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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.

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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
A Familiar Point
v4913   52
N a few seconds ago by SimplisticFormulas
Source: EGMO 2023/6
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 the circle that is tangent to $AB$ and internally tangent to $\Omega$ at $S_b$, and let $\omega_c$ be the circle that is tangent to $AC$ and internally tangent to $\Omega$ at $S_c$. Show that the line $IN_a$, and the lines through the intersections of $\omega_b$ and $\omega_c$, meet on $\Omega$.
52 replies
v4913
Apr 16, 2023
SimplisticFormulas
a few seconds ago
Tangential quadrilateral and 8 lengths
popcorn1   72
N 4 minutes ago by cj13609517288
Source: IMO 2021 P4
Let $\Gamma$ be a circle with centre $I$, and $A B C D$ a convex quadrilateral such that each of the segments $A B, B C, C D$ and $D A$ is tangent to $\Gamma$. Let $\Omega$ be the circumcircle of the triangle $A I C$. The extension of $B A$ beyond $A$ meets $\Omega$ at $X$, and the extension of $B C$ beyond $C$ meets $\Omega$ at $Z$. The extensions of $A D$ and $C D$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that \[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\]
Proposed by Dominik Burek, Poland and Tomasz Ciesla, Poland
72 replies
popcorn1
Jul 20, 2021
cj13609517288
4 minutes ago
An algorithm for discovering prime numbers?
Lukaluce   3
N 22 minutes ago by TopGbulliedU
Source: 2025 Junior Macedonian Mathematical Olympiad P3
Is there an infinite sequence of prime numbers $p_1, p_2, ..., p_n, ...,$ such that for every $i \in \mathbb{N}, p_{i + 1} \in \{2p_i - 1, 2p_i + 1\}$ is satisfied? Explain the answer.
3 replies
Lukaluce
May 18, 2025
TopGbulliedU
22 minutes ago
Random concyclicity in a square config
Maths_VC   5
N 22 minutes ago by Royal_mhyasd
Source: Serbia JBMO TST 2025, Problem 1
Let $M$ be a random point on the smaller arc $AB$ of the circumcircle of square $ABCD$, and let $N$ be the intersection point of segments $AC$ and $DM$. The feet of the tangents from point $D$ to the circumcircle of the triangle $OMN$ are $P$ and $Q$ , where $O$ is the center of the square. Prove that points $A$, $C$, $P$ and $Q$ lie on a single circle.
5 replies
Maths_VC
Tuesday at 7:38 PM
Royal_mhyasd
22 minutes ago
Basic ideas in junior diophantine equations
Maths_VC   3
N 36 minutes ago by Royal_mhyasd
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
3 replies
Maths_VC
Tuesday at 7:54 PM
Royal_mhyasd
36 minutes ago
Prime number theory
giangtruong13   2
N an hour ago by RagvaloD
Find all prime numbers $p,q$ such that: $p^2-pq-q^3=1$
2 replies
giangtruong13
2 hours ago
RagvaloD
an hour ago
Problem 2
delegat   147
N 2 hours ago by math-olympiad-clown
Source: 0
Let $n\ge 3$ be an integer, and let $a_2,a_3,\ldots ,a_n$ be positive real numbers such that $a_{2}a_{3}\cdots a_{n}=1$. Prove that
\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]

Proposed by Angelo Di Pasquale, Australia
147 replies
delegat
Jul 10, 2012
math-olympiad-clown
2 hours ago
Coloring points of a square, finding a monochromatic hexagon
goodar2006   6
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part 2-P1
Prove that for each coloring of the points inside or on the boundary of a square with $1391$ colors, there exists a monochromatic regular hexagon.
6 replies
goodar2006
Sep 15, 2012
quantam13
2 hours ago
Van der Warden Theorem!
goodar2006   7
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part 2-P2
Suppose $W(k,2)$ is the smallest number such that if $n\ge W(k,2)$, for each coloring of the set $\{1,2,...,n\}$ with two colors there exists a monochromatic arithmetic progression of length $k$. Prove that


$W(k,2)=\Omega (2^{\frac{k}{2}})$.
7 replies
goodar2006
Sep 15, 2012
quantam13
2 hours ago
Maxi-inequality
giangtruong13   0
2 hours ago
Let $a,b,c >0$ and $a+b+c=2abc$. Find max: $$P= \sum_{cyc} \frac{a+2}{\sqrt{6(a^2+2)}}$$
0 replies
giangtruong13
2 hours ago
0 replies
Isosceles triangles among a group of points
goodar2006   2
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part1-P2
Consider a set of $n$ points in plane. Prove that the number of isosceles triangles having their vertices among these $n$ points is $\mathcal O (n^{\frac{7}{3}})$. Find a configuration of $n$ points in plane such that the number of equilateral triangles with vertices among these $n$ points is $\Omega (n^2)$.
2 replies
goodar2006
Jul 27, 2012
quantam13
2 hours ago
APMO Number Theory
somebodyyouusedtoknow   12
N 2 hours ago by math-olympiad-clown
Source: APMO 2023 Problem 2
Find all integers $n$ satisfying $n \geq 2$ and $\dfrac{\sigma(n)}{p(n)-1} = n$, in which $\sigma(n)$ denotes the sum of all positive divisors of $n$, and $p(n)$ denotes the largest prime divisor of $n$.
12 replies
somebodyyouusedtoknow
Jul 5, 2023
math-olympiad-clown
2 hours ago
My Unsolved Problem
ZeltaQN2008   0
2 hours ago
Source: IDK
Let \( P(x) = x^{2024} + a_{2023}x^{2023} + \cdots + a_1x + a_0 \) be a polynomial with real coefficients.

(a) Suppose that \( 2023a_{2023}^2 - 4048a_{2022} < 0 \). Prove that the polynomial \( P(x) \) cannot have 2024 real roots.

(b) Suppose that \( a_0 = 1 \) and \( 2023(a_1^2 + a_2^2 + \cdots + a_{2023}^2) \leq 4 \). Prove that \( P(x) \geq 0 \) for all real numbers \( x \).
0 replies
ZeltaQN2008
2 hours ago
0 replies
Points of a grid
goodar2006   2
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part1-P4
Prove that from an $n\times n$ grid, one can find $\Omega (n^{\frac{5}{3}})$ points such that no four of them are vertices of a square with sides parallel to lines of the grid. Imagine yourself as Erdos (!) and guess what is the best exponent instead of $\frac{5}{3}$!
2 replies
goodar2006
Jul 27, 2012
quantam13
2 hours ago
Similarity through arc midpoint in right triangle
cjquines0   11
N Apr 28, 2025 by ItsBesi
Source: Iranian Geometry Olympiad 2016 Medium 4
Let $\omega$ be the circumcircle of right-angled triangle $ABC$ ($\angle A = 90^{\circ}$). The tangent to $\omega$ at point $A$ intersects the line $BC$ at point $P$. Suppose that $M$ is the midpoint of the minor arc $AB$, and $PM$ intersects $\omega$ for the second time in $Q$. The tangent to $\omega$ at point $Q$ intersects $AC$ at $K$. Prove that $\angle PKC = 90^{\circ}$.

Proposed by Davood Vakili
11 replies
cjquines0
May 26, 2017
ItsBesi
Apr 28, 2025
Similarity through arc midpoint in right triangle
G H J
G H BBookmark kLocked kLocked NReply
Source: Iranian Geometry Olympiad 2016 Medium 4
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cjquines0
510 posts
#1 • 5 Y
Y by itslumi, tony88, Adventure10, Rounak_iitr, ItsBesi
Let $\omega$ be the circumcircle of right-angled triangle $ABC$ ($\angle A = 90^{\circ}$). The tangent to $\omega$ at point $A$ intersects the line $BC$ at point $P$. Suppose that $M$ is the midpoint of the minor arc $AB$, and $PM$ intersects $\omega$ for the second time in $Q$. The tangent to $\omega$ at point $Q$ intersects $AC$ at $K$. Prove that $\angle PKC = 90^{\circ}$.

Proposed by Davood Vakili
This post has been edited 1 time. Last edited by cjquines0, May 26, 2017, 11:15 AM
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iliyash
600 posts
#2 • 2 Y
Y by Adventure10, Mango247
Here's the full problems and solution link:
http://igo-official.ir/wp-content/uploads/2016/12/igo_2016_-_problems__solutions_-_english.pdf
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aleksam
101 posts
#3 • 5 Y
Y by Pluto1708, GeoKing, Adventure10, Mango247, redred123
We will use the well-known fact that the tangent to the circumcircle of the triangle $ABC$ divides the side $BC$ in ratio $-\frac {AB^2}{AC^2}$. Before applying it, let note a Lemma.
Lemma:$\frac{AQ}{CQ}=\frac{AB}{AC}$.
Proof: This can be proven considering the inversion with center $P$ and radius $PA$, and as the circle of inversion is orthogonal to $\omega$, $(M, Q)$ and $(B,C)$ are excanging places. So, by applying the inverison distance formula, we get $\frac{AQ}{CQ}=\frac{AM\times BP}{BM\times AP}=\frac{BP}{AP}=\frac{AB}{AC}$.
Further, we easily compute the ratios $\frac{AK}{KC}=\frac{AQ^2}{CQ^2}=\frac{AB^2}{AC^2}=\frac{BP}{CP}$, and by Thales, we have the conclusion.
This post has been edited 2 times. Last edited by aleksam, May 26, 2017, 11:47 AM
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Euler365
143 posts
#4 • 2 Y
Y by tony88, Adventure10
I too have inversion solution but it is different.
Let $K'$ be on $AC$ such that $\angle PK'C = 90^{\circ}$. Let $N$ be midpoint of $BC$. We shall prove that $K'Q$ is tangent to $\omega$.
Perform an inversion with centre $P$ and radius $PA$.
Then $\omega$ gets mapped to $\omega$ itself. $K'$ goes to $PK' \cap AN$ say $L$ and $Q$ goes to $M$. So line $K'Q$ gets mapped to $\odot (PLM)$.
Now note that $MN$ is perpendicular bisector of $AB$ and hence $MN$ is perpendicular bisector of $PL$.
However circumcentre of $\odot (PLM)$ lies on perpendicular bisector of $PL$.
Also $N$ is circumcentre of $\omega$.
$\therefore$ The centres of $\omega , \odot (PLM)$ and their intersection pts. are collinear.
So $\omega$ and $\odot (PLM)$ are tangent to each other. $\therefore$ before inversion $K'Q$ and $\omega$ must have been tangent to each other as desired.
This post has been edited 4 times. Last edited by Euler365, Sep 15, 2019, 9:29 AM
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ShinyDitto
63 posts
#5 • 5 Y
Y by Jafarly8097, itslumi, shalomrav, popdit, Professor33
I am not sure if this works for $AB>AC$ but I hope it does.

Let $AQ$ and $MC$ meet at $R$. Pascal on $AQQMCA$ shows that $R$, $K$ and $P$ are collinear. Pascal on $ABCMMQ$ shows that $PR$ is parallel to $AB$. Thus $PK$ is parallel to $AB$ and $\angle PKC=90^{\circ}$.
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JustKeepRunning
2958 posts
#6
Y by
@above, how can you use pascal if the lines $AB$ and $MM$ do not even intersect? I have not learned about this use of pascal before, so could you explain?
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Mahdi_Mashayekhi
697 posts
#7
Y by
we have to prove PK || AB or PB/PC = KA/KC.
first note that PMA and PAQ are similar. PMB and PCQ are similar. PBA and PAC are similar.

PB/PC = (AB/AC)^2 and KA/KC = (QA/QC)^2
so we have to prove AB/AC = QA/QC.
AB/AC = PB/PA
PA/PQ = AM/AQ and PQ/PB = QC/BM so PB/PA = QA/QC.
we're Done.
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tony88
5 posts
#8
Y by
Euler365 wrote:
I too have inversion solution but it is different.

Thank you for your very good inversion solution!!
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lian_the_noob12
173 posts
#9 • 2 Y
Y by Rounak_iitr, LeYohan
Finally Solved With $\textbf{Pascal's Theorem}$ :)

$\color{red} \boxed{\textbf{SOLUTION}}$

Let, $AQ \cap MC \equiv X, AB \cap AC \equiv Y$

By $\textbf{Pascal's Theorem}$ on $Q,Q,A,A,C,M$ We get $QQ \cap AC \equiv K, QA \cap CM \equiv X, AA \cap MQ \equiv P$
So, $P,X,K$ are collinear.

We need to show $PK \equiv PX \parallel AB$
As $M$ is the midpoint of minor arc $AB, CM$ is the angle bisector of $\angle ACB$
Let, $\angle ACM=\angle BCM=\alpha \implies \angle ABC=90-2\alpha$
Now,
$$\angle PQX=\angle AQM=\angle ACM=\angle BCM=\angle PCX$$$\implies PQCX$ cyclic
Therefore,
$$\angle CXK=\angle PQC=\angle PQX + \angle XQC = \alpha + \angle ABC=\alpha + 90- 2\alpha=90-\alpha$$So,
$$\angle  AYC=90- \alpha= \angle CXK=\angle YXK \implies AY \parallel PX \implies AB \parallel PX \implies AB \parallel PK \implies \angle PKC=\angle BAC=90 \blacksquare$$
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ACGNmath
327 posts
#11
Y by
It suffices to show that $PK\parallel AB$. This can be done with 3 applications of Pascal's Theorem. Let $T=MA\cap QB$ and $X=QA\cap CM$.

Pascal's on $QQMACB$ yields $K, P, T$ are collinear.

Pascal's on $ABCMMQ$ yields $PX\parallel AB$ (since $MM$ intersects $AB$ at the point at infinity along that line)

Pascal's on $QQAACM$ yields $K,X,P$ are collinear.

Therefore, $K,P,T,X$ are collinear and these lie on a line parallel to $AB$, so we are done.
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Tuguldur
19 posts
#12
Y by
It's just angle chasing
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ItsBesi
147 posts
#13 • 1 Y
Y by sami1618
Nice problem!
From Pascal Theorem on $(Q,Q,M,C,A,A)$ we get that the intersections of these lines are collinear:
$$QQ \cap AC=\{K\}$$$$QM \cap AA=\{P\}$$$$MC \cap AQ=\{X\}$$
Hence: Points $\overline{K-P-X}$ are collinear

Claim Points $P$,$C$,$Q$ and $X$ are concyclic
Claim
$\angle CXQ \equiv \angle CXA \stackrel{\triangle CXA}{=} 180-\angle XAC-\angle XCA=\angle CAQ-\angle MCA \stackrel{\omega}{=} \angle CMQ-\angle MCA=\angle CMQ-\angle BCM=$
$=180-\angle PMQ-\angle BCM \equiv 180-\angle PMQ-\angle PCM \stackrel{\triangle PMC}{=} \angle MPC \equiv \angle QPC \implies \angle CXQ=\angle CPQ \implies$
Points $P$,$C$,$Q$ and $X$ are concyclic $\square$

Claim $\angle PKC=90^{\circ}$
Claim
$\angle PKC \equiv \angle XKA \stackrel{\triangle XKA}{=} 180-\angle KXA-\angle KAX=\angle PXA-\angle QAC \equiv \angle PXQ-\angle QAC \stackrel{\odot(PCQX)}{=} 180-\angle PCQ-\angle QAC$
$ \equiv 180-\angle BCQ-\angle QAC \stackrel{\omega}{=} \angle QAB-\angle QAC=\angle BAC=90^{\circ} \implies$
$$\angle PKC=90^{\circ} \blacksquare$$
Nice problem!
From Pascal Theorem on $(Q,Q,M,C,A,A)$ we get that the intersections of these lines are collinear:
$$QQ \cap AC=\{K\}$$$$QM \cap AA=\{P\}$$$$MC \cap AQ=\{X\}$$
Hence: Points $\overline{K-P-X}$ are collinear

Claim Points $P$,$C$,$Q$ and $X$ are concyclic
Claim
$\angle CXQ \equiv \angle CXA \stackrel{\triangle CXA}{=} 180-\angle XAC-\angle XCA=\angle CAQ-\angle MCA \stackrel{\omega}{=} \angle CMQ-\angle MCA=\angle CMQ-\angle BCM=$
$=180-\angle PMQ-\angle BCM \equiv 180-\angle PMQ-\angle PCM \stackrel{\triangle PMC}{=} \angle MPC \equiv \angle QPC \implies \angle CXQ=\angle CPQ \implies$
Points $P$,$C$,$Q$ and $X$ are concyclic $\square$

Claim $\angle PKC=90^{\circ}$
Claim
$\angle PKC \equiv \angle XKA \stackrel{\triangle XKA}{=} 180-\angle KXA-\angle KAX=\angle PXA-\angle QAC \equiv \angle PXQ-\angle QAC \stackrel{\odot(PCQX)}{=} 180-\angle PCQ-\angle QAC$
$ \equiv 180-\angle BCQ-\angle QAC \stackrel{\omega}{=} \angle QAB-\angle QAC=\angle BAC=90^{\circ} \implies$
$$\angle PKC=90^{\circ} \blacksquare$$

Edit: I just relaized that you can finish this way easier.
After you prove that points $P$,$C$,$Q$ and $X$ are concyclic, we also have points $B$,$A$,$Q$ and $C$ are concyclic combining with $\overline{Q-A-X}$ and $\overline{C-B-P}$ are collinear by Reim's Theorem you get that: $BA \parallel PX \implies BA \parallel PK \implies \angle PKC=\angle BAC=90 \implies \angle PKC=90^{\circ} $
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This post has been edited 1 time. Last edited by ItsBesi, Apr 30, 2025, 8:30 PM
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