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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
ISI UGB 2025 P7
SomeonecoolLovesMaths   2
N 5 minutes ago by Primeniyazidayi
Source: ISI UGB 2025 P7
Consider a ball that moves inside an acute-angled triangle along a straight line, unit it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence = angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.

IMAGE
2 replies
1 viewing
SomeonecoolLovesMaths
an hour ago
Primeniyazidayi
5 minutes ago
Knights NOT crowded on the chessboard
mshtand1   1
N 6 minutes ago by sarjinius
Source: Ukrainian Mathematical Olympiad 2025. Day 1, Problem 8.4
What is the maximum number of knights that can be placed on a chessboard of size \(8 \times 8\) such that any knight, after making 1 or 2 arbitrary moves, does not land on a square occupied by another knight?

Proposed by Bogdan Rublov
1 reply
mshtand1
Mar 13, 2025
sarjinius
6 minutes ago
ISI UGB 2025 P5
SomeonecoolLovesMaths   2
N 19 minutes ago by SomeonecoolLovesMaths
Source: ISI UGB 2025 P5
Let $a,b,c$ be nonzero real numbers such that $a+b+c \neq 0$. Assume that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}$$Show that for any odd integer $k$, $$\frac{1}{a^k} + \frac{1}{b^k} + \frac{1}{c^k} = \frac{1}{a^k+b^k+c^k}.$$
2 replies
1 viewing
SomeonecoolLovesMaths
an hour ago
SomeonecoolLovesMaths
19 minutes ago
A bit tricky invariant with 98 numbers on the board.
Nuran2010   0
29 minutes ago
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
The numbers $\frac{50}{1},\frac{50}{2},...\frac{50}{97},\frac{50}{98}$ are written on the board.In each step,two random numbers $a$ and $b$ are chosen and deleted.Then,the number $2ab-a-b-1$ is written instead.What will be the number remained on the board after the last step.
0 replies
Nuran2010
29 minutes ago
0 replies
Looks like power mean, but it is not
Nuran2010   0
35 minutes ago
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
For $a,b,c$ positive real numbers satisfying $a^2+b^2+c^2 \geq 3$,show that:

$\sqrt[3]{\frac{a^3+b^3+c^3}{3}}+\frac{a+b+c}{9} \geq \frac{4}{3}$.
0 replies
+1 w
Nuran2010
35 minutes ago
0 replies
Taking antipode on isosceles triangle's circumcenter
Nuran2010   0
39 minutes ago
Source: Azerbaijan Al-Khwarizmi IJMO TST 2025
In isosceles triangle, the condition $AB=AC>BC$ is satisfied. Point $D$ is taken on the circumcircle of $ABC$ such that $\angle CAD=90^{\circ}$.A line parallel to $AC$ which passes from $D$ intersects $AB$ and $BC$ respectively at $E$ and $F$.Show that circumcircle of $ADE$ passes from circumcenter of $DFC$.
0 replies
Nuran2010
39 minutes ago
0 replies
R to R, with x+f(xy)=f(1+f(y))x
NicoN9   3
N 40 minutes ago by EeEeRUT
Source: Own.
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that\[
x+f(xy)=f(1+f(y))x
\]for all $x, y\in \mathbb{R}$.
3 replies
NicoN9
4 hours ago
EeEeRUT
40 minutes ago
find angle
TBazar   7
N 44 minutes ago by TBazar
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
7 replies
TBazar
May 8, 2025
TBazar
44 minutes ago
ISI UGB 2025 P4
SomeonecoolLovesMaths   0
an hour ago
Source: ISI UGB 2025 P4
Let $S^1 = \{ z \in \mathbb{C} \mid |z| =1 \}$ be the unit circle in the complex plane. Let $f \colon S^1 \longrightarrow S^2$ be the map given by $f(z) = z^2$. We define $f^{(1)} \colon = f$ and $f^{(k+1)} \colon = f \circ f^{(k)}$ for $k \geq 1$. The smallest positive integer $n$ such that $f^{(n)}(z) = z$ is called the period of $z$. Determine the total number of points in $S^1$ of period $2025$.
(Hint : $2025 = 3^4 \times 5^2$)
0 replies
SomeonecoolLovesMaths
an hour ago
0 replies
ISI UGB 2025 P8
SomeonecoolLovesMaths   0
an hour ago
Source: ISI UGB 2025 P8
Let $n \geq 2$ and let $a_1 \leq a_2 \leq \cdots \leq a_n$ be positive integers such that $\sum_{i=1}^{n} a_i = \prod_{i=1}^{n} a_i$. Prove that $\sum_{i=1}^{n} a_i \leq 2n$ and determine when equality holds.
0 replies
SomeonecoolLovesMaths
an hour ago
0 replies
ISI UGB 2025 P6
SomeonecoolLovesMaths   0
an hour ago
Source: ISI UGB 2025 P6
Let $\mathbb{N}$ denote the set of natural numbers, and let $\left( a_i, b_i \right)$, $1 \leq i \leq 9$, be nine distinct tuples in $\mathbb{N} \times \mathbb{N}$. Show that there are three distinct elements in the set $\{ 2^{a_i} 3^{b_i} \colon 1 \leq i \leq 9 \}$ whose product is a perfect cube.
0 replies
SomeonecoolLovesMaths
an hour ago
0 replies
ISI UGB 2025 P2
SomeonecoolLovesMaths   0
an hour ago
Source: ISI UGB 2025 P2
If the interior angles of a triangle $ABC$ satisfy the equality, $$\sin ^2 A + \sin ^2 B + \sin^2  C = 2 \left( \cos ^2 A + \cos ^2 B + \cos ^2 C \right),$$prove that the triangle must have a right angle.
0 replies
SomeonecoolLovesMaths
an hour ago
0 replies
Six variables
Nguyenhuyen_AG   1
N 2 hours ago by TNKT
Let $a,\,b,\,c,\,x,\,y,\,z$ be six positive real numbers. Prove that
$$\frac{a}{b+c} \cdot \frac{y+z}{x} + \frac{b}{c+a} \cdot \frac{z+x}{y} + \frac{c}{a+b} \cdot \frac{x+y}{z} \geqslant 2+\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}.$$
1 reply
Nguyenhuyen_AG
Today at 5:09 AM
TNKT
2 hours ago
Anything real in this system must be integer
Assassino9931   3
N 2 hours ago by Sardor_lil
Source: Al-Khwarizmi International Junior Olympiad 2025 P1
Determine the largest integer $c$ for which the following statement holds: there exists at least one triple $(x,y,z)$ of integers such that
\begin{align*} x^2 + 4(y + z) = y^2 + 4(z + x) = z^2 + 4(x + y) = c \end{align*}and all triples $(x,y,z)$ of real numbers, satisfying the equations, are such that $x,y,z$ are integers.

Marek Maruin, Slovakia
3 replies
Assassino9931
May 9, 2025
Sardor_lil
2 hours ago
Triangle centres
shobber   4
N May 27, 2014 by Sardor
Source: China TST 2005
In acute angled triangle $ABC$, $BC=a$,$CA=b$,$AB=c$, and $a>b>c$. $I,O,H$ are the incentre, circumcentre and orthocentre of $\triangle{ABC}$ respectively. Point $D \in BC$, $E \in CA$ and $AE=BD$, $CD+CE=AB$. Let the intersectionf of $BE$ and $AD$ be $K$. Prove that $KH \parallel IO$ and $KH = 2IO$.
4 replies
shobber
Jun 27, 2006
Sardor
May 27, 2014
Triangle centres
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G H BBookmark kLocked kLocked NReply
Source: China TST 2005
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shobber
3498 posts
#1 • 2 Y
Y by Adventure10, Mango247
In acute angled triangle $ABC$, $BC=a$,$CA=b$,$AB=c$, and $a>b>c$. $I,O,H$ are the incentre, circumcentre and orthocentre of $\triangle{ABC}$ respectively. Point $D \in BC$, $E \in CA$ and $AE=BD$, $CD+CE=AB$. Let the intersectionf of $BE$ and $AD$ be $K$. Prove that $KH \parallel IO$ and $KH = 2IO$.
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cefer
294 posts
#2 • 2 Y
Y by Adventure10, Mango247
Let $CK$ meet $AB$ at $F$ then from Seva teorem
$\frac{AE}{EC}\frac{CD}{DB}\frac{BF}{FA}=1 \Longrightarrow AE=BD \Longrightarrow\frac{CD}{EC}=\frac{FA}{FB}\Longrightarrow \frac{CD+CE}{EC}=\frac{AB}{FB}$ $\Longrightarrow EC=FB$ and $AF=CD$
So $AF=CD=p-b, BF=CE=p-a, BD=AE=p-c$
Let $A_1,B_1,C_1$ be the feet of altitudes from $A,B,C$, respectively.Let $K_1,K_2$ be feet of perpendiculars from $K$to the side $BC$ and to the altitude $AA_1$ and let $S,T$ and $R$ be the feet of perpendiculars from $I$ and $O$ to the side $BC$, and foot of altitude $O$ to $IS$, respectively.
From the Seva teorem we get $\frac{AD}{KD}=\frac{p}{p-a}=\frac{AA_1}{KK_1} \Longrightarrow KK_1=\frac{p-a}{p}AA_1$
$|HK_2|=|HA_1-KK_1|=|AA_1-AH-AA_1\frac{p-a}{p}|=|\frac{a}{p}AA_1-AH|=|\frac{2S}{p}-AH|=|2r-2OT|$

$\Longrightarrow HK_2=2IR$ $(1)$
$KK_2=A_1K_1=A_1D-K_1D=A_1D-A_1D \frac{KD}{AD}=A_1D-A_1D\frac{p-a}{p}=\frac{a}{p}A_1D=\frac{a}{p}(CA_1-CD)=\frac{a}{p}(\frac{a^2+b^2-c^2}{2a}-(p-b))=b-c$
$\Longrightarrow KK_2=2(p-c-\frac{a}{2})=2ST$
$\Longrightarrow KK_2=2OR$$(2)$
Using $(1)$,$(2)$ and $\angle KK_2H=\angle ORI=\frac{\pi}{2}$
we get $\bigtriangleup KK_2H \sim \bigtriangleup ORI$ with ratio $2$.
So we get $KH \Vert OI$ and $KH=2OI$
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N.T.TUAN
3595 posts
#3 • 2 Y
Y by Adventure10, Mango247
We have $(p-a)\overrightarrow{KA}+(p-b)\overrightarrow{KB}+(p-c)\overrightarrow{KC}=\overrightarrow{0}$, $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OH}$, $a\overrightarrow{IA}+b\overrightarrow{IB}+c\overrightarrow{IC}=\overrightarrow{0}$. Now, it is easy ! :D
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Sardor
804 posts
#4 • 2 Y
Y by Adventure10, Mango247
It's very easy problem.We have $ K $ is Nagel point of the triangle $ ABC $ and by Nagel line theorem $ K,I,G $ are collinear and $ GK=2IG $, on the other hand $ H,G,O $ are collinear ( Euler line ) and $ HO=2OG $ ,so the triangle $ IGO $ and the triangle $ HGK $ similar, so $ IO $ parallel to $ HK $ and $ HK=2IO $ .
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Sardor
804 posts
#5 • 1 Y
Y by Adventure10
Where $ G $ is centroid of the trianle $ ABC $.
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N Quick Reply
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