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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
D1032 : A general result on polynomial 2
Dattier   1
N a few seconds ago by Dattier
Source: les dattes à Dattier
Let $P \in \mathbb Q[x,y]$ with $\max(\deg_x(P),\deg_y(P)) \leq d$ and $\forall (a,b) \in \mathbb Z^2 \cap [0,d]^2, P(a,b) \in \mathbb Z$.

Is it true that $\forall (a,b) \in\mathbb Z^2, P(a,b) \in \mathbb Z$?
1 reply
Dattier
Yesterday at 5:19 PM
Dattier
a few seconds ago
greatest volume
hzbrl   2
N 12 minutes ago by hzbrl
Source: purple comet
A large sphere with radius 7 contains three smaller balls each with radius 3 . The three balls are each externally tangent to the other two balls and internally tangent to the large sphere. There are four right circular cones that can be inscribed in the large sphere in such a way that the bases of the cones are tangent to all three balls. Of these four cones, the one with the greatest volume has volume $n \pi$. Find $n$.
2 replies
hzbrl
May 8, 2025
hzbrl
12 minutes ago
inequality
danilorj   2
N 12 minutes ago by danilorj
Let $a, b, c$ be nonnegative real numbers such that $a + b + c = 3$. Prove that
\[
\frac{a}{4 - b} + \frac{b}{4 - c} + \frac{c}{4 - a} + \frac{1}{16}(1 - a)^2(1 - b)^2(1 - c)^2 \leq 1,
\]and determine all such triples $(a, b, c)$ where the equality holds.
2 replies
danilorj
Yesterday at 9:08 PM
danilorj
12 minutes ago
2010 Japan MO Finals
parkjungmin   2
N 20 minutes ago by egxa
Is there anyone who can solve question problem 5?
2 replies
parkjungmin
an hour ago
egxa
20 minutes ago
Functional Equation!
EthanWYX2009   3
N 35 minutes ago by liyufish
Source: 2025 TST 24
Find all functions $f:\mathbb Z\to\mathbb Z$ such that $f$ is unbounded and
\[2f(m)f(n)-f(n-m)-1\]is a perfect square for all integer $m,n.$
3 replies
EthanWYX2009
Mar 29, 2025
liyufish
35 minutes ago
Three concurrent circles
jayme   1
N 36 minutes ago by jayme
Source: own?
Dear Mathlinkers,

1. ABC a triangle
2. 0 the circumcircle
3. Tb, Tc the tangents to 0 wrt. B, C
4. D the point of intersection of Tb and Tc
5. B', C' the symmetrics of B, C wrt AC, AB
6. 1b, 1c the circumcircles of the triangles BB'D, CC'D.

Prove : 1b, 1c and 0 are concurrents.

Sincerely
Jean-Louis
1 reply
1 viewing
jayme
Yesterday at 3:08 PM
jayme
36 minutes ago
All-Russian Olympiad
ABCD1728   3
N an hour ago by RagvaloD
When did the first ARMO occur? 2025 is the 51-st, but ARMO on AoPS starts from 1993, there are only 33 years.
3 replies
ABCD1728
2 hours ago
RagvaloD
an hour ago
Hard geometry
Lukariman   6
N 2 hours ago by Lukariman
Given circle (O) and chord AB with different diameters. The tangents of circle (O) at A and B intersect at point P. On the small arc AB, take point C so that triangle CAB is not isosceles. The lines CA and BP intersect at D, BC and AP intersect at E. Prove that the centers of the circles circumscribing triangles ACE, BCD and OPC are collinear.
6 replies
Lukariman
Yesterday at 4:28 AM
Lukariman
2 hours ago
Simple but hard
TUAN2k8   1
N 2 hours ago by Funcshun840
Source: Own
I need synthetic solution:
Given an acute triangle $ABC$ with orthocenter $H$.Let $AD,BE$ and $CF$ be the altitudes of triangle.Let $X$ and $Y$ be reflections of points $E,F$ across the line $AD$, respectively.Let $M$ and $N$ be the midpoints of $BH$ and $CH$, respectively.Let $K=YM \cap AB$ and $L=XN \cap AC$.Prove that $K,D$ and $L$ are collinear.
1 reply
1 viewing
TUAN2k8
4 hours ago
Funcshun840
2 hours ago
Iran geometry
Dadgarnia   23
N 3 hours ago by Ilikeminecraft
Source: Iranian TST 2018, first exam, day1, problem 3
In triangle $ABC$ let $M$ be the midpoint of $BC$. Let $\omega$ be a circle inside of $ABC$ and is tangent to $AB,AC$ at $E,F$, respectively. The tangents from $M$ to $\omega$ meet $\omega$ at $P,Q$ such that $P$ and $B$ lie on the same side of $AM$. Let $X \equiv PM \cap BF $ and $Y \equiv QM \cap CE $. If $2PM=BC$ prove that $XY$ is tangent to $\omega$.

Proposed by Iman Maghsoudi
23 replies
Dadgarnia
Apr 7, 2018
Ilikeminecraft
3 hours ago
Dou Fang Geometry in Taiwan TST
Li4   9
N 3 hours ago by WLOGQED1729
Source: 2025 Taiwan TST Round 3 Mock P2
Let $\omega$ and $\Omega$ be the incircle and circumcircle of the acute triangle $ABC$, respectively. Draw a square $WXYZ$ so that all of its sides are tangent to $\omega$, and $X$, $Y$ are both on $BC$. Extend $AW$ and $AZ$, intersecting $\Omega$ at $P$ and $Q$, respectively. Prove that $PX$ and $QY$ intersects on $\Omega$.

Proposed by kyou46, Li4, Revolilol.
9 replies
Li4
Apr 26, 2025
WLOGQED1729
3 hours ago
A4 BMO SHL 2024
mihaig   0
3 hours ago
Source: Someone known
Let $a\ge b\ge c\ge0$ be real numbers such that $ab+bc+ca=3.$
Prove
$$3+\left(2-\sqrt 3\right)\cdot\frac{\left(b-c\right)^2}{b+\left(\sqrt 3-1\right)c}\leq a+b+c.$$Prove that if $k<\sqrt 3-1$ is a positive constant, then
$$3+\left(2-\sqrt 3\right)\cdot\frac{\left(b-c\right)^2}{b+kc}\leq a+b+c$$is not always true
0 replies
mihaig
3 hours ago
0 replies
Nice one
imnotgoodatmathsorry   5
N 4 hours ago by arqady
Source: Own
With $x,y,z >0$.Prove that: $\frac{xy}{4y+4z+x} + \frac{yz}{4z+4x+y} +\frac{zx}{4x+4y+z} \le \frac{x+y+z}{9}$
5 replies
imnotgoodatmathsorry
May 2, 2025
arqady
4 hours ago
Equal segments in a cyclic quadrilateral
a_507_bc   4
N 4 hours ago by AylyGayypow009
Source: Greece JBMO TST 2023 P2
Consider a cyclic quadrilateral $ABCD$ in which $BC = CD$ and $AB < AD$. Let $E$ be a point on the side $AD$ and $F$ a point on the line $BC$ such that $AE = AB = AF$. Prove that $EF \parallel BD$.
4 replies
a_507_bc
Jul 29, 2023
AylyGayypow009
4 hours ago
image under a rotation center through an angle of 120°
orl   9
N Mar 6, 2024 by AshAuktober
Source: IMO 1986, Day 1, Problem 2
Given a point $P_0$ in the plane of the triangle $A_1A_2A_3$. Define $A_s=A_{s-3}$ for all $s\ge4$. Construct a set of points $P_1,P_2,P_3,\ldots$ such that $P_{k+1}$ is the image of $P_k$ under a rotation center $A_{k+1}$ through an angle $120^o$ clockwise for $k=0,1,2,\ldots$. Prove that if $P_{1986}=P_0$, then the triangle $A_1A_2A_3$ is equilateral.
9 replies
orl
Nov 11, 2005
AshAuktober
Mar 6, 2024
image under a rotation center through an angle of 120°
G H J
Source: IMO 1986, Day 1, Problem 2
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orl
3647 posts
#1 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Given a point $P_0$ in the plane of the triangle $A_1A_2A_3$. Define $A_s=A_{s-3}$ for all $s\ge4$. Construct a set of points $P_1,P_2,P_3,\ldots$ such that $P_{k+1}$ is the image of $P_k$ under a rotation center $A_{k+1}$ through an angle $120^o$ clockwise for $k=0,1,2,\ldots$. Prove that if $P_{1986}=P_0$, then the triangle $A_1A_2A_3$ is equilateral.
Z K Y
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ZetaX
7579 posts
#2 • 5 Y
Y by Sskkrr, Adventure10, Mango247, Mango247, and 1 other user
Write $x= A_1-A_2 , y= A_2-A_3 , z= A_3-A_1$.
The mappings are (with $\zeta$ being the 'first' third root of unity):
$P \mapsto \zeta (P-A_1)+A_1$
$P \mapsto \zeta (P-A_2)+A_2$
$P \mapsto \zeta (P-A_3)+A_3$
and using them in a row gives (easy to calculate out):
$P \mapsto P + \zeta^2 x + \zeta y + z$
Using the chain of mappings $1986= 3 \cdot 662$ times is the same as using that mapping $662$ times, thus the mapping
$P \mapsto P+ 662(\zeta^2 x + \zeta y + z)$
Now the last shall equal $P$, so $\zeta^2 x + \zeta y + z = 0$, which means (by subtracting $x+y+z=0$) that $(\zeta^2-1) x +(\zeta-1) y =0 \iff x= \zeta y$.
Similary, $z=\zeta x$, thus $|x|=|y|=|z|$, thus the triangle is equilateral.
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archimedes1
1482 posts
#3 • 2 Y
Y by Adventure10, Mango247
Does anyone have a synthetic solution or know if a nice one exists?
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FOURRIER
727 posts
#4 • 2 Y
Y by Adventure10, Mango247
There's one here, but in french
p.20

http://www.animath.fr/cours/deho_geo/deho_geo.pdf
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archimedes1
1482 posts
#5 • 3 Y
Y by Adventure10, Mango247, and 1 other user
FOURRIER wrote:
There's one here, but in french
p.20

http://www.animath.fr/cours/deho_geo/deho_geo.pdf
Thank you very much! Luckily "Math French" is a little similar to "math English." So I think I have understood the solution, and this is not a literal translation from that file.
Synthetic
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FOURRIER
727 posts
#6 • 3 Y
Y by Vietnamisalwaysinmyheart, Adventure10, Mango247
I think the solution found by Zetax is shorter and better (I also did like him).
Z K Y
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sqing
42171 posts
#7 • 1 Y
Y by Adventure10
orl wrote:
Given a point $P_0$ in the plane of the triangle $A_1A_2A_3$. Define $A_s=A_{s-3}$ for all $s\ge4$. Construct a set of points $P_1,P_2,P_3,\ldots$ such that $P_{k+1}$ is the image of $P_k$ under a rotation center $A_{k+1}$ through an angle $120^o$ clockwise for $k=0,1,2,\ldots$. Prove that if $P_{1986}=P_0$, then the triangle $A_1A_2A_3$ is equilateral.
Professor Chang Gengzhe, one of the propositional person of this problem, died in Beijing on November 18, 2018.
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Sskkrr
1 post
#8 • 2 Y
Y by Adventure10, Mango247
FOURRIER wrote:
There's one here, but in french
p.20

http://www.animath.fr/cours/deho_geo/deho_geo.pdf

Sir can you please repost this link because this link is dead when you open this there is no solution of above problem.kindly give me other link for this solution
Z K Y
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HamstPan38825
8866 posts
#9
Y by
Let $a_1, a_2, a_3, p_0$ be the complex numbers corresponding to their points in the complex plane, and set $\omega = e^{2i \pi/3}$. We can compute via a series of rotations that$$p_3 = p_0 - a_1 + \omega^2 a_1 + \omega a_2 - \omega a_3 - \omega^2 a_2 + a_3.$$In other words, we must have$$a_3-a_1+\omega^2(a_1-a_2) + \omega(a_2-a_3) = 0$$in order to have $P_{1986} = P_0$ because $3 \mid 1986$. Now, consider the polynomial$$x^2(a_1-a_2) + x(a_2 - a_3) + (a_3 - a_1) = 0.$$It has roots $\omega$ and $1$, so by Vieta's,$$\frac{a_3 - a_1}{a_1 - a_2} = \omega,$$which implies that $a_1, a_2, a_3$ are the vertices of an equilateral triangle, as required.
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AshAuktober
1007 posts
#10
Y by
Observe that by the theory of rotation composition, we have that $P_{k+3}$ is a translation of $P_k \forall k \ge 0$. But since $P_{1986}$ is reached by performing $662$ such translations, and $P_0 = P_{1986}$, we have that $P_k = P_{k+3} \forall k \ge 0$.
Now construct an equilateral triangle $A_1A_2B$ such that $B$ is on the same side of line $A_1A_2$ as $A_3$. Then clearly if $P_0 = B$, we also have $P_2  = B$.Therefore $P_2 = P_3 = B$, but this is only possible if $B = A_3$. Since $\Delta A_1A_2B$ is equilateral by definition, we are done. $\square$
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