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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.

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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
Inequality with xy+yz+zx=1
Kimchiks926   14
N 14 minutes ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 4
The positive real numbers $x,y,z$ satisfy $xy+yz+zx=1$. Prove that:
$$ 2(x^2+y^2+z^2)+\frac{4}{3}\bigg (\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\bigg) \ge 5 $$
14 replies
Kimchiks926
Nov 12, 2022
math-olympiad-clown
14 minutes ago
Problem 7
SlovEcience   7
N 19 minutes ago by GreekIdiot
Consider the sequence \((u_n)\) defined by \(u_0 = 5\) and
\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]a) Prove that there exist infinitely many positive integers \(n\) such that \(u_n > 2020n\).

b) Compute
\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]
7 replies
SlovEcience
May 14, 2025
GreekIdiot
19 minutes ago
D1033 : A problem of probability for dominoes 3*1
Dattier   1
N an hour ago by Dattier
Source: les dattes à Dattier
Let $G$ a grid of 9*9, we choose a little square in $G$ of this grid three times, we can choose three times the same.

What the probability of cover with 3*1 dominoes this grid removed by theses little squares (one, two or three) ?
1 reply
Dattier
May 15, 2025
Dattier
an hour ago
Inequality
lgx57   1
N an hour ago by sqing
Source: Own
$a,b,c \in \mathbb{R}^{+}$,$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$. Prove that
$$a^abc+b^bac+c^cab \ge 27(ab+bc+ca)$$
1 reply
lgx57
2 hours ago
sqing
an hour ago
Easy but Nice 12
TelvCohl   2
N an hour ago by AuroralMoss
Source: Own
Given a $ \triangle ABC $ with orthocenter $ H $ and a point $ P $ lying on the Euler line of $ \triangle ABC. $ Prove that the midpoint of $ PH $ lies on the Thomson cubic of the pedal triangle of $ P $ WRT $ \triangle ABC. $
2 replies
TelvCohl
Mar 8, 2025
AuroralMoss
an hour ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   8
N an hour ago by atdaotlohbh
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
8 replies
OgnjenTesic
May 22, 2025
atdaotlohbh
an hour ago
Balkan Mathematical Olympiad
ABCD1728   1
N an hour ago by ABCD1728
Can anyone provide the PDF version of the book "Balkan Mathematical Olympiads" by Mircea Becheanu and Bogdan Enescu (published by XYZ press in 2014), thanks!
1 reply
ABCD1728
May 24, 2025
ABCD1728
an hour ago
Unexpecredly Quick-Solve Inequality
Primeniyazidayi   2
N an hour ago by sqing
Source: German MO 2025,Round 4,Grade 11/12 Day 2 P1
If $a, b, c>0$, prove that $$\frac{a^5}{b^2}+\frac{b}{c}+\frac{c^3}{a^2}>2a$$
2 replies
Primeniyazidayi
4 hours ago
sqing
an hour ago
cute geo
Royal_mhyasd   0
an hour ago
Source: own(?)
Let $\triangle ABC$ be an acute triangle and $I$ it's incenter. Let $A'$, $B'$ and $C'$ be the projections of $I$ onto $BC$, $AC$ and $AB$ respectively. $BC \cap B'C' = \{K\}$ and $Y$ is the projection of $A'$ onto $KI$. Let $M$ be the middle of the arc $BC$ not containing $A$ and $T$ the second intersection of $A'M$ and the circumcircle of $ABC$. If $N$ is the midpoint of $AI$, $TY \cap IA' = \{P\}$, $BN \cap PC' = \{D\}$ and $CN \cap PB' =\{E\}$, prove that $NEPD$ is cyclic.
PS i'm not sure if this problem is actually original so if it isn't someone please tell me so i can change the source (if that's possible)
0 replies
1 viewing
Royal_mhyasd
an hour ago
0 replies
Nice inequality
TUAN2k8   2
N an hour ago by TUAN2k8
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
2 replies
TUAN2k8
Today at 2:03 AM
TUAN2k8
an hour ago
An NT for a break
reni_wee   1
N 2 hours ago by reni_wee
Source: ONTCP 2.4.1
Prove that there are no positive integers $x,k$ and $n \geq 2$ such that $x^2+1 = k(2^n -1)$.
1 reply
reni_wee
2 hours ago
reni_wee
2 hours ago
p divides x^x-c
mistakesinsolutions   6
N 2 hours ago by reni_wee
Show that for integer c and a prime p, $ p |x^x-c $ has a solution
6 replies
mistakesinsolutions
Jun 13, 2023
reni_wee
2 hours ago
exponential diophantine in integers
skellyrah   1
N 2 hours ago by skellyrah
find all integers x,y,z such that $$ 45^x = 5^y + 2000^z $$
1 reply
skellyrah
Yesterday at 7:04 PM
skellyrah
2 hours ago
IMO 2017 Problem 4
Amir Hossein   117
N 2 hours ago by ezpotd
Source: IMO 2017, Day 2, P4
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Proposed by Charles Leytem, Luxembourg
117 replies
Amir Hossein
Jul 19, 2017
ezpotd
2 hours ago
Lines from vertices to some point are perpendicular
chaotic_iak   4
N Sep 10, 2023 by ismayilzadei1387
Source: 2012 Indonesia Round 2.5 TST 1 Problem 3
The incircle of a triangle $ABC$ is tangent to the sides $AB,AC$ at $M,N$ respectively. Suppose $P$ is the intersection between $MN$ and the bisector of $\angle ABC$. Prove that $BP$ and $CP$ are perpendicular.
4 replies
chaotic_iak
May 10, 2012
ismayilzadei1387
Sep 10, 2023
Lines from vertices to some point are perpendicular
G H J
Source: 2012 Indonesia Round 2.5 TST 1 Problem 3
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chaotic_iak
2932 posts
#1 • 2 Y
Y by Adventure10, Mango247
The incircle of a triangle $ABC$ is tangent to the sides $AB,AC$ at $M,N$ respectively. Suppose $P$ is the intersection between $MN$ and the bisector of $\angle ABC$. Prove that $BP$ and $CP$ are perpendicular.
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hadikh
36 posts
#2 • 2 Y
Y by Adventure10, Mango247
chaotic_iak wrote:
The incircle of a triangle $ABC$ is tangent to the sides $AB,AC$ at $M,N$ respectively. Suppose $P$ is the intersection between $MN$ and the bisector of $\angle ABC$. Prove that $BP$ and $CP$ are perpendicular.
Hint
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chaotic_iak
2932 posts
#3 • 1 Y
Y by Adventure10
There's a reason why this is in Proposed & Own Problems (or otherwise known as "solved by poster, but looking for alternate solutions").

I didn't solve this in the test because I tried on the Algebra one, and when I see this, I immediately think of analytic geometry (hey, one circle, one bisector; use the bisector as an axis and we have a good fact that the gradients of the two sides of the bisected angle are opposites!)...which leads to something messy. So I gave up. Without thinking of simple angle chasing. :headbash:

(Also, I get the one-liner solution probably precisely as you want to say from a friend.)
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hectorleo123
347 posts
#4
Y by
chaotic_iak wrote:
The incircle of a triangle $ABC$ is tangent to the sides $AB,AC$ at $M,N$ respectively. Suppose $P$ is the intersection between $MN$ and the bisector of $\angle ABC$. Prove that $BP$ and $CP$ are perpendicular.
$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
$\color{green}\boxed{\textbf{Solution 1:}}$
$\color{green}\rule{24cm}{0.3pt}$
Let $I$ be the incenter of $\triangle ABC$
Let $\angle MBI=\angle IBC=\alpha, \angle NCI=\angle ICB=\beta$
$$\Rightarrow \angle ABC=2\alpha, \angle ACP=2\beta$$$$\Rightarrow \angle BAC=180-2\alpha -2\beta$$$$\Rightarrow \angle NAI=90-\alpha -\beta...(I)$$Since $\angle IMA=\angle INA=90$
$$\Rightarrow ANIM \text{ is cyclic}$$$$\Rightarrow \angle NMI=\angle NAI$$By $(I):$
$$\Rightarrow \angle NMI=90-\alpha-\beta$$In $\triangle BMP:$
$$\alpha +90+90-\alpha -\beta +\angle MPB=180$$$$\Rightarrow \angle MPB=\beta=\angle NCI$$$$\Rightarrow PNIC \text{ is cyclic}$$$$\Rightarrow \angle IPC=\angle INC=90$$$$\Rightarrow \boxed{BP\perp CP}_\blacksquare$$$\color{green}\rule{24cm}{0.3pt}$
$\color{green}\boxed{\textbf{Solution 2:}}$
$\color{green}\rule{24cm}{0.3pt}$
Let $T$ be the point of tangency of the incircle to the side $BC$
Working in complex numbers, where $\odot (MNP)$ is the unit circle, let $m=m$, $n=n$, $t=1$ and $i=0$
$A$ is the intersection of the tangents to the incircle that pass through $M$ and $N:$
$$\Rightarrow a=\frac{2mn}{m+n}$$Analogously we have to:
$$b=\frac{2m}{m+1}$$$$c=\frac{2n}{n+1}$$$P, B \text{ and }O$ are collinears:
$$\Rightarrow \frac{p}{\overline{p}}=\frac{b}{\overline{b}}$$$$\Rightarrow \frac{p}{\overline{p}}=\frac{\frac{2m}{m+1}}{\frac{2}{m+1}}=m$$$$\Rightarrow \overline{p}=\frac{p}{m}...(\blacktriangle)$$$M, N \text{ and }P$ are collinears:
$$\Rightarrow \frac{p-m}{\overline{p}-\overline{m}}=\frac{m-n}{\overline{m}-\overline{n}}=-mn$$$$\Rightarrow \overline{p}=\frac{m+n-p}{mn}...(\blacksquare)$$By $(\blacktriangle)$ and $(\blacksquare):$
$$\Rightarrow \frac{p}{m}=\frac{m+n-p}{mn}$$$$\Rightarrow p=\frac{m+n}{n+1}...(\star)$$$BP\perp CP:$
$$\Leftrightarrow \frac{b-p}{\overline{b}-\overline{p}}=-\frac{c-p}{\overline{c}-\overline{p}}$$By $(\star):$
$$\Leftrightarrow \frac{\frac{2m}{m+1}-\frac{m+n}{n+1}}{\frac{2}{m+1}-\frac{m+n}{m(n+1)}}=-\frac{\frac{2n}{n+1}-\frac{m+n}{n+1}}{\frac{2}{n+1}-\frac{m+n}{m(n+1)}}$$$$\Leftrightarrow \frac{2m(n+1)-(m+n)(m+1)}{\frac{2m(n+1)-(m+n)(m+1)}{m}}=-\frac{n-m}{\frac{m-n}{m}}$$$$\Leftrightarrow m=m\checkmark$$$$\Rightarrow \boxed{BP\perp CP}_\blacksquare$$$\color{green}\rule{24cm}{0.3pt}$
$\color{blue}\rule{24cm}{0.3pt}$
This post has been edited 1 time. Last edited by hectorleo123, Jul 25, 2023, 1:10 AM
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ismayilzadei1387
219 posts
#5 • 1 Y
Y by FriIzi
It's just Iran Lemma
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N Quick Reply
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