problem string | solution string | candidates list | tags list | metadata dict |
|---|---|---|---|---|
Given is a tree $G$ with $2023$ vertices. The longest path in the graph has length $2n$ . A vertex is called good if it has degree at most $6$ . Find the smallest possible value of $n$ if there doesn't exist a vertex having $6$ good neighbors. | [] | [
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"2023 Contests",
"2023 Bulgarian Spring Mathematical Competition"
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"path": "Contest Collections/2023 Contests/2023 Bulgarian Spring Mathematical Competition/3039503.json"
} | |
A positive integer $b$ is called good if there exist positive integers $1=a_1, a_2, \ldots, a_{2023}=b$ such that $|a_{i+1}-a_i|=2^i$ . Find the number of the good integers. | I claim the answer is $2^{2021}$ .
Note that $b$ is of form
\[
b=1+\epsilon_1 2^1 + \epsilon_2 2^2 + \cdots + \epsilon_{2022}2^{2022},\quad\text{where}\quad \epsilon_i \in\{-1,1\}.
\]
Note that as $b>0$ , we must have $\epsilon_{2022}=1$ . Next I show any choice of $\epsilon_1,\dots,\epsilon_{2021}$ induces a d... | [] | [
"origin:aops",
"2023 Contests",
"2023 Bulgarian Spring Mathematical Competition"
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"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Bulgarian Spring Mathematical Competition/3039506.json"
} |
Given is a cyclic quadrilateral $ABCD$ and a point $E$ lies on the segment $DA$ such that $2\angle EBD = \angle ABC$ . Prove that $DE= \frac {AC.BD}{AB+BC}$ . | Let $F$ be the point on ray $DA$ such that $DF=DC$ . Let $G$ be the intersection between $BD$ and $CF$ .**Claim 1:** $BEFG$ is cyclic.**Proof:** By angle chasing, $$ \angle GFD=\frac{180^o-\angle FDC}{2}=\frac{\angle ABC}{2}=\angle EBG \square $$ **Claim 2:** The intersection $H$ of $BE$ and $CF$ li... | [
"Let $BE\\cap (ABCD)=F$ , the angle bisector of $CBA$ meet $AC$ at $G$ , and $CF$ meet $AD$ at $T$ .\nBy angle chasing, $\\angle DCT=\\angle DBF=\\frac12\\angle CBA=90-\\frac12 \\angle CDT$ , hence $CD=TD$ .\nAlso, $\\angle GAE=\\angle CAD=\\angle CBD=\\angle GBE$ , so $BAGE$ is cyclic. \nThus, $\\... | [
"origin:aops",
"2023 Contests",
"2023 Bulgarian Spring Mathematical Competition"
] | {
"answer_score": 164,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Bulgarian Spring Mathematical Competition/3039510.json"
} |
Find all real $a$ such that the equation $3^{\cos (2x)+1}-(a-5)3^{\cos^2(2x)}=7$ has a real root.
<details><summary>Remark</summary>This was the statement given at the contest, but there was actually a typo and the intended equation was $3^{\cos (2x)+1}-(a-5)3^{\cos^2(x)}=7$ , which is much easier.</details> | <blockquote>Find all real $a$ such that the equation $3^{\cos (2x)+1}-(a-5)3^{\cos^2(2x)}=7$ has a real root.</blockquote>
So we are looking for the range of $a=5+\frac{3^{1+\cos 2x}-7}{3^{\cos^22x}}$ over $\mathbb R$ And so the range of $f(x)=5+\frac{3^{1+x}-7}{3^{x^2}}$ over $[-1,+1]$ $f(x)=5+3^{1+x-x^2}-7... | [] | [
"origin:aops",
"2023 Contests",
"2023 Bulgarian Spring Mathematical Competition"
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"answer_score": 1046,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Bulgarian Spring Mathematical Competition/3039515.json"
} |
Find all positive integers $n$ , such that there exists a positive integer $m$ and primes $1<p<q$ such that $q-p \mid m$ and $p, q \mid n^m+1$ . | Well claim that the answer is all odd $n>1$ . Clearly, $n=1$ fails. We first handle evens.
By way of contradiction assume that there exists an even integer $n$ such that $q-p \mid m$ and $p,q\mid n^m+1$ . If $2 \mid n$ then, $ 2 \nmid n^m+1$ . But then, both $p$ and $q$ are odd primes, which implies tha... | [
"Very nice. I claim the answer are all odd $n>1$ . Clearly $n\\ne 1$ , and let $n>1$ be an odd number. We take $p=2$ and $q$ such that $q\\mid n^2+1$ and $q$ is prime. As $n^2+1\\equiv 2\\pmod{4}$ and $n^2+1>2$ , such a $q$ indeed exists. With these choices, we take $m$ such that $m\\equiv 0\\pm... | [
"origin:aops",
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"answer_score": 76,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Bulgarian Spring Mathematical Competition/3039518.json"
} |
Given is a convex octagon $A_1A_2 \ldots A_8$ . Given a triangulation $T$ , one can take two triangles $\triangle A_iA_jA_k$ and $\triangle A_iA_kA_l$ and replace them with $\triangle A_iA_jA_l$ and $\triangle A_jA_lA_k$ . Find the minimal number of operations $k$ we have to do so that for any pair of triang... | [] | [
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"2023 Contests",
"2023 Bulgarian Spring Mathematical Competition"
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"answer_score": 0,
"boxed": false,
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"n_reply": 0,
"path": "Contest Collections/2023 Contests/2023 Bulgarian Spring Mathematical Competition/3039523.json"
} | |
An isosceles $\triangle ABC$ has $\angle BAC =\angle ABC =72^{o}$ . The angle bisector $AL$ meets the line through $C$ parallel to $AB$ at $D$ . $a)$ Prove that the circumcenter of $\triangle ADC$ lies on $BD$ . $b)$ Prove that $\frac {BE} {BL}$ is irrational. | <details><summary>bash too much for a</summary>WLOG $A(0,0) , B(2,0) ,C(1,n)$ notice CAB = 72 then by angle inclination $Grad_{AC}=\tan(72)$ , $Grad_{AD}=\tan(36)$ thus $C(1,n) , C(1,\frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1})$ equation line AD $ \frac{\sqrt{10+2\sqrt{5}}}{\sqrt{5}-1} = \frac{\sqrt{10-2\sqrt{5}}}{\s... | [
"For this solution you would have gotten 5/6 cause you haven't explicitly proven y is irrational. "
] | [
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"answer_score": 30,
"boxed": false,
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"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Bulgarian Spring Mathematical Competition/3039525.json"
} |
Given a prime $p$ , find $\gcd(\binom{2^pp}{1},\binom{2^pp}{3},\ldots, \binom{2^pp}{2^pp-1}) $ . | If $p=2$ then $\gcd( \binom{8}{1},..., \binom{8}{7})= 8$ If $p>2$ $\binom{2^pp}{1}=2^p *p$ For odd $n \neq k$ and $n \leq 2^p p-1 $ we have that $n\not | 2^p* p $ so $\frac{\binom{2^pp}{n}}{2^p* p}$ is integer
So enough to find $\gcd( 2^p p, \binom{2^pp}{p})$ and easy to show that $2^p | \binom{2^pp}{p... | [
"What does k mean in this solution?"
] | [
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"answer_score": 24,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Bulgarian Spring Mathematical Competition/3039528.json"
} |
Given is a directed graph with $28$ vertices, such that there do not exist vertices $u, v$ , such that $u \rightarrow v$ and $v \rightarrow u$ . Every $16$ vertices form a directed cycle. Prove that among any $17$ vertices, we can choose $15$ which form a directed cycle. | [] | [
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"path": "Contest Collections/2023 Contests/2023 Bulgarian Spring Mathematical Competition/3039531.json"
} | |
Given is triangle $ABC$ with angle bisector $CL$ and the $C-$ median meets the circumcircle $\Gamma$ at $D$ . If $K$ is the midpoint of arc $ACB$ and $P$ is the symmetric point of $L$ with respect to the tangent at $K$ to $\Gamma$ , then prove that $DLCP$ is cyclic. | Nice problem,
Let $CD\cap AB=S$ , $KP\cap AB=T$ ,
Note that $2SK=LP$ , $LP\bot AB$ ,so $TS=SL$ ,
We have $LC\bot CK$ ,so $SLCK$ is cyclic
Which means that $\angle TKS = \angle SKL =\angle SCL=\angle SCB- \dfrac{1}{2}\angle C$ Since $\measuredangle (DK,AB)=90^{\circ}- \dfrac{1}{2}\angle C+\angle SCB$ So $\meas... | [] | [
"origin:aops",
"2023 Contests",
"2023 Bulgarian Spring Mathematical Competition"
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"answer_score": 24,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Bulgarian Spring Mathematical Competition/3039534.json"
} |
Tick, Trick and Track have 20, 23 and 25 tickets respectively for the carousel at the fair in Duckburg. They agree that they will only ride all three together, for which they must each give up one of their tickets. Also, before a ride, if they want, they can redistribute tickets among themselves as many times as they w... | [
"No solution?"
] | [
"origin:aops",
"2023 Bundeswettbewerb Mathematik",
"2023 Contests"
] | {
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Bundeswettbewerb Mathematik/3028087.json"
} | |
Given two parallelograms $ABCD$ and $AECF$ with common diagonal $AC$ , where $E$ and $F$ lie inside parallelogram $ABCD$ . Show: The circumcircles of the triangles $AEB$ , $BFC$ , $CED$ and $DFA$ have one point in common. | <details><summary>solution 1</summary>Write $C=-A$ , $D=-B$ and $F=-E$ (so their common midpoint is $0$ ) and define
\begin{align*}
P&=+A+E-B
Q&=-A+E+B
R&=-A-E-B
S&=+A-E+B
\end{align*}
It is easy to check that $A$ , $B$ , $C$ , $D$ , $E$ und $F$ are the midpoints of $PS$ , $QS$ , $QR$ ... | [
"**Idea:** Inversion at circle with midpoint $C$ ."
] | [
"origin:aops",
"2023 Bundeswettbewerb Mathematik",
"2023 Contests"
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"answer_score": 110,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Bundeswettbewerb Mathematik/3028088.json"
} |
Given a real number $\alpha$ in whose decimal representation $\alpha=0,a_1a_2a_3\dots$ each decimal digit $a_i$ $(i=1,2,3,\dots)$ is a prime number. The decimal digits are arranged along the path indicated by arrows in the accompanying figure, which can be thought of as continuing infinitely to the right and do... | [] | [
"origin:aops",
"2023 Bundeswettbewerb Mathematik",
"2023 Contests"
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"path": "Contest Collections/2023 Contests/2023 Bundeswettbewerb Mathematik/3029115.json"
} | |
Determine all triples $(x, y, z)$ of integers that satisfy the equation $x^2+ y^2+ z^2 - xy - yz - zx = 3$ | we multiply with two and observe that it comes down as $(x-y)^2+(y-z)^2+(z-x)^2=6$ which gives us $x-y=2 ,x-z=1$ and $y-z=-1$ and $x-y=-2,x-z=-1$ and $y-z=-1$ which gives $(x,y,z) \in \boxed{\{(k,k+1,k+2),(k-2,k-1,k)\} , k \in \mathbb{Z}}$ and it's permutations $\blacksquare$ | [
"Sol.\n\n<details><summary>Click to expand</summary>$\\Longrightarrow 2x^2+2y^2+2z^2-2xy-2xz-2yz= (x-y)^2+(x-z)^2+(y-z)^2=6=1+1+4$ $\\Longrightarrow (x,y,z)=(t,t+1,t+2), (t,t-1,t-2),~~~ t\\in Z_0$ and permutations thereof.</details>",
"<details><summary>alt sol</summary>$x^2+y^2+z^2-xy-yz-zx=(y-x)^2+(z-y)(z-x)=... | [
"origin:aops",
"2023 Bundeswettbewerb Mathematik",
"2023 Contests"
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"answer_score": 1114,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Bundeswettbewerb Mathematik/3029135.json"
} |
Determine the greatest common divisor of the numbers $p^6-7p^2+6$ where $p$ runs through the prime numbers $p \ge 11$ . | My try:
<details><summary>Click to expand</summary>$f(p)=p^6-7p^2+6$ $f(\pm1)=0\Rightarrow p^6-7p^2+6=(p+1)(p-1)(p^4+p^2-6)$ $p^2=X\rightarrow p^4+p^2-6=X^2+X-6=(X+3)(X-2)$ $\Longrightarrow p^6-7p^2+6=(p-1)(p+1)(p^2+3)(p^2-2)$ $p=4k+1\Longrightarrow (p-1)(p+1)(p^2+3)(p^2-2)=32k(2k+1)(4k^2+2k+1)(16k^2+8k-1)$ $p=4k... | [
"Factoring is a good idea, but shouldn't it be <details><summary>factorization</summary>$(p-1)(p+1)(p^2+3)(p^2-2)$</details>?",
"<blockquote>Factoring is a good idea, but shouldn't it be <details><summary>factorization</summary>$(p-1)(p+1)(p^2+3)(p^2-2)$</details>?</blockquote>\n\nYes, then observe that $(p-1)(p... | [
"origin:aops",
"2023 Bundeswettbewerb Mathematik",
"2023 Contests"
] | {
"answer_score": 22,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 Bundeswettbewerb Mathematik/3151632.json"
} |
A hilly island has $2023$ lookouts. It is known that each of them is in line of sight with at least $42$ of the other lookouts. For any two distinct lookouts $X$ and $Y$ there is a positive integer $n$ and lookouts $A_1,A_2,\dots,A_{n+1}$ such that $A_1=X$ and $A_{n+1}=Y$ and $A_1$ is in line of sight... | I think, the correct condition is $A_{n+1}=Y$ instead $A_2=Y$ .
Let be $M$ the maximum possible value of the viewing distance between two lookouts and $A_1, A_2,...,A_M, A_{M+1}$ a sequence of lookouts with the property $A_1$ is in line of sight with $A_2$ , $A_2$ with $A_3$ , $\dots$ and $A_M$ with ... | [
"Answer is here: [ https://www.sciencedirect.com/science/article/pii/009589568990066X?ref=pdf_download&fr=RR-2&rr=7f15e1919aca90e0]( https://www.sciencedirect.com/science/article/pii/009589568990066X?ref=pdf_download&fr=RR-2&rr=7f15e1919aca90e0)"
] | [
"origin:aops",
"2023 Bundeswettbewerb Mathematik",
"2023 Contests"
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"answer_score": 156,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Bundeswettbewerb Mathematik/3151633.json"
} |
Let $ABC$ be a triangle with incenter $I$ . Let $M_b$ and $M_a$ be the midpoints of $AC$ and $BC$ , respectively. Let $B'$ be the point of intersection of lines $M_bI$ and $BC$ , and let $A'$ be the point of intersection of lines $M_aI$ and $AC$ .
If triangles $ABC$ and $A'B'C$ have the same ar... | <blockquote>Let $ABC$ be a triangle with incenter $I$ . Let $M_b$ and $M_a$ be the midpoints of $AC$ and $BC$ , respectively. Let $B'$ be the point of intersection of lines $M_bI$ and $BC$ , and let $A'$ be the point of intersection of lines $M_aI$ and $AC$ .
If triangles $ABC$ and $A'B'C$ have... | [
"We use barycentric coordinates: $A = (1:0:0), \\ B = (0:1:0), \\ M_a = (0:1:1), \\ M_b = (1:0:1)$ . Let $CA' = t$ , then $A'A = b-t$ , therefore $A' = (t:0:b-t)$ .\n\nNow, $I = (a:b:c)$ is the incentreand we have that $M_a, I, A'$ are collinear, therefore, $$ \\begin{vmatrix}\na & b &c\n0 & 1 & 1\nt & ... | [
"origin:aops",
"2023 Bundeswettbewerb Mathematik",
"2023 Contests"
] | {
"answer_score": 36,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Bundeswettbewerb Mathematik/3151634.json"
} |
Exactly $n$ chords (i.e. diagonals and edges) of a regular $2n$ -gon are coloured red, satisfying the following two conditions:
(1) Each of the $2n$ vertices occurs exactly once as the endpoint of a red chord.
(2) No two red chords have the same length.
For which positive integers $n \ge 2$ is this possible? | <blockquote>Exactly $n$ chords (i.e. diagonals and edges) of a regular $2n$ -gon are coloured red, satisfying the following two conditions:
(1) Each of the $2n$ vertices occurs exactly once as the endpoint of a red chord.
(2) No two red chords have the same length.
For which positive integers $n \ge 2$ is this... | [
"That's correct. :)",
"Was that apparent to the PSC? Because that problem (among many more) with its solution was sent to almost everyone preceeding the last Bundesrunde for preparation. So anyone that received that file would have been capable to just copy the solution. :(",
"I am pretty sure that it was not o... | [
"origin:aops",
"2023 Bundeswettbewerb Mathematik",
"2023 Contests"
] | {
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Bundeswettbewerb Mathematik/3151635.json"
} |
William is thinking of an integer between 1 and 50, inclusive. Victor can choose a positive integer $m$ and ask William: "does $m$ divide your number?", to which William must answer truthfully. Victor continues asking these questions until he determines William's number. What is the minimum number of questions that... | <blockquote><blockquote>Because $2^6 > 50$ , the sum of the exponents can't be larger than $5$ , so he won't need to test all powers (for example if $2$ doesn't divide $m$ , he won't test $2^2, 2^3$ , etc.
Also, if $2$ divides, he doesn't need to test prime numbers larger than $23$ , or else $m>50$ .</blockq... | [
" $15$ , basically ask prime powers as follows:\n\n - if currently $p^k$ divides, ask $p^{k+1}$ - if does not divide, ask next prime\nstart with $2$ Worst case w/ this algorithm is when number equals $1$ , with $15$ questions asked",
"Induct to show that answer with $50$ replaced with $n$ is $\\pi(n)$... | [
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"2023 Canada National Olympiad",
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"answer_score": 34,
"boxed": false,
"end_of_proof": false,
"n_reply": 19,
"path": "Contest Collections/2023 Contests/2023 Canada National Olympiad/3029959.json"
} |
There are 20 students in a high school class, and each student has exactly three close friends in the class. Five of the students have bought tickets to an upcoming concert. If any student sees that at least two of their close friends have bought tickets, then they will buy a ticket too.
Is it possible that the entire... | *<span style="color:#f0f">**<u>(Answer)= No it's not possible</u>**</span>*
Denote set of all students who don't have ticket to be $\mathcal{F}$ . Call friend pair $(\Omega , \mho)$ <span style="color:#f00">**partial induced substructure**</span> if $\Omega$ is having ticket and $\mho$ not. And for a person ... | [
"No.\n\nColor vertex if buying ticket. Connect edge if friends. Note quantity (colored vertices) + (edge connecting a colored with a noncolored vertex) remains nonincreasing while updating color of vertices. Quantity starts at $20$ , must end at $22$ right before the last guy gets updated (if all can be colored)... | [
"origin:aops",
"2023 Canada National Olympiad",
"2023 Contests"
] | {
"answer_score": 138,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2023 Contests/2023 Canada National Olympiad/3029960.json"
} |
An acute triangle is a triangle that has all angles less than $90^{\circ}$ ( $90^{\circ}$ is a Right Angle). Let $ABC$ be an acute triangle with altitudes $AD$ , $BE$ , and $CF$ meeting at $H$ . The circle passing through points $D$ , $E$ , and $F$ meets $AD$ , $BE$ , and $CF$ again at $X$ , $Y$ , a... | Solution using ptolemy theorem :)
Let $A'=AH \cap (ABC)$ and define $B',C'$ similarly
Note $\triangle A'B'C' \sim \triangle DEF$ with ratio $2$ .
As $X,Y,Z$ are midpoint of $AH,BH,CH,$ (As they lie on Nine point circle) we need to prove $$ \frac{AH}{AA'}+\frac{BH}{BB'}+\frac{CH}{CC'} \geq \frac{3}{2} ... | [
" $\\frac{AH}{DX} = \\frac{2AH}{AH_A} = \\frac{2H_BH_C}{H_A H_B + H_AH_B}$ and cyclic variants. Result easily follows",
"<blockquote>An acute triangle is a triangle that has all angles less than $90^{\\circ}$ ( $90^{\\circ}$ is a Right Angle). Let $ABC$ be an acute triangle with altitudes $AD$ , $BE$ , an... | [
"origin:aops",
"2023 Canada National Olympiad",
"2023 Contests"
] | {
"answer_score": 136,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2023 Contests/2023 Canada National Olympiad/3029963.json"
} |
Let $f(x)$ be a non-constant polynomial with integer coefficients such that $f(1) \neq 1$ . For a positive integer $n$ , define $\text{divs}(n)$ to be the set of positive divisors of $n$ .
A positive integer $m$ is $f$ -cool if there exists a positive integer $n$ for which $$ f[\text{divs}(m)]=\text{divs... | Some people might get a -1 by not considering $f(1)<0$ (in particular $f(1)=-1$ )?
We will prove that there are finitely many cool pairs of positive integers $(m,n)$ such that $f(\text{divs}(m))=\text{divs}(n)$ . Throwing out the finitely many pairs with $m=1$ or $n=1$ , The key is to consider the least prime... | [
"Take prime $p$ dividing $f(1)$ and $m >> N$ where $f$ is strictly increasing past $N$ . $f$ of largest factors of $m$ must correspond to largest factors of $n$ , and size argument (relying on the fact that $l^n$ cannot equal $p$ ever for n > 1) wins",
"alternatively: there are finitely many $a... | [
"origin:aops",
"2023 Canada National Olympiad",
"2023 Contests"
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"answer_score": 208,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2023 Contests/2023 Canada National Olympiad/3029964.json"
} |
A country with $n$ cities has some two-way roads connecting certain pairs of cities. Someone notices that if the country is split into two parts in any way, then there would be at most $kn$ roads between the two parts (where $k$ is a fixed positive integer). What is the largest integer $m$ (in terms of $n$ an... | I claim that the answer is largest $m$ such that there is guaranteed to be a set of $m$ vertices with no edge between them is $\boxed{\lceil \tfrac{n}{4k}\rceil}.$ I will show that $\text{max}(m) \ge \lceil \tfrac{n}{4k}\rceil$ and that $\text{max}(m) \le \lceil \tfrac{n}{4k}\rceil.$
First, I will show that... | [
"<details><summary>Solution</summary>The answer is $\\lceil \\frac{n}{4k}\\rceil$ . The construction is as follows: let $n=4kq+r$ where $1\\le r\\le 4k$ . Then the graph \n\n $$ \\underbrace{K_{4k} \\sqcup \\cdots \\sqcup K_{4k}}_q \\sqcup K_r $$ works (gives answer of $q+1=\\lceil \\frac{n}{4k}\\rce... | [
"origin:aops",
"2023 Canada National Olympiad",
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"answer_score": 1186,
"boxed": true,
"end_of_proof": true,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 Canada National Olympiad/3030002.json"
} |
Given are reals $a, b$ . Prove that at least one of the equations $x^4-2b^3x+a^4=0$ and $x^4-2a^3x+b^4=0$ has a real root.
Proposed by N. Agakhanov | If $a = b$ take $x=a$ .
If $a = -b$ take $x = a$ in the second equation.
If $a = 0$ , then take $x=0$ for the first equation.
Suppose wlog that $|a| > |b|$ . Let $Q(x) = x^4 - 2a^3x + b^4$ be the second polynomial.
We see that $Q(a) = b^4 - a^4 < 0$ and $Q(-a) = 3a^4 + b^4 > 0$ and $0\neq a \neq -a$ ... | [
"first take a=b it shows (P(a)=0) P(x)=x^4-2a^3x+b^4, Q(x)=x^4-2b^3x+a^4) \na don;t equal to b P(0) and Q(0) are big than 0 ((((( (P(a) = - Q(b) )))))) if they dont equal to 0 one of them will be smal than 0 after P(a) or Q(b) will be small than 0 after that then using polynomial is continious one of them w... | [
"origin:aops",
"2023 Caucasus Mathematical Olympiad",
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"boxed": false,
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"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3029749.json"
} |
Let $n>k>1$ be positive integers and let $G$ be a graph with $n$ vertices such that among any $k$ vertices, there is a vertex connected to the rest $k-1$ vertices. Find the minimal possible number of edges of $G$ .
Proposed by V. Dolnikov | Consider the complementary graph. $\textbf{Answer:}$ $\frac{n(n-1)}{2}-\frac{(k-1)(k-2)}{2}$ for even $k$ and $\frac{n(n-1)}{2}-max(\lfloor \frac{n}{2} \rfloor,\frac{(k-1)(k-2)}{2})$ for odd $k$ . $\textbf{Construction:}$ Draw all the edges on an complete graph on $k-1$ vertices. Any pairing works for odd $... | [] | [
"origin:aops",
"2023 Caucasus Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 76,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3029752.json"
} |
Let $ABC$ be an acute-angled triangle, and let $AA_1, BB_1, CC_1$ be its altitudes. Points $A', B', C'$ are chosen on the segments $AA_1, BB_1, CC_1$ , respectively, so that $\angle BA'C = \angle AC'B = \angle CB'A = 90^{o}$ . Let segments $AC'$ and $CA'$ intersect at $B"$ ; points $A", C"$ are defined s... | Note that $AB'^2=AB_1\cdot AC=AC_1\cdot AB=AC'^2$ , so we must have $AB'=AC'$ . This means that $AB'A''C'$ is a kite, since $\angle AB'A''=\angle AC'A''=90^{\circ}$ , and it must have an incircle, $\omega_1$ . Define $\omega_2$ and $\omega_3$ similarly.
If $\omega_1$ , $\omega_2$ , and $\omega_3$ were pa... | [
"you need to prove that the points are cyclic",
"<blockquote>you need to prove that the points are cyclic</blockquote>\n\nThe definition of circumscribed is that it has an incircle: see [here](https://en.wikipedia.org/wiki/Tangential_quadrilateral). For further measure:**Claim:** $A'B\"C'A\"B'C\"$ cannot be insc... | [
"origin:aops",
"2023 Caucasus Mathematical Olympiad",
"2023 Contests"
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"boxed": false,
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"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3030717.json"
} |
Determine the least positive integer $n{}$ for which the following statement is true: the product of any $n{}$ odd consecutive positive integers is divisible by $45$ . | By Pigeonhole we have that for $n=6$ we have at least two multiples of $3$ and at least one multiple of $5$ , which is an answer.Just find contradicting cases for $n \leq 5$ and we are done $\blacksquare$ | [
"The answer is $6$ .\n\nFor $n=5$ , just consider $45\\nmid17\\times19\\times21\\times23\\times25$ .\n\nSince for any consecutive $6$ odd positive integers, at least $2(=\\lfloor\\frac63\\rfloor)$ of which are divisible by $3$ , and at least $1=(=\\lfloor\\frac65\\rfloor)$ of which is divisible by $5$ , ... | [
"origin:aops",
"2023 Caucasus Mathematical Olympiad",
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"answer_score": 110,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3112241.json"
} |
In a convex hexagon the value of each angle is $120^{\circ}$ . The perimeter of the hexagon equals $2$ . Prove that this hexagon can be covered by a triangle with perimeter at most $3$ .
| let's name the 6 sides of the hexagon $a,b,c,d,e,f$ .
let's make an equilateral triangle that is made by extending the sides of $a,c,e$ and $b,d,f$ .
the perimeter of the two equilateral triangles are $a+c+e+2(b+d+f)$ or $2(a+c+e)+b+d+f$ .
Because $a+b+c+d+e+f=2$ , one of $a+c+e$ and $b+d+f$ are smaller tha... | [
"If the hexagon is ABCDEF, then either AB+CD+EF or BC+DE+FA is at most 1; WLOG the former. Then extend BC, DE, FA, to form three equilateral triangles on sides AB, CD, EF. If the value of AB+CD+EF was x<=1, then the perimeter of the triangle formed by the extensions of BC, DE, FA is precisely (2-x)+2x=2+x<=3"
] | [
"origin:aops",
"2023 Caucasus Mathematical Olympiad",
"2023 Contests"
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"answer_score": 22,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3112242.json"
} |
The numbers $1, 2, 3,\ldots, 2\underbrace{00\ldots0}_{100 \text{ zeroes}}2$ are written on the board. Is it possible to paint half of them red and remaining ones blue, so that the sum of red numbers is divisible by the sum of blue ones?
| The sum of all the numbers are $200 \cdot\cdot\cdot03 \times 1000\cdot\cdot\cdot01$ .
Let's look in a bizzare way. Let's make the time when red/blue is the biggest.
It is when $1+2+\cdot\cdot\cdot+1000\cdot\cdot\cdot01$ and $1000\cdot\cdot\cdot02+\cdot\cdot\cdot+200\cdot\cdot\cdot02$ The red/blue is smaller than 3,... | [
":):):):)"
] | [
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"2023 Caucasus Mathematical Olympiad",
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"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3112244.json"
} |
Pasha and Vova play the game crossing out the cells of the $3\times 101$ board by turns. At the start, the central cell is crossed out. By one move the player chooses the diagonal (there can be $1, 2$ or $3$ cells in the diagonal) and crosses out cells of this diagonal which are still uncrossed. At least one new ... | I didn't dwell on it much, but for $k\geq2$ , $4(k-1)$ moves seem to be made in the most probable game in a $3\times (2k+1)$ table. Since the number of moves will be even, the $2$ nd player wins the game. | [
"Im not sure but if first player crosses a 2 cell diagonal. And then if 2nd crosses 1 cell 1st crosses 2 cell, if 1 then 2, if 3 then 3 to make the number divisible by 3",
"<blockquote>Im not sure but if first player crosses a 2 cell diagonal. And then if 2nd crosses 1 cell 1st crosses 2 cell, if 1 then 2, if 3... | [
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"2023 Caucasus Mathematical Olympiad",
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"boxed": false,
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"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3112245.json"
} |
Let $n{}$ and $m$ be positive integers, $n>m>1$ . Let $n{}$ divided by $m$ have partial quotient $q$ and remainder $r$ (so that $n = qm + r$ , where $r\in\{0,1,...,m-1\}$ ). Let $n-1$ divided by $m$ have partial quotient $q^{'}$ and remainder $r^{'}$ .
a) It appears that $q+q^{'} =r +r^{'} = 99$ ... | a) If $r\geq 1, q=q' \Longrightarrow 2q=99$ which forces $q$ to be a non integer. If $r=0, n=qm \Longrightarrow n-1=m(q-1)+(m-1)$ so $q'=q-1, r=m-1$ . Now $2q-1=m-1=99 \Longrightarrow q=50, m=100$ so $n=5000$ .
b) Established from before, $n=5000, 2n=10000=100^2$ . | [
"You can just do q', no need for q^{'}"
] | [
"origin:aops",
"2023 Caucasus Mathematical Olympiad",
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"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3112246.json"
} |
a) Determine if there exists a convex hexagon $ABCDEF$ with $$ \angle ABD + \angle AED > 180^{\circ}, $$ $$ \angle BCE + \angle BFE > 180^{\circ}, $$ $$ \angle CDF + \angle CAF > 180^{\circ}. $$ b) The same question, with additional condition, that diagonals $AD, BE,$ and $CF$ are concurrent. | [] | [
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"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3112250.json"
} | |
Is it possible to fill a table $1\times n$ with pairwise distinct integers such that for any $k = 1, 2,\ldots, n$ one can find a rectangle $1\times k$ in which the sum of the numbers equals $0$ if
a) $n= 11$ ;
b) $n= 12$ ? | The answer to both of (a) and (b) are yes.
Construction:
(a): $\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}\hline
0&1&-1&2&-2&3&-3&4&-4&5&-5\hline
\end{array}$ (b): $\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}\hline
0&5&6&-11&1&-1&2&-2&3&-3&4&-4\hline
\end{array}$ | [
"My construction : $ 5,-5,4,-4,3,-3,2,-2,1,-1,0$ $0,1,2,-3,4,-4,5,-5,6,-6,7,-7$ Pretty much similar to sn6dh's but you know what i mean ;;"
] | [
"origin:aops",
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"answer_score": 4,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3112267.json"
} |
Let $a, b, c$ be positive integers such that $$ \gcd(a, b) + \text{lcm}(a, b) = \gcd(a, c) + \text{lcm}(a, c). $$ Does it follow from this that $b = c$ ? | $lcm(a, b)$ and $lcm(a, c)$ are both divisible by $a$ . Taking modulo $a$ on both sides, we see $\gcd(a, b)\equiv\gcd(a, c)\pmod a$ . As both are positive and $\le a$ , we have that $\gcd(a, b)=\gcd(a, c)$ . Thus, $a\cdot b/\gcd(a, b)=lcm(a, b)=lcm(a, c)=a\cdot c/\gcd(a, c)$ . Cancelling out $a$ and the gc... | [
"@all can we do it with Local-Global Principle?\nSuch that $v_p(a) =x$ $v_p(b) =y$ $v_p(c) =z$ Where p is prime\nWlog x>=y>=z\nAnd we get y=z the can we say b=c(?)\n",
"So let's start. First i accepted a=dx b=dy and c=dz.\nAfter that gcd(a,b)=d\nlcm(a,b)=dxy\ngcm(a,c)=d and finally\nlcm(a,c)=dxz\nLet's use it ... | [
"origin:aops",
"2023 Caucasus Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 1020,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3112272.json"
} |
Sasha has $10$ cards with numbers $1, 2, 4, 8,\ldots, 512$ . He writes the number $0$ on the board and invites Dima to play a game. Dima tells the integer $0 < p < 10, p$ can vary from round to round. Sasha chooses $p$ cards before which he puts a “ $+$ ” sign, and before the other cards he puts a “ $-$ " sign... | The answer is $256$ . Let the current number be negative. We can do this because for the other way we can swap the $+$ and $-$ signs in our solution.
Let $1\leq p \leq8$ :
Let Sasha put a $+$ sign before $512$ and put $-$ signs for $256,128$ ,Sasha can put the other signs arbitrarily. This way the number ... | [
"<blockquote>Sasha has $10$ cards with numbers $1, 2, 4, 8,\\ldots, 512$ . He writes the number $0$ on the board and invites Dima to play a game. Dima tells the integer $0 < p < 10, p$ can vary from round to round. Sasha chooses $p$ cards before which he puts a “ $+$ ” sign, and before the other cards he p... | [
"origin:aops",
"2023 Caucasus Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 44,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3112274.json"
} |
Let $ABC$ be an equilateral triangle with the side length equals $a+ b+ c$ . On the side $AB{}$ of the triangle $ABC$ points $C_1$ and $C_2$ are chosen, on the side $BC$ points $A_1$ and $A_2$ , arc chosen, and on the side $CA$ points $B_1$ and $B_2$ are chosen such that $A_1A_2 = CB_1 = BC_2 = a... | Note that $A_1B_2, C_1A_2, B_1C_2$ concur at some point $P$ in $\triangle{ABC}$ , and these three lines meet each other at $60^{\circ}$ angles. By angle chasing, $C_1B_2A'P$ , etc, are cyclic, and since $PA'$ meets $PC_1, PB_2$ at $60^{\circ}$ angles, $A' \in B_1C_2$ , and similarly $B' \in C_1A_2, C' \... | [] | [
"origin:aops",
"2023 Caucasus Mathematical Olympiad",
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"answer_score": 134,
"boxed": false,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3112280.json"
} |
Let $n \leq 100$ be an integer. Hare puts real numbers in the cells of a $100 \times 100$ table. By asking Hare one question, Wolf can find out the sum of all numbers of a square $n \times n$ , or the sum of all numbers of a rectangle $1 \times (n - 1)$ (or $(n - 1) \times 1$ ). Find the greatest $n{}$ such t... | The answer is $51$ .
If a cell can be a corner of an $n\times n$ square, then one can find the number in that cell.
Since for $n=51$ , every cell can be a corner of an $n\times n$ square, Wolf can find the number in all cells.
For $n\geq52$ , for all $n\times n$ squares and $(n-1)\times1, 1\times(n-1)$ rec... | [] | [
"origin:aops",
"2023 Caucasus Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 26,
"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3112283.json"
} |
Numbers $1, 2,\ldots, n$ are written on the board. By one move, we replace some two numbers $ a, b$ with the number $a^2-b{}$ . Find all $n{}$ such that after $n-1$ moves it is possible to obtain $0$ . | <blockquote>Numbers $1, 2,\ldots, n$ are written on the board. By one move, we replace some two numbers $ a, b$ with the number $a^2-b{}$ . Find all $n{}$ such that after $n-1$ moves it is possible to obtain $0$ .</blockquote> $\color{blue}\boxed{\textbf{Answer: n}\equiv \textbf{0(mod 4) and n}\equiv \textbf{... | [] | [
"origin:aops",
"2023 Caucasus Mathematical Olympiad",
"2023 Contests"
] | {
"answer_score": 1262,
"boxed": true,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 Caucasus Mathematical Olympiad/3112284.json"
} |
Define the sequences $(a_n),(b_n)$ by
\begin{align*}
& a_n, b_n > 0, \forall n\in\mathbb{N_+}
& a_{n+1} = a_n - \frac{1}{1+\sum_{i=1}^n\frac{1}{a_i}}
& b_{n+1} = b_n + \frac{1}{1+\sum_{i=1}^n\frac{1}{b_i}}
\end{align*}
1) If $a_{100}b_{100} = a_{101}b_{101}$ , find the value of $a_1-b_1$ ;
2) If $a_{100} = b_{... | Now $a_{n+1}(1+\sum\limits_{i=1}^n\frac1{a_i})=a_n(1+\sum\limits_{i=1}^n\frac1{a_i})-1=a_n(1+\sum\limits_{i=1}^{n-1}\frac1{a_i})=\cdots=a_1$ So $\frac{a_1}{a_{n+2}}=1+\sum\limits_{i=1}^{n+1}\frac1{a_i}=1+\sum\limits_{i=1}^n\frac1{a_i}+\frac1{a_{n+1}}=\frac{a_1+1}{a_{n+1}}$ $a_{n+2}=\dfrac{a_1}{a_1+1}a_{n+1}$ Which ... | [
"<blockquote>Now $a_{n+1}(1+\\sum\\limits_{i=1}^n\\frac1{a_i})=a_n(1+\\sum\\limits_{i=1}^n\\frac1{a_i})-1=a_n(1+\\sum\\limits_{i=1}^{n-1}\\frac1{a_i})=\\cdots=a_1$ So $\\frac{a_1}{a_{n+2}}=1+\\sum\\limits_{i=1}^{n+1}\\frac1{a_i}=1+\\sum\\limits_{i=1}^n\\frac1{a_i}+\\frac1{a_{n+1}}=\\frac{a_1+1}{a_{n+1}}$ $a_{n+2... | [
"origin:aops",
"2023 China MO",
"2023 Contests"
] | {
"answer_score": 1120,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 China MO/2985920.json"
} |
Given positive integer $m,n$ , color the points of the regular $(2m+2n)$ -gon in black and white, $2m$ in black and $2n$ in white.
The *coloring distance* $d(B,C) $ of two black points $B,C$ is defined as the smaller number of white points in the two paths linking the two black points.
The *coloring distance*... | I am a bit unsure about this, but <details><summary>here</summary>**Terms.** An edge of a matching refers to one of the $A_iB_i$ -s or $C_iD_i$ -s. The good side of an edge in a black matching (white respectively) refers to the side in which there are less white points than the other.**Lemma.** For any white matching... | [
"Let $M$ be the maximum number of intersections by drawing $m+n$ chords,each connecting to points with the same color,and the chords connecting the same color do not intersect.\nClearly for any black chord,it has at most the minimum number of white points on either side of this chord;summing up we get $M \\le ... | [
"origin:aops",
"2023 China MO",
"2023 Contests"
] | {
"answer_score": 1048,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 China MO/2985922.json"
} |
Let $\triangle ABC$ be an equilateral triangle of side length 1. Let $D,E,F$ be points on $BC,AC,AB$ respectively, such that $\frac{DE}{20} = \frac{EF}{22} = \frac{FD}{38}$ . Let $X,Y,Z$ be on lines $BC,CA,AB$ respectively, such that $XY\perp DE, YZ\perp EF, ZX\perp FD$ . Find all possible values of $\frac... | Consider the center $K$ of the spiral similarity $\Phi:\triangle DEF\to\triangle XYZ$ . Simple angle chasing concludes $K$ is in fact the Miquel point of $D,E,F$ wrt $\triangle ABC$ . $\Phi$ rotates the whole plane by $90^{\circ}$ so $KF\perp KZ$ etc.
Let $KP\perp BC,KQ\perp CA,KR\perp AB$ and $\theta... | [
"What A STRANGE geometry problem.",
"这道问题就是纯粹的狗屎",
"<blockquote>这道问题就是纯粹的狗屎</blockquote>\n\nI think the geometry part of this problem is nice.\nBut yyj is exactly like what you said",
"<blockquote><blockquote>这道问题就是纯粹的狗屎</blockquote>\n\nI think the geometry part of this problem is nice.\nBut yyj is exactly li... | [
"origin:aops",
"2023 China MO",
"2023 Contests"
] | {
"answer_score": 1054,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 China MO/2985926.json"
} |
Find the minimum positive integer $n\ge 3$ , such that there exist $n$ points $A_1,A_2,\cdots, A_n$ satisfying no three points are collinear and for any $1\le i\le n$ , there exist $1\le j \le n (j\neq i)$ , segment $A_jA_{j+1}$ pass through the midpoint of segment $A_iA_{i+1}$ , where $A_{n+1}=A_1$ | <details><summary>Another kind of construction</summary>[asy]
pair a1=dir(60);pair a2=-a1;pair a3=dir(120);pair a6=-a3;pair a4=E;pair a5=-a4;
draw(a1--a2--a3--a4--a5--a6--cycle);label(" $A_1$ ",a1,NE);label(" $A_2$ ",a2,SW);label(" $A_3$ ",a3,NW);label(" $A_4$ ",a4,E);label(" $A_5$ ",a5,W);label(" $A_6$ ",a6,SE);
[/asy... | [
"The answer is $6$ , which can be achieved by selecting the points $(0,0),(1,1),(2,0),(0,3),(1,-2),(2,3)$ , which works. To show that this is minimal we can easily see that $n=3,4$ fail because we can't have every edge intersected by another. For $n=5$ casework bash."
] | [
"origin:aops",
"2023 China MO",
"2023 Contests"
] | {
"answer_score": 12,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 China MO/2986332.json"
} |
There are $n(n\ge 8)$ airports, some of which have one-way direct routes between them. For any two airports $a$ and $b$ , there is at most one one-way direct route from $a$ to $b$ (there may be both one-way direct routes from $a$ to $b$ and from $b$ to $a$ ). For any set $A$ composed of airports $(1\... | <blockquote>Let $x$ be the given vertex. Let $a_{i}$ be the number of vertices distance $i$ away from $x$ . So first $a_0 = 1$ ( $x$ itself). As above states, we want to prove $a_0+ a_1 + \dots + a_{\lfloor \sqrt{n/2} \rfloor} > \frac{n}{2}.$ Suppose FTSOC $a_0+a_1+ \dots +a_{i} \le \frac{n}{2}$ for all ... | [
"Proving we can get to more than $\\frac n2$ places in $\\sqrt{\\dfrac n2}$ steps suffices.",
"Let $x$ be the given vertex. Let $a_{i}$ be the number of vertices distance $i$ away from $x$ . So first $a_0 = 1$ ( $x$ itself). As above states, we want to prove $a_0+ a_1 + \\dots + a_{\\lfloor \\sqrt{... | [
"origin:aops",
"2023 China MO",
"2023 Contests"
] | {
"answer_score": 150,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 China MO/2986343.json"
} |
Prove that there exist $C>0$ , which satisfies the following conclusion:
For any infinite positive arithmetic integer sequence $a_1, a_2, a_3,\cdots$ , if the greatest common divisor of $a_1$ and $a_2$ is squarefree, then there exists a positive integer $m\le C\cdot {a_2}^2$ , such that $a_m$ is squarefree.
No... | Classify primes into three types:
- If a prime $p$ has $p^2 \mid a_2 - a_1$ , or if $p \mid a_2 - a_1$ but not $\gcd(a_1, a_2)$ , then the prime is *completely harmless*; no term of the sequence is divisible by $p^2$
- If $p \nmid a_2 - a_1$ , then we say $p$ is \alert{mostly harmless}.
- Otherwise, if $p ... | [
"<blockquote>Prove there exist $C>0$ , which satisfies the following conclusion:\nFor any infinite positive integer sequence $a_1, a_2, a_3,\\cdots$ , if the greatest common divisor of $a_1$ and $a_2$ Is squarefree, then there exists a positive integer $m\\le C\\cdot {a_2}^2$ , such that $a_m$ is squarefre... | [
"origin:aops",
"2023 China MO",
"2023 Contests"
] | {
"answer_score": 176,
"boxed": false,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2023 Contests/2023 China MO/2986348.json"
} |
Given $m,n\in\mathbb N_+,$ define $$ S(m,n)=\left\{(a,b)\in\mathbb N_+^2\mid 1\leq a\leq m,1\leq b\leq n,\gcd (a,b)=1\right\}. $$ Prove that: for $\forall d,r\in\mathbb N_+,$ there exists $m,n\in\mathbb N_+,m,n\geq d$ and $\left|S(m,n)\right|\equiv r\pmod d.$ | <details><summary>Solution</summary>$m=r+d, n=1+d \prod\limits_{p<m} p^2$ .
It is straightforward to check that $|S(m,1)| = r+d$ and (via mobius / PIE) $$ S(m,n) = \sum_{k \text{ squarefree }, k\le m} (-1)^{\omega(k)} \lfloor m/k \rfloor\lfloor n/k \rfloor $$ Thus, $$ S(m,n)-S(m,1) = \sum_{k \text{ squarefree },... | [
"Easier than P1...I suppose.",
" $m=kd+r$ and $n=1+ld\\prod_{p<m}p$ .( $k,l\\in \\mathbb{N}^+$ )"
] | [
"origin:aops",
"2023 Contests",
"2023 China Team Selection Test"
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"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3032002.json"
} |
Given an integer $n \geqslant 2$ . Suppose there is a point $P$ inside a convex cyclic $2n$ -gon $A_1 \ldots A_{2n}$ satisfying $$ \angle PA_1A_2 = \angle PA_2A_3 = \ldots = \angle PA_{2n}A_1, $$ prove that $$ \prod_{i=1}^{n} \left|A_{2i - 1}A_{2i} \right| = \prod_{i=1}^{n} \left|A_{2i}A_{2i+1} \right|, $$ ... | Awesome problem! I realised my solution was a bit different from the ones posted above, so here it is:
By inversion through $P$ as above we find $P$ has an isogonal conjugate with respect to $A_1A_2\dots A_{2n}$ - let this be $Q$ . Now let $|PA_i|=p_i$ and $|QA_i|=q_i$ . Also let $A_iA_{i+1}=m_i$ . Now trian... | [
"I proved the case n=2 and made a futile 3-hour attempt to generalise it. This is not a very hard problem but I had a very hard time solving it. ",
"Very slick.\n\nExtend $PA_{i}$ to meet the circle again at $B_{i+1} \\ne A_{i},$ now the key observation is that $B_{1} \\dots B_{2n}$ is a rotated version of... | [
"origin:aops",
"2023 Contests",
"2023 China Team Selection Test"
] | {
"answer_score": 132,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3032062.json"
} |
Prove that: (1) In the complex plane, each line (except for the real axis) that crosses the origin has at most one point ${z}$ , satisfy $$ \frac {1+z^{23}}{z^{64}}\in\mathbb R. $$ (2) For any non-zero complex number ${a}$ and any real number $\theta$ , the equation $1+z^{23}+az^{64}=0$ has roots in $$ S_{\t... | <details><summary>Reposted Solution</summary>We will show that for any $a\in \mathbb{C}$ , there exists at most one point on a line of this form such that $\frac{1+z^{23}}{az^{64}}\in \mathbb{R}$ . Let $z=cr$ .
This is equivalent to
\begin{align*}
&\frac{1+z^{23}}{az^{64}} = \overline{\frac{1+z^{2... | [
"<blockquote>Prove that:**(1)** In the complex plane, each line (except for the real axis) that crosses the origin has at most one point ${z}$ , satisfy $\\frac {1+z^{23}}{z^{64}}\\in\\mathbb R$ .</blockquote>\nSetting $z=\\rho e^{i\\theta}$ with $\\rho\\ge 0$ and $\\theta\\in[0,2\\pi)$ : $\\frac{1+z^{23}}{... | [
"origin:aops",
"2023 Contests",
"2023 China Team Selection Test"
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"answer_score": 128,
"boxed": false,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3032064.json"
} |
$n$ people attend a party. There are no more than $n$ pairs of friends among them.
Two people shake hands if and only if they have at least $1$ common friend.
Given integer $m\ge 3$ such that $n\leq m^3$ .
Prove that there exists a person $A$ , the number of people that shake hands with $A$ is no more than ... | Take the obvious graph theory interpretation, letting the graph be $G$ . Evidently we may assume $G$ is connected, else we can find a connected component with $|E| \geq |V|$ and win by inductive hypothesis, since any valid $m$ for $G$ will also be valid for this connected component alone. Thus $G$ is either ... | [
"Some progress, will finish later:\nConvert the problem to the obvious graph theory interpretation. If the graph isn't a tree, take a connected component that is (or an isolated vertices). \\newline\nAssume $v_z$ is the vertice with the highest degree x and let $a_1, a_2, ... a_x$ be the degrees of the adjacent... | [
"origin:aops",
"2023 Contests",
"2023 China Team Selection Test"
] | {
"answer_score": 206,
"boxed": false,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3032071.json"
} |
(1) Let $a,b$ be coprime positive integers. Prove that there exists constants $\lambda$ and $\beta$ such that for all integers $m$ ,
$$ \left| \sum\limits_{k=1}^{m-1} \left\{ \frac{ak}{m} \right\}\left\{ \frac{bk}{m} \right\} - \lambda m \right| \le \beta $$ (2) Prove that there exists $N$ such tha... | This problem is similar to modern art, in the sense that I have no clue what the point of this problem even is.
1) Trivial by combining Lagrange interpolation on each of the finite cases of m modulo ab.
2) We proceed by forgetting part 1 and using the power of combinatorics instead.
First, notice that we can rewrite... | [
"<details><summary>Sketch</summary>(1) can be done in a couple of ways; summing $\\lfloor \\frac{ak}{m} \\rfloor \\lfloor \\frac{bk}{m} \\rfloor $ or computing $\\int_0^1 \\{ax\\}\\{bx\\}$ works.\n\nFor (2), we prove that <details><summary>Hint 1</summary>$\\int_0^1 \\{ax\\}\\{bx\\} = \\frac 14 + \\frac{1}{12ab... | [
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"2023 Contests",
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"answer_score": 186,
"boxed": false,
"end_of_proof": true,
"n_reply": 20,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3032437.json"
} |
Let $\triangle ABC$ be a triangle, and let $P_1,\cdots,P_n$ be points inside where no three given points are collinear. Prove that we can partition $\triangle ABC$ into $2n+1$ triangles such that their vertices are among $A,B,C,P_1,\cdots,P_n$ , and at least $n+\sqrt{n}+1$ of them contain at least one of $A... | <details><summary>Wrong</summary>We sort $P_1,\cdots,P_n$ such that $$ \angle BAP_1 < \angle BAP_2 < \cdots < \angle BAP_n $$ We first use $n+1$ triangles: $\triangle BAP_1$ , $\triangle P_jAP_{j+1}$ for $1\le j\le n-1$ , and $\triangle P_nAC$ .
Let $\theta_j=\angle P_jBC$ . Let $m$ be the length of th... | [
"Was it given that any three points cannot lie on the same line?",
"I believe your solution is incomplete, because having sqrt(n) points in increasing order left to right doesn’t necessarily mean that you can connect them like you did in your example.",
"<blockquote>I believe your solution is incomplete, becaus... | [
"origin:aops",
"2023 Contests",
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] | {
"answer_score": 88,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3032550.json"
} |
Find the largest positive integer $m$ which makes it possible to color several cells of a $70\times 70$ table red such that
- There are no two red cells satisfying: the two rows in which they are have the same number of red cells, while the two columns in which they are also have the same number of red cells;
- T... | I just spent eleven hours on this. Bruh...
Remove columns and rows with no red cells, and suppose there are $m$ columns in the original grid.
First, suppose there are exactly $s$ rows having the same number of red cells $k$ , then none of their cells can share a column since otherwise, those two cells fail. Now,... | [
"Nice Problem, Last writeup of 2024! The answer is $2k+\\lceil\\left(\\log_2 (k+1)\\right)\\rceil$ for $n=2k$ We local spam. My construction is the same as in #2.\n<details><summary>Bound</summary>Merge $R$ rows with same sum, call $S$ , let the merged row be called $\\mathcal{R}$ , define $f({\\mathcal{R}... | [
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"answer_score": 1100,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3034668.json"
} |
Given the integer $n\geq 2$ and a integer ${a}$ , which is coprime with ${n}$ . A country has ${n}$ islands $D_1$ , $D_2$ , $\cdots$ , $D_n$ . For any $1\leq i\neq j\leq n$ , there is a one-way ferry $D_i$ to $D_j$ if and only if $ij\equiv ia\pmod n$ . A tourist can initially fly to any of the islands, ... | The answer is $\boxed{1+\gamma (n)}$ .
Note:Assume $n=2^{\alpha}p_1p_2\cdots p_ip_{i+1}^{\alpha _{i+1}}\cdots p_m^{\alpha _m}$ ( $\alpha _{i+1},\ldots ,\alpha_m\geqslant 2$ ),then $$ \gamma (n)=\begin{cases}3m-i\quad \alpha =03m-i+1\quad \alpha =13m-i+3\quad \alpha \geqslant 2\end{cases} $$ For example: $\gamma (4)... | [] | [
"origin:aops",
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"answer_score": 1012,
"boxed": true,
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"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3034672.json"
} |
In non-isosceles acute ${}{\triangle ABC}$ , $AP$ , $BQ$ , $CR$ is the height of the triangle. $A_1$ is the midpoint of $BC$ , $AA_1$ intersects $QR$ at $K$ , $QR$ intersects a straight line that crosses ${A}$ and is parallel to $BC$ at point ${D}$ , the line connecting the midpoint of $AH$ and $... | Solved with **Mogmog8**, **Ritwin**, and **v4913**- a new addition to DMW-anti !
[asy]
//certain anti prob
//setup
size(9cm); defaultpen(fontsize(10pt));
pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0);
blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(234,218,255);// blu1 lighter
... | [
"The main claim: $\\omega$ is the nine-point circle of $\\triangle ABC$ , then we notice the incircle satisfy $(1)$ all at once so use inversion at a vertice to construct the corresponding circle. The rest can be done by bashing.\n",
"Here is a proof of the main claim mentioned above.\n\nLet $M_a$ be the m... | [
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"answer_score": 164,
"boxed": false,
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"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3034708.json"
} |
Let $n\in\mathbb N_+.$ For $1\leq i,j,k\leq n,a_{ijk}\in\{ -1,1\} .$ Prove that: $\exists x_1,x_2,\cdots ,x_n,y_1,y_2,\cdots ,y_n,z_1,z_2,\cdots ,z_n\in \{-1,1\} ,$ satisfy $$ \left| \sum\limits_{i=1}^n\sum\limits_{j=1}^n\sum\limits_{k=1}^na_{ijk}x_iy_jz_k\right| >\frac {n^2}3. $$ *Created by Yu Deng* | Let $$ S_l=\sum_{1\le j,k\le n} a_{j,k,l}x_jy_k $$ By picking $z_l$ s appropriately the answer is at least $\sum_{l=1}^n |S_l|$ Note $$ E[S_l^2]=\sum_{1\le j_1,k_1,j_2,k_2\le n} a_{j_1,k_1,l}a_{j_2,k_2,l} E[x_{j_1}y_{k_1}x_{j_2}y_{k_2}] =\sum_{1\le j,k\le n} a_{j,k,l}^2x_j^2y_k^2=n^2 $$ We compute $E[S_l^4]$ ... | [
"This is the stuff you see in nightmares",
"What the...",
"@EthanWYX2009, can you post P10, please? ",
"Looks interesting!",
"Splendid! but rather unfriendly to those who haven't seen this technique, I guess?",
"I guess. This is a rather standard problem. My first instinct is when the constant 1/3 is repl... | [
"origin:aops",
"2023 Contests",
"2023 China Team Selection Test"
] | {
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 16,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3034743.json"
} |
The set of nonempty integers $A$ is said to be "elegant" if it is for any $a\in A,$ $1\leq k\leq 2023,$ $$ \left| \left\{ b\in A:\left\lfloor\frac b{3^k}\right\rfloor =\left\lfloor\frac a{3^k}\right\rfloor\right\}\right| =2^k. $$
Prove that if the intersection of the integer set $S$ and any "elegant" set i... | <details><summary>sol</summary>Denote $I_k^l=[k\cdot3^l,(k+1)\cdot3^l)\cap\mathbb{Z}$ , call a set $A\subset\mathbb{Z}$ " $l$ -good" if it satisfy the condition inthe question with $2023\to l$ .
Lemma 1: Suppose $A\subset I_k^l\subset I_{k'}^{l+1}$ and $B\subset I_t^l\subset I_{k'}^{l+1}$ are $l$ -good sets ... | [] | [
"origin:aops",
"2023 Contests",
"2023 China Team Selection Test"
] | {
"answer_score": 38,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3035052.json"
} |
Prove that there exists some positive real number $\lambda$ such that for any $D_{>1}\in\mathbb{R}$ , one can always find an acute triangle $\triangle ABC$ in the Cartesian plane such that [list] [*] $A, B, C$ lie on lattice points; [*] $AB, BC, CA>D$ ; [*] $S_{\triangle ABC}<\frac{\sqrt 3}{4}D^2+\lambda\cdot ... | <blockquote><details><summary>Solution</summary>The problem is equivalent to finding integers $x,y$ such that $$ D^2/4 \le x^2+y^2 \le D^2/4 +10D^{4/5} $$ and $$ \{\sqrt{3}y\}, 1-\{\sqrt{3}x\}< 100D^{-1/5} $$ For I can let (0,0) be the first vertex, (2x,2y) be the second vertex and $(x-\sqrt{3}y+\{\sqrt{3}y\}, ... | [
"The fourth line should have side lengths greater than D right？",
"Thank you @above and @below for pointing out my mistake!",
"<blockquote> [list] [*] $A, B, C$ lie on lattice points; [*] ${\\color{red}AB,BC,CA}>D$ ; [*] $S_{\\triangle ABC}<\\frac{\\sqrt 3}{4}D^2+\\lambda\\cdot D^{4/5}$ .</blockquote>\n\n",... | [
"origin:aops",
"2023 Contests",
"2023 China Team Selection Test"
] | {
"answer_score": 56,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3035077.json"
} |
Does there exists a positive irrational number ${x},$ such that there are at most finite positive integers ${n},$ satisfy that for any integer $1\leq k\leq n,$ $\{kx\}\geq\frac 1{n+1}?$ | <details><summary>Solution</summary>The answer is no. Indeed, for any positive irrational number $x$ , there are infinitely many $n\in\mathbb{Z}^+$ such that $\{kx\}\ge\frac{1}{n+1}$ for $1\le k\le n,$ where $k\in\mathbb{Z}^+$ .
For a positive irrational number $x$ , consider the set $$ S_x\coloneq\{n: \{k... | [
"<details><summary>Proof that infinitely many integers do not satisfy the property in the problem</summary>Assume not. Then there exists $N$ such that for all $n>N$ , we have $\\{nx\\} > \\frac{1}{n+1}$ , since that implies any $n' > \\frac{1}{\\min_{1\\le j\\le N} \\{jx\\}}$ satisfy $\\min_{1\\le k\\le n'} ... | [
"origin:aops",
"2023 Contests",
"2023 China Team Selection Test"
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"answer_score": 190,
"boxed": false,
"end_of_proof": true,
"n_reply": 7,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3040414.json"
} |
For any nonempty, finite set $B$ and real $x$ , define $$ d_B(x) = \min_{b\in B} |x-b| $$ (1) Given positive integer $m$ . Find the smallest real number $\lambda$ (possibly depending on $m$ ) such that for any positive integer $n$ and any reals $x_1,\cdots,x_n \in [0,1]$ , there exists an $m$ -element set ... | Thanks for Baby cabbage helping me to solve this problem!
For(1),it's similar to above solving by Canbankan.
For(2),we let $ k>2^{2\times\left\lceil\log_{1/2}{\varepsilon}\right\rceil+5m}$ For $S_k=\left \{ 0,1,2,…,k \right \} $
We set $n=2^{k+1}-1$ and $ x_i=2^{-1-\left \lfloor \log_{2}{i} \right \rfloor }$ We supp... | [
"sketch for a solution for (2):\nLet N be very big. Let $1<a_{1}<a_{2}<...<a_{N}<2$ . Now let $t$ be very big. Now take the $x_{i}$ 's to be all whole numbers inside the intervalls $[t^{a_{j}}, t^{a_{j}} + t^{2-a_{j}}]$ . To prove that this works, we use that these intervalls have almost the same sums, and tha... | [
"origin:aops",
"2023 Contests",
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"answer_score": 128,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3040594.json"
} |
For a convex quadrilateral $ABCD$ , call a point in the interior of $ABCD$ **balanced**, if
(1) $P$ is not on $AC,BD$ (2) Let $AP,BP,CP,DP$ intersect the boundaries of $ABCD$ at $A', B', C', D'$ , respectively, then $$ AP \cdot PA' = BP \cdot PB' = CP \cdot PC' = DP \cdot PD' $$ Find the maximum possible ... | [
"Not so difficult as a CTST 3. \nScratch: The answer is 3. The key is to prove that if a quad ABCD has a balance point not on AC or BD , then AB//CD or AD//BC. Then some doing casework (trapezoid or parallelogram) finishes it(the former is at most 3 and the latter at most 2 balance points). We can achieve 3 when it... | [
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"2023 Contests",
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"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3040596.json"
} | |
Whether there are integers $a_1$ , $a_2$ , $\cdots$ , that are different from each other, satisfying:
(1) For $\forall k\in\mathbb N_+$ , $a_{k^2}>0$ and $a_{k^2+k}<0$ ;
(2) For $\forall n\in\mathbb N_+$ , $\left| a_{n+1}-a_n\right|\leqslant 2023\sqrt n$ ? | Great problem.
We prove the
For any positive constant $C$
There do not exist integers $a_1$ , $a_2$ , $\cdots$ , that are different from each other, satisfying:
(1) For $\forall k\in\mathbb N_+$ , $a_{k^2}>0$ and $a_{k^2+k}<0$ ;
(2) For $\forall n\in\mathbb N_+$ , $\left| a_{n+1}-a_n\right|\leqslant C\l... | [
"<details><summary>Solution</summary>Nice troll problem. The answer is no.\nLet $f(k)$ denote a (fixed) integer in $[k^2,k^2+k-1]$ satisfying $a_{f(k)} > 0 > a_{f(k)+1}$ . Define $g(k)$ similarly for $[k^2+k, (k+1)^2-1]$ Then by triangular inequality, when $C<k$ then $a_{f(k)\\pm C} \\le a_{f(k)+1}+2023(... | [
"origin:aops",
"2023 Contests",
"2023 China Team Selection Test"
] | {
"answer_score": 68,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3041821.json"
} |
Let $\Gamma, \Gamma_1, \Gamma_2$ be mutually tangent circles. The three circles are also tangent to a line $l$ . Let $\Gamma, \Gamma_1$ be tangent to each other at $B_1$ , $\Gamma, \Gamma_2$ be tangent to each other at $B_2$ , $\Gamma_1, \Gamma_2$ be tangent to each other at $C$ . $\Gamma, \Gamma_1, \Gamma... | Here is a solution by 几何大神杨皓哲. Since he is a new user, he can't post images and latex, so I will post for him.
<details><summary>God's solution</summary>Here is an alternative solution that I smh did not notice to use Ceva.
Define $A'$ , $X_1$ , $X_2$ as the point diametrically of $A$ in $\Gamma$ , $A_1$ in ... | [
"<details><summary>solution</summary>Let $A_1B_1 \\cap A_2B_2 = X$ . By homotheties we may see that $X$ is the top point of $\\Gamma$ . By Ceva it suffices to prove that $CX$ is the median of $CA_1A_2$ , which is readily equivalent to $X$ being on the radical axis of $\\Gamma_1, \\Gamma_2$ . But since $A... | [
"origin:aops",
"2023 Contests",
"2023 China Team Selection Test"
] | {
"answer_score": 48,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3042024.json"
} |
Find the greatest constant $\lambda$ such that for any doubly stochastic matrix of order 100, we can pick $150$ entries such that if the other $9850$ entries were replaced by $0$ , the sum of entries in each row and each column is at least $\lambda$ .
Note: A doubly stochastic matrix of order $n$ is a $n\ti... | <details><summary>Solution</summary>Claim: if I construct a bipartite with vertices $R=\{r_1,\cdots,r_{100}\}$ representing rows, $C=\{c_1,\cdots,c_{100}\}$ representing columns. Draw an edge between $r_j$ and $c_k$ if $x_{j,k} \ge l$ , then $l$ works if and only if the graph has a matching involving at leas... | [
"50 that was a little difficult",
"<blockquote>50 that was a little difficult</blockquote>\nNot really since this is TST 3, I believe the difficulty this day is a lot easier because the previous day is too hard.\n------\nThe answer is $\\lambda_{\\max}=\\frac{17}{1900}=\\frac{51}{75\\times 76}. $ Construct a bip... | [
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"answer_score": 172,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3042026.json"
} |
Let $A,B$ be two fixed points on the unit circle $\omega$ , satisfying $\sqrt{2} < AB < 2$ . Let $P$ be a point that can move on the unit circle, and it can move to anywhere on the unit circle satisfying $\triangle ABP$ is acute and $AP>AB>BP$ . Let $H$ be the orthocenter of $\triangle ABP$ and $S$ be a ... | [
"The fixed point is the midpoint of AB, regardless of where P and B lie.\n@CANBANKAN I'll send you my solution again soon",
"Basically complex bash, with A, B, P, S, T lying on the unit circle. One may compute that q=p(b+p)/(a+p), and then compute angle QMH is a right angle"
] | [
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"answer_score": 0,
"boxed": false,
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"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3043352.json"
} | |
Find all functions $f:\mathbb {Z}\to\mathbb Z$ , satisfy that for any integer ${a}$ , ${b}$ , ${c}$ , $$ 2f(a^2+b^2+c^2)-2f(ab+bc+ca)=f(a-b)^2+f(b-c)^2+f(c-a)^2 $$ | The answers are $f \equiv 0$ and $f \equiv x$ . Swapping $b$ and $c$ yields
\[f(a-b)^2+f(b-c)^2+f(c-a)^2 = f(a-c)^2+f(c-b)^2+f(b-a)^2.\]
So for any $x, y$ , we have
\[f(x)^2+f(y)^2 + f(-x-y)^2 = f(-x)^2+f(-y)^2 + f(x+y)^2.\]
For $g(x) =f(x)^2 - f(-x)^2$ , we have $g(x)+g(y)=-g(-x-y)$ . Setting $y=0$ , $g(x)... | [
"obviously this problem just need to get f(0),f(1),f(-1),f(2),f(-2).and using Mathematical induction we can get the answer is f=0andf(x)=x.",
"<blockquote>obviously </blockquote>\nYou are welcome to show us more details of your solution to this certainly entirely trivial problem. :)\n",
"The solutions are zero ... | [
"origin:aops",
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"answer_score": 174,
"boxed": false,
"end_of_proof": false,
"n_reply": 10,
"path": "Contest Collections/2023 Contests/2023 China Team Selection Test/3043535.json"
} |
Given integer $n\geq 2$ . Find the minimum value of $\lambda {}$ , satisfy that for any real numbers $a_1$ , $a_2$ , $\cdots$ , ${a_n}$ and ${b}$ , $$ \lambda\sum\limits_{i=1}^n\sqrt{|a_i-b|}+\sqrt{n\left|\sum\limits_{i=1}^na_i\right|}\geqslant\sum\limits_{i=1}^n\sqrt{|a_i|}. $$ | Here is a smoothing solution.
The answer is $\frac{n-1+\sqrt{n-1}}{\sqrt{n}}$ . The construction is $(a_1,\cdots,a_n)=(1,\cdots,1,-(n-1))$ and $b=1$ .
Step 1: We may assume $\sum a_j = 0$ .
Proof: I will maximize $$ f(c) = \sum\limits_{i=1}^n\sqrt{|a_i+c|} - \sqrt{n\left|\sum\limits_{i=1}^n (a_i+c)\ri... | [
"the answer should be (n-1+sqrt(n-1)/sqrtn.However I can't prove it...",
"<details><summary>Answer</summary>$\\lambda = \\frac{n-1+\\sqrt{n-1}}{\\sqrt{n}}$ .</details>\n<details><summary>Solution</summary>Note that $\\sqrt{\\left|a\\right|}+\\sqrt{\\left|b\\right|}\\geq \\sqrt{\\left|a\\pm b\\right|}$ holds for... | [
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Let $n$ be a positive integer. Initially, a $2n \times 2n$ grid has $k$ black cells and the rest white cells. The following two operations are allowed :
(1) If a $2\times 2$ square has exactly three black cells, the fourth is changed to a black cell;
(2) If there are exactly two black cells in a $2 \times 2$ ... | <details><summary>Solution</summary>The answer is $n^2+n+1$ .
Proof that $n^2+n$ can fail: I divide the board into $n^2$ 2x2 squares. For squares above the main diagonal of 2x2 squares, paint the upper right square black and for squares below the main diagonal only paint the lower left square black.
Proof that ... | [
"Doesn’t 4 just work for n>= 2?",
"Yea I agree. The answer is 4 for n>=2 and 3 for n=1\n\nBasically have an L shape, then it can be turned into a 3*2 black grid. Shift the top 2*2 square right, we can create a 3*3 grid, and repeat. To see k=3 doesn't work: either we are never able to add a new square, or we can a... | [
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Let $a,b,d$ be integers such that $\left|a\right| \geqslant 2$ , $d \geqslant 0$ and $b \geqslant \left( \left|a\right| + 1\right)^{d + 1}$ . For a real coefficient polynomial $f$ of degree $d$ and integer $n$ , let $r_n$ denote the residue of $\left[ f(n) \cdot a^n \right]$ mod $b$ . If $\left \{ r_n ... | <details><summary>My Sketch</summary><details><summary>Part 1</summary>For the sake of contradiction, assume $f$ doesn't belong to $\mathbb{Q} [x]$ .
Assume for all $x \in \mathbb{R}$ , $x (mod b)$ is the smallest non-negative real number $y$ so that $b|(x-y)$ .
Also assume that under mod $b$ or mod $1$ , w... | [
"I believe that the statement is still true even if we let $b>(|a|+1)^d$ ",
"Post for storage.\n\nLet $u_n=f(n)\\cdot a^n$ . After trying $d=0,1$ , the idea become clear: Try to find some relation between $u_n,u_{n+1},\\dots,u_{n+d+1}$ . \n\nFirst, by Lagrange interpolation, we have that: $$ \\sum_{k=0}^{d+1... | [
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Given a prime $p$ and a real number $\lambda \in (0,1)$ . Let $s$ and $t$ be positive integers such that $s \leqslant t < \frac{\lambda p}{12}$ . $S$ and $T$ are sets of $s$ and $t$ consecutive positive integers respectively, which satisfy $$ \left| \left\{ (x,y) \in S \times T : kx \equiv y \pmod p \... | This is secretly a geometry problem. This is a solution by codecode.
Let $A = \{ (x,y) \in S\times T \mid kx\equiv y(\bmod\; p), 0\le x,y\le p-1\}$ Label the points $A_j = (x_j,y_j)$ .
The key claim is that all $A_j$ are collinear; if that's the case, observe that the points form an arithmetic sequence. Note $\... | [
"REDACTED",
"<blockquote>The mod d should be mod p.</blockquote>\n\nSorry for the typo. It is fixed now.",
"REDACTED"
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In a field, $2023$ friends are standing in such a way that all distances between them are distinct. Each of them fires a water pistol at the friend that stands closest. Prove that at least one person does not get wet. | We find the two people who're the closest. They'll clearly spray each other. Among the rest of the people, we can still find two people who're the closest, and who're going to spray each other. After $1011$ turns, the person left must be dry.
-----
This is a very old combinatorics problem, I wonder why it is still be... | [] | [
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A frog jumps around on the integers on the number line. If it lands on an even number $n$ , it jumps to the number $\frac{n}{2}$ . If it lands on an odd number $n$ , it jumps to the number $n + 5$ . At some point it lands on the number $25$ . At which numbers may it have been three jumps ago? | $$ 25\to50\begin{cases}45\to90100\begin{cases}95200\end{cases}\end{cases} $$ | [
"Just reverse the process. The only possible numbers are $90, 95, 200$ ."
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The numbers $1, 2, 3, . . . , 16$ must be placed in the $16$ squares in such a way that the sum of the numbers in each of the four rows and columns is the same. What is the smallest possible sum of the four numbers in the corner squares?
![Image](https://cdn.artofproblemsolving.com/attachments/c/2/fad1837625fd71e8e... | Let the sum of the numbers in four corners be $k$ and the sum of each column/row be $x$ . Then $1+2+\cdots+16=4x-k$ . Which implies that $4\mid k$ . So $k\ge12$ . Construction, $$ \begin{matrix}
1&7&13&14&2
6&&&&8
11&&&&12
15&&&&10
4&3&9&16&5
\end{matrix} $$ | [] | [
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Georg has a circular game board with 100 squares labelled $1, 2, . . . , 100$ . Georg chooses three numbers $a, b, c$ among the numbers $1, 2, . . . , 99$ . The numbers need not be distinct. Initially there is a piece on the square labelled $100$ . First, Georg moves the piece $a$ squares forward $33$ times an... | [] | [
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In the $9$ -gon $ABCDEFGHI$ , all sides have equal lengths and all angles are equal. Prove that $|AB| + |AC| = |AE|$ .
 | Also an algebra problem: Put the problem on the complex plane with $A$ at $1,$ $B$ at $z,$ $C$ at $z^2,$ etc., where $z=e^{\frac{2\pi i}{9}}.$ Then $|AB|+|AC|=|AE|$ becomes $\left|1-z\right|+\left|1-z^2\right|=\left|1-z^4\right|.$ Let's factor both sides: $\left|1-z\right|+\left|1-z\right|\left|1+z\r... | [
"Better yet, we can simply use the fact that $AH+AB=AE$ , which follows from that fact that $\\triangle HBE$ is equilateral and $A \\in (HBE)$ . (If you are unfamiliar with the result, apply Ptolemy to $AHEB$ .)",
"So we can make up a trigonometry problem: Prove that $\\sin20^\\circ+\\sin40^\\circ=\\sin80^... | [
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Consider the triangle $ABC{}$ and let $I_A{}$ be its $A{}$ -excenter. Let $M,N$ and $P{}$ be the projections of $I_A{}$ onto the lines $AC,BC{}$ and $AB{}$ respectively. Prove that if $\overrightarrow{I_AM}+\overrightarrow{I_AP}=\overrightarrow{I_AN}$ then $ABC{}$ is an equilateral triangle. | [] | [
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[list=a]
[*]Determine all real numbers $x{}$ satisfying $\lfloor x\rfloor^2-x=-0.99$ .
[*]Prove that if $a\leqslant -1$ , the equation $\lfloor x\rfloor^2-x=a$ does not have real solutions.
[/list] | $a$ Clearly if $x < -1$ then no solution and if $x > 1$ then also no solution. It's easy to show that $x=0.99$ and $x=-0.01$ are the only solutions because only one solution in $]-1,0[$ and one in $]0,1[$ . | [
"@above: $\\lfloor x \\rfloor \\le x,$ not the other way around.\n\nb. We have that $\\lfloor x \\rfloor > x - 1,$ hence $-1 \\ge a = \\lfloor x \\rfloor^2 - x > (x - 1)^2 - x,$ and so $x^2 - 3x + 2 < 0.$ Therefore, $(x - 1)(x - 2) < 0,$ so $x \\in (1, 2).$ But then $a = \\lfloor x \\rfloor^2 - x = 1 -... | [
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} |
Let $x,y{}$ and $z{}$ be positive real numbers satisfying $x+y+z=1$ . Prove that
[list=a]
[*]\[1-\frac{x^2-yz}{x^2+x}=\frac{(1-y)(1-z)}{x^2+x};\]
[*]\[\frac{x^2-yz}{x^2+x}+\frac{y^2-zx}{y^2+y}+\frac{z^2-xy}{z^2+z}\leqslant 0.\]
[/list] | $a$ is just multiplying things out $b$ is proving that $\sum_{cyc}\frac{(1-y)(1-z)}{x^2+x} \ge 3$ Now we have that $(1-y)(1-z) \ge \frac{(2-y-z)^2}{4} = \frac{(x+1)^2}{4}$ by AM-GM.
so $$ \sum_{cyc}\frac{(1-y)(1-z)}{x^2+x} \ge \frac{(x+1)}{4x} = \frac{3}{4} + \frac{1}{4} \cdot (\frac{1}{x} + \frac{1}{y} + \frac{1}... | [
"Let $x,y{}$ and $z{}$ be positive real numbers satisfying $x+y+z=1$ . Prove that $$ \\frac{x^2-yz}{x^2+kx}+\\frac{y^2-zx}{y^2+ky}+\\frac{z^2-xy}{z^2+kz}\\leqslant 0 $$ Where $k\\ge 0.$ ",
"Let $x,y{}$ and $z{}$ be positive real numbers satisfying $x+y+z=1$ . Prove that $$ \\frac{x^2-yz}{x^2+xy}+... | [
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Determine all strictly increasing functions $f:\mathbb{N}_0\to\mathbb{N}_0$ which satisfy \[f(x)\cdot f(y)\mid (1+2x)\cdot f(y)+(1+2y)\cdot f(x)\]for all non-negative integers $x{}$ and $y{}$ . | Joy Bangla! The following proof is what @straight has pointed out.
The answer is $f(x)=2x+1$ and $f(x)=4x+2$ which can be easily verified.
Let $P(x,y)$ denote the given assertion.
Also consider $2<p_1<p_2< \cdots$ to be the infinite sequence of primes. $P(0,0): f(0) \in \{1,2\}$ .**<span style="color:#00f">Cas... | [
" $f(x) = 2x+1$ or $f(x) = 4x+2$ .\n\nYou can prove for primes $2a+1$ that $f(a) = 2a+1$ or $f(a) = 4a+2$ . Then use the strict increasing fact to show that in both cases we need every number between to primes to satisfy $f(b) = 2b+1$ or $f(b) = 4b+2$ ",
"arnab da orz "
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Determine all real numbers $x{}$ satisfying $2^{x-1}+2^{1/\sqrt{x}}=3$ . | <blockquote>Determine all real numbers $x{}$ satisfying $2^{x-1}+2^{1/\sqrt{x}}=3$ .</blockquote>
Let $f(x)=2^{x-1}+2^{\frac 1{\sqrt x}}$ $f'(x)=\frac{\ln 2}{2\sqrt x}\left(\sqrt x2^{x}-2^{\frac 1{\sqrt x}}\right))$ $\sqrt x2^{x}-2^{\frac 1{\sqrt x}}$ is continuous increasing and is zero when $x=1$ So $f(x)$ ... | [
"Note that $x=1$ works. It suffices to show that $\\frac{d}{dx} 2^{x-1}+2^{1/\\sqrt{x}} = \\ln(2)\\left( 2^{x-1} -\\frac{1}{2}x^{-\\frac{3}{2}}2^\\frac{1}{\\sqrt x} \\right)< 0$ when $x < 1$ and $> 0$ when $x >1$ . When $x < 1$ , $2^x < 2^\\frac{1}{\\sqrt x} < x^{-\\frac{3}{2}}2^\\frac{1}{\\sqrt x}$ , an... | [
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Let $ABC$ be an equilateral triangle. On the small arc $AB{}$ of its circumcircle $\Omega$ , consider the point $N{}$ such that the small arc $NB$ measures $30^\circ{}$ . The perpendiculars from $N{}$ onto $AC$ and $AB$ intersect $\Omega$ again at $P{}$ and $Q{}$ respectively. Let $H_1,H_2$ and ... | [] | [
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Let $n\geqslant 2$ be an integer. Determine all complex numbers $z{}$ which satisfy \[|z^{n+1}-z^n|\geqslant|z^{n+1}-1|+|z^{n+1}-z|.\] | Wow, nice.
------
The Triangle Inequality applied to the right hand side yields
\begin{align*}
|z|^n|z-1| &\geqslant |z^{n+1} - 1| + |z^{n+1} - z|
&= |z^{n+1} - 1| + |z - z^{n+1}| \geqslant |z - 1|.\qquad(\dagger)
\end{align*}
This inequality is true when either $|z - 1| = 0$ or $|z|^n \geqslant 1$ ; either way,... | [] | [
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Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that any real numbers $x{}$ and $y{}$ satisfy \[f(xf(x)+f(y))=f(f(x^2))+y.\] | <blockquote>Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that any real numbers $x{}$ and $y{}$ satisfy \[f(xf(x)+f(y))=f(f(x^2))+y.\]</blockquote> $\color{blue}\boxed{\textbf{Answer:}f(x)\equiv x, -x}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$ f(xf(x)+f(y))=f(f(x^2)... | [
"Clearly $f$ is injective and surjective.\n\nAlso, $f(0) = 0$ , because $\\exists \\alpha \\colon f(\\alpha) = 0$ and $P(\\alpha,y)$ gives $f(f(y)) = f(f(\\alpha^2)) + y$ but $P(0,y)$ gives $f(f(y)) = f(f(0)) + y$ ,\nso we get $f(f(\\alpha)) = f(f(0))$ and thus $\\alpha = 0$ . \n\nNow, subbing in $P(0... | [
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Determine all continuous functions $f:\mathbb{R}\to\mathbb{R}$ for which $f(1)=e$ and \[f(x+y)=e^{3xy}\cdot f(x)f(y),\]for all real numbers $x{}$ and $y{}$ . | The answer is of course as ysharifi pointed out i.e
\[ f(x)=e^{\frac{x(3x-1)}{2}}\]
Here is the calculation that leads to the apparently weird looking $\frac{x(3x-1)}{2}$ for naturals.
Put $y=1$ to get that $f(n+1)=e^{3n}f(n)f(1)=e^{3n+1}f(n)$ .
Now taking log on both sides we get $\log (f(n+1))=3n+1+\log(f(n)... | [
"The answer should be $f(x)=e^{\\frac{x(3x-1)}{2}}.$ The proof is standard: use $f(1)=e$ to first find $f(n)$ for integers $n,$ then find $f(q)$ for rational numbers $q,$ and finally for all real numbers using sequences of rational numbers and continuity of $f.$ "
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Let $A{}$ and $B$ be invertible $n\times n$ matrices with real entries. Suppose that the inverse of $A+B^{-1}$ is $A^{-1}+B$ . Prove that $\det(AB)=1$ . Does this property hold for $2\times 2$ matrices with complex entries? | Similarly to the above solution, we'll take $C = AB.$ Since $(A + B^{-1})(A^{-1} + B) = I_n,$ we get that $AB + B^{-1}A^{1} + I_n = O_n.$ But $B^{-1}A^{-1} = (AB)^{-1},$ so by substituting $C = AB$ we get $C + C^{-1} + I_n = O_n,$ or equivalently $C^2 + C + I_n = O_n.$
On one hand, we have that $\det(C ... | [
"C=AB real entries \nI+C+C^2=0\nC is diagonalisable in M(n,C) since x^2+x+1 has simple root in C \nIt eigenvalue are $j, j^2$ with the same multiplicity since entries of C are reals $det(C)=j^a.j^{2a}=1$ ",
"What do you think about this statement ?\nWe consider the set $S=\\{A,B\\in M_3(\\mathbb{R}); (A+B^{-... | [
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Let $f:[a,b]\to[a,b]$ be a continuous function. It is known that there exist $\alpha,\beta\in (a,b)$ such that $f(\alpha)=a$ and $f(\beta)=b$ . Prove that the function $f\circ f$ has at least three fixed points. | <blockquote>Let $f:[a,b]\to[a,b]$ be a continuous function. It is known that there exist $\alpha,\beta\in (a,b)$ such that $f(\alpha)=a$ and $f(\beta)=b$ . Prove that the function $f\circ f$ has at least three fixed points.</blockquote>
We may normalise to assume $a = -b.$ Replacing $f$ by the function $... | [
"<details><summary>Sol</summary>We work with the interval $[0,1]$ for easier calculations.\nWe know that $f(c)=0$ and $f(d)=1$ for some $(c,d) \\in (0,1)$ .\nWe also know that $f$ has atleast one fixed point say $k$ .\nIf $f(0)=0, f(1)=1$ we are already done!\nDefine $g(x)=f(f(x))-x$ . We have $g(0)=f... | [
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Let $A{}$ and $B{}$ be $3\times 3{}$ matrices with complex entries, satisfying $A^2=B^2=O_3$ . Prove that if $A{}$ and $B{}$ commute, then $AB=O_3$ . Is the converse true? | My friend [Arde](https://artofproblemsolving.com/community/user/1031921) asked me to post this here:
<blockquote>My less interesting solution using spaces: $A^2 = O_3$ implies $\text{col}(A) \subseteq \text{ker}(A)$ . Also, \[\text{dim}(\text{col}(A)) + \text{dim}(\text{ker}(A)) = 3\]implies $\text{dim}(\text{col}... | [
"Here's a neat three-line solution: firstly, apply Sylvester's inequality on the pair $(A,A)$ to infer that \\[2\\text{rk}(A)\\leqslant \\text{rk}(A^2)+3=\\text{rk}(O_3)+3=3,\\]hence $\\text{rk}(A)\\leqslant 1$ . Now, apply Frobenius' inequality on the triple $(B,A,B)$ and use the fact that $AB=BA$ to get th... | [
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Let $f:[-\pi/2,\pi/2]\to\mathbb{R}$ be a twice differentiable function which satisfies \[\left(f''(x)-f(x)\right)\cdot\tan(x)+2f'(x)\geqslant 1,\]for all $x\in(-\pi/2,\pi/2)$ . Prove that \[\int_{-\pi/2}^{\pi/2}f(x)\cdot \sin(x) \ dx\geqslant \pi-2.\] | A similar solution :
As $\cos x $ is non-negative in the interval $[-\pi/2,\pi/2]$ hence by multiplying with $\cos x$ on both sides we get that the given inequality is equivalent to $ [f(x)\sin x +\cos x]'' \ge 0 $ which means that $g(x)= f(x)\sin x+ \cos x$ is a convex function in $(-\pi/2,\pi/2)$ . So $g... | [
"Let $g(x)=f(x)\\sin(x)+\\cos(x)$ . The inequality in the hypothesis is equivalent to $g''\\geq 0$ , i.e. $g$ is convex. We therefore have by Jensen $$ 2+\\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} f(x)\\sin(x) \\ dx = \\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}} g(x) \\ dx = \\int_{-\\frac{\\pi}{2}}^{\\frac{\\... | [
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"path": "Contest Collections/2023 Contests/2023 District Olympiad/3030266.json"
} |
Let $(G,\cdot)$ be a grup with neutral element $e{}$ , and let $H{}$ and $K$ be proper subgroups of $G$ , satisfying $H\cap K=\{e\}$ . It is known that $(G\setminus(H\cup K))\cup\{e\}$ is closed under the operation of $G$ . Prove that $x^2=e$ for all the elements $x{}$ of $G{}$ . | I think we can prove more strongly that, assuming $H$ and $K$ are both not the trivial group, $H \cong K \cong \mathbb Z/2\mathbb Z$ and $G \cong (\mathbb Z/2\mathbb Z)^2$ .
Let $P := G\setminus(H\cup K)$ for ease of typesetting; then $G = P \cup H\cup K$ . Pick $x\in P$ , which must exist (e.g. $x = hk$ ... | [
"Let $L=(G\\setminus (H\\cup K))\\cup \\{e\\}$ . Quite clearly, $L$ is also closed by taking inverses, so $L$ is also a proper subgroup of $G$ , giving $H\\cup K\\cup L$ a partition (in the sense of trivial intersection) of $G$ .\n\nNotice that is general, if $h\\in H\\setminus\\{e\\}$ and $k\\in K\\set... | [
"origin:aops",
"2023 Contests",
"2023 District Olympiad"
] | {
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2023 Contests/2023 District Olympiad/3030267.json"
} |
Let $f:[0,1]\to\mathbb{R}$ be a continuous function. Prove that \[\lim_{n\to\infty}\int_0^1 f(x^n) \ dx=f(0).\]Furthermore, if $f(0)=0$ and $f$ is right-differentiable in $0{}$ , prove that the limits \[\lim_{\varepsilon\to0}\int_\varepsilon^1\frac{f(x)}{x} \ dx\quad\text{and}\quad\lim_{n\to\infty}\left(n\int_0^... | This proof is a bit of an elementary nature.
a) We know that for every $x \in [0,1)$ we have $\lim_{n \to \infty} f(x^n) =f(0)$ . Since $f$ is a continuous function on a compact set we can say that $\exists M>0$ such that $|f(x)|< M$ $\forall x \in [0,1]$ . Now for small enough $\varepsilon>0$ choose $a= ... | [
"From the continuity of $f$ , the sequence of functions $(f(x^n))_{n\\geq 1}$ converges pointwise to the almost-continuous function $$ \\overline{f}(x)=\\begin{cases} f(0) & x\\in [0,1) f(1) & x=1\\end{cases}. $$ By Dominated Convergence Theorem and the fact that $\\overline{f}$ is both Riemann and Lebesgu... | [
"origin:aops",
"2023 Contests",
"2023 District Olympiad"
] | {
"answer_score": 66,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 District Olympiad/3030271.json"
} |
Consider the functions $f,g,h:\mathbb{R}_{\geqslant 0}\to\mathbb{R}_{\geqslant 0}$ and the binary operation $*:\mathbb{R}_{\geqslant 0}\times \mathbb{R}_{\geqslant 0}\to \mathbb{R}_{\geqslant 0}$ defined as \[x*y=f(x)+g(y)+h(x)\cdot|x-y|,\]for all $x,y\in\mathbb{R}_{\geqslant 0}$ . Suppose that $(\mathbb{R}_{\geq... | Easy to see that $e=0$ .
It follows that $f(x)=x(1-h(x))$ and $g(x)=x(1-h(0))$ which implies that $\text{Im } h \subseteq [0,1]$ since $f$ and $ g$ only take nonnegative values.
We have $$ x*y=x+y-xh(x)- yh(0) + h(x)\cdot |x-y| $$ Since $"*"$ is commutative we get $$ xh(x)-yh(y)=(x-y)h(0)+|x-y|\cdot (h(... | [] | [
"origin:aops",
"2023 Contests",
"2023 District Olympiad"
] | {
"answer_score": 64,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2023 Contests/2023 District Olympiad/3030276.json"
} |
There are $n \ge 3$ positive real numbers $a_1, a_2, \dots, a_n$ . For each $1 \le i \le n$ we let $b_i = \frac{a_{i-1} + a_{i+1}}{a_i}$ (here we define $a_0$ to be $a_n$ and $a_{n+1}$ to be $a_1$ ). Assume that for all $i$ and $j$ in the range $1$ to $n$ , we have $a_i \le a_j$ if and only if $... | Let $m=\min(a_i)$ , and $M=\max(a_i)$ . The key is the following observation.**Lemma**: If $a_i=m$ and $a_j=M$ , then $a_{i-1}=a_{i+1}=m$ and $a_{j-1}=a_{j+1}=M$ .
*Proof*: Since $a_i=m\leq M=a_j$ , we know that $b_i\leq b_j$ . But then
$$ 2=\frac{m+m}{m}\leq \frac{a_{i-1}+a_{i+1}}{a_i}\leq \frac{a_{j-1... | [
"Let $m=\\min(a_1,\\ldots,a_n)$ and $M=\\max(a_1,\\ldots,a_n)$ . Assume FTSOC that $M>m$ . Then, there must exist an index $i$ such that $a_i=m$ and $\\max(a_{i-1},a_{i+1})>m$ , giving $b_i>2$ . Similarly, there must exist an index $j$ such that $a_j=M$ and $\\min(a_{j-1},a_{j+1})<M$ , giving $b_j<2... | [
"origin:aops",
"2023 Contests",
"2023 EGMO"
] | {
"answer_score": 130,
"boxed": false,
"end_of_proof": true,
"n_reply": 22,
"path": "Contest Collections/2023 Contests/2023 EGMO/3054398.json"
} |
Let $ABC$ be a triangle with circumcircle $\Omega$ . Let $S_b$ and $S_c$ respectively denote the midpoints of the arcs $AC$ and $AB$ that do not contain the third vertex. Let $N_a$ denote the midpoint of arc $BAC$ (the arc $BC$ including $A$ ). Let $I$ be the incenter of $ABC$ . Let $\omega_b$ be... | <details><summary>Bary</summary>We use barycentric coordinates. Let $T$ be the intersection of $NI$ with $\Omega$ , and let $E$ be the point where the $A$ -excircle is tangent to $BC$ . We will show that the radical axis of $\omega_b$ and $\omega_c$ is the line through $A$ isogonal to $AE$ , which is $... | [
"Suppose that $\\omega_b$ is tangent to $\\overline{AB}$ at $D$ and $\\omega_c$ is tangent to $\\overline{AC}$ at $E$ . The homothety at $S_b$ sending $\\omega_b$ to $\\Omega$ sends $D$ to $S_c$ , so $D$ lies on $\\overline{S_bS_c}$ . Similarly, $E$ lies on $\\overline{S_bS_c}$ . Since $S... | [
"origin:aops",
"2023 Contests",
"2023 EGMO"
] | {
"answer_score": 354,
"boxed": false,
"end_of_proof": true,
"n_reply": 50,
"path": "Contest Collections/2023 Contests/2023 EGMO/3054399.json"
} |
Let $k$ be a positive integer. Lexi has a dictionary $\mathbb{D}$ consisting of some $k$ -letter strings containing only the letters $A$ and $B$ . Lexi would like to write either the letter $A$ or the letter $B$ in each cell of a $k \times k$ grid so that each column contains a string from $\mathbb{D}$ ... | $\mathfrak{The \;Twenty-Forth\; Of\; August,\; 2025}$ ʕ•ᴥ•ʔ
<details><summary>Solution - grinding Russian Combinatorics</summary>$$ \color{red} \spadesuit\color{red} \boxed{\textbf{Answer \& Approach.}}\color{red} \spadesuit $$ We claim that the answer is $m=2^{k-1}$ . To prove this, we will show that it is always ... | [
"Replace $A$ with $0$ and $B$ with $1$ .\n\nThe answer is $m=2^{k-1}$ . To show that $m=2^{k-1}-1$ does not work, take $\\mathcal{D}$ containing all strings other than $00 \\cdots 0$ starting with $0$ . Then, the top row of the grid must contain a $1$ , but no string in $\\mathcal{D}$ starts with ... | [
"origin:aops",
"2023 Contests",
"2023 EGMO"
] | {
"answer_score": 1120,
"boxed": false,
"end_of_proof": false,
"n_reply": 25,
"path": "Contest Collections/2023 Contests/2023 EGMO/3054400.json"
} |
Turbo the snail sits on a point on a circle with circumference $1$ . Given an infinite sequence of positive real numbers $c_1, c_2, c_3, \dots$ , Turbo successively crawls distances $c_1, c_2, c_3, \dots$ around the circle, each time choosing to crawl either clockwise or counterclockwise.
Determine the largest cons... | The answer is $C = \boxed{\frac12}$ .
Notice that $C = \frac12$ works because there never exists a time where Turbo is forced to go to the antipode of his starting point. There always exists a direction in which he can travel away from it.
To show $C > \frac12$ does not work, consider an arbitrarily small $\e... | [
"We claim that $C=\\tfrac{1}{2}$ . If $C>\\tfrac{1}{2}$ , then let $c$ be a real number in $(\\tfrac{1}{2},C)$ . Let $a_i=c$ for odd $i$ and $a_i=\\tfrac{1}{2}$ for even $i$ . If Turbo crawls twice in the same direction, it will travel a distance of $c+\\tfrac{1}{2}>1$ in these two moves, covering the... | [
"origin:aops",
"2023 Contests",
"2023 EGMO"
] | {
"answer_score": 1030,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2023 Contests/2023 EGMO/3054404.json"
} |
We are given a positive integer $s \ge 2$ . For each positive integer $k$ , we define its *twist* $k’$ as follows: write $k$ as $as+b$ , where $a, b$ are non-negative integers and $b < s$ , then $k’ = bs+a$ . For the positive integer $n$ , consider the infinite sequence $d_1, d_2, \dots$ where $d_1=n$ a... | We first prove the following key lemma:**Lemma**: If $d_1=n$ , then the sequence $\pmod{s^2-1}$ is precisely $k, ks, ks^2, ks^3, \dots$ *Proof*: Unconditionally,
$$ as+b\equiv k \pmod{s^2-1}\implies ks\equiv s(as+b)\equiv bs+as^2\equiv bs+a \pmod{s^2-1} $$ so the conclusion follows. $\square$ ... | [
"For any nonnegative integers $a$ and $b$ , we have \\[s(bs+a)=bs^2+as \\equiv as+b \\pmod{s^2-1}.\\] If the sequence starting with $n$ results in $1$ after $m$ twists, we have $n \\equiv sn' \\equiv s^2n'' \\equiv \\cdots \\equiv s^m \\pmod{s^2-1}$ . The powers of $s$ cycle between $1$ and $s \\pmod... | [
"origin:aops",
"2023 Contests",
"2023 EGMO"
] | {
"answer_score": 168,
"boxed": false,
"end_of_proof": true,
"n_reply": 9,
"path": "Contest Collections/2023 Contests/2023 EGMO/3054412.json"
} |
We are given an acute triangle $ABC$ . Let $D$ be the point on its circumcircle such that $AD$ is a diameter. Suppose that points $K$ and $L$ lie on segments $AB$ and $AC$ , respectively, and that $DK$ and $DL$ are tangent to circle $AKL$ .
Show that line $KL$ passes through the orthocenter of trian... | $\textbf{Problem: }$ We are given an acute triangle $ABC$ . Let $D$ be the point on its circumcircle such that $AD$ is a diameter. Suppose that points $K$ and $L$ lie on segments $AB$ and $AC$ , respectively, and that $DK$ and $DL$ are tangent to circle $AKL$ .
Show that line $KL$ passes through the... | [
"Let $H$ be the orthocenter of $ABC$ , and let $M$ be the midpoint of $\\overline{KL}$ . Notice that \\[\\angle DBK=\\angle DMK=\\angle DCL=\\angle DML=90^\\circ,\\] so $BDMK$ and $CDML$ are cyclic. We have \\[\\angle ABM=\\angle KDM=90^\\circ-\\angle DKM=90^\\circ-\\angle BAC=\\angle ABH,\\] and similarl... | [
"origin:aops",
"2023 Contests",
"2023 EGMO"
] | {
"answer_score": 226,
"boxed": false,
"end_of_proof": false,
"n_reply": 62,
"path": "Contest Collections/2023 Contests/2023 EGMO/3054414.json"
} |
Let $a, b, c, d$ be positive reals strictly smaller than $1$ , such that $a+b+c+d=2$ . Prove that $$ \sqrt{(1-a)(1-b)(1-c)(1-d)} \leq \frac{ac+bd}{2}. $$ | $\color{blue} \boxed{\textbf{SOLUTION}}$ By $\textbf{AM-GM Inequality,}$ $$ \sqrt{(1-a)(1-b)(1-c)(1-d)} \leq \frac{(1-a)(1-c)+(1-b)(1-d)}{2}=\frac{ac+bd+2-(a+b+c+d)}{2}=\dfrac{ac+bd}{2} \blacksquare $$ | [
"By AM-GM, $\\sqrt{(1-a)(1-b)(1-c)(1-d)} \\leq \\dfrac{(1-a)(1-c)+(1-b)(1-d)}{2}=\\dfrac{ac+bd+2-(a+b+c+d)}{2}=\\dfrac{ac+bd}{2},$ as desired.",
"<u>**1st solution:**</u> $\\sqrt{(1-a)(1-b)(1-c)(1-d)}\\overset{\\text{AM-GM}}{\\le}\\frac{(1-b)(1-d)+(1-a)(1-c)}{2}=\\frac{ac+bd+2-\\sum_{cyc}a}{2}=\\frac{ac+bd}{2}$ ... | [
"origin:aops",
"2023 Federal Competition For Advanced Students, P1",
"2023 Contests"
] | {
"answer_score": 1108,
"boxed": false,
"end_of_proof": false,
"n_reply": 11,
"path": "Contest Collections/2023 Contests/2023 Federal Competition For Advanced Students, P1/3065879.json"
} |
Given is a triangle $ABC$ . The points $P, Q$ lie on the extensions of $BC$ beyond $B, C$ , respectively, such that $BP=BA$ and $CQ=CA$ . Prove that the circumcenter of triangle $APQ$ lies on the angle bisector of $\angle BAC$ . | $\angle CAQ = \angle CQA \implies \angle OAC = \angle OQC = \angle OQP = \angle OPQ = \angle OPB = \angle OAB$ . $\blacksquare$ <details><summary>lol</summary>(first attempt at nonchalant solve)</details> | [
"Let $O$ be the circumcenter of triangle $APQ$ . Then, $\\angle BAO=\\angle PAO-\\angle PAB=(90^\\circ-\\angle AQP)-\\angle PAB=90^\\circ-\\dfrac{\\angle B+\\angle C}{2}=\\dfrac{\\angle A}{2},$ as desired.",
"The perpendicular bisectors of $AP$ and $AQ$ are the bisectors of the exterior angles of $\\wideh... | [
"origin:aops",
"2023 Federal Competition For Advanced Students, P1",
"2023 Contests"
] | {
"answer_score": 104,
"boxed": false,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Federal Competition For Advanced Students, P1/3065881.json"
} |
Given a positive integer $n$ , find the proportion of the subsets of $\{1,2, \ldots, 2n\}$ such that their smallest element is odd. | The ratio is $\frac{2}{3}$ . We proceed by induction. The case $n=1$ is obvious with the subsets in question being $\{ 1 \}, \{ 2 \}, \{ 1, 2 \}$ .
We now consider the case for $n+1$ . Let $S$ be the set for $n$ . We note that the set of subsets including $2n+1$ but not $2n+2$ has a bijection to the powerse... | [
"Simple counting exercise. The number of subsets of $\\{1,2,\\ldots ,2n \\}$ such that the smallest element is equal to $2i+1$ is $2^{2n-2i-1}$ , and so in total we have $2^1+2^3+\\ldots+2^{2n-1}=\\dfrac{2(4^n-1)}{3}$ such subsets, and so the answer is $\\dfrac{2(4^n-1)}{3 \\cdot 4^n}$ "
] | [
"origin:aops",
"2023 Federal Competition For Advanced Students, P1",
"2023 Contests"
] | {
"answer_score": 26,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2023 Contests/2023 Federal Competition For Advanced Students, P1/3065882.json"
} |
Find all pairs of positive integers $(n, k)$ satisfying the equation $$ n!+n=n^k. $$ | Notice that: $(n-1)!=n^{k-1}-1\Longrightarrow (n-1)! \equiv -1 \pmod n$ however this is only possible when $n \in \mathbb{P}$ Furthermore notice that for for $n\le7$ the only solutions are: $(n,k)=(2,2), (3,2), (5,3)$ **<span style="color:#00f">Claim:</span>** There are no solution for $n\ge11$ *Proof:* $(n-1)!... | [
"what exactly are we solving for? A function $f(n)=k$ that gives a satisfactory $k$ for every $n$ ? A specific $n$ and $k?$ ",
"Find all pairs of positive integers $(n, k)$ satisfying the equation. ",
"This problem has certainly appeared before. Firstly, we may assume $n,k \\geq 1$ . If $k=1$ then ... | [
"origin:aops",
"2023 Federal Competition For Advanced Students, P1",
"2023 Contests"
] | {
"answer_score": 1128,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2023 Contests/2023 Federal Competition For Advanced Students, P1/3065883.json"
} |
Determine all pairs $(m,n)$ of integers with $n \ge m$ satisfying the equation
\[n^3+m^3-nm(n+m)=2023.\] | Notice that $(n+m)(m^2 -2mn +n^2) = 2023$ .
You can write this as $(m+n)(m - n)^2 = 2023$ .
The positive integer divisors of 2023 are $ 1, 7, 17, 119, 289, 2023$ . There are two possibilities : $ 1) : (m-n)^2 = 1$ and $(n+m) = 2023$ . In this case, due to $ n \ge m $ , $ (m-n)= -1$ and $ (m+n)= 2023$ . If y... | [
" $(n-m)^2(n+m)=2023=17^2\\cdot7$ . Thus $n-m|17$ .\nIf $n-m=1$ then $n+m=2023$ and $(n,m)=(1012,1011)$ .\nIf $n-m=17$ then $n+m=7$ and $(n,m)=(12,-5)$ .",
"<blockquote> $(n-m)^2(n+m)=2023=17^2\\cdot7$ . Thus $n-m|17$ .\nIf $n-m=1$ then $n+m=2023$ and $(n,m)=(1012,1011)$ .\nIf $n-m=17$ then $n... | [
"origin:aops",
"2023 German National Olympiad",
"2023 Contests"
] | {
"answer_score": 132,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2023 Contests/2023 German National Olympiad/3092050.json"
} |
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