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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)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
Very easy case of a folklore polynomial equation
Assassino9931   4
N 2 minutes ago by luutrongphuc
Source: Bulgaria EGMO TST 2025 P6
Determine all polynomials $P(x)$ of odd degree with real coefficients such that $P(x^2 + 2025) = P(x)^2 + 2025$.
4 replies
Assassino9931
Friday at 11:51 PM
luutrongphuc
2 minutes ago
<HKG = 90^o if AG = GH
parmenides51   15
N 9 minutes ago by complex2math
Source: 2020 Balkan MO shortlist G2
Let $G, H$ be the centroid and orthocentre of $\vartriangle ABC$ which has an obtuse angle at $\angle B$. Let $\omega$ be the circle with diameter $AG$. $\omega$ intersects $\odot(ABC)$ again at $L \ne A$. The tangent to $\omega$ at $L$ intersects $\odot(ABC)$ at $K \ne L$. Given that $AG = GH$, prove $\angle HKG = 90^o$
.
Sam Bealing, United Kingdom
15 replies
parmenides51
Sep 14, 2021
complex2math
9 minutes ago
Old problem
kwin   0
20 minutes ago
Let $a, b, c \ge 0$ and $ ab+bc+ca>0$. Prove that:
$$ \frac{1}{(a+b)^2} + \frac{1}{(b+c)^2} + \frac{1}{(c+a)^2} + \frac{15}{(a+b+c)^2} \ge \frac{6}{ab+bc+ca}$$Is there any generalizations?
0 replies
kwin
20 minutes ago
0 replies
Interesting inequalities
sqing   1
N 24 minutes ago by sqing
Source: Own
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
$$ a+b+kc\geq\sqrt{k}$$$$ a+b+kc^2\geq\frac{3\sqrt[3]{k}}{4}$$Where $ k>0. $
$$ a+b+c\geq1$$$$ a+b+4c\geq2$$$$ a+b+c^2\geq\frac{3}{4}$$$$ a+b+8c^2\geq\frac{3}{2}$$
1 reply
sqing
an hour ago
sqing
24 minutes ago
IMO 2011 Problem 5
orl   84
N 36 minutes ago by alexanderhamilton124
Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m- n)$. Prove that, for all integers $m$ and $n$ with $f(m) \leq f(n)$, the number $f(n)$ is divisible by $f(m)$.

Proposed by Mahyar Sefidgaran, Iran
84 replies
orl
Jul 19, 2011
alexanderhamilton124
36 minutes ago
n^k + mn^l + 1 divides n^(k+1) - 1
cjquines0   34
N 38 minutes ago by SimplisticFormulas
Source: 2016 IMO Shortlist N4
Let $n, m, k$ and $l$ be positive integers with $n \neq 1$ such that $n^k + mn^l + 1$ divides $n^{k+l} - 1$. Prove that
[list]
[*]$m = 1$ and $l = 2k$; or
[*]$l|k$ and $m = \frac{n^{k-l}-1}{n^l-1}$.
[/list]
34 replies
cjquines0
Jul 19, 2017
SimplisticFormulas
38 minutes ago
Geometry with orthocenter config
thdnder   5
N 41 minutes ago by thdnder
Source: Own
Let $ABC$ be a triangle, and let $AD, BE, CF$ be its altitudes. Let $H$ be its orthocenter, and let $O_B$ and $O_C$ be the circumcenters of triangles $AHC$ and $AHB$. Let $G$ be the second intersection of the circumcircles of triangles $FDO_B$ and $EDO_C$. Prove that the lines $DG$, $EF$, and $A$-median of $\triangle ABC$ are concurrent.
5 replies
thdnder
Apr 29, 2025
thdnder
41 minutes ago
Coefficient Problem
P162008   1
N an hour ago by maromex
Consider the polynomial $g(x) = \prod_{i=1}^{7} \left(1 + x^{i!} + x^{2i!} + x^{3i!} + \cdots + x^{(i-1)i!} + x^{ii!}\right)$
Find the coefficient of $x^{2025}$ in the expansion of $g(x).$
1 reply
P162008
an hour ago
maromex
an hour ago
A coincidence about triangles with common incenter
flower417477   7
N an hour ago by flower417477
$\triangle ABC,\triangle ADE$ have the same incenter $I$.Prove that $BCDE$ is concyclic iff $BC,DE,AI$ is concurrent
7 replies
flower417477
Apr 30, 2025
flower417477
an hour ago
Number theory
gggzul   1
N an hour ago by Pal702004
Is the number
$$10^{32}+10^{28}+...+10^4+1$$a perfect square?
1 reply
gggzul
3 hours ago
Pal702004
an hour ago
Find all b's
TheMathBob   3
N 2 hours ago by Jupiterballs
Source: Polish Math Olympiad 2023 2nd stage P1
Find all positive integers $b$ with the following property: there exists positive integers $a,k,l$ such that $a^k + b^l$ and $a^l + b^k$ are divisible by $b^{k+l}$ where $k \neq l$.
3 replies
TheMathBob
Feb 10, 2023
Jupiterballs
2 hours ago
Sintetic geometry problem
ICE_CNME_4   7
N 2 hours ago by Bryan0224
Source: Math Gazette Contest 2025
Let there be the triangle ABC and the points E ∈ (AC), F ∈ (AB), such that BE and CF are concurrent in O.
If {L} = AO ∩ EF and K ∈ BC, such that LK ⊥ BC, show that EKL = FKL.
7 replies
ICE_CNME_4
Yesterday at 9:30 PM
Bryan0224
2 hours ago
Number theory
falantrng   39
N 2 hours ago by Jupiterballs
Source: RMM 2018 D2 P4
Let $a,b,c,d$ be positive integers such that $ad \neq bc$ and $gcd(a,b,c,d)=1$. Let $S$ be the set of values attained by $\gcd(an+b,cn+d)$ as $n$ runs through the positive integers. Show that $S$ is the set of all positive divisors of some positive integer.
39 replies
falantrng
Feb 25, 2018
Jupiterballs
2 hours ago
2020 EGMO P6: m such that 3-term linear recurrence is always a square
alifenix-   23
N 2 hours ago by Jupiterballs
Source: 2020 EGMO P6
Let $m > 1$ be an integer. A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = a_2 = 1$, $a_3 = 4$, and for all $n \ge 4$, $$a_n = m(a_{n - 1} + a_{n - 2}) - a_{n - 3}.$$
Determine all integers $m$ such that every term of the sequence is a square.
23 replies
1 viewing
alifenix-
Apr 18, 2020
Jupiterballs
2 hours ago
AMO 2020 Day 1 Problem 3
mikestro   7
N Jun 16, 2020 by stroller
Let $ABC$ be a triangle with $\angle ACB=90^{\circ}$. Suppose that the tangent line at $C$ to the circle passing through $A,B,C$ intersects the line $AB$ at $D$. Let $E$ be the midpoint of $CD$ and let $F$ be a point on $EB$ such that $AF$ is parallel to $CD$.
Prove that the lines $AB$ and $CF$ are perpendicular.
7 replies
mikestro
Jun 16, 2020
stroller
Jun 16, 2020
AMO 2020 Day 1 Problem 3
G H J
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mikestro
73 posts
#1 • 1 Y
Y by Mango247
Let $ABC$ be a triangle with $\angle ACB=90^{\circ}$. Suppose that the tangent line at $C$ to the circle passing through $A,B,C$ intersects the line $AB$ at $D$. Let $E$ be the midpoint of $CD$ and let $F$ be a point on $EB$ such that $AF$ is parallel to $CD$.
Prove that the lines $AB$ and $CF$ are perpendicular.
Attachments:
This post has been edited 6 times. Last edited by mikestro, Jul 4, 2020, 4:13 AM
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brainiacmaniac31
2170 posts
#2 • 3 Y
Y by Math-wiz, Mango247, Mango247
We use coordinates.

[asy]
import geometry;
import olympiad;
size(250);
pair a, b, c, d, e, f;
a = dir(0);
b = dir(180);
c = dir(125);
d = (sec(125*pi/180),0);
e = (d+c)/2;
f = extension(e, b, a, a+d-c);

draw(unitcircle);
draw(a--b--c--a);
draw(b--d--c);
draw(e--f--a);
draw(c--f, dashed);

dot("$A$",a,dir(a));
dot("$B$",b,dir(225));
dot("$C$",c,dir(c));
dot("$D$",d,dir(d));
dot("$E$",e,dir(e));
dot("$F$",f,dir(f));


[/asy]
Let $A=(1,0),B=(-1,0),C=(\cos\theta,\sin\theta)$. Then, since the slope of the tangent at $C$ is $-\cot\theta$, we find that $D=(\sec\theta,0)$. From here it is easy to see that $E=\left(\frac{\cos^2\theta+1}{2\cos\theta},\frac{\sin\theta}{2}\right)$. Using Point-Slope form, we know that
$$\overleftrightarrow{AF}\equiv y=-\cot\theta(x-1)$$and
$$\overleftrightarrow{EB}\equiv y=\frac{\sin\theta\cos\theta}{(\cos\theta+1)^2}(x+1).$$The intersection of these lines is $F$, and if $CF\perp AB$ then the $x$-coordinate of $F$ must be $\cos\theta$. SInce $\overleftrightarrow{AF}\cap\overleftrightarrow{BE}=F$, we can plug in $\cos\theta$ into our two line equations and show that $y$ is equal.
$$\-\cot\theta(\cos\theta-1)=\frac{\sin\theta\cos\theta}{(\cos\theta+1)^2}(\cos\theta+1)$$$$\iff -\frac{\cos\theta(\cos\theta-1)}{\sin\theta}=\frac{\sin\theta\cos\theta}{\cos\theta+1}$$$$\iff -(\cos\theta-1)(\cos\theta+1)=\sin^2\theta$$$$\iff 1-\cos^2\theta=\sin^2\theta,$$which is obvious.
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tworigami
844 posts
#3 • 2 Y
Y by KillerOrca2015, porkemon2
Pretty easy for an AMO #3.

Note that $$-1 = (C,D;\infty_{CD}, E) \stackrel{A}{=} (AC \cap BE, B; F, E) \stackrel{C}{=} (A, B; CF \cap AB, D) $$so $CF$ is a symmedian in $\triangle CAB$ and since $\angle C = 90^\circ$ we get $CF \perp AB$ as desired.
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stroller
894 posts
#4
Y by
tworigami wrote:
Pretty easy for an AMO #3.

Well, it's Australian MO :roll:
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parmenides51
30651 posts
#5
Y by
Let $ABC$ be a triangle with $\angle ACB = 90^o$ . Suppose that the tangent line at $C$ to the circle passing through $A, B, C$ intersects the line AB at $D$. Let $E$ be the midpoint of $CD$ and let $F$ be the point on the line $EB$ such that $AF$ is parallel to $CD$. Prove that the lines $AB$ and $CF$ are perpendicular.
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Jjesus
508 posts
#6
Y by
Be: $G=AB\cap CF$ and $\angle BAC=\alpha$, $\angle BCG=\beta$
$\frac{DB}{BG}=\frac{DB}{BC}.\frac{BC}{BG}=\frac{sin\alpha}{cos2\alpha}.\frac{cos(\alpha-\beta)}{sin\beta}$
$\frac{FC}{FG}=\frac{FC}{FB}.\frac{FB}{FG}=\frac{cos\alpha}{cos\beta}.\frac{cos(\alpha-\beta)}{cos2\alpha}$
By Menelaus Theorem $EC.DB.FG=DE.BG.FC$
$\Rightarrow \frac{DB}{BG}=\frac{FC}{FG}\Rightarrow \frac{sin\alpha}{sin\beta}=\frac{cos\alpha}{cos\beta}$
$\Rightarrow tan\alpha=tan\beta \Rightarrow \alpha=\beta$
$\angle GCA=90^{\circ}-\angle BCG=90^{\circ}-\beta=90^{\circ}-\alpha=90^{\circ}-\angle BAC$ QED
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OlympusHero
17020 posts
#7 • 1 Y
Y by amar_04
Wait, the USAMO already happened? :o
This post has been edited 1 time. Last edited by OlympusHero, Jun 16, 2020, 9:31 PM
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stroller
894 posts
#8 • 1 Y
Y by amar_04
stroller wrote:
tworigami wrote:
Pretty easy for an AMO #3.

Well, it's Australian MO :roll:
OlympusHero wrote:
Wait, the USAMO already happened? :o

:(
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