}} = \frac{DM^2}{CM^2}$ . The rest is uses of the Sine Law. | [] | [
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In a chess tournament participated $ 100 $ players. Every player played one game with every other player. For a win $1$ point is given, for loss $ 0 $ and for a draw both players get $0,5$ points. Ion got more points than every other player. Mihai lost only one game, but got less points than every other player.... | [] | [
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Every point on a circle is coloured in blue or yellow such there is at least a point of each color.
Prove that for every colouring of the circle there is always an isosceles triangle inscribed inside the circle
1) with all vertexes of the same colour.
2) with vertexes of both colours.
For every colouring of the circle... | [] | [
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Outside of the trapezoid $ABCD$ with the smaller base $AB$ are constructed the squares $ADEF$ and $BCGH$ . Prove that the perpendicular bisector of $AB$ passes through the midpoint of $FH$ . | Trivial by coordinate bash.
Choose coordinates so that $A(-1,0)$ , $B(1,0)$ , $C(c,a)$ , $D(d,a)$ - then the $y$ -axis is the perpendicular bisector of $AB$ . It suffices to compute the $x$ -coordinates of $F$ and $H$ and to check that their sum (and hence their arithmetic mean) is $0$ .
This can be don... | [
"Complex numbers is very clean. Note $h-b=(b-c)i$ , $f-a=(a-d)i$ so the midpoint of $FH$ is $\\frac{a+b}{2}+\\frac{(a+b)-(c+d)}{2}i. $ Orienting so that $\\Im(a)=\\Im(b)$ now solves the problem. This is because the perpendicular bisector of $AB$ is $\\Re(z)=\\frac{\\Re(a)+\\Re(b)}{2}$ and $(a+b)-(c+d)... | [
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} |
Determine the smallest natural number $n$ for which there exist distinct nonzero naturals $a, b, c$ , such that $n=a+b+c$ and $(a + b)(b + c)(c + a)$ is a perfect cube. | Let $gcd(a, b, c, n) = d$ then the first equation turns into $n' = x + y + z$ where $x, y, z$ are coprime. Having $(a + b)(b + c)(c + a) = m^3$ we can turn it into $(x + y)(y + z)(z + x) = m'^3 \implies (n' - z)(n' - x)(n' - y) = m'^3$ . Then we have $$ n' - z = k $$ $$ n' - x = k^2 $$ $$ n' - y = h^... | [
"if $n=9$ we have to check $(4,3,2),(5,3,1),(6,2,1)$ if $n=8$ we have to check $(4,3,1),(5,2,1)$ if $n=7$ we have to check $(4,2,1)$ if $n=6$ we have to check $(3,2,1)$ None of them work\nFor $n=10$ take $(7,2,1)$ "
] | [
"origin:aops",
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"answer_score": 1076,
"boxed": true,
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"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Romania/3052828.json"
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Given is a triangle $ABC$ . Let the points $P$ and $Q$ be on the sides $AB, AC$ , respectively, so that $AP=AQ$ , and $PQ$ passes through the incenter $I$ . Let $(BPI)$ meet $(CQI)$ at $M$ , $PM$ meets $BI$ at $D$ and $QM$ meets $CI$ at $E$ . Prove that the line $MI$ passes through the midpo... | $I$ is the midpoint of $PQ$ so its enough to show that $PQ$ and $DE$ are parallel, because of homothety. Angle chasing gives $BPI, BIC, IQC$ similar, then $\angle BIC+\angle PMQ=\pi$ , so $DIEM$ is cyclic. From this we have that $BIC$ and $EID$ are similar. Next angle chase to prove that $MBP$ and $... | [
"interesting to see Mixtilinear at Junior\n\n1) $M,D,I,E$ lie on a circle\n2)Sinse $DE//PQ$ we have $M,K,I$ collinear where $K$ is the midpoint of $DE$ **Attachments:**\n\n[geogebra-export.pdf](https://cdn.artofproblemsolving.com/attachments/b/b/d7a3aed1195290a0fb97d8537436583cd986bd.pdf)",
"Let $IM \\c... | [
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"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Romania/3052837.json"
} |
Initially the numbers $i^3-i$ for $i=2,3 \ldots 2n+1$ are written on a blackboard, where $n\geq 2$ is a positive integer. On one move we can delete three numbers $a, b, c$ and write the number $\frac{abc} {ab+bc+ca}$ . Prove that when two numbers remain on the blackboard, their sum will be greater than $16$ . | As in post #2, notice that $\frac{ab+bc+ca}{abc}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Thus, if the set of numbers on the board is $\mathcal{S}$ , $\sum_{n\in \mathcal{S}}\frac{1}{n}$ remains invariant. Notice that at the beginning this quantity is $$ \sum_{i=2}^{2n+1}\frac{1}{i^3-i}=\sum_{i=2}^{2n+1}\frac{1}{2}\... | [
"1/a + 1/b + 1/c invariant....",
"Sniped,\n\nCall $a_i = i^3-i$ for all $i = 2,3,\\ldots, 2n+1$ . $$ X = \\sum_{i=2}^{2n+1}\\frac{1}{a_i} $$ is an invariant. \nMore precise, $$ X = \\frac{2n^2+3n}{8n^2+12n+4} < \\frac{1}{4} $$ .\n\nSuppose $b$ and $c$ are the last two numbers on the board. Suppose ... | [
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Given is a cube $3 \times 3 \times 3$ with $27$ unit cubes. In each such cube a positive integer is written. Call a $\textit {strip}$ a block $1 \times 1 \times 3$ of $3$ cubes. The numbers are written so that for each cube, its number is the sum of three other numbers, one from each of the three strips it is... | If all the numbers are equal to 1, there is nothing to prove. If the cube contains numbers greater than 1, we assume that there are even numbers written in the cube, and let n be the smallest of them. Then n must be the sum of 3 odd numbers, which is impossible, as the sum of three odd numbers is an odd number. Therefo... | [
"How can this be possible? Let $a_1, a_2$ and $a_3$ be on a same strip. Then $a_1 = a_2 + a_i + a_m$ which gives $a_1 > a_2$ Implying this we get $a_2 = a_3 + a_n + a_p$ which gives $a_2 > a_3$ then how can $a_3$ be equal to one of $a_1$ and $a_2$ plus something else when it is smaller than both o... | [
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"boxed": false,
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} |
Determine the real numbers $x$ , $y$ , $z > 0$ for which $xyz \leq \min\left\{4(x - \frac{1}{y}), 4(y - \frac{1}{z}), 4(z - \frac{1}{x})\right\}$ | Note that $xyz \leq 4\left(x - \frac{1}{y}\right)$ requires $4\sqrt{xz} \leq xyz + \frac{4}{y} \leq 4x$ by AM-GM, thus $z\leq x$ . Analogously from the other two we get $x \leq y$ and $y \leq z$ , so necessarily $x=y=z$ . Now the requirement holds if and only if $x^3 \leq 4\left(x - \frac{1}{x}\right)$ , i.e.... | [
"<blockquote>Determine the real numbers $x$ , $y$ , $z > 0$ for which $xyz \\leq \\min\\left\\{\\frac{y}{x-1}, \\frac{z}{y-1}, \\frac{x}{z-1}\\right\\}$ </blockquote>\nUhhh ?\nThere are infinitely many such $x,y,z$ (for example any $x,y,z\\in(1,\\frac 43)$ )\n",
"<blockquote><blockquote>Determine the real ... | [
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Let $ABC$ be an acute-angled triangle with $BC > AB$ , such that the points $A$ , $H$ , $I$ and $C$ are concyclic (where $H$ is the orthocenter and $I$ is the incenter of triangle $ABC$ ). The line $AC$ intersects the circumcircle of triangle $BHC$ at point $T$ , and the line $BC$ intersects the ci... | The answer is $\angle A=80^\circ, \angle B=60^\circ, \angle C=40^\circ$ .
Note that since $AHIC$ is cyclic, $180^\circ-\angle B=\angle AHC=\angle AIC=90^\circ+\dfrac{\angle B}{2},$ and so $\angle B=60^\circ$ .
Moreover, note that $AP=BT=AB$ , since $\angle BTA=180^\circ-\angle BTC=180^\circ-\angle BHC=\angle... | [] | [
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Let $ABCDEF$ be a regular hexagon of side length $2$ . Let us construct parallels to its sides passing through its vertices and midpoints, which divide the hexagon into $24$ congruent equilateral triangles, whose vertices are called nodes. For each node $X$ , we define its trio as the figure formed by three adjac... | [] | [
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"2023 Contests"
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"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Romania/3063071.json"
} | |
Let $M \geq 1$ be a real number. Determine all natural numbers $n$ for which there exist distinct natural numbers $a$ , $b$ , $c > M$ , such that $n = (a,b) \cdot (b,c) + (b,c) \cdot (c,a) + (c,a) \cdot (a,b)$ (where $(x,y)$ denotes the greatest common divisor of natural numbers $x$ and $y$ ). | Call such a positive integer $n$ *good*. I claim that the good positive integers $n$ are only the ones such that $v_2(n)$ is even and $n$ is not a power of $4$ (including $1$ ). Let $K=\gcd(a,b)\gcd(b,c)+\gcd(b,c)\gcd(c,a)+\gcd(c,a)\gcd(a,b)$ .
We have the following 2 crucial Claims:**Claim 1:** $n$ is ... | [
"motivation:- <details><summary>Click to expand</summary>playing around helps to see that odd numbers work, then we can see that some even work but not all, which helps in guessing the answer now to deal with proving the other side, we observe the claim-1 of @above's solution. This ends the problem with some finish... | [
"origin:aops",
"2023 Junior Balkan Team Selection Tests - Romania",
"2023 Contests"
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"answer_score": 240,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Romania/3063072.json"
} |
Determine all natural numbers $n \geq 2$ with at most four natural divisors, which have the property that for any two distinct proper divisors $d_1$ and $d_2$ of $n$ , the positive integer $d_1-d_2$ divides $n$ . | I claim $n=4$ or $n=p$ , where $p$ is an arbitrary prime are the only answers. Clearly $n=4$ works, whereas the statement is vacuous as $n=p$ .
Now let $d(n)$ denotes the number of positive integer divisors of $n$ . We inspect $d(n)\in\{3,4\}$ . If $d(n)=3$ then $n=p^2$ for some prime $p$ . As $p-1\... | [
"<blockquote>The former implies $p-1\\mid q$ and the latter gives $q-1\\mid p$ ; this is a contradiction as $p\\ge q-1$ and $q\\ge p-1$ cannot simultaneously hold.</blockquote>\nActually, this can happen in a special case when $p=q-1$ so $p=2, q=3$ . This gives us solution $n=6$ as well, otherwise your ... | [
"origin:aops",
"2023 Junior Balkan Team Selection Tests - Romania",
"2023 Contests"
] | {
"answer_score": 60,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Romania/3075782.json"
} |
Let the equilateral triangles $ABC$ and $DEF$ be congruent with the centers $O_1$ , respectively $O_2$ , so that segment $AB$ intersects segments $DE$ and $DF$ at $M, N$ , and the segment $AC$ intersects the segments $DF$ and $EF$ at $P$ and $Q$ , respectively. We denote by $I$ the intersection ... | If $T$ is the intersection of the bisectors of $\angle{NMD}, \angle{NPA}$ then $MDTAP$ is cyclic and $I$ is the antipode of $T$ wrt its circumcircle. Also, $ID = IA$ . Since $AO_1 = DO_2$ , $\triangle{IO_1A} \cong \triangle{IO_2D} \implies IO_1 = IO_2$ . Similarly, $JO_1 = JO_2$ , so we are done. | [] | [
"origin:aops",
"2023 Junior Balkan Team Selection Tests - Romania",
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"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Romania/3075789.json"
} |
Let be $a$ be positive real number. Prove that there are no real numbers $b$ and $c$ , with $b < c$ , so that for any distinct numbers $x, y \in (b, c)$ we have $|\frac{x+y} {x-y}| \leq a$ . | This is not suited at all for Romanian JBMO TST P4 because it is an analysis problem and it is trivial.
We have to show that for every constant $a$ and open interval $I$ there exists $x,y\in I$ such that $\left|\frac{x+y}{x-y}\right|>a$ .
For this we can just take any $y\in I$ and since $\lim_{x\to y} \lef... | [
"Can anyone translate the official solutions and post it here? I think it is too easy or the wording is not correct…",
"The wording is correct - the problem is indeed easy. \nBelow is a translation of the official solution by ChatGpt, checked by me.\n<details><summary>Official solution</summary>Assume that the nu... | [
"origin:aops",
"2023 Junior Balkan Team Selection Tests - Romania",
"2023 Contests"
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"answer_score": 16,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Romania/3075791.json"
} |
Given is a positive integer $n \geq 2$ and three pairwise disjoint sets $A, B, C$ , each of $n$ distinct real numbers. Denote by $a$ the number of triples $(x, y, z) \in A \times B \times C$ satisfying $x<y<z$ and let $b$ denote the number of triples $(x, y, z) \in A \times B \times C$ such that $x>y>z$... | Let $B=\{y_1,\dots,y_n\}$ . For any $j$ , denote by $a_j$ the number of elements of $A$ below $y_j$ . Likewise, denote by $b_j$ the number of elements of $C$ above $y_j$ . Using the fact $A\cap B = B\cap C = C\cap A = \varnothing$ , we obtain
\[
a = \sum_j a_j b_j \quad\text{and}\quad b = \sum_j (n-a_j)(n... | [] | [
"origin:aops",
"2023 Junior Balkan Team Selection Tests - Romania",
"2023 Contests"
] | {
"answer_score": 22,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Romania/3075797.json"
} |
Suppose that $a, b,$ and $c$ are positive real numbers such that $$ a + b + c \ge \frac{1}{a} + \frac{1}{b} + \frac{1}{c}. $$ Find the largest possible value of the expression $$ \frac{a + b - c}{a^3 + b^3 + abc} + \frac{b + c - a}{b^3 + c^3 + abc} + \frac{c + a - b}{c^3 + a^3 + abc}. $$ | Note that if $a+b-c,b+c-a,c+a-b$ are all non-negative, then in light of $x^3+y^3 \geq xy(x+y)$ we obtain $\displaystyle \sum \dfrac{a+b-c}{a^3+b^3+abc} \leq \dfrac{1}{a+b+c} \cdot \dfrac{\displaystyle 2 \sum ab-\sum a^2}{abc} \leq \dfrac{\displaystyle \sum ab}{abc(a+b+c)}=\dfrac{\displaystyle \sum 1/a}{\displaystyl... | [
"Let $a, b,$ and $c$ are positive real numbers such that $a + b + c \\ge \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}.$ Prove that $$ \\frac{a + b - c}{a^3 + b^3 + abc} + \\frac{b + c - a}{b^3 + c^3 + abc} + \\frac{c + a - b}{c^3 + a^3 + abc}\\leq 1 $$ ",
"notice we have $a^2bc+ab^2c+abc^2 \\geq ab+bc+c... | [
"origin:aops",
"2023 Junior Balkan Team Selection Tests - Romania",
"2023 Contests"
] | {
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Romania/3099021.json"
} |
Let $a$ and $b$ be two distinct positive integers with the same parity. Prove that the fraction $\frac{a!+b!}{2^a}$ is not an integer. | First we can say $b\geq a$ because if $a > b$ it is so clear that given condition is not satisfied. Then if we factorize the numerator we get, \[\frac{a!+b!}{2^a} = \frac{a!(1+\frac{b!}{a!})}{2^a}\] If $b\geq a+2$ then $\frac{b!}{a!}$ is even so $1+\frac{b!}{a!}$ is odd and this is not possible because we get... | [
" $\\nu_2(a!)\\neq\\nu_2(b!)$ so $\\nu_2(a!+b!)\\leq\\min(\\nu_2(a!),\\nu_2(b!))<a$ .",
"We know that $v_2(a!)<\\frac a2+\\frac a4+\\frac a8+\\cdots=a$ If $b>a$ then $v_2(b!+a!)=v_2(a!)<a$ If $b<a$ then $v_2(b!+a!)=v_2(b!)<v_2(a!)<a$ ",
"We know that $v_2(a!) = \\left\\lfloor \\frac{a}{2} \\right\\rflo... | [
"origin:aops",
"2023 Junior Balkan Team Selection Tests - Romania",
"2023 Contests"
] | {
"answer_score": 130,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Romania/3172659.json"
} |
Consider a grid with $n{}$ lines and $m{}$ columns $(n,m\in\mathbb{N},m,n\ge2)$ made of $n\cdot m \; 1\times1$ squares called ${cells}$ . A ${snake}$ is a sequence of cells with the following properties: the first cell is on the first line of the grid and the last cell is on the last line of the grid, starti... | Perhaps the mean implies that expected value is very useful here...
<details><summary>elaboration of my idea</summary>If we 'drop down' to a line/row, then we can continue in one of two directions until we reach an edge. When we 'drop down', let's break it down by case.
If we drop at the leftmost square of that line... | [
"See [here](https://pregatirematematicaolimpiadejuniori.wordpress.com/2023/04/14/2023ro/).\nAll solutions from this site are in romanian, but I think you can handle :) ."
] | [
"origin:aops",
"2023 Junior Balkan Team Selection Tests - Romania",
"2023 Contests"
] | {
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Romania/3172668.json"
} |
Let $ABC$ be an acute triangle with $\angle B > \angle C$ . On the circle $\mathcal{C}(O, R)$ circumscribed to this triangle points $D, E, J, K, S$ are chosen such that $A, E, J$ and $K$ are on the same side of the line $BC$ , the diameter $DE$ is perpendicular on the chord $BC$ , $S\in \overarc{EK},\ov... | $\angle AFB=2\angle ACB=\angle AOB$ implies $AFOB$ is cyclic. In fact, we will show the $A=60$ , $B=75$ , and $C=45$ , which proves the problem since $\angle AMB=90=\angle AOB$ and $\angle BAM=30=\angle SMK$ .
Since $\angle BCJ=\angle CBK$ , we get $BCKJ$ is an isosceles trapezoid, so $PQ\parallel BC$ . ... | [] | [
"origin:aops",
"2023 Junior Balkan Team Selection Tests - Romania",
"2023 Contests"
] | {
"answer_score": 60,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Junior Balkan Team Selection Tests - Romania/3172682.json"
} |
$a,b,c$ are positive real numbers such that $a+b+c\ge 3$ and $a^2+b^2+c^2=2abc+1$ . Prove that $$ a+b+c\le 2\sqrt{abc}+1 $$ | $(a+b+c-1)^2\leq 4abc $ $\Rightarrow$ $4abc+2(a+b+c)\geq 2(ab+bc+ca)+1+a^2+b^2+c^2$ $abc+a+b+c\geq ab+bc+ca +1$ $ \Rightarrow$ $(a-1)(b-1)(c-1)\geq 0 $ lets prove all numbers $\geq 1$ or two of them $<1 $ Let all but we have $a+b+c>3$ Lets only one $<1$ let is $a$ $\Rightarrow$ $(a^2-1)(b^2-1)=(ab-c)^... | [
"<blockquote> $a,b,c$ are positive real numbers such that $a+b+c\\ge 3$ and $a^2+b^2+c^2=2abc+1$ . Prove that $$ a+b+c\\le 2\\sqrt{abc}+1 $$ </blockquote>\n2023?\nThanks.\nLet $a,b,c$ be nonnegative real numbers such that $a^2+b^2+c^2=3.$ [Prove that](https://artofproblemsolving.com/community/c4h1583160p... | [
"origin:aops",
"2023 Contests",
"2023 Kazakhstan National Olympiad"
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"answer_score": 132,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 Kazakhstan National Olympiad/3036571.json"
} |
$a,b,c$ are positive real numbers such that $\max\{\frac{a(b+c)}{a^2+bc},\frac{b(c+a)}{b^2+ca},\frac{c(a+b)}{c^2+ab}\}\le \frac{5}{2}$ . Prove inequality $$ \frac{a(b+c)}{a^2+bc}+\frac{b(c+a)}{b^2+ca}+\frac{c(a+b)}{c^2+ab}\le 3 $$ | I will prove it for triangle sides $a,b,c$ and same thing maybe works for the main problem, I am not sure. $\sum \frac{(a-b)(a-c)}{a^2+bc} \ge 0 \iff \sum (b-c)^2(2ac+2ac-bc-a^2)(a^2+bc) \ge 0 \leftarrow \sum (b-c)^2(ab+ac-bc)(a^2+bc) \overset{*} \ge 0$ .
Assume that $a\ge b\ge c$ . $$ (a-b)^2(ac+bc-ab)(c^2+ab)+(a... | [
"Let $ a,b,c$ be the side-lengths of a triangle. Prove that: $$ \\frac {a(b + c)}{a^2 + bc} + \\frac {b(c + a)}{b^2 + ca} + \\frac {c(a + b)}{c^2 + ab}\\leq 3 $$ $$ \\frac{a(b+c)}{a^2+2bc}+\\frac{b(c+a)}{b^2+2ca}+\\frac{c(a+b)}{c^2+2ab} \\leq 2 $$ ",
"This problem was proposed by Mudok (probably)",
"?... | [
"origin:aops",
"2023 Contests",
"2023 Kazakhstan National Olympiad"
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"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Kazakhstan National Olympiad/3036573.json"
} |
The $C$ -excircle of a triangle $ABC$ touches $AB, AC, BC$ at $M, N, K$ . The points $P, Q$ lie on $NK$ so that $AN=AP, BK=BQ$ . Prove that the circumradius of $\triangle MPQ$ is equal to the inradius of $\triangle ABC$ . | I think its incorrect?
<details><summary>:)</summary>Let line $AP$ and $BQ$ intersects at $R$ .And incircle of $\triangle ABC$ tangent to sides $AB,BC,CA$ at $C_1,A_1,B_1$ respectively.
We have $AN=AP$ the $\angle ANP=\angle APN=\angle CNK$ and because $CN=CK$ then we get $\angle NAP=\angle NCK$ .That ... | [
"<details><summary>Solution</summary>We claim that $MPQ$ is congruent to the triangle $DEF$ of points where the incircle of $ABC$ touches the sides.\nIndeed, we claim that upon reflection over the midpoint of $AB$ , one triangle is mapped to the other.\nIt is clear that $M$ is mapped to $F$ . By symmetry,... | [
"origin:aops",
"2023 Contests",
"2023 Kazakhstan National Olympiad"
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"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Kazakhstan National Olympiad/3036885.json"
} |
A triangle $ABC$ with obtuse angle $C$ and $AC>BC$ has center $O$ of its circumcircle $\omega$ . The tangent at $C$ to $\omega$ meets $AB$ at $D$ . Let $\Omega$ be the circumcircle of $AOB$ . Let $OD, AC$ meet $\Omega$ at $E, F$ and let $OF \cap CE=T$ , $OD \cap BC=K$ . Prove that $OTBK$ is... | An alternative solution:
Let $X$ be the reflection of $C$ over $OD$ . If $E'$ is the midpoint of $XC$ on $OD$ , then
\[\vert DE'\vert \cdot \vert DO\vert=\vert DC\vert^2=\vert DB\vert \cdot \vert DA,\]
hence $ABE'O$ is cyclic and hence $E=E'$ . We claim that $X$ also lies on $(OBTK)$ .
Indeed,
\[\angle ... | [] | [
"origin:aops",
"2023 Contests",
"2023 Kazakhstan National Olympiad"
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"answer_score": 30,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Kazakhstan National Olympiad/3036892.json"
} |
Let $n>100$ be an integer. The numbers $1,2 \ldots, 4n$ are split into $n$ groups of $4$ . Prove that there are at least $\frac{(n-6)^2}{2}$ quadruples $(a, b, c, d)$ such that they are all in different groups, $a<b<c<d$ and $c-b \leq |ad-bc|\leq d-a$ . | Observe that for natural numbers $a,b$ with $a+2\leq b$ , we have\[b-(a+1)<|a(b+1)-(a+1)b|<(b+1)-a.\]It suffices to show that there are at least $\frac{(n-6)^2}2$ pairs $(a,b)$ with the four numbers $a$ , $a+1$ , $b$ , $b+1$ in separate groups.
Assume the groups are numbered. Let there be $k$ pairs of co... | [
"Any solution?",
"<details><summary>Hint</summary>count the number of $(x, x+1, y, y+1)$ quadruples for $y > x+1.$</details>",
"any solution?",
"<blockquote>\n\nNotice that the number of pairs $(a,b)$ with the four numbers $a$ , $a+1$ , $b$ , $b+1$ in separate groups is at least \\[{k\\choose2}-\\le... | [
"origin:aops",
"2023 Contests",
"2023 Kazakhstan National Olympiad"
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"answer_score": 50,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Kazakhstan National Olympiad/3036896.json"
} |
Solve the given equation in prime numbers $$ p^3+q^3+r^3=p^2qr $$ | <blockquote>Solve the given equation in prime numbers $$ p^3+q^3+r^3=p^2qr $$ </blockquote>
This problem took me a long time as I didn't think it would be necessary to use the Legendre symbol :coolspeak:
Solved with **complex_afi** $\color{blue}\boxed{\textbf{Answer: (3,3,3)}}$ $\color{blue}\boxed{\textbf{Proof:}... | [
"A similar approach as in 2023 RMM/1 should work, I think.",
"<details><summary>wrong</summary>$(p+q+r)(p^2+q^2+r^2-pq-qr-rp)=pqr(p-3)$ . $p=2$ is impossible since the right hand side is negative but the left hand side is positive. $q=2$ implies $p^3+r^3+8=2p^2r$ , or $8=(r-p)(p^2+pr-r^2)$ . By casework, ther... | [
"origin:aops",
"2023 Contests",
"2023 Kazakhstan National Olympiad"
] | {
"answer_score": 1320,
"boxed": false,
"end_of_proof": false,
"n_reply": 12,
"path": "Contest Collections/2023 Contests/2023 Kazakhstan National Olympiad/3037324.json"
} |
Given $x,y>0$ such that $x^2y^2+2x^3y=1$ . Find the minimum value of sum $x+y$ | <blockquote>Given $x,y>0$ such that $x^2y^2+2x^3y=1$ . Find the minimum value of sum $x+y$ </blockquote> $(x+y)^2=x^2+\frac{2x^3y+x^2y^2}{x^2}=x^2+\frac 1{x^2}\ge 2$ And so $\boxed{x+y\ge\sqrt 2}$ , reached for example when $(x,y)=(1,\sqrt 2-1)$ | [
"<details><summary>Solution</summary>For $(1,\\sqrt{2}-1)$ we get sum $\\sqrt{2}$ . To prove that this is minimal, it suffices to show that\n\\[(x+y)^4 \\ge 4(x^2y^2+2x^3y).\\]\nBy homogenity it suffices to prove this for $x=1$ i.e.\n\\[(y+1)^4 \\ge 4(y^2+2y).\\]\nWriting $T=y^2+2y$ , this reduces to\n\\[(T+1... | [
"origin:aops",
"2023 Contests",
"2023 Kazakhstan National Olympiad"
] | {
"answer_score": 1012,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Kazakhstan National Olympiad/3037325.json"
} |
The altitudes of an acute triangle $ABC$ intersect at $H$ . The tangent line at $H$ to the circumcircle of triangle $BHC$ intersects the lines $AB$ and $AC$ at points $Q$ and $P$ respectively. The circumcircles of triangles $ABC$ and $APQ$ intersect at point $K$ ( $K\neq A$ ). The tangent lines at ... | Let $A'$ be the antipode of $A$ on $(ABC)$ . $(AH)\cap (ABC)=\{A,N\},PQ\cap BC=S$ Let the altitudes from $A,B,C$ to $BC,CA,AB$ be $D,E,F$ respectively. $AD\cap (ABC)=\{A,G\},PQ\cap AN=L,AN\cap BC=R$ $\textbf{Claim:} \ SN$ is tangent to $(ABC)$ . $A'N\perp LA$ and $LH\perp AA'$ since $\angle PAA'=9... | [
"<blockquote>Outline of Proof:\n1. $S=AK\\cap QP\\cap BC,$ by radical axis on $(ABC), (APQ)$ and $(BCPQ)$ . \n2. $AF$ diameter of $\\odot AQP,$ since $TA \\parallel BC.$ 3. $AA’$ diameter of $\\odot ABC,$ then $A’ = HM\\cap KF.$ 4. $SH^2 = SB\\cdot SC = SK\\cdot SA,$ which implies $QP \\parallel E... | [
"origin:aops",
"2023 Contests",
"2023 Kazakhstan National Olympiad"
] | {
"answer_score": 200,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Kazakhstan National Olympiad/3037333.json"
} |
Inside an equilateral triangle with side $3$ there are two rhombuses with sides $1,061$ and acute angles $60^\circ$ . Prove that these two rhombuses intersect each other. (The vertices of the rhombus are strictly inside the triangle.) | [
"Bump one\n"
] | [
"origin:aops",
"2023 Contests",
"2023 Kazakhstan National Olympiad"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Kazakhstan National Olympiad/3037337.json"
} | |
Given are positive integers $a, b, m, k$ with $k \geq 2$ . Prove that there exist infinitely many $n$ , such that $\gcd (\varphi_m(n), \lfloor \sqrt[k] {an+b} \rfloor)=1$ , where $\varphi_m(n)$ is the $m$ -th iteration of $\varphi(n)$ . | <details><summary>Solution</summary>We will find infinitely many $n$ , such that $\lfloor \sqrt[k] {an+b} \rfloor$ is a prime, say $p$ , and $n$ satisfies $\nu_p(n)=1$ and $p$ is the greatest prime factor of $n$ . Notice that then $p \nmid \varphi_m(n)$ for any positive integer $m$ , so this way we will h... | [] | [
"origin:aops",
"2023 Contests",
"2023 Kazakhstan National Olympiad"
] | {
"answer_score": 48,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Kazakhstan National Olympiad/3037584.json"
} |
Let $p$ be an odd prime. Let $A(n)$ be the number of subsets of $\{1,2,...,n\}$ such that the sum of elements of the subset is a multiple of $p$ . Prove that if $2^{p-1}-1$ is not a multiple of $p^2$ , there exists infinitely many positive integer $m$ for any integer $k$ that satisfies the following. (The... | How sad, this is just routine complex combinatorics. See, for instance, PFTB chapter 4, example 7.**Lemma:** The number of subsets of $\{1,2,\ldots,an\}$ whose sum of elements is divisible by $n{}$ is equal to \[\frac{1}{n}\sum_{\substack{d\mid n 2\nmid d}}\varphi(d)\cdot2^{an/d}.\]**Proof:** Consider the polynomi... | [
"Do you mean $2^{p-1}-1$ is not a multiple of $p^2$ ?",
"<blockquote>Do you mean $2^{p-1}-1$ is not a multiple of $p^2$ ?</blockquote>\nYes. My bad.\n",
"I solved this problem using the generating function, but I've heard there's a simpler solution than this :oops: \n\nDef) for $1\\le i\\le p-1$ , $a_i... | [
"origin:aops",
"2023 Korea - Final Round",
"2023 Contests"
] | {
"answer_score": 150,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Korea - Final Round/3039229.json"
} |
In a triangle $ABC ~(\overline{AB} < \overline{AC})$ , points $D (\neq A, B)$ and $E (\neq A, C)$ lies on side $AB$ and $AC$ respectively. Point $P$ satisfies $\overline{PB}=\overline{PD}, \overline{PC}=\overline{PE}$ . $X (\neq A, C)$ is on the arc $AC$ of the circumcircle of triangle $ABC$ not inclu... | Construct parallelograms $YDXD'$ and $YEXE'$ . Note that $D,E,D',E'$ are trivially collinear. Moreover, let $M,S,T$ be the midpoints of $AX,BD,CE$ respectively. It suffices to prove that $A,M,S,T$ are concyclic, or equivalently that $\angle SMT=\angle A$ . Since $BD' \parallel SM$ and $CE' \parallel MT$ ... | [
"Let $G$ be the miquel point of $BDEC$ , let $M=(D+B)/2$ and $N=(C+E)/2$ , since $GDM \\sim GEN$ hence $G$ lies on $(AMN)$ as well similarly if $K=(X+Y)/2$ then $K \\in (GAMN)$ and because $AP$ is the diameter we are done.",
"Let $(ADE) \\cap (ABC) = K$ Let the midpoint of $BD,CE$ as $M,N$ L... | [
"origin:aops",
"2023 Korea - Final Round",
"2023 Contests"
] | {
"answer_score": 148,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2023 Contests/2023 Korea - Final Round/3039230.json"
} |
Function $f : \mathbb{R^+} \rightarrow \mathbb{R^+}$ satisfies the following condition.
(Condition) For each positive real number $x$ , there exists a positive real number $y$ such that $(x + f(y))(y + f(x)) \leq 4$ , and the number of $y$ is finite.
Prove $f(x) > f(y)$ for any positive real numbers $x < ... | This looks like an IMO 2022/2 reboot. We prove something slightly stronger:
<blockquote>Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be a function such that for any positivr eal number $x{}$ there exists at least one, but <span style="color:#f00">countably many</span> positive real numbers $y{}$ for which \[(x+f(y))(y+f(... | [
"Not even an hour and a half past since FKMO Day 1 ended, and all three problems has been posted. Nice.",
"for $x$ , denote the set of $y$ satisfing the condition as $A(x)$ and the set of $y$ satisfing $y\\le x$ , $f(y)\\le f(x)$ as $E(x)$ Claim) $E(x)$ is a finite set.\n<details><summary>pf</summary... | [
"origin:aops",
"2023 Korea - Final Round",
"2023 Contests"
] | {
"answer_score": 190,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Korea - Final Round/3039232.json"
} |
Find all positive integers $n$ satisfying the following. $$ 2^n-1 \text{ doesn't have a prime factor larger than } 7 $$ | Almost the same as Argentina MO 2023 P2 :D
It's clear that $n = 5$ doesn't work, and that $2 \nmid 2^n - 1$ . Notice that: $$ 2^1 - 1 = 1 $$ $$ 2^2 - 1 = 3 $$ $$ 2^3 - 1 = 7 $$ $$ 2^4 - 1 = 15 = 3 \cdot 5 $$ $$ 2^6 - 1 = 63 = 3^2 \cdot 7 $$ By Zsigmondy's Theorem, there is a prime number $p$ that ... | [
"I took the test and I solved this problem. It needs some long calculations but they are not that hard. The answer seems to be 1,2,3,4,6",
"<details><summary>Solution</summary>Note that $3 \\mid 2^n-1$ iff $n$ is even, $5 \\mid 2^n-1$ iff $4 \\mid n$ and $7 \\mid 2^n-1$ iff $3 \\mid n$ .\nNow if $n$ ... | [
"origin:aops",
"2023 Korea - Final Round",
"2023 Contests"
] | {
"answer_score": 1040,
"boxed": false,
"end_of_proof": false,
"n_reply": 29,
"path": "Contest Collections/2023 Contests/2023 Korea - Final Round/3039731.json"
} |
For positive integer $n\geq 3$ and real numbers $a_1,...,a_n,b_1,...,b_n$ , prove the following. $$ \sum_{i=1}^n a_i(b_i-b_{i+3})\leq\frac{3n}{8}\sum_{i=1}^n((a_i-a_{i+1})^2+(b_i-b_{i+1})^2) $$ ( $a_{n+1}=a_1$ , and for $i=1,2,3$ $b_{n+i}=b_i$ .) | Since $\frac{3n}{8}$ is definitely not the best bound, we’ll try to find a better bound $\lambda_n<\frac{3n}{8}$
Some preparations:
Lemma $1$ : (Ky Fan) If $\sum_{i=1}^{n}a_i=0$ , then $cos\frac{2\pi}{n}\cdot\sum_{i=1}^{n} a_i^2\ge\sum_{i=1}^{n}a_ia_{i+1}$
It’s a well-known inequality that can be proved by F... | [
"Claim) $\\Sigma (a_i -a_{i+k})^2 \\le k^2 \\Sigma (a_i -a_{i+1})^2$ <details><summary>pf</summary>trivial by Cauchy Inequality</details>\n\nNote that the inequality doesnt change by adding the constant $c$ to all $ a_i (1\\le i \\le n)$ .\nTherefore, we can assume that $\\Sigma a_i =0$ .\n\nNow, firstly divid... | [
"origin:aops",
"2023 Korea - Final Round",
"2023 Contests"
] | {
"answer_score": 62,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2023 Contests/2023 Korea - Final Round/3039736.json"
} |
Given a positive integer $n$ , there are $n$ boxes $B_1,...,B_n$ . The following procedure can be used to add balls. $$ \text{(Procedure) Chosen two positive integers }n\geq i\geq j\geq 1\text{, we add one ball each to the boxes }B_k\text{ that }i\geq k\geq j. $$ For positive integers $x_1,...,x_n$ let $f(x_1,... | Consider the numbers $\pmod3$ .
Suppose the maximum number of operations we need is $k$ . The answer is as follows:
For $n=3m, k=2m+1$ .
For $n=3m+1, k=2m+2$ .
For $n=3m+2, k=2m+2$ .
<span style="color:#f00">**Construction:**</span>Suppose the number of $0$ s, $1$ s, $2$ s be $A$ , $B$ , $C$ . We consider... | [
"<blockquote> $$ \\text{(Procedure) Chosen two positive integers }1\\geq i\\geq j\\geq n\\text{, we add one ball each to the boxes }B_k\\text{ that }i\\geq k\\geq j. $$ </blockquote>\nDo you mean this? $$ \\text{(Procedure) Chosen two positive integers }1\\leqslant i\\leqslant j\\leqslant n\\text{, we add one ba... | [
"origin:aops",
"2023 Korea - Final Round",
"2023 Contests"
] | {
"answer_score": 1210,
"boxed": false,
"end_of_proof": true,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 Korea - Final Round/3039737.json"
} |
Let $ABC$ be a triangle such that $AB<AC$ . Let $D$ be a point on the segment $BC$ such that $BD<CD$ . The angle bisectors of $\angle ADB$ and $\angle ADC$ meet the segments $AB$ and $AC$ at $E$ and $F$ respectively. Let $\omega$ be the circumcircle of $AEF$ and $M$ be the midpoint of $EF$ . ... | Take $EF \cap BC$ at $X$ and $AX \cap (AEF)$ at $T'$ .It is known that, $(X,D;B,C)=-1$ take pencil from $A$ to $(AEF)$ $\implies$ $ET'FX$ is harmonic.So, $TX$ is symmedian in $\triangle ETF$ .Thus, $\angle ET'X=\angle MT'F=\angle XFE=\angle YEF$ it means $T-M-Y$ .The $\angle XFE=\angle YEF$ get... | [
"First note that $DE$ and $DF$ are orthogonal and $DE$ bisects $\\angle BDA$ , so $-1=(DA,DB;DE,DF)\\overset{D}{=}(DA\\cap EF, BC\\cap EF,E,F)\\overset{A}{=}(D,BC\\cap EF; B,C)=(B,C;D,BC\\cap EF)$ .\n\nNow we have $$ (B,C;D,BC\\cap EF)=-1=(E,F;\\infty_{EF},M)\\overset{Y}{=}(E,F;X,T)\\overset{A}{=}(B,C;D,BC... | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Balkan MO TST"
] | {
"answer_score": 30,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 Macedonian Balkan MO TST/3058740.json"
} |
Let $f$ be a non-zero function from the set of positive integers to the set of non-negative integers such that for all positive integers $a$ and $b$ we have $$ 2f(ab)=(b+1)f(a)+(a+1)f(b). $$ Prove that for every prime number $p$ there exists a prime $q$ and positive integers $x_{1}$ , ..., $x_{n}$ and $... | <blockquote>
A related question: is it possible to find all functions which satisfy the equation?</blockquote>
Yes, it is.
Let $\mathcal{A}$ be set of primes for which $f(p)\ne 0$ and $\mathcal{B}=\text{set of remaining primes}$ . As @grupyorum wrote, for all $q\in \mathcal{A}$ $f(q^{n}) = (q^{n-1}+\cdots+1)f(... | [
"First, I think one must assume $f$ is not identically zero: $f\\equiv 0$ enjoys the given condition, while $f(q^p)/f(q)$ is not defined for this function. Note that if $f$ is not identically zero, then it is clear there is a prime $q$ so that $f(q)>0$ . \n\nNext, we show for any $q\\in\\mathbb{N}$ and... | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Balkan MO TST"
] | {
"answer_score": 70,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Macedonian Balkan MO TST/3058742.json"
} |
At a chess tournament, every pair of contestants played each other at most once. If any two con-
testants, $A$ and $B$ , failed to play each other, then exactly two other contestants, $C$ and $D$ , played
against both $A$ and $B$ during the tournament. Moreover, no $4$ contestants played exactly $5$ games... | Make a graph where two contestants are connected with an edge if they $\textbf{did not}$ play each other. Now we have that whenever there is an edge $AB$ in the graph, there is also an edge $CD$ such that $AC$ , $AD$ , $BC$ and $BD$ aren't connected with edges. Call such pairs of edges $\textit{conjugates}... | [] | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Balkan MO TST"
] | {
"answer_score": 44,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Macedonian Balkan MO TST/3058751.json"
} |
Let $(a_{n})_{n=1}^{\infty}$ be a sequence of positive real numbers defined by $a_{1}=1$ , $a_{2}=2$ and $$ \frac{a_{n+1}^{4}}{a_{n}^3} = 2a_{n+2}-a_{n+1}. $$ Prove that the following inequality holds for every positive integer $N>1$ : $$ \sum_{k=1}^{N}\frac{a_{k}^{2}}{a_{k+1}}<3. $$ *Note: The bound is not s... | After rearranging the conditions we get: $a_{n+2}=\frac{a_{n+1}}{2}+\frac{a_{n+1}^4}{2a_n^3}$ thus the sequence is strictly increasing.
Furthermore notice that: $\frac{a_n^2}{a_{n+1}}\le \bigg(\frac{4}{9}\bigg)^{n-1}$ for $n\ge 2$ **Claim:** $\frac{1}{n^2}\ge \bigg(\frac{4}{9}\bigg)^{n-1}$ *Proof:* $\Longrightarr... | [
"Let $b_n = a_{n+1}/a_n$ for $n\\ge 1$ . Then $b_1=2$ and $1+b_n^3 = 2b_{n+1}$ using the given recursion. From here, we get $2(b_{n+1}-1)=b_n^3-1$ , yielding\n\\[\n\\frac{b_{n+1}-1}{b_n-1} = \\frac{b_n^2+b_n+1}{2} \\Rightarrow b_{n+1}-1 = 2^{-n}\\prod_{1\\le i\\le n}\\left(b_i^2+b_i+1\\right).\n\\]\nThe las... | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Balkan MO TST"
] | {
"answer_score": 122,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Macedonian Balkan MO TST/3058761.json"
} |
Determine all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ we have: $$ xf(x+y)+yf(y-x) = f(x^2+y^2)\,. $$ *Authored by Nikola Velov* | <blockquote>Determine all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for all $x,y \in \mathbb{R}$ we have: $$ xf(x+y)+yf(y-x) = f(x^2+y^2)\,. $$ *Authored by Nikola Velov*</blockquote>
My solution
Easy to see that $f\equiv 0$ . If $f \not \equiv 0$ Let $P(x;y)$ be the assertion $P(0;0) \Righta... | [
"I claim all such functions are form $f(x)=cx$ , where $c\\in\\mathbb{R}$ is arbitrary.\n\nDenote the given assertion by $P(x,y)$ . Note that $P(0,y)$ gives $f(y^2)=yf(y)$ . In particular $f(0)=0$ and $f((-y)^2)=-yf(-y)$ so that $f(-y)=-f(y)$ for any $y$ . Now,\n\\begin{align*}\nP(x,y)&\\Rightarrow xf... | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Mathematical Olympiad"
] | {
"answer_score": 1050,
"boxed": true,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 Macedonian Mathematical Olympiad/3049237.json"
} |
Let $p$ and $q$ be odd prime numbers and $a$ a positive integer so that $p|a^q+1$ and $q|a^p+1$ . Show that $p|a+1$ or $q|a+1$ .
*Authored by Nikola Velov* | First of all let $p=q$ , this yields $a^p+1\overset{\text{FLT}}{\equiv}a+1\pmod p$ however since $p\mid a^p+1$ we have that $p\mid a+1$ So from now on $\text{ WLOG }$ assume that $p<q$ Rewriting the first condition yields $a^q+1\equiv 0\pmod p\Longrightarrow a^q\equiv-1\pmod p\text{ thus }a^{2q}\equiv1\pmod ... | [
"First, let $p=q$ , so $p\\mid a^p+1$ . By Fermat, $a^p+1\\equiv a+1\\pmod{p}$ , so $p\\mid a+1$ . Hence assume $p<q$ without loss. Let $d$ be the smallest positive integer such that $p\\mid a^d-1$ . Using $a^q\\equiv -1\\pmod{p}$ , $a^{2q}\\equiv 1\\pmod{p}$ and $p>2$ , we get that $d\\nmid q$ and ... | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Mathematical Olympiad"
] | {
"answer_score": 158,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Macedonian Mathematical Olympiad/3049239.json"
} |
In a city of gnomes there are $1000$ identical towers, each of which has $1000$ stories, with exactly one gnome living on each story. Every gnome in the city wears a hat colored in one of $1000$ possible colors and any two gnomes in the same tower have different hats. A pair of gnomes are friends if they wear hat... | Replace $1000$ with any even integer $n$ . We will prove that the answer is $\dfrac{n^3}{4}$ . Let $a_i$ be the $n$ colours, and consider the $n$ graphs $G_i$ such that the vertices of each $G_i$ are the $n$ gnomes of colour $i$ , and we connect two gnomes with an edge if and only if they are friends.*... | [
"Claim) for an arbitrary colour $i$ , there are at most ${n^2} \\over 4$ pairs of friends that have the colour $i$ .\nPf) Enough to prove for colour 1.\n Let there are $a_i$ gnomes in the $i$ th floor that has the colour 1.\nNote that $\\sum _{i=1} ^n a_i =n$ .\nLet #(pair of friends that has the colour 1) ... | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Mathematical Olympiad"
] | {
"answer_score": 154,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Macedonian Mathematical Olympiad/3049242.json"
} |
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