problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
Let $ABC$ be a scalene acute triangle with orthocenter $H$ . The circle with center $A$ and radius $AH$ meets the circumcircle of $BHC$ at $T_{a} \neq H$ . Define $T_{b}$ and $T_{c}$ similarly. Show that $H$ lies on the circumcircle of $T_{a}T_{b}T_{c}$ .
*Authored by Nikola Velov* | Let $A_1, B_1, C_1$ be the feet of the altitudes through $A,B,C$ . After applying the negative inversion centered at $H$ , with radius $\sqrt{HA\cdot HA_1}=\sqrt{HB\cdot HB_1}=\sqrt{HC\cdot HC_1}$ . The problem becomes: Let $I$ be the incenter of a triangle $\triangle DEF$ . Let $T_1$ be the intersection of t... | [] | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Mathematical Olympiad"
] | {
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Macedonian Mathematical Olympiad/3049245.json"
} |
There are $n$ boys and $n$ girls sitting around a circular table, where $n>3$ . In every move, we are allowed to swap
the places of $2$ adjacent children. The **entropy** of a configuration is the minimal number of moves
such that at the end of them each child has at least one neighbor of the same gender.
Find ... | This was a cruel problem to actually give on the test, but alright. $\color{black} \rule{7.4cm}{1pt}$ We claim that the desired maximum is $\boxed{n-2}$ .
To achieve this entropy, consecutively place around the table: $1$ boy (Bob), $\left\lfloor\frac{n-1}{2}\right\rfloor$ girls (block G1), $\left\lceil\frac{n-1... | [
"Bumping this thread",
"# redacted",
"@above this isn't correct, as it turns out the worst possible distribution is not the alternating one. The worst config i've found is\n<details><summary>Click to expand</summary>Taking a girl, $\\frac{n-1}{2}$ boys, $\\frac{n-1}{2}$ girls, a boy, $\\frac{n-1}{2}$ girl... | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Mathematical Olympiad"
] | {
"answer_score": 1264,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Macedonian Mathematical Olympiad/3049248.json"
} |
Let $s(n)$ denote the smallest prime divisor and $d(n)$ denote the number of positive divisors of a positive integer $n>1$ . Is it possible to choose $2023$ positive integers $a_{1},a_{2},...,a_{2023}$ with $a_{1}<a_{2}-1<...<a_{2023}-2022$ such that for all $k=1,...,2022$ we have $d(a_{k+1}-a_{k}-1)>2023... | <details><summary>Solution</summary>Take sufficiently large constant $M$ ; choose $a_1$ arbitrarily and select $a_2, a_3, \ldots, a_{2023}$ so that $a_{k+1}-a_k=M!+1$ for all integers $k \in [1;2022]$ . Observe that $s(a_{k+1}-a_k)=s(M!+1)>M$ and that $\tau (a_{k+1}-a_k-1)=\tau (M!)>M$ , so we are done.
Com... | [
"**S**olved with **L567, mueller.25**\n\n<details><summary>solution</summary>Yes, it is possible. Inductively construct such a sequence by selecting $a_1 = 1$ and for each $k > 1$ we choose $a_k-a_{k-1}$ to be $1$ plus the product of all the first $2023^{2023^{2023^k}}$ primes.</details>",
"The answer i... | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Team Selection Test"
] | {
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Macedonian Team Selection Test/3076279.json"
} |
Let $ABC$ be an acute triangle such that $AB<AC$ and $AB<BC$ . Let $P$ be a point on the segment $BC$ such that $\angle APB = \angle BAC$ . The tangent to the circumcircle of triangle $ABC$ at $A$ meets the circumcircle of triangle $APB$ at $Q \neq A$ . Let $Q'$ be the reflection of $Q$ with respec... | Let $AB =c, BC=a, CA=b$ . We have $\angle BAC= \angle APB = \alpha$ . Clearly, $\triangle BAC \sim \triangle BPA \Rightarrow BA^2=BP\cdot BC \Rightarrow BP = \frac{c^2}{a}$ .
Angle chasing gives: $\angle ACB = \angle QAB =\angle QPB= \angle BAP = \gamma$ . Therefore, $QP\| AC$ .
Now $\angle BAQ^{'}= \angle QBA... | [
"<details><summary>Solution</summary>We have $\\angle BPS=\\angle QAB=\\angle ACB$ , so $PS \\parallel AC$ . Moreover, $\\angle BAS=\\angle ABQ=\\alpha-\\gamma$ , so $\\angle SAC=\\angle ACB$ , which means that $ASPC$ is a cyclic isosceles trapezoid. Hence $\\angle ACS=\\angle PAC=\\alpha-\\gamma$ and thus ... | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Team Selection Test"
] | {
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Macedonian Team Selection Test/3076280.json"
} |
Let $f:\mathbb{N} \rightarrow \mathbb{N}$ be a monotonically increasing function over the natural numbers, such that $f(f(n))=n^{2}$ . What is the smallest, and what is the largest value that $f(2023)$ can take?
*Proposed by Ilija Jovcheski* | Still has no solution so i'll post a sketch: $f$ is injective so $f(n)\ge n$ . Show $f(1)=1,f(2)=3,f(3)=4, f(4)=9$ .
We claim all functions are of the following form:
1. $f(1)=1,f(2)=3,f(3)=4,f(4)=9$ 2. We keep a pointer $a$ , which starts at $3$ .
3. Assign $f$ of $f(a)$ to be $a^2$ , $f$ of $f(a+1)$ t... | [] | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Team Selection Test"
] | {
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Macedonian Team Selection Test/3076281.json"
} |
Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a function satisfying the following property: If $A, B, C \in \mathbb{R}^2$ are
the vertices of an equilateral triangle with sides of length $1$ , then
$$ f(A) + f(B) + f(C) = 0. $$ Show that $f(x) = 0$ for all $x \in \mathbb{R}^2$ .
*Proposed by Ilir Snopce* | **<span style="color:#f00">Claim 1</span>** : If $A,B \in \mathbb{R}^2$ such that $\text{dist}(A,B) = \sqrt{3}$ , then $f(A)=f(B)$ .
*<span style="color:#00f">Proof </span>*: Construct distinct $C,D$ on opposite sides of $AB$ , such that $CA=CB=DA=DB=1$ , which clearly exist by triangle inequality. As $AB=\sqr... | [
"Let $A,B,C\\in \\mathbb R^2$ such that they form an equilateral triangle with side length $1$ . Reflect $A$ across $BC$ to get the rhombus $ABCA’$ . Note that $f(A)+f(B)+f(C)=f(A’) +f(B)+f(C) =0$ so $f(A)=f(A’)$ and we get the following : For any two points $A,A’$ with $\\operatorname{dist}(A,A’)=\... | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Team Selection Test"
] | {
"answer_score": 152,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Macedonian Team Selection Test/3076282.json"
} |
Let $Q(x) = a_{2023}x^{2023}+a_{2022}x^{2022}+\dots+a_{1}x+a_{0} \in \mathbb{Z}[x]$ be a polynomial with integer coefficients. For an odd prime number $p$ we define the polynomial $Q_{p}(x) = a_{2023}^{p-2}x^{2023}+a_{2022}^{p-2}x^{2022}+\dots+a_{1}^{p-2}x+a_{0}^{p-2}.$
Assume that there exist infinitely primes ... | **S**olved with **starchan, AdhityaMV, Siddharth03, mueller.25**
The maximum value of $Q(2023)$ is $\frac{2023^{2024}-1}{2022}$ , achieved when $Q(x) = x^{2023}+x^{2022}+\dots + 1$ . We show that each $a_i \in \{-1, 0, 1\}$ after which it is clear that we cannot do better than the value prescribed above. Constru... | [
"NOTE: The above guy is trying to copy my solution. Beware the 567**S**olved with **L567, AdhityaMV, Siddharth03, mueller.25**\n<details><summary>solution</summary>The maximum value of $Q(2023)$ is $\\frac{2023^{2024}-1}{2022}$ , achieved when $Q(x) = x^{2023}+x^{2022}+\\dots + 1$ . We show that each $a_i \\in... | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Team Selection Test"
] | {
"answer_score": 24,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Macedonian Team Selection Test/3076286.json"
} |
Lucky and Jinx were given a paper with $2023$ points arranged as the vertices of a regular polygon.
They were then tasked to color all the segments connecting these points such that no triangle formed
with these points has all edges in the same color, nor in three different colors and no quadrilateral
(not necessaril... | <details><summary>Answer and short sketch of the solution</summary>$2022$ and $2020$ . **<span style="font-size:100%">Claim. $1$ :</span>** If we colour the edges of $K_n$ with at least $n$ colours, there gonna be a triangle which has all edges in different colours.**<span style="font-size:100%">Proof:</span>** D... | [] | [
"origin:aops",
"2023 Contests",
"2023 Macedonian Team Selection Test"
] | {
"answer_score": 24,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Macedonian Team Selection Test/3076288.json"
} |
For which $n\ge 3$ does there exist positive integers $a_1<a_2<\cdots <a_n$ , such that: $$ a_n=a_1+...+a_{n-1}, \hspace{0.5cm} \frac{1}{a_1}=\frac{1}{a_2}+...+\frac{1}{a_n} $$ are both true?
*Proposed by Ivan Chan Kai Chin* | Official Solution:
Answer: All $n\ge 4$ .
Solution: For $n=3$ this is impossible, because if $a<b<c$ are integers such that $c=a+b$ and $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$ , then we can rewrite it as $$ \frac{1}{a}=\frac{1}{b}+\frac{1}{a+b} $$ $$ \iff b^2-ab-a^2=0 $$ This implies the discriminant ... | [
"Answer: All naturals bigger than $3$ .\nWe are going to construct inductively,\nFor $n=4$ take $(a_1,a_2,a_3,a_4)=(1,2,3,6)$ For $n=5$ take $(a_1,a_2,a_3,a_4,a_5)=(9,25,30,36,100)$ Use the identity below for the induction step: $$ \\frac{1}{x}=\\frac{1}{x+2}+\\frac{1}{x^2+x}+\\frac{1}{x^2+3x+2} $$ thus we... | [
"origin:aops",
"2023 Malaysian APMO Camp Selection Test",
"2023 Contests"
] | {
"answer_score": 178,
"boxed": false,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Malaysian APMO Camp Selection Test/3018270.json"
} |
Ivan is playing Lego with $4n^2$ $1 \times 2$ blocks. First, he places $2n^2$ $1 \times 2$ blocks to fit a $2n \times 2n$ square as the bottom layer. Then he builds the top layer on top of the bottom layer using the remaining $2n^2$ $1 \times 2$ blocks. Note that the blocks in the bottom layer are connect... | For each of the $4n^2$ cells, draw a line to two other cells, the partner cells on the top and bottom layer (i.e. belonging to the same lego). Since the building is connected, the lines form a hamiltonian cycle. At the corners this cycle must form an "L-shape". Note you have to visit corners in a cycle, by this I mea... | [] | [
"origin:aops",
"2023 Malaysian APMO Camp Selection Test",
"2023 Contests"
] | {
"answer_score": 2,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Malaysian APMO Camp Selection Test/3018271.json"
} |
Let triangle $ABC$ with $AB<AC$ has orthocenter $H$ , and let the midpoint of $BC$ be $M$ . The internal angle bisector of $\angle BAC$ meet $CH$ at $X$ , and the external angle bisector of $\angle BAC$ meet $BH$ at $Y$ . The circles $(BHX)$ and $(CHY)$ meet again at $Z$ .
Prove that $\angle H... | Other solutions (from the official packet):
Solution 2: In this solution, all angles used are directed angles.
Let $BE$ and $CF$ be altitudes, and let $K, L$ be the midpoints of $BE$ and $CF$ respectively. Then $MK\parallel CE$ so $\angle HKM=90^{\circ}$ . Likewise $L$ lies on the circle with diameter... | [
"Note that $Z$ is the center of the spiral similarity taking $BY$ to $XC$ . If $M_B, M_C$ are the midpoints of the altitudes from $B, C$ respectively then $M_B, M_C$ lie on $(HM)$ , so it suffices to show that the spiral similarity taking $BY \\rightarrow XC$ also takes $M_B \\rightarrow M_C$ , which... | [
"origin:aops",
"2023 Malaysian APMO Camp Selection Test",
"2023 Contests"
] | {
"answer_score": 220,
"boxed": false,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Malaysian APMO Camp Selection Test/3018272.json"
} |
Let $k$ be a fixed integer. In the town of Ivanland, there are at least $k+1$ citizens standing on a plane such that the distances between any two citizens are distinct. An election is to be held such that every citizen votes the $k$ -th closest citizen to be the president. What is the maximal number of votes a ci... | Official solution (Credits to Anzo for the write up!)
We first provide an example where a center point, $O$ , has exactly $5k$ "votes''. Indeed, consider a regular pentagon $P_1\cdots P_5$ with center $O$ and radius 1, and consider a circle with radius $\epsilon > 0$ (but small) centered around (each) $P_i$... | [
"The answer is $5k$ .\n\n<span style=\"color:#f00\">Lemma 1:</span> Given $n$ points (not necessary distinct) $P_1, \\dots, P_n$ on the complex plane, for any $\\epsilon>0$ , $\\exists P_1', \\dots, P_n'$ s.t. $\\forall i,\\ |P_i'-P_i|<\\epsilon$ and the distances between any $2$ points are not $0$ an... | [
"origin:aops",
"2023 Malaysian APMO Camp Selection Test",
"2023 Contests"
] | {
"answer_score": 228,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Malaysian APMO Camp Selection Test/3018273.json"
} |
Let $n\ge 3$ , $d$ be positive integers. For an integer $x$ , denote $r(x)$ be the remainder of $x$ when divided by $n$ such that $0\le r(x)\le n-1$ . Let $c$ be a positive integer with $1<c<n$ and $\gcd(c,n)=1$ , and suppose $a_1, \cdots, a_d$ are positive integers with $a_1+\cdots+a_d\le n-1$ .
(a... | Denote, for each $n$ , the minimum $d$ for which the conclusion holds as $d(n)$ .
We claim the following:
\[
d(n) =
\begin{cases}
\lceil\frac n2\rceil & 4\mid n\text{ or }n\text{ odd}
2\cdot\lceil\frac n6\rceil & 4\mid n - 2
\end{cases}
\]
We first construct a counterexample with $d... | [
"Solution that solves only part (a): (From the official packet) \n\nWithout loss of generality, suppose $a_1 \\le a_2 \\le \\ldots \\le a_d$ . Furthermore, let $k$ be the largest integer for which $a_k \\le \\frac{n-1}{c}$ . Let $S:=a_1+a_2+\\ldots+a_k$ .\n\nFirst, observe that $$ S+(d-k) \\cdot \\frac{n}{c}... | [
"origin:aops",
"2023 Malaysian APMO Camp Selection Test",
"2023 Contests"
] | {
"answer_score": 364,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Malaysian APMO Camp Selection Test/3018274.json"
} |
Let $P$ be a cyclic polygon with circumcenter $O$ that does not lie on any diagonal, and let $S$ be the set of points on 2D plane containing $P$ and $O$ .
The $\textit{Matcha Sweep Game}$ is a game between two players $A$ and $B$ , with $A$ going first, such that each choosing a nonempty subset $T$ ... | Official Solution:
Answer: All polygons $P$ such that $P$ is not an acute-angled triangle.
Solution: If $P$ is an acute-angled triangle, then $O$ lies inside $P$ , so $A$ cannot choose $T=S$ .
Regardless of the $T$ chosen by $A$ , there are at most 3 points left, so $B$ can just choose all the rem... | [
"We claim that the answer is all polygons except for acute triangles.\n\nFirst, we demonstrate that $A$ cannot win with an acute triangle. In this case, $O$ lies within $P$ . Therefore, $A$ cannot select all four points at once. Since he selects at least one point, $B$ is left with at most 3 points to sele... | [
"origin:aops",
"2023 Malaysian IMO Team Selection Test",
"2023 Contests"
] | {
"answer_score": 156,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Malaysian IMO Team Selection Test/3062606.json"
} |
Let $a_1, a_2, \cdots, a_n$ be a sequence of real numbers with $a_1+a_2+\cdots+a_n=0$ . Define the score $S(\sigma)$ of a permutation $\sigma=(b_1, \cdots b_n)$ of $(a_1, \cdots a_n)$ to be the minima of the sum $$ (x_1-b_1)^2+\cdots+(x_n-b_n)^2 $$ over all real numbers $x_1\le \cdots \le x_n$ .
Prove th... | Note that: $$ \sum (x_i - b_i)^2 = \sum b_i^2 + \sum x_i^2 - 2 \sum x_ib_i $$ Where if we call $S_i = b_1 + \cdots + b_i$ (being $S_0 = 0$ ) we have: $$ \sum_{i=1}^n x_ib_i = \sum x_i (S_i - S_{i-1}) = \sum_{i=1}^{n-1} S_i(x_i - x_{i+1}) + x_n (b_1 + \cdots + b_n) = \sum_{i=1}^{n-1} S_i(x_i - x_{i+1}) $$ So: $... | [
"Proposer's note: this is inspired by the property of isotonic regression (monotone smoothing), cf. https://link.springer.com/article/10.1007/BF01580873. \n\n",
"There exists $y_1$ and $y_2,\\cdots y_n\\ge 0$ such that $x_k=\\sum\\limits_{i=1}^k y_i$ . Let $\\sum b_i^2=\\sum a_i^2=T$ .\nThe expression becom... | [
"origin:aops",
"2023 Malaysian IMO Team Selection Test",
"2023 Contests"
] | {
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Malaysian IMO Team Selection Test/3062607.json"
} |
Let $ABC$ be an acute triangle with $AB\neq AC$ . Let $D, E, F$ be the midpoints of the sides $BC$ , $CA$ , and $AB$ respectively, and $M, N$ be the midpoints of minor arc $BC$ not containing $A$ and major arc $BAC$ respectively. Suppose $W, X, Y, Z$ are the incenter, $D$ -excenter, $E$ -excenter, ... | Denote by $\mathcal{K}$ the Kiepert hyperbola of $ABC$ . Suppose $P$ is the antipode of $N$ in $(NWX)$ . Let $H$ be the orthocenter of $\triangle ABC$ . Notice that $W$ and $X$ also lies on $\mathcal{K}$ .
Claim: $H$ is also the orthocenter of $\triangle XWP$ .
Proof:
Since $\angle NXP= \angle NWP=... | [
"Let $W'$ be an arbitrary point on the angle bisector of $EDF$ , and $X'Y'Z'$ be the anti-cevian triangle of $W'$ wrt $DEF$ . We claim that the circumcircles of triangle $ABC,W'NX',Y'MZ'$ meet at a point. Let the second intersection of $(ABC)$ and $(W'NX')$ be $L_1$ , $(ABC)$ and $(Y'MZ')$ be $L... | [
"origin:aops",
"2023 Malaysian IMO Team Selection Test",
"2023 Contests"
] | {
"answer_score": 224,
"boxed": false,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Malaysian IMO Team Selection Test/3062608.json"
} |
Do there exist infinitely many triples of positive integers $(a, b, c)$ such that $a$ , $b$ , $c$ are pairwise coprime, and $a! + b! + c!$ is divisible by $a^2 + b^2 + c^2$ ?
*Proposed by Anzo Teh Zhao Yang* | Let $a=x^3+18, b=x^3, c=x^3-18$ . I will prove that these triple work for infinitely many values of $x$ .
In fact take $x=(k^4-6k^2+36)/4-k$ where $k$ is a positive integer. Let $k=2m$ . Then $x=4m^4-6m^2-2m+9$ . Take also $m \equiv 1 (mod 3)$ .
Then it is trivial to show that $gcd(x,2)=gcd(x,3)=1$ , hence ... | [
"This is just an idea. There are infinitely many primitive pythagorean triplets, write $a = m^2-n^2$ , $b = 2mn$ and $c = m^2+n^2$ , with $m,n$ positive integers with $m,n$ coprime and one even. Take $c$ a prime number $p$ , there are infinitely many prime numbers that can be written as a sum of two squa... | [
"origin:aops",
"2023 Malaysian IMO Team Selection Test",
"2023 Contests"
] | {
"answer_score": 110,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Malaysian IMO Team Selection Test/3063074.json"
} |
Let $ABCD$ be a cyclic quadrilateral, with circumcircle $\omega$ and circumcenter $O$ . Let $AB$ intersect $CD$ at $E$ , $AD$ intersect $BC$ at $F$ , and $AC$ intersect $BD$ at $G$ .
The points $A_1, B_1, C_1, D_1$ are chosen on rays $GA$ , $GB$ , $GC$ , $GD$ such that: $\bullet$ $\displ... | Here is my solution that doesn't require ratio bash. Pretty much all solutions I have read so far (internally, and in AoPS here) involves some form of bashing @_@
Let ray $OG$ intersect line $EF$ and $\omega$ at $X$ and $Y$ respectively. Let $Z$ be the diametrically opposite point of $Y$ in $\omega$ . ... | [
"It is well known that the image of $G$ the inversion wrt $\\omega$ is the foot from $G$ to $EF$ , call it $P$ . Let the center of $\\omega_1=(A_1B_1C_1D_1)$ be $O_1$ , then $GO_1O$ are collinear by homothety. Therefore the image of $\\omega_1$ is a line $\\ell$ perpendicular to $OO_1$ , i.e. perp... | [
"origin:aops",
"2023 Malaysian IMO Team Selection Test",
"2023 Contests"
] | {
"answer_score": 272,
"boxed": false,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Malaysian IMO Team Selection Test/3063075.json"
} |
Suppose there are $n$ points on the plane, no three of which are collinear. Draw $n-1$ non-intersecting segments (except possibly at endpoints) between pairs of points, such that it is possible to travel between any two points by travelling along the segments. Such a configuration of points and segments is called ... | Official Solution: **Answer.** All odd $n$ .**Solution.**We interpret the network as a tree and root it at an arbitrary vertex. For a permutation $p_1,p_2,\cdots,p_n$ of $\{1,2,\cdots,n\}$ , we say that it is even if it can be transformed into the identity permutation with an even number of transpositions and even ... | [
"Is this ISL? ",
"No, its not",
"<details><summary>partial progress</summary>For even integers $n=2k\\geq 6$ , consider the tree with two connected nodes $a$ and $b$ , with $a$ connected to $2$ other nodes and $b$ connected to $2k-2$ other nodes. In this way, each clockwise shift induces a permutati... | [
"origin:aops",
"2023 Malaysian IMO Team Selection Test",
"2023 Contests"
] | {
"answer_score": 464,
"boxed": false,
"end_of_proof": true,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 Malaysian IMO Team Selection Test/3063077.json"
} |
Integers $a, b, c, d$ satisfy $a+b+c+d=0$ . Show that $$ n=(ab-cd)\cdot(bc-ad)\cdot(ca-bd) $$ is a perfect square. | $d=-(a+b+c)$
So $ab-cd=ab+c(a+b+c)=ab+ac+c^2+bc=(a+c)(b+c)$ $bc-ad=bc+a(a+b+c)=bc+a^2+ba+ac=(a+c)(a+b)$ $ac-bd=ac+b(a+b+c)=ac+ab+b^2+bc=(a+b)(b+c)$ So, $n=(a+c)(b+c)(a+c)(a+b)(a+b)(b+c)=[(a+b)(b+c)(a+c)]^2$ , a perfect sqaure as $a,b,c$ integers | [
"Waaaaaayyyyyy tooooooo trivial for EGMO TST. What were they thinking??"
] | [
"origin:aops",
"2023 Contests",
"2023 Moldova EGMO TST"
] | {
"answer_score": 12,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Moldova EGMO TST/3008611.json"
} |
Show that for every integer $n\geq2$ there are two distinct powers of $n$ such that their sum is greater than $10^{2023}$ and their positive difference is divisible with $2023$ . | [] | [
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Let there be a quadrilateral $ABCD$ such that $\angle CAD=45, \angle ACD=30, \angle BAC=\angle BCA=15$ . Find $\angle DBC$ . | As $\angle ABC = 180-(2.15)=150=2(180-105)=2(180-\angle ADC)$ , $\angle ADC$ obtuse and $AB=BC$ , $B$ is the circumcenter of $\triangle ADC$ . So , $BD=BC$ . So $\angle DBC=180-2\angle BCD = 180-2.45=\fbox{90}$ | [] | [
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Find all triplets of prime numbers $(m, n, p)$ , that satisfy the system of equations: $$ \left\{\begin{matrix} 2m-n+13p=2072,3m+11n+13p=2961.\end{matrix}\right. $$ | $m+12n=889 \implies n \leq 74$ , so trying $n=73$ , we get $m=13$ , $p=163$ .
Now, we take mod $13$ .
Then,
\begin{align*}
2m-n &\equiv 5 \pmod{13}
3m-2n &\equiv 10 \pmod{13}
\therefore 4m-2n &\equiv 10 \pmod{13} \implies 13 \mid m
\end{align*}
Thus, since $m$ is prime, $m=13 \implies n=73 \impli... | [
"Only one such prime triplet exists: $\\fbox{m =13, n = 73, p = 163}$ ."
] | [
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Find all pairs of real numbers $(x, y)$ , that satisfy the system of equations: $$ \left\{\begin{matrix} 6(1-x)^2=\dfrac{1}{y} 6(1-y)^2=\dfrac{1}{x}.\end{matrix}\right. $$ | We have $6y(1-x)^2=1=6x(1-y)^2$ $\implies$ $x^2y-2xy+y=xy^2-2xy+x$ $\implies$ $xy(x-y)=x-y$ So, if $xy=1$ or $x=\frac{1}{y}$ , by pluggin it into the first relation we get $6x^2-13x+6=0$ and this is equivalent to $(x-\frac{2}{3})(x-\frac{3}{2})=0$ So $(x,y)=(\frac{2}{3},\frac{3}{2}),(\frac{3}{2},\frac{2... | [] | [
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Let there be a square $ABCD$ . Points $E$ and $F$ are on sides $(BC)$ and $(AB)$ such that $BF=CE$ . LInes $AE$ and $CF$ intersect in point $G$ . Prove that $EF$ and $DG$ are perpendicular. | analytic geometry solution
<details><summary>Click to expand</summary>wlog let $ABCD$ be the unit square $A (0,1)$ , $B (1,1)$ , $C(1,0)$ , $D(0,0)$ $E(1,a)$ with $0<a<1$ and $F(1-a,1) $ gradient of $EF$ is $\ell_{EF} = \frac{a-1}{1-(1-a)}=\frac{a-1}{a}$ $(CF): y-0=\frac{ 0-1}{1-(1-a)}(x-1)=\frac{1}{... | [] | [
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} |
Find all triplets of integers $(a, b, c)$ , that verify the equation $$ |a+3|+b^2+4\cdot c^2-14\cdot b-12\cdot c+55=0. $$ | <blockquote>This can be rewritten as $|a+3|+(b-7)^2+(2c-3)^2=2$ Thus either one of the terms on the LHS is $0$ and the others equal $1$ or $|a+3|=2$ and the squares are $0$ . However this cannot happen as $2c-3$ , with $c$ an integer, can never equal 0.So it must be that one of the terms on the LHS is $0$ ... | [
"This can be rewritten as $|a+3|+(b-7)^2+(2c-3)^2=2$ Thus either one of the terms on the LHS is $0$ and the others equal $1$ or $|a+3|=2$ and the squares are $0$ . However this cannot happen as $2c-3$ , with $c$ an integer, can never equal 0.So it must be that one of the terms on the LHS is $0$ and th... | [
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} |
Prove that the number $1$ can be written as a sum of $2023$ fractions of the form $\frac{1}{k_i}$ , where all nonnegative integers $k_i (1\leq i\leq 2023)$ are distinct. | too easy for TST $1=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{2020}}+\frac{1}{2^{2021}}+\frac{1}{3\times2^{2020}}+\frac{1}{6\times2^{2020}}$ | [] | [
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} |
Solve the equation $$ \left[\frac{x^2+1}{x}\right]-\left[\frac{x}{x^2+1}\right]=3. $$ | <blockquote>x in positive integer?</blockquote> $x$ is real | [
"x in positive integer? ",
"if in positive integer answer: x--> 1,2,3"
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Cirlce $\Omega$ is inscribed in triangle $ABC$ with $\angle BAC=40$ . Point $D$ is inside the angle $BAC$ and is the intersection of exterior bisectors of angles $B$ and $C$ with the common side $BC$ . Tangent form $D$ touches $\Omega$ in $E$ . FInd $\angle BEC$ . | $D$ is the A-excenter of $\triangle BAC$ . Let $I$ be the incenter of $\triangle BAC$ or simply center of $\Omega$ . Now we know that $\angle IBD=\angle ICD=90$ . As $\angle ICD+\angle IBD=180$ , $ I,C,B,D$ concylic. Now as $DE$ tangent to $\Omega$ so $\angle IED=90$ . As $\angle IED=90=\angle IBD$ ... | [] | [
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Find all three digit positive integers that have distinct digits and after their greatest digit is switched to $1$ become multiples of $30$ . | write such number as $N=\overline{abc}$ with $a,b,c$ being digits we observe that $c$ cannot be maximum out of the $3$ , as then the number would end in $1$ , which is clearly absurd (it needs to be a multiple of 10)So $c=0$ and the number is $\overline{ab0}$ If $b$ is bigger than $a$ , after the tranfor... | [] | [
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Let there be an integer $n\geq2$ . In a chess tournament $n$ players play between each other one game. No game ended in a draw. Show that after the end of the tournament the players can be arranged in a list: $P_1, P_2, P_3,\ldots,P_n$ such that for every $i (1\leq i\leq n-1)$ the player $P_i$ won against play... | Prove by induction
n=2 can be easily checked
Suppose we already have a list of n players $P_1, P_2, P_3,\ldots,P_n$ If $P_{n+1}$ won against $P_1$ then we're done (I)
If there is a number $i (1\leq i\leq n-1)$ such that $P_i$ won against $P_{n+1}$ and $P_{n+1}$ won against $P_{i+1}$ then we're done (II)... | [] | [
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Let $a,b,c$ be distinct positive integers and let $r,s,t$ be positive integers such that: $ab+1=r^2,ac+1=s^2,bc+1=t^2$ Prove that it is not possible that all three fractions $ \frac{rt}{s}, \frac{rs}{t}, \frac{st}{r}$ are integers. | Let $a>b>c$ $\frac{st}{r}$ is integer , so $\frac{s^2t^2}{r^2}$ is integer too. $\frac{s^2t^2}{r^2}=\frac{(ac+1)(bc+1)}{ab+1}=\frac{abc^2+c(a+b)+1}{ab+1} =\frac{(ab+1)(c^2+1)+(ac+bc-c^2-ab)}{ab+1} = c^2+1-\frac{(a-c)(b-c)}{ab+1}$ $(a-c)(b-c)>0$ and $ab+1|(a-c)(b-c)$ so $(a-c)(b-c)\geq ab+1$ $ab-ac-bc+c^2 \ge... | [
"https://www.artofproblemsolving.com/community/c3046h1184227_the_system_is_an_arithmetic_progression",
"Any ideas?",
"Here's a similar solution to the above.\n\nAssume otherwise. Note that if $\\dfrac{rt}{s}$ is an integer, then $\\dfrac{r^2s^2}{t^2}$ must be too. Therefore, $\\dfrac{(ab+1)(bc+1)}{ac+1}$ i... | [
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} |
Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$ . Let $M$ and $N$ be the midpoints of $AB$ and $AC$ , respectively; let $AD$ be an altitude in this triangle. A point $K$ is chosen on the segment $MN$ so that $BK=CK$ . The ray $KD$ meets the circumcircle $\Omega$ of $ABC$ at $Q$ . Pr... | Ack this took 40 minutes
[asy]
size(250);
defaultpen(linewidth(0.7)+fontsize(11));
pair A = dir(125), B = dir(210), C = dir(330), G = centroid(A,B,C), D = foot(A,B,C), O = circumcenter(A,B,C), M = (A+B)/2, N = (A+C)/2, K = foot(O,M,N), P=(A+D)/2;
pair X = extension(A,K,B,C), Ap = A+X-D, Y = 2*D-G, Q = intersectionpoin... | [
"<blockquote>Let $\\triangle ABC$ be an acute-angled triangle with $AB<AC$ . Let $M$ and $N$ be the midpoints of $AB$ and $AC$ , respectively; let $AD$ be an altitude in this triangle. A point $K$ is chosen on the segment $MN$ so that $BK=CK$ . The ray $KD$ meets the circumcircle $\\Omega$ of ... | [
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"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2023 Contests/2023 Moldova Team Selection Test/1632766.json"
} |
The sequence $\left(a_n \right)$ is defined by $a_1=1, \ a_2=2$ and $$ a_{n+2} = 2a_{n+1}-pa_n, \ \forall n \ge 1, $$ for some prime $p.$ Find all $p$ for which there exists $m$ such that $a_m=-3.$ | The following sollution works for any integer p (not compulsory prime or even positive), and it consists in doing casework based on p mod 3: $1.\; p \equiv 0 \mod{3} $ This is the easiest one as $a_{n} $ is never divisible by 3. $2. \; p \equiv -1 \mod{3} $ Note that $ a_{n+2} \equiv 2a_{n+1}-pa_n \equiv 2a_{n+1}+a_... | [
"Same trick as for this classical problem from [Mathematical Excalibur](https://artofproblemsolving.com/community/c6h1183114) (from where it is surely copied).\n<details><summary>Solution</summary>Working modulo $p$ we see that $a_n \\equiv 2^{n-1} \\mod p$ .\nWorking modulo $p-1$ we see that $a_n \\equiv n \... | [
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} |
Polynomials $(P_n(X))_{n\in\mathbb{N}}$ are defined as: $$ P_0(X)=0, \quad P_1(X)=X+2, $$ $$ P_n(X)=P_{n-1}(X)+3P_{n-1}(X)\cdot P_{n-2}(X)+P_{n-2}(X), \quad (\forall) n\geq2. $$ Show that if $ k $ divides $m$ then $P_k(X)$ divides $P_m(X).$ | <blockquote><blockquote>Polynomials $(P_n(X))_{n\in\mathbb{N}}$ are defined as: $$ P_0(X)=0, \quad P_1(X)=X+2, $$ $$ P_n(X)=P_{n-1}(X)+3P_{n-1}(X)\cdot P_{n-2}(X)+P_{n-2}(X), \quad (\forall) n\geq2. $$ Show that if $ k $ divides $m$ then $P_k(X)$ divides $P_m(X).$ </blockquote>
Easy to find (and easier... | [
"<blockquote>Polynomials $(P_n(X))_{n\\in\\mathbb{N}}$ are defined as: $$ P_0(X)=0, \\quad P_1(X)=X+2, $$ $$ P_n(X)=P_{n-1}(X)+3P_{n-1}(X)\\cdot P_{n-2}(X)+P_{n-2}(X), \\quad (\\forall) n\\geq2. $$ Show that if $ k $ divides $m$ then $P_k(X)$ divides $P_m(X).$ </blockquote>\nEasy to find (and easie... | [
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Let $ n $ be a positive integer. A sequence $(a_1,a_2,\ldots,a_n)$ of length is called $balanced$ if for every $ k $ $(1\leq k\leq n)$ the term $ a_k $ is equal with the number of distinct numbers from the subsequence $(a_1,a_2,\ldots,a_k).$ a) How many balanced sequences $(a_1,a_2,\ldots,a_n)$ of length... | <blockquote><blockquote>the definition is equivalent to $a_1=1$ and $a_n=a_{n-1}$ or $a_{n-1}-1$ for $n>1$ the problem is trivial from that</blockquote>
shouldn't it be $a_n = a_{n-1}$ or $a_n = a_{n-1} + 1$ for $n>1$ ?</blockquote>
Yes, that is my fault.
<details><summary>Answer</summary>(a) is $2^{n-1... | [
"the definition is equivalent to $a_1=1$ and $a_n=a_{n-1}$ or $a_{n-1}+1$ for $n>1$ the problem is trivial from that",
"<blockquote>the definition is equivalent to $a_1=1$ and $a_n=a_{n-1}$ or $a_{n-1}-1$ for $n>1$ the problem is trivial from that</blockquote>\n\nshouldn't it be $a_n = a_{n-1}$ or... | [
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} |
Find all pairs of positive integers $(n,k)$ for which the number $m=1^{2k+1}+2^{2k+1}+\cdots+n^{2k+1}$ is divisible by $n+2.$ | Case 1: $n$ is odd
Then $m=1^{2k+1}+(2^{2k+1}+n^{2k+1})+(3^{2k+1}+(n-1)^{2k+1})+... \equiv 1 \pmod {n+2}$ so $n+2 \not | m$ Case 2: $n=2t$ is even
Then $m=m=1^{2k+1}+(2^{2k+1}+n^{2k+1})+(3^{2k+1}+(n-1)^{2k+1})+...\equiv 1+(\frac{n+2}{2})^{2k+1}=1+(t+1)^{2k+1} \pmod {2t+2}$ So $n+2 \not | m$ | [] | [
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Show that if $2023$ real numbers $x_1,x_2,\dots,x_{2023}$ satisfy $x_1\geq x_2\geq\dots\geq x_{2023}\geq0,$ then $$ x_1^2+3x_2^2+5x_3^2+\cdots+(2\cdot2023-1)\cdot x^2_{2023}\leq(x_1+x_2+\cdots+x_{2023})^2. $$ When does the equality take place? | $(x_1+x_2+\cdots+x_{2023})^2-(x_1^2+3x_2^2+5x_3^2+\cdots+(2\cdot2023-1)\cdot x^2_{2023})=2 \sum_{i=2}^{2023} x_i( \sum_{j=1}^{i-1} x_j-(i-1)x_i)
\geq 0$ because $ \sum_{j=1}^{i-1} x_j-(i-1)x_i \geq 0$ And equality if all $x_i$ are equal.
Or we can prove it by induction using $(x_1+x_2+...+x_n)^2-(x_1^2+3x_2^2+... | [
"Actually\nEquality can be when some are 0, and the others are equal."
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} |
Find all integers $ n $ $(n\geq2)$ with the property: for every $ n $ distinct disks in a plane with at least a common point one of the disks contains the center of another disk. | [
"Its trivial"
] | [
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"boxed": false,
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Let $ABC$ be an acute triangle with orthocenter $ H $ and $AB<AC.$ Let $\Omega_1$ be a circle with diameter $AC$ and $\Omega_2$ a circle with diameter $ AB.$ Line $BH$ intersects $\Omega_1$ in points $ D $ and $E$ such that $E$ is not on segment $BH.$ Line $ CH $ intersects $\Omega_2$ in ... | I think this is quite well known.**Claim 1:** Points $D,E,F,G$ lie on a circle $\omega$ centered at point $A$ .
*Proof:* Let $Y,Z$ be the feet of the altitudes from $B,C$ respectively. Note that $AG^2=AF^2=AZ \cdot AB=AY \cdot AC=AE^2=AD^2,$ and the claim follows $\blacksquare$ **Claim 2:** $BC$ is the po... | [
"Let $B_1$ and $C_1$ be the feet from $B$ and $C$ to the opposite sides, respectively. Note that $D, E, F, G$ are invariant under inversion in circle with center $A$ and radius $\\sqrt{AC_1\\cdot AB}=\\sqrt{AB_1\\cdot AC}$ . Hence $DEFG$ is concyclic with center $A$ .\n\nNow $CE$ is tangent to $\... | [
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"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2023 Contests/2023 Moldova Team Selection Test/3048285.json"
} |
Let $ n $ $(n\geq2)$ be an integer. Find the greatest possible value of the expression $$ E=\frac{a_1}{1+a_1^2}+\frac{a_2}{1+a_2^2}+\ldots+\frac{a_n}{1+a_n^2} $$ if the positive real numbers $a_1,a_2,\ldots,a_n$ satisfy $a_1+a_2+\ldots+a_n=\frac{n}{2}.$ What are the values of $a_1,a_2,\ldots,a_n$ when the... | <blockquote>Let $ n $ $(n\geq2)$ be an integer. Find the greatest possible value of the expression $$ E=\frac{a_1}{1+a_1^2}+\frac{a_2}{1+a_2^2}+\ldots+\frac{a_n}{1+a_n^2} $$ if the positive real numbers $a_1,a_2,\ldots,a_n$ satisfy $a_1+a_2+\ldots+a_n=\frac{n}{2}.$ What are the values of $a_1,a_2,\ldots,a_... | [
"For $x\\ge0$ we have\n\\begin{align}\\frac x{1+x^2}\\le\\frac{12x+4}{25}\\iff(2x-1)^2(3x+4)\\ge0,\\tag1\\end{align}which is true. So\n\\[E\\le\\sum_{k=1}^n\\frac{12a_k+4}{25}=\\frac{12\\times\\frac n2+4n}{25}=\\frac{2n}5.\\]\nThe greatest value is achieved when $(1)$ reaches equality, or\n\\[a_1=a_2=\\cdots=a_... | [
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"boxed": false,
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} |
Let $ABC$ be a triangle with $\angle ACB=90$ and $AC>BC.$ Let $\Omega$ be the circumcircle of $ABC.$ Point $ D $ is the midpoint of small arc $AC$ of $\Omega.$ Point $ M $ is symmetric with $ A$ with respect to $D.$ Point $ N$ is the midpoint of $MC.$ Line $AN$ intersects $\Omega$ in point... | $1.$ See that $DN \parallel AC$ , so if $QM$ passes through the midpoint of $AC$ , then $Q$ is the midpoint $DN$ . $2.$ $M,C,B$ are collinear, because $AD=DC=DM$ , so $\angle ACM = 90^\circ = \angle ACB$ . $3.$ $DN$ is tangent to $\Omega$ , because if $O$ is the center of $\Omega$ , then $DN \para... | [
"It suffices to prove that $Q$ is the midpoint of $ND$ . Let $K$ be the midpoint of $BC$ . By the Ratio Lemma, this reduces to $\\dfrac{\\sin \\angle NBQ}{\\sin \\angle QBD}=\\dfrac{BD}{BN}$ .\n\nNote that $\\dfrac{\\sin \\angle NBQ}{\\sin \\angle QBD}=\\dfrac{\\sin \\angle CBN}{\\sin \\angle NAD}=\\dfrac{AD}... | [
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"answer_score": 70,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Moldova Team Selection Test/3048862.json"
} |
Find all sets $ A$ of nonnegative integers with the property: if for the nonnegative intergers $m$ and $ n $ we have $m+n\in A$ then $m\cdot n\in A.$ | $A=\emptyset$ , $A=\{0\}$ , $A=\{0,1\}$ , $A=\{0,1,2\}$ , $A=\{0,1,2,3\}$ , $A=\{0,1,2,3,4\}$ , $A=\mathbb Z_{\ge0}$ work, now assume $A$ is something else.
If $n\in A$ then we can conclude $\{0,1,\ldots,n\}\subseteq A$ by inductively applying $(n-1)+1\in A\Rightarrow n-1\in A$ .
If $A$ is unbounded ab... | [
" $A=\\emptyset$ works. Suppose $A$ is not empty. $A=\\left \\{0\\right \\}$ works, now assume that $A$ has at least one non zero element.\nIf $a\\in A,a\\neq 0$ then $1+(a-1)\\in A$ so $1\\cdot(a-1)=a-1\\in A$ .\nAs a result, if $A$ is finite then is on the form $A=\\left \\{0,1,..,n\\right \\}$ , if... | [
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"boxed": false,
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"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Moldova Team Selection Test/3048865.json"
} |
Let $u, v$ be arbitrary positive real numbers. Prove that \[\min{(u, \frac{100}{v}, v+\frac{2023}{u})} \leq \sqrt{2123}.\] | <blockquote>Let $u, v$ be arbitrary positive real numbers. Prove that \[\min{(u, \frac{100}{v}, v+\frac{2023}{u})} \leq \sqrt{2123}.\]</blockquote>**Solution.**if $u>\sqrt{2123}$ and $\frac{100}{v}>\sqrt{2123}$ , then
\begin{align*}& v+ \frac{2023}{u}<\frac{100}{\sqrt{2123}}+ \frac{2023}{\sqrt{2123}} =\sqrt{2123... | [
"<blockquote>Let $u, v$ be arbitrary positive real numbers. Prove that \\[\\min{(u, \\frac{100}{v}, v+\\frac{2023}{u})} \\leq \\sqrt{2123}.\\]</blockquote>\nIf $u\\le\\sqrt{2123}$ : $\\min(u,\\frac {100}v,v+\\frac{2023}u)\\le u\\le\\sqrt{2123}$ If $v\\ge\\frac{100}{\\sqrt{2123}}$ : $\\min(u,\\frac {100}v,v... | [
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} |
In an acute triangle $ABC$ the points $D, E$ are the feet of the altitudes through $B, C$ respectively. Let $L$ be the point on segment $BD$ such that $AD=DL$ . Similarly, let $K$ be the point on segment $CE$ such that $AE=EK$ . Let $M$ be the midpoint of $KL$ . The circumcircle of $ABC$ intersect ... | By the Ratio Lemma on $\triangle{AKL}$ , $\frac{\sin \angle{KAM}}{\sin \angle{LAM}} = \frac{AL}{AK} = \frac{AD}{AE} = \frac{AB}{AC}$ . If $U = BS \cap CT$ , then by the Law of Sines on $\triangle{ASU}, \triangle{ATU}$ , $\frac{\sin \angle{SAU}}{\sin \angle{TAU}} = \frac{\sin \angle{ASU} \cdot \frac{SU}{AU}}{\sin \... | [] | [
"origin:aops",
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"boxed": false,
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"path": "Contest Collections/2023 Contests/2023 Mongolian Mathematical Olympiad/3245672.json"
} |
Five girls and five boys took part in a competition. Suppose that we can number the boys and girls $1, 2, 3, 4, 5$ such that for each $1 \leq i,j \leq 5$ , there are exactly $|i-j|$ contestants that the girl numbered $i$ and the boy numbered $j$ both know. Let $a_i$ and $b_i$ be the number of contestants ... | [] | [
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} | |
Find all functions $f : \mathbb{R} \to \mathbb{R}$ and $h : \mathbb{R}^2 \to \mathbb{R}$ such that \[f(x+y-z)^2=f(xy)+h(x+y+z, xy+yz+zx)\] for all real numbers $x,y,z$ . | <blockquote>Find all functions $f : \mathbb{R} \to \mathbb{R}$ and $h : \mathbb{R}^2 \to \mathbb{R}$ such that \[f(x+y-z)^2=f(xy)+h(x+y+z, xy+yz+zx)\] for all real numbers $x,y,z$ .</blockquote>
Let $P(x,y,z)$ be the assertion $f(x+y-z)^2=f(xy)+h(x+y+z,xy+yz+zx)$ Note that $(x+y+z)^2\ge 3(xy+yz+zx)$ and so we... | [] | [
"origin:aops",
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"answer_score": 1102,
"boxed": true,
"end_of_proof": false,
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"path": "Contest Collections/2023 Contests/2023 Mongolian Mathematical Olympiad/3245678.json"
} |
There are $n$ students in a class, and some pairs of these students are friends. Among any six students, there are two of them that are not friends, and for any pair of students that are not friends there is a student among the remaining four that is friends with both of them. Find the maximum value of $n$ . | Answer: $25$ .
Example: Take $5$ subgraphs $A,B,C,D,E$ . Let each of them has $5$ vertices and any two vertices in the same subgraph are not connected. Any two vertices in different subgraphs are connected.
Proof: $\rule{15cm}{0.2mm}$ Claim: Everyone has at least $n-5$ friends.
Proof: Let $A$ be a vertex. If... | [
"This was also problem 2 of the Lithuanian Grand Duchy Olympiad 2023. The answer is 25.",
"Oh, that reminds my nostalgic memories. If I had solved it during the contest, my life would have been different. I hate this problem."
] | [
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"path": "Contest Collections/2023 Contests/2023 Mongolian Mathematical Olympiad/3245679.json"
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Let $m$ be a positive integer. We say that a sequence of positive integers written on a circle is good, if the sum of any $m$ consecutive numbers on this circle is a power of $m$ .
1. Let $n \geq 2$ be a positive integer. Prove that for any good sequence with $mn$ numbers, we can remove $m$ numbers so the r... | Deceivingly short.
[color = #7851A9]**Part 1:**[/color] Let $a_1, a_2, \dots, a_{mn}$ be the numbers with cyclic indices, and let $a_i$ be the maximum. Then, for every $k$ such that $i - m \leq k < i$ , we have
\[a_i < a_{k + 1} + a_{k + 2} + \dots + a_{k + m} \leq ma_i\]
and hence all of these sums correspond ... | [] | [
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} |
Let points $M$ and $N$ lie on sides $AB$ and $BC$ of triangle $ABC$ in such a way that $MN||AC$ . Points $M'$ and $N'$ are the reflections of $M$ and $N$ about $BC$ and $AB$ respectively. Let $M'A$ meet $BC$ at $X$ , and let $N'C$ meet $AB$ at $Y$ . Prove that $A,C,X,Y$ are concyclic. | I believe there are some typos (which I edited below).
<blockquote>Let points $M$ and $N$ lie on sides $AB$ and $BC$ of triangle $ABC$ in such a way that $MN||AC$ . Points $M'$ and $N'$ are the reflections of $M$ and $N$ about $BC$ and $AB$ respectively. Let $M'A$ meet $BC$ at $X$ , and let... | [
"How can MN be parallel to BC when N lies on the side BC?",
"<blockquote>Let points $M$ and $N$ lie on sides $AB$ and $BC$ of triangle $ABC$ in such a way that $MN||AC$ . Points $M'$ and $N'$ are the reflections of $M$ and $N$ about $BC$ and $AB$ respectively. Let $M'A$ meet $BC$ at $X... | [
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} |
Let $n$ be a positive integer and let $p$ be a prime number. Prove that if $a$ , $b$ , $c$ are integers (not necessarily positive) satisfying the equations \[ a^n + pb = b^n + pc = c^n + pa\] then $a = b = c$ .
*Proposed by Angelo Di Pasquale, Australia* | After trying to derive relations that allows applying LTE, I found this solution which do not use it at all.
Assume for contradiction that $a,b,c$ are distinct. We can easily see that if $a,b,c$ are not pairwise distinct, then they must be all equal. Then, we must have $a \ne b, b \ne c, c\ne a.$ If $n=1,$ we w... | [
"<blockquote>Let $ n$ be a positive integer and let $ p$ be a prime number. Prove that if $ a$ , $ b$ , $ c$ are integers (not necessarily positive) satisfying the equations\n\\[ a^n \\plus{} pb \\equal{} b^n \\plus{} pc \\equal{} c^n \\plus{} pa\\]\nthen $ a \\equal{} b \\equal{} c$ .</blockquote>\r\n\r\n... | [
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"path": "Contest Collections/2023 Contests/2023 Myanmar IMO Training/287869.json"
} |
Let $\triangle ABC$ be a triangle such that $AB = AC$ , and let its circumcircle be $\Gamma$ . Let $\omega$ be a circle which is tangent to $AB$ and $AC$ at $B$ and $C$ . Point $P$ belongs to $\omega$ , and lines $PB$ and $PC$ intersect $\Gamma$ again at $Q$ and $R$ . $X$ and $Y$ are points... | Let $O$ be the circumcenter of the circle $(ABC)$ , and let the midpoints of $AB$ and $AC$ be $M$ and $N$ respectively. Denote by $T$ the arc midpoint of $BC$ not containing $A$ , i.e. the center of $\omega$ . $\textbf{Claim.} OX \cdot OY = R^2,$ where $R$ is the circumradius. $\textit{Proof.}$ Let... | [] | [
"origin:aops",
"2023 Contests",
"2023 Myanmar IMO Training"
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"boxed": false,
"end_of_proof": true,
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} |
Some cells of an infinite chessboard (infinite in all directions) are coloured blue so that at least one of the $100$ cells in any $10 \times 10$ rectangular grid is blue. Prove that, for any positive integer $n$ , it is possible to select $n$ rows and $n$ columns so that all of the $n^2$ cells in their inte... | I'll solve this problem by solving the following:
Let $m$ be a positive integer. There is an infinite chessboard. Each cell of the chessboard contains an integer in $\{1, 2, \dots, m\}$ . Prove that for any positive integer $n$ , one can always find $n$ rows and $n$ columns whose $n^2$ intersections contain ... | [] | [
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"path": "Contest Collections/2023 Contests/2023 Myanmar IMO Training/3095451.json"
} |
Let $n \geq 2$ be a positive integer. A total of $2n$ balls are coloured with $n$ colours so that there are two balls of each colour. These balls are put inside $n$ cylindrical boxes with two balls in each box, one on top of the other. Phoe Wa Lone has an empty cylindrical box and his goal is to sort the balls ... | The answer is $\left\lfloor\frac{3n}2\right\rfloor$ .**Bound:** Take two colours, say $a,b$ , and consider the two boxes $\dbinom a b$ $\dbinom b a$ .
It takes one move to move a ball to a different box, and then two more moves to complete the boxes (i.e. have the same colours).
If $n$ is even, repeating this wi... | [
"The answer is n \nProof:\nBase case:\nn=2\nWe have two colours and four balls\nThere at least two balls in a box and at least two balls have a common colour.\nTake,the same colour balls in the same boxes.\nTherefore,we can do this move for n moves.\nn=3\nWe have three colours and six balls. \nBy repeating the same... | [
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"boxed": false,
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} |
Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that $$ m+f(n) \mid f(m)^2 - nf(n) $$ for all positive integers $m$ and $n$ .
(Here, $f(m)^2$ denotes $\left(f(m)\right)^2$ .) | I just realized that this is in $\mathbb{N}$ but a cool problem though , i will not post the correct solution $\textbf{Answer:}$ $f(n)=n \forall n \in \mathbb{N}$ . Clearly this works $\textbf{Solution:}$ Let $Q(m,n)$ -denote the given asseriton.
First we claim that $f(0)=0$ Proof: $Q(0.1)\implies f(1) \mid f(0)... | [
"Let $P(m,n)$ denote the assertion, note that $m + f(n) \\mid f(m)^2 + mn$ , $P(1,1)$ gives size contradiction if $f(1) > 1$ so $f(1)=1$ and $P(1,n) \\implies 1 + f(n) \\mid 1+ n \\implies f(a)=a$ for $a = \\text{prime} - 1$ , finally note that $P(m,a) \\implies m + a \\mid f(m)^2 - a^2 \\implies m + ... | [
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"boxed": false,
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"path": "Contest Collections/2023 Contests/2023 Myanmar IMO Training/3111848.json"
} |
Find all real numbers $a, b, c$ that satisfy $$ 2a - b =a^2b, \qquad 2b-c = b^2 c, \qquad 2c-a= c^2 a. $$ | So, here is the solution by the author.
It is obvious that the triplets $(0, 0, 0), (1, 1, 1), (-1, -1, -1)$ satisfy the given system. Now, we will prove that these are the only solutions.
Firstly, one can easily verify that if one of $a, b, c$ is zero, then all are zero. Thus, we look for non-zero solutions. We ca... | [
"Note that $a=f(f(f(a)))$ where $f(x)=\\frac{2x}{1+x^2}$ .\n\nFixed points: $\\frac{2x}{1+x^2}=x\\implies x^3-x=0\\implies x(x+1)(x-1)=0\\implies x\\in\\{-1,0,1\\}$ .\n\nThree-cycle points:\nSince $\\tan 2\\theta=\\frac{2\\tan\\theta}{1-\\tan^2\\theta}$ , we have $f(i\\tan\\theta)=i\\tan 2\\theta$ and so we ... | [
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For a real number $x$ , let $\lfloor x\rfloor$ stand for the largest integer that is less than or equal to $x$ . Prove that
\[ \left\lfloor{(n-1)!\over n(n+1)}\right\rfloor \]
is even for every positive integer $n$ . | <details><summary>Solution</summary>Let $\left\lfloor{(n-1)!\over n(n+1)}\right\rfloor = F(n)$ for convenience.
Check that $n \le 10$ works. In practice, you can really check up to as many as you wish, but the purpose is to eliminate the small cases that could be nuisances to the general procedure.
For $n > 10$ ,... | [
"It's easy to know $n=1,2,3,4$ , $\\left\\lfloor\\frac{(n-1)!}{n(n+1)}\\right\\rfloor$ is even.\r\nNow assume $n\\geq5$ :\r\n(1) $n$ is odd prime,By Wilson: $(n-1)!+1\\equiv0\\pmod{n}$ \r\nAnd $n+1$ is composite,suppose $n+1=ab,2\\leq a\\leq b\\leq n-1$ \r\n(i) $a\\neq b$ , $a,b\\in\\{1,2,\\ldots,n-1\\},\\th... | [
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"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2023 Contests/2023 Myanmar IMO Training/82831.json"
} |
Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that $$ \gcd(f(x),y)f(xy)=f(x)f(y) $$ for all positive integers $x, y$ . | Swap $x,y$ to get $\gcd(f(x),y)=\gcd(f(y),x)$ , and so for $y=f(x)$ we obtain that $f(x)=\gcd(f(f(x)),x)$ . Hence, this implies that $f(x) \mid x$ for all $x$ . Therefore, $f(p) \mid p$ for all primes $p$ , i.e. $f(p) \in \{1, p \}$ for all primes $p$ . Let $A=\{p \,\, {\rm prime} : f(p)=1 \}$ and $B=... | [
"<details><summary>pls someone check this</summary>denotes P(x,y) as the assertion $P(x,1) = f(x) = f(x)*f(1)$ $f(1) = 1 $ $P(x,p) = \\gcd(f(x),p) *f(xp) = f(x)*f(p)$ if $p \\mid f(x)$ for all x however contradiction to f(1) $pf(xp) = f(x)*f(p)$ so $f(xp) = f(x)*f(p)$ now \nif x prime $P(p,y)=\\gcd(f(p),y) *f... | [
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"2023 Contests",
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"answer_score": 298,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 Nordic/3057566.json"
} |
Find all functions $f:\mathbb{N}_0 \to \mathbb{Z}$ such that $$ f(k)-f(l) \mid k^2-l^2 $$ for all integers $k, l \geq 0$ . | If $f$ works so does $f+c$ for $c \in \mathbb Z$ so wlog $f(0)=0$ . Also $f$ is injective else LHS makes no sense.
Let $P(k,l)$ the assertion of the given F.E. $P(p,0)$ where $p$ is a prime number $$ f(p) \mid p^2 \implies f(p) \in \{-p^2, -p, -1, 1, p, p^2 \} $$ Also note that if $f$ works so does $... | [
"A bit confusing. If $f(k)=f(l)$ then the denominator doesn’t make sense so are we going to assume $f$ is injective?",
"<blockquote>If $f$ works so does $f+c$ for $c \\in \\mathbb Z$ so wlog $f(0)=0$ . Also $f$ is injective else LHS makes no sense.\nLet $P(k,l)$ the assertion of the given F.E. $P(p... | [
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"boxed": false,
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"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Nordic/3057567.json"
} |
Alice and Bianca have one hundred marbles. At the start of the game they split these hundred marbles into two piles. Thereafter, a move consists of choosing a pile, then choosing a positive integer not larger than half of the number of marbles in that pile, and finally removing that number of marbles from the chosen pi... | A slightly more symmetric way to write the solution is to say that $(a,b)$ is a losing position iff $\frac{a+1}{b+1}$ is a power by $2$ (which could be less than $1$ ). This also makes the proof quite obvious: We always change $a+1$ or $b+1$ by a factor of less than $2$ , so from a losing position we must m... | [
"<details><summary>sketch</summary>Bianca wins if the pile sizes are $(k, 2^nk+2^n-1)$ for any $k \\in Z^+$ and $n \\in Z ^{ \\ge 0}$ .\n\nIf Bianca performs a move on $(k, 2^nk+2^n-1)$ then Alice can make a move to get piles $(k, 2^{n-1}k+2^{n-1}-1)$ . Moreover if Alice does not start with $(k, 2^nk+2^n-1... | [
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"boxed": false,
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} |
Let $ABC$ be a triangle, and $M$ the midpoint of the side $BC$ . Let $E$ and $F$ be points on the sides $AC$ and $AB$ , respectively, so that $ME=MF$ . Let $D$ be the second intersection of the circumcircle of $MEF$ and the side $BC$ . Consider the lines $\ell_D$ , $\ell_E$ and $\ell_F$ through ... | suppose $\ell_D, \ell_E$ insect at J， and J' is on $\ell_D$ to use the sin form of Ceva's theorem ，we only need to prove $\begin{aligned} & \frac{\sin \angle \mathrm{FEJ}}{\sin \angle \mathrm{DEJ}} \cdot \frac{\sin \angle \mathrm{EDJ'}}{\sin \angle \mathrm{FDJ'}} \cdot \frac{\sin \angle \mathrm{DFJ}}{\sin \angle ... | [
"Let the perpenediculars from $B$ to $FD$ and from $C$ to $ED$ intersect $FD$ at $P$ and $ED$ at $Q$ , respectively, and intersect each other at $X$ . **Claim:** Points $P,Q,X,D,M$ are concyclic.\n*Proof:* Note that $\\angle XBC=90^\\circ-\\angle FDB=90^\\circ-\\angle FEM=90^\\circ-\\angle EFM=90... | [
"origin:aops",
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"2023 Nordic"
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"answer_score": 184,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 Nordic/3057570.json"
} |
Let $ABC$ be a triangle and $H$ and $D$ be the feet of the height and bisector relative to $A$ in $BC$ , respectively. Let $E$ be the intersection of the tangent to the circumcircle of $ABC$ by $A$ with $BC$ and $M$ be the midpoint of $AD$ . Finally, let $r$ be the line perpendicular to $BC$ tha... | Let $2\alpha,2\beta,2\gamma$ be the angles $\angle BAC, \angle CBA, \angle ACB,$ respectively. We first prove that $EA = ED.$ We will do so by showing that $\angle EAD = \angle EDA.$ WLOG assume $AB < AC;$ the case when $AB > AC$ is symmetrical to that of $AB < AC,$ and $AB = AC$ is impossible.
Let $X... | [
"Because $E$ is the center of the Apollonius Circle of $A$ relative to $BC$ and $AB/AC = DB/DC$ , it follows that $EA=ED$ , from which we get that $\\angle AME = 90^{\\circ} = \\angle AHE$ , thus $AMHE$ is inscribed in a circumference whose center is the midpoint of $AE$ ; call that center $N$ .\nNow, ... | [
"origin:aops",
"2023 Contests",
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"answer_score": 66,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Olimphíada/3107113.json"
} |
The Fibonacci sequence is defined by $F_1 = F_2 = 1$ and $F_{n+2} = F_{n+1}+F_n$ for every integer $n$ . A sequence $(a_n)$ of integers is said to be $\textit{phirme}$ if there is a fixed integer $k$ such that $a_n + a_{n+1} = F_{n+k}$ for all $n \geq 1$ . Show that if $(a_n)$ is a $\textit{phirme}$ s... | <span style="color:#f00">Lemma:</span> $\frac{F_{n+k-2} - a_n}{(-1)^{n-1}}$ is constant $\forall$ $n \geq 1$ .
*Proof:* It's like to prove that $\frac{F_{n+k-2} - a_n}{(-1)^{n-1}}$ $= \frac{F_{n+k-1} - a_{n+1}}{(-1)^n}$ $\Longleftrightarrow$ $F_{n+k-2} - a_n = $ $\frac{F_{n+k-1} - a_{n+1}}{(-1)}$ $\Long... | [
"Let $a_1=F_{k-1}-c$ for some integer $c$ $a_2=F_{k+1}-a_1=F_k+c$ $a_3=F_{k+2}-a_2=F_{k+1}-c$ And by induction we can prove $a_n = F_{n+k-2} + (-1)^nc$ "
] | [
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"answer_score": 26,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Olimphíada/3107116.json"
} |
Let $n$ be a positive integer. On a blackboard is a circle, and around it $n$ non-negative integers are written. Raphinha plays a game in which an operation consists of erasing a number $a$ neighboring $b$ and $c$ , with $b \geq c$ , and writing in its place $b + c$ if $b + c \leq 5a/4$ and $b - c$ othe... | [
"BUMP! BUMP! BUMP!\n"
] | [
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"boxed": false,
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"path": "Contest Collections/2023 Contests/2023 Olimphíada/3107128.json"
} | |
We say that a prime $p$ is $\textit{philé}$ if there is a polynomial $P$ of non-negative integer coefficients smaller than $p$ and with degree $3$ , that is, $P(x) = ax^3 + bx^2 + cx + d$ where $a, b, c, d < p$ , such that $$ \{P(n) | 1 \leq n \leq p\} $$ is a complete residue system modulo $p$ . Find ... | [
"If p=3k+2 or p=3, then P(x)=x^3 works; \n\nIf p=3k+1, consider S=P(1)^k+P(2)^k+...+P(p)^k. \nNote that p|1^d+2^d+...+p^d for all d=1,2,...,p-2\nSo, on one hand, S=1^k+2^k+...+p^k= 0 (modp)\nOn the other hand, by expanding P(x), S=a(1^p-1+2^p-1+...+p^p-1)= -a (modp)\nwhich contradicts with the assumption that 0<a<p... | [
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} | |
The Fibonacci sequence is defined by $F_1 = F_2 = 1$ and $F_{n+2} = F_{n+1}+F_n$ for every integer $n$ . Let $k$ be a fixed integer. A sequence $(a_n)$ of integers is said to be $\textit{phirme}$ if $a_n + a_{n+1} = F_{n+k}$ for all $n \geq 1$ . Find all $\textit{phirme}$ sequences in terms of $n$ and... | $a_n+a_{n+1}=F_{n+k}=F_{n+k-2}+F_{n+k-1}$ $\Rightarrow (a_n-F_{n+k-2})+(a_{n+1}-F_{n+k-1})=0$ Let $(b_n)$ be a sequence such that $b_n=a_n-F_{n+k-2}$ $\Rightarrow b_n+b_{n+1}=0$ $\Rightarrow b_{n+1}=-b_n$ Let $b_1=d$ then we have: $b_{2i+1}=d, b_{2i}=-d$ $\Rightarrow a_{2i+1}=F_{2i+k-1} +d,a_{2i}=F_{2i+k-1}-d... | [] | [
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} |
Let $ABCD$ be a quadrilateral circumscribed around a circle $\omega$ with center $I$ . Assume $P$ and $Q$ are distinct points and are isogonal conjugates such that $P, Q$ , and $I$ are collinear. Show that $ABCD$ is a kite, that is, it has two disjoint pairs of consecutive equal sides. | [] | [
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We say that a prime $p$ is $n$ - $\textit{rephinado}$ if $n | p - 1$ and all $1, 2, \ldots , \lfloor \sqrt[\delta]{p}\rfloor$ are $n$ -th residuals modulo $p$ , where $\delta = \varphi+1$ . Are there infinitely many $n$ for which there are infinitely many $n$ - $\textit{rephinado}$ primes?
Notes: $\var... | The answer is no. Suppose, FTSoC that there exists a $n$ satisfying the condition and let $p$ be a $n$ - $\textit{rephinado}$ prime. It's known that we have $\frac{(p-1)(n-1)}{n}$ no $n$ -th residues modulo $p$ . Since $(\sqrt[\delta]{p})^2>p$ , any integer in the interval $[1,p-1]$ has at most one prime g... | [
"This is somewhat funny. The answer is no.\nIf, for certain $n$ , there are infinitely many prime $p$ which is n-rephinado, then all the positive integer smaller than $p$ which is the product of some positive integers smaller than $p^{\\frac{1}{\\delta}}$ is a $n$ -th power residue $\\mathrm{mod}p$ .\nIt i... | [
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"2023 Contests",
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"answer_score": 194,
"boxed": false,
"end_of_proof": true,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Olimphíada/3107204.json"
} |
Let $n \geq 2023$ be an integer. For each real $x$ , we say that $\lfloor x \rceil$ is the closest integer to $x$ , and if there are two closest integers then it is the greater of the two. Suppose there is a positive real $a$ such that $$ \lfloor an \rceil = n + \bigg\lfloor\frac{n}{a} \bigg\rceil. $$ Show t... | [] | [
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We all know the Fibonacci sequence. However, a slightly less known sequence is the $k$ -bonacci sequence. In it, we have $F_1^{(k)} = F_2^{(k)} = \cdots = F_{k-1}^{(k)} = 0, F_k^{(k)} = 1$ and $$ F^{(k)}_{n+k} = F^{(k)}_{n+k-1} + F^{(k)}_{n+k-2} + \cdots + F^{(k)}_n, $$ for all $n \geq 1$ . Find all positive int... | [] | [
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"boxed": false,
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} | |
Find all $f:\mathbb{R}\rightarrow \mathbb{R}$ continuous functions such that $\lim_{x\rightarrow \infty} f(x) =\infty$ and $\forall x,y\in \mathbb{R}, |x-y|>\varphi, \exists n<\varphi^{2023}, n\in \mathbb{N}$ such that $$ f^n(x)+f^n(y)=x+y $$ | Solved with **mxlcv**
<details><summary>solution</summary>We show that the only such continous function is $f(x) = x$ for all $x \in \mathbb{R}$ . Observe that this obviously satisfies the problem's constraints. Let us now show this is the only valid function. Write $N = k!$ where $k = \lfloor \varphi^{2023} \rfl... | [
"<blockquote>Find all $f:\\mathbb{R}\\rightarrow \\mathbb{R}$ continuous functions such that $lim_{x\\rightarrow \\infty}=\\infty$ \\newline and $\\forall x,y\\in \\mathbb{R}, |x-y|>\\varphi, \\exists n<\\varphi^{2023}, n\\in \\mathbb{N}$ such that $$ f^n(x)+f^n(y)=x+y $$ </blockquote>\nWhat is $f^n(x)$ ... | [
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"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Olympic Revenge/3041154.json"
} |
Find all triples ( $a$ , $b$ , $n$ ) of positive integers such that $$ a^3=b^2+2^n $$ | @above unfair, you missed out details and now forever you'll be the first solver despite not actually finishing it, oh well
on an unrelated note, rip my sanity
<details><summary>solution which actually has all solutions</summary>First, we solve for odd $b$ and deal with the annoying case of $b$ even later. We do ca... | [
"Note that if $a^3-b^2=2^n$ then $(4a)^3-(8b)^2=2^{n+6}$ so we have infinitely many solutions.",
"the problem requires you to find all natural solutions of the equation, so simply saying that there are infinitely many of them (if there is one) is not enough!",
"<blockquote>the problem requires you to find a... | [
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"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 Olympic Revenge/3041156.json"
} |
Define a $\emph{big circle}$ in a sphere as a circle that has two diametrically oposite points of the sphere in it. Suppose $(AB)$ as the big circle that passes through $A$ and $B$ . Also, let a $\emph{Spheric Triangle}$ be $3$ connected by big circles. The angle between two circles that intersect is defined... | [] | [
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} | |
Let $S=\{(x,y,z)\in \mathbb{Z}^3\}$ the set of points with integer coordinates in the space. Gugu has infinitely many solid spheres. All with radii $\ge (\frac{\pi}2)^3$ . Is it possible for Gugu to cover all points of $S$ with his spheres? | [
"could anyone post a way of thinking?it's too strangefor me. thanks :)"
] | [
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Let $ABCD$ be a circumscribed quadrilateral and $T=AC\cap BD$ . Let $I_1$ , $I_2$ , $I_3$ , $I_4$ the incenters of $\Delta TAB$ , $\Delta TBC$ , $TCD$ , $TDA$ , respectively, and $J_1$ , $J_2$ , $J_3$ , $J_4$ the incenters of $\Delta ABC$ , $\Delta BCD$ , $\Delta CDA$ , $\Delta DAB$ . Show that $I... | [
"The problem appears [here](c6h2343104) and [here](c6h2588021).",
"@above Indeed. It is an old problem. I used much the same of its proof when doing this a year ago. \n<blockquote>P79: Let ABCD has both an incircle and a circumcircle. AC cross BD at P. Prove that the incenters of ABC,BCD,CDA,DAB,ABP,BCP,CDP,DAP a... | [
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} | |
We say that $H$ permeates $G$ if $G$ and $H$ are finite groups and for all subgroup $F$ of $G$ there is $H'\cong H$ with $H'\le F$ or $F\le H'\le G$ . Suppose that a non-abelian group $H$ permeates $G$ and let $S=\langle H'\le G | H'\cong H\rangle$ . Show that $$ |\bigcap_{H'\in S} H'|>1 $$ | [] | [
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Points $C_1$ and $C_2$ lie on side $AB$ of triangle $ABC$ , where the point $C_1$ belongs to the segment $AC_2$ and $\angle ACC_1= \angle BCC_2$ . On segments $CC_1$ and $CC_2$ points $A'$ and $B'$ are taken such that $\angle CAA'= \angle CBB' = \angle C_1CC_2$ . Prove that the center of the circ... | Invert around $C$ with factor $\sqrt{CA\cdot CB}$ , together with symmetry w.r.t. the angle bisector of $\angle C$ .
<blockquote> Points $C_1$ and $C_2$ lie on arc $BC$ of the circumcircle of triangle $\triangle ABC$ , where the point $C_1$ belongs to the arc $AC_2$ and $\angle ACC_1= \angle BCC_2$ . ... | [] | [
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In an acute-angled triangle $ABC$ with orthocenter $H$ , the line $AH$ cuts $BC$ at point $A_1$ . Let $\Gamma$ be a circle centered on side $AB$ tangent to $AA_1$ at point $H$ . Prove that $\Gamma$ is tangent to the circumscribed circle of triangle $AMA_1$ , where $M$ is the midpoint of $AC$ . | Let $B_1,C_1$ be the foots of the altitudes from $B,C$ respectively.
Invert around $A$ with factor $AH\cdot AA_1=AB_1\cdot AC=AC_1\cdot AB$ , so that $B\leftrightarrow C_1, C\leftrightarrow B_1, H\leftrightarrow A_1$ . $M$ goes to a point $M'$ , such that $B_1$ is the midpoint of $AM'$ . $\Gamma$ goes ... | [] | [
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} |
In an acute triangle $ABC$ the line $OI$ is parallel to side $BC$ . Prove that the center of the nine-point circle of triangle $ABC$ lies on the line $MI$ , where $M$ is the midpoint of $BC$ . | Let $N,L$ be the midpoints of arc $BC$ with $L\in BAC$ , $H$ the orthocenter, and $K$ the midpoint of $AH$ .
We know that $KM$ is a diameter of the nine-point circle, so we just need to prove that $I$ lies on $KM$ .
Note that $KM\parallel AO$ as $AK=KH=OM$ ( $KHMO$ is a parallelogram), so we need t... | [
"Without loss of generality assume $AB < AC$ . Let $W$ be the intersection of the angle bisector $AL$ with the circumcircle of $\\Delta ABC$ . Then $MW = OW-OM = R-r = \\sqrt{OI^2 + r^2} = MI$ and $\\measuredangle BMI = 90^\\circ - 2\\measuredangle AWO = \\angle A + 2\\angle C - 90^\\circ$ . Let $AH_A$ a... | [
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} |
Points $X_1$ and $X_2$ move along fixed circles with centers $O_1$ and $O_2$ , respectively, so that $O_1X_1 \parallel O_2X_2$ . Find the locus of the intersection point of lines $O_1X_2$ and $O_2X_1$ . | [] | [
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In triangle ABC $\angle ABC=60^{o}$ and $O$ is the center of the circumscribed circle. The bisector $BL$ intersects the circumscribed circle at the point $W$ . Prove that $OW$ is tangent to $(BOL)$ | The inversion around $W$ fixing $A$ , $C$ and $O$ maps $L$ to $B$ , hence $WL \cdot WB = WO^2$ . | [] | [
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Given a circle $\Omega$ tangent to side $AB$ of angle $\angle BAC$ and lying outside this angle. We consider circles $w$ inscribed in angle $BAC$ . The internal tangent of $\Omega$ and $w$ , different from $AB$ , touches $w$ at a point $K$ . Let $L$ be the point of tangency of $w$ and $AC$ . Prove... | [] | [
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Altitudes $BB_1$ and $CC_1$ of acute triangle $ABC$ intersect at $H$ , and $\angle A = 60^{o}$ , $AB < AC$ . The median $AM$ intersects the circumcircle of $ABC$ at point $K$ ; $L$ is the midpoint of the arc $BC$ of the circumcircle that does not contain point $A$ ; lines $B_1C_1$ and $BC$ inte... | Did you mean $\angle EHL = \angle ABK + 30^\circ$ ?
Let $O$ be the circumcenter of $\Delta ABC$ and $A_1$ be the foot of the remaining altitude. Then $L$ is the center of the circle containing points $B$ , $H$ , $O$ , $C$ in this order. Since $HA_1 \parallel OL \perp BC$ , by trivial angle chase this im... | [] | [
"origin:aops",
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"boxed": false,
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Let $I$ be the incenter of triangle $ABC$ , tangent to sides $AB$ and $AC$ at points $E$ and $F$ , respectively. The lines through $E$ and $F$ parallel to $AI$ intersect lines $BI$ and $CI$ at points $P$ and $Q$ , respectively. Prove that the center of the circumcircle of triangle $IPQ$ lies o... | Cool problem.
<details><summary>solution</summary>Rephrasing in terms of $IBC$ and after a few initial observations, the problem becomes as follows:
<blockquote>
Let $ABC$ be a triangle with circumcenter $O$ and foot $D$ from $A$ to $BC$ . Say $X \in AB, Y \in AC$ such that $DX // OB, DY // OC$ . Prove t... | [
"By counting angles, it is easy to prove that the triangles $EIP$ and $FIQ$ are similar. Let $D$ be the tangent point of the inscribed circle with $BC.$ By counting, it can be shown that the $DPEI$ and $DIFQ$ quadrilaterals differ in a rotary homotety at point $D,$ therefore, with this rotary homotety... | [
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"boxed": false,
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"path": "Contest Collections/2023 Contests/2023 Oral Moscow Geometry Olympiad/3057105.json"
} |
Given is a triangle $ABC$ and $M$ is the midpoint of the minor arc $BC$ . Let $M_1$ be the reflection of $M$ with respect to side $BC$ . Prove that the nine-point circle bisects $AM_1$ . | My solution: Let $H$ be the orthocenter of $ABC$ . Then, with symmetry with respect to $BC$ , the circle $(ABC)$ turns into $(BHC),$ so $BHM_1C$ is inscribed. Now the statement of the problem is obviously obtained by a homothetic with the center $A$ and the coefficient $1/2$ . In this case, the circle $(B... | [] | [
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"boxed": false,
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} |
In trapezoid $ABCD$ with bases $AD, BC$ , $AD = 2BC$ and $M$ is midpoint of $AB$ . Prove that line $BD$ passes through the midpoint of segment $CM$ . | Continue $AB$ and $DC$ to their intersection at point $E$ and use Menelaus for $\Delta ECM$ and the points $B$ , $D$ , $N$ , where $N$ is the midpoint of $MC$ . | [] | [
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There is a square sheet of paper. How to get a rectangular sheet of paper with an aspect ratio equal to $\sqrt2$ ? (There are no tools, the sheet can only be bent.) | [] | [
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Given isosceles tetrahedron $PABC$ (faces are equal triangles). Let $A_0$ , $B_0$ and $C_0$ be the touchpoints of the circle inscribed in the triangle $ABC$ with sides $BC$ , $AC$ and $AB$ respectively, $A_1$ , $B_1$ and $C_1$ are the touchpoints of the excircles of triangles $PCA$ , $PAB$ and $P... | [] | [
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In a triangle $ABC$ with $AB<AC$ , $D$ is a point on segment $AC$ such that $BD = CD$ .
A line parallel to $BD$ meets segment $BC$ at $E$ and line $AB$ at $F$ . Point $G$ is the intersection of $AE$ and $BD$ .
Show that $\angle BCG = \angle BCF$ . *(Côte d’Ivoire)*
| another solution.
let $X$ be a point on $BC$ such that $\measuredangle CYF=\measuredangle DBC$ , let $Y$ be a point on line $XF$ such that $\measuredangle XCY=\measuredangle CXY$ $$ \boxed{\textbf{Claim:}\triangle FYC\sim\triangle AGD} $$ by doing some work we find that the claim is true | [
"Ok, this is unnecessarily long. In the actual contest, I found a solution using moving points, had some time left, scratched a figure on a napkin, and found this solution :\n\nConsider $f$ the reflection over line $BC$ .\n\nWhat we want to show is: $f(CF)=CG$ . Let $f(X)=X’$ . We have $F’ \\in (BA’)$ and $... | [
"origin:aops",
"2023 Pan-African",
"2023 Contests"
] | {
"answer_score": 1016,
"boxed": false,
"end_of_proof": true,
"n_reply": 13,
"path": "Contest Collections/2023 Contests/2023 Pan-African/3073901.json"
} |
Find all positive integers $m$ and $n$ with no common divisor greater than 1 such that $m^3 + n^3$ divides $m^2 + 20mn + n^2$ . *(Professor Yongjin Song)* | <blockquote> $m^3 + n^3 = (m+n)(m^2 - mn + n^2)$ , so $m^2 - mn + n^2 \mid m^2 + 20mn + n^2$ , which implies $m^2 - mn + n^2 \mid m^2 + 20mn + n^2 - (m^2 - mn + n^2) = 21mn$ .
Now notice that $m^2 - mn + n^2$ is relatively prime to $mn$ , so it must divide $21$ </blockquote>
Indeed but even though it is triv... | [
"Yeah, no. $m+n$ divides $18$ , and do ugly case work. Solutions are $(1,1);(1,2);(1,5)$ ",
"<blockquote>Yeah, no. $m+n$ divides $18$ , and do ugly case work. Solutions are $(1,1);(1,2);(1,5)$ </blockquote>\n\nYou can also add that $m^2-mn+n^2$ divides $21$ , To reduce number of cases",
"Can you expl... | [
"origin:aops",
"2023 Pan-African",
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"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 Pan-African/3073906.json"
} |
Consider a sequence of real numbers defined by:
\begin{align*}
x_{1} & = c
x_{n+1} & = cx_{n} + \sqrt{c^{2} - 1}\sqrt{x_{n}^{2} - 1} \quad \text{for all } n \geq 1.
\end{align*}
Show that if $c$ is a positive integer, then $x_{n}$ is an integer for all $n \geq 1$ . *(South Africa)* | <blockquote>Consider a sequence of real numbers defined by:
\begin{align*}
x_{1} & = c
x_{n+1} & = cx_{n} + \sqrt{c^{2} - 1}\sqrt{x_{n}^{2} - 1} \quad \text{for all } n \geq 1.
\end{align*}
Show that if $c$ is a positive integer, then $x_{n}$ is an integer for all $n \geq 1$ . *(South ... | [
"By induction, $x_k = T_k(x_1)$ where $T_n(x)$ is the $n$ th Chebyshev polynomial. Since $x_k$ is an integer polynomial in $x_1 = c$ , and $c$ is an integer, so is $x_k$ .",
"Nice solution! With this solution, using the characteristical equation we can find the exact form of the general therm of this s... | [
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"2023 Pan-African",
"2023 Contests"
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"answer_score": 30,
"boxed": false,
"end_of_proof": false,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 Pan-African/3073909.json"
} |
Let $a, b$ be reals with $a \neq 0$ and let $$ P(x)=ax^4-4ax^3+(5a+b)x^2-4bx+b. $$ Show that all roots of $P(x)$ are real and positive if and only if $a=b$ . | <blockquote>Let $a, b$ be reals with $a \neq 0$ and let $$ P(x)=ax^4-4ax^3+(5a+b)x^2-4bx+b. $$ Show that all roots of $P(x)$ are real and positive if and only if $a=b$ .</blockquote>
Let $k_1,k_2,k_3,k_4$ be the solutions of the polynomial $P(x)$ $$ a(x-k_1)(x-k_2)(x-k_3)(x-k_4)=(x)=ax^4-4ax^3+(5a+b)x^... | [
"<blockquote>Let $a, b$ be reals with $a \\neq 0$ and let $$ P(x)=ax^4-4ax^3+(5a+b)x^2-4bx+b. $$ Show that all roots of $P(x)$ are real and positive if and only if $a=b$ .</blockquote>\n\nWe will rather consider $P(x)=x^4-4x^3+(5+c)x^2-4cx+c$ .\nIf $c=1$ , then $P(x)=(x-1)^4$ which has only positive ... | [
"origin:aops",
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] | {
"answer_score": 1162,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 Pan-African/3077077.json"
} |
Manzi has $n$ stamps and an album with $10$ pages. He distributes the $n$ stamps in the album such that each page has a distinct number of stamps. He finds that, no matter how he does this, there is always a set of $4$ pages such that the total number of stamps in these $4$ pages is at least $\frac{n}{2}$ . ... | I’ll vent here to bump this. I did P5 in 10 minutes, then spent 3 effing hours on this cursed problem guessing bounds all of which turned out to be incorrect, didn’t even try P6 (or as I’ve learned some call Q6) which was way easier, and missed gold by 1 point. Talk about luck.
After contest, I found out the what I wa... | [
"<blockquote>I’ll vent here to bump this. I did P5 in 10 minutes, then spent 3 effing hours on this cursed problem guessing bounds all of which turned out to be incorrect, didn’t even try P6 (or as I’ve learned some call Q6) which was way easier, and missed gold by 1 point. Talk about luck.\n\nAfter contest, I foun... | [
"origin:aops",
"2023 Pan-African",
"2023 Contests"
] | {
"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Pan-African/3077078.json"
} |
Let $ABC$ be an acute triangle with $AB<AC$ . Let $D, E,$ and $F$ be the feet of the perpendiculars from $A, B,$ and $C$ to the opposite sides, respectively. Let $P$ be the foot of the perpendicular from $F$ to line $DE$ . Line $FP$ and the circumcircle of triangle $BDF$ meet again at $Q$ . Show th... | Let $AD$ and $FP$ intersect at $G$ . Let $DE$ intersect $AB$ at a point $S$ . It stands that $\mathcal{H}(A,F,B,S)$ , so $AF/BF = AT/BT$ . Since $\angle FPT = 90^{\circ}$ , we know that $AP/BP = AF/BF$ so by the internal angle bisector theorem we have that $\angle APG = \angle BPQ \quad ( \star )$ .
We ... | [
"repeated sol>>",
"<blockquote>Let $AD$ and $FP$ intersect at $G$ . Let $DE$ intersect $AB$ at a point $S$ . It stands that $\\mathcal{H}(A,F,B,S)$ , so $AF/BF = AT/BT$ . Since $\\angle FPT = 90^{\\circ}$ , we know that $AP/BP = AF/BF$ so by the internal angle bisector theorem we have that $\\angl... | [
"origin:aops",
"2023 Pan-African",
"2023 Contests"
] | {
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Pan-African/3077080.json"
} |
Find all ordered pairs $(a, b)$ of positive integers such that $a^2 + b^2 + 25 = 15ab$ and $a^2 + ab + b^2$ is prime. | <blockquote>Find all ordered pairs $(a, b)$ of positive integers such that $a^2 + b^2 + 25 = 15ab$ and $a^2 + ab + b^2$ is prime.</blockquote> $17(a^2+b^2+ab)=(4a+4b+5)(4a+4b-5)+a^2+b^2+25-15ab$ $=(4a+4b+5)(4a+4b-5)$ And so $(4a+4b+5)(4a+4b-5)=17p$ for some prime $p$ and so :
Either $4a+4b-5=1$ , and so $... | [
"<blockquote>Find all ordered pairs $(a, b)$ of positive integers such that $a^2 + b^2 + 25 = 15ab$ and $a^2 + ab + b^2$ is prime.</blockquote>\nAre you sure this is the exact problem ?\n\nFirst equation $a^2 + b^2 + 25 = 15ab$ is a Pell equation with infinitely many solutions (starting with the trivial $(... | [
"origin:aops",
"2023 Philippine MO",
"2023 Contests"
] | {
"answer_score": 1032,
"boxed": true,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 Philippine MO/3035165.json"
} |
Find all primes $p$ such that $\dfrac{2^{p+1}-4}{p}$ is a perfect square. | The answer is $\boxed{p=3,7}$ . | [
"If $p=2$ , it is not true. now we let $p\\geq 3$ . $\\frac{2^{p+1}-4}{p}=\\frac{4(2^{\\frac {p-1}2}-1)(2^{\\frac {p-1}2}+1)}{p}\\Rightarrow 2^{\\frac {p-1}2}-1$ or $2^{\\frac {p-1}2}+1$ is a perfect square.\n------\nIf $2^{\\frac {p-1}2}-1$ is a perfect square, it is obviously true when $p=3$ .\nNow let $... | [
"origin:aops",
"2023 Philippine MO",
"2023 Contests"
] | {
"answer_score": 1002,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Philippine MO/3035166.json"
} |
In $\triangle ABC$ , $AB > AC$ . Point $P$ is on line $BC$ such that $AP$ is tangent to its circumcircle. Let $M$ be the midpoint of $AB$ , and suppose the circumcircle of $\triangle PMA$ meets line $AC$ again at $N$ . Point $Q$ is the reflection of $P$ with respect to the midpoint of segment $BC$ ... | [] | [
"origin:aops",
"2023 Philippine MO",
"2023 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 0,
"path": "Contest Collections/2023 Contests/2023 Philippine MO/3035167.json"
} | |
In chess, a knight placed on a chess board can move by jumping to an adjacent square in one direction (up, down, left, or right) then jumping to the next two squares in a perpendicular direction. We then say that a square in a chess board *can be attacked* by a knight if the knight can end up on that square after a mov... | <details><summary>Solution</summary>The answer is $\boxed{16}$ . $16$ knights can be placed as follows, where a blue square represents a square with a knight:
[asy]
size(150);
int n = 10; // Number of rows/columns
real sq_size = 1;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
// Draw grid squar... | [
"<span style=\"color:#f00\">Incorrect conjecture</span>"
] | [
"origin:aops",
"2023 Philippine MO",
"2023 Contests"
] | {
"answer_score": 1062,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Philippine MO/3035168.json"
} |
Silverio is very happy for the 25th year of the PMO. In his jubilation, he ends up writing a finite sequence of As and Gs on a nearby blackboard. He then performs the following operation: if he finds at least one occurrence of the string "AG", he chooses one at random and replaces it with "GAAA". He performs this opera... | Fix the number $n$ of $G$ ’s in the string; this is an invariant during the entirety of the operation. In any intermediate string, label the letters from left to right by $1,2,\dots,m$ , and let
\[
1 \le p_1 < p_2 < \cdots < p_n \le m
\]
be the positions of the $G$ ’s. Consider the sequence $(p_1,p_2,\dots,p_n)$ ... | [
"Sketch: note the number of Gs don't change, label them $G_1,\\cdots G_n$ from left to right. Note their order never swaps. We prove by induction that the number of moves on each G is bounded. For $G_1$ it is obvious as it would keep moving towards the front. Suppose it is true for $G_{i-1}$ . Then, it never c... | [
"origin:aops",
"2023 Philippine MO",
"2023 Contests"
] | {
"answer_score": 34,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Philippine MO/3035169.json"
} |
Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$ f(2f(x)) = f(x - f(y)) + f(x) + y $$ for all $x, y \in \mathbb{R}$ . | <blockquote>Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$ f(2f(x)) = f(x - f(y)) + f(x) + y $$ for all $x, y \in \mathbb{R}$ .</blockquote>
Let $P(x,y)$ be the assertion $f(2f(x))=f(x-f(y))+f(x)+y$ $f(x)$ is injective (look at $y$ ) $P(x,-f(0)-f(x)+f(2f(x)))$ $\implies$ $f(x-f... | [
"<details><summary>solution</summary>Injective by plugging $P(x,a),P(x,b)$ . $P(x,-f(x)) \\implies f(-f(x))=x-2f(x)$ . $P(0,y) \\implies 2f(y)=2y+f(0)-f(2f(0)) \\implies f(x)=x+c$ for all $x$ . Plugging back gives $c=0$ .</details>",
"<details><summary>Hopefully right</summary>We can prove function is surje... | [
"origin:aops",
"2023 Philippine MO",
"2023 Contests"
] | {
"answer_score": 1046,
"boxed": true,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 Philippine MO/3035170.json"
} |
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