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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

Our full course list for upcoming classes is below:
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0 replies
jwelsh
Jul 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
IMO 2025 Medal Cutoffs Prediction
GreenTea2593   42
N 9 minutes ago by swynca
What are your prediction for IMO 2025 medal cutoffs?
42 replies
GreenTea2593
Today at 4:44 AM
swynca
9 minutes ago
two sequences of positive integers and inequalities
rmtf1111   54
N 12 minutes ago by Luckylily
Source: EGMO 2019 P5
Let $n\ge 2$ be an integer, and let $a_1, a_2, \cdots , a_n$ be positive integers. Show that there exist positive integers $b_1, b_2, \cdots, b_n$ satisfying the following three conditions:

$\text{(A)} \ a_i\le b_i$ for $i=1, 2, \cdots , n;$

$\text{(B)} \ $ the remainders of $b_1, b_2, \cdots, b_n$ on division by $n$ are pairwise different; and

$\text{(C)} \ $ $b_1+b_2+\cdots b_n \le n\left(\frac{n-1}{2}+\left\lfloor \frac{a_1+a_2+\cdots a_n}{n}\right \rfloor \right)$

(Here, $\lfloor x \rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)
54 replies
rmtf1111
Apr 10, 2019
Luckylily
12 minutes ago
Conic geo for the win
AlephG_64   0
12 minutes ago
Source: knamprihodilinoneseichas
Let $\mathcal{P}$ be a parabola that passes through the vertices of a triangle $ABC$. Suppose there exists a circle $\Theta$ that is tangent to $\mathcal{P}$ at two points and tangent to the circumcircle of $ABC$. Prove $\Theta$ is tangent to an excircle or incircle of $ABC$.
0 replies
AlephG_64
12 minutes ago
0 replies
IMO 2025 P2
sarjinius   63
N 16 minutes ago by lksb
Source: 2025 IMO P2
Let $\Omega$ and $\Gamma$ be circles with centres $M$ and $N$, respectively, such that the radius of $\Omega$ is less than the radius of $\Gamma$. Suppose $\Omega$ and $\Gamma$ intersect at two distinct points $A$ and $B$. Line $MN$ intersects $\Omega$ at $C$ and $\Gamma$ at $D$, so that $C, M, N, D$ lie on $MN$ in that order. Let $P$ be the circumcentre of triangle $ACD$. Line $AP$ meets $\Omega$ again at $E\neq A$ and meets $\Gamma$ again at $F\neq A$. Let $H$ be the orthocentre of triangle $PMN$.

Prove that the line through $H$ parallel to $AP$ is tangent to the circumcircle of triangle $BEF$.
63 replies
sarjinius
Yesterday at 3:38 AM
lksb
16 minutes ago
Greatest algebra ever
EpicBird08   13
N 18 minutes ago by Assassino9931
Source: ISL 2024/A2
Let $n$ be a positive integer. Find the minimum possible value of
\[
S = 2^0 x_0^2 + 2^1 x_1^2 + \dots + 2^n x_n^2,
\]where $x_0, x_1, \dots, x_n$ are nonnegative integers such that $x_0 + x_1 + \dots + x_n = n$.
13 replies
EpicBird08
Today at 3:00 AM
Assassino9931
18 minutes ago
Hard sequence
straight   0
19 minutes ago
Source: Own
Consider a sequence $(a_n)_n, n \rightarrow \infty$ of real numbers.

Consider an infinite $\mathbb{N} \times \mathbb{N}$ grid $a_{i,j}$. In the first row of this grid, we place $a_0$ in every square ($a_{0,n} = a_0)$. In the first column of this grid, we place $a_n$ in the $n$-th square ($a_{n,0} = a_n)$.
Next, fill up the grid according to the following rule: $a_{i,j} = a_{i-1,j} + a_{i,j-1}$.

If $\lim_{i \rightarrow \infty} a_{i,j} = \infty$ for all $j = 0,1,...$, does this mean that $a_n = 0$ for all $n$?

Hint?
0 replies
straight
19 minutes ago
0 replies
The inekoalaty game
sarjinius   26
N 40 minutes ago by straight
Source: 2025 IMO P5
Alice and Bazza are playing the inekoalaty game, a two‑player game whose rules depend on a positive real number $\lambda$ which is known to both players. On the $n$th turn of the game (starting with $n=1$) the following happens:
[list]
[*] If $n$ is odd, Alice chooses a nonnegative real number $x_n$ such that
\[
    x_1 + x_2 + \cdots + x_n \le \lambda n.
  \][*]If $n$ is even, Bazza chooses a nonnegative real number $x_n$ such that
\[
    x_1^2 + x_2^2 + \cdots + x_n^2 \le n.
  \][/list]
If a player cannot choose a suitable $x_n$, the game ends and the other player wins. If the game goes on forever, neither player wins. All chosen numbers are known to both players.

Determine all values of $\lambda$ for which Alice has a winning strategy and all those for which Bazza has a winning strategy.

Proposed by Massimiliano Foschi and Leonardo Franchi, Italy
26 replies
+1 w
sarjinius
Today at 3:18 AM
straight
40 minutes ago
BMO Shortlist 2021 G6
Lukaluce   23
N an hour ago by Rayvhs
Source: BMO Shortlist 2021
Let $ABC$ be an acute triangle such that $AB < AC$. Let $\omega$ be the circumcircle of $ABC$
and assume that the tangent to $\omega$ at $A$ intersects the line $BC$ at $D$. Let $\Omega$ be the circle with
center $D$ and radius $AD$. Denote by $E$ the second intersection point of $\omega$ and $\Omega$. Let $M$ be the
midpoint of $BC$. If the line $BE$ meets $\Omega$ again at $X$, and the line $CX$ meets $\Omega$ for the second
time at $Y$, show that $A, Y$, and $M$ are collinear.

Proposed by Nikola Velov, North Macedonia
23 replies
Lukaluce
May 8, 2022
Rayvhs
an hour ago
quadrilateral geo with length conditions
OronSH   9
N an hour ago by player-019
Source: IMO Shortlist 2024 G1
Let $ABCD$ be a cyclic quadrilateral such that $AC<BD<AD$ and $\angle DBA<90^\circ$. Point $E$ lies on the line through $D$ parallel to $AB$ such that $E$ and $C$ lie on opposite sides of line $AD$, and $AC=DE$. Point $F$ lies on the line through $A$ parallel to $CD$ such that $F$ and $C$ lie on opposite sides of line $AD$, and $BD=AF$.

Prove that the perpendicular bisectors of segments $BC$ and $EF$ intersect on the circumcircle of $ABCD$.

Proposed by Mykhailo Shtandenko, Ukraine
9 replies
OronSH
Today at 3:13 AM
player-019
an hour ago
k What happened to 2025 IMO P4 post?
sarjinius   3
N an hour ago by AltruisticApe
Contest is over, why deleted?
3 replies
1 viewing
sarjinius
Today at 4:30 PM
AltruisticApe
an hour ago
Periodic sequence
EeEeRUT   5
N an hour ago by dangerousliri
Source: Isl 2024 A5
Find all periodic sequence $a_1,a_2,\dots$ of real numbers such that the following conditions hold for all $n\geqslant 1$:$$a_{n+2}+a_{n}^2=a_n+a_{n+1}^2\quad\text{and}\quad |a_{n+1}-a_n|\leqslant 1.$$
Proposed by Dorlir Ahmeti, Kosovo
5 replies
EeEeRUT
Today at 3:01 AM
dangerousliri
an hour ago
Bonza functions
KevinYang2.71   51
N an hour ago by Oznerol1
Source: 2025 IMO P3
Let $\mathbb{N}$ denote the set of positive integers. A function $f\colon\mathbb{N}\to\mathbb{N}$ is said to be bonza if
\[
f(a)~~\text{divides}~~b^a-f(b)^{f(a)}
\]for all positive integers $a$ and $b$.

Determine the smallest real constant $c$ such that $f(n)\leqslant cn$ for all bonza functions $f$ and all positive integers $n$.

Proposed by Lorenzo Sarria, Colombia
51 replies
+1 w
KevinYang2.71
Yesterday at 3:38 AM
Oznerol1
an hour ago
Next term is sum of three largest proper divisors
vsamc   3
N an hour ago by KevinYang2.71
Source: 2025 IMO P4
A proper divisor of a positive integer $N$ is a positive divisor of $N$ other than $N$ itself.

The infinite sequence $a_1, a_2, \cdots$ consists of positive integers, each of which has at least three proper divisors. For each $n\geq 1$, the integer $a_{n+1}$ is the sum of the three largest proper divisors of $a_n$.

Determine all possible values of $a_1$.
3 replies
vsamc
an hour ago
KevinYang2.71
an hour ago
2024 International Math Olympiad Number Theory Shortlist, Problem 3
brainfertilzer   10
N 2 hours ago by vsamc
Source: 2024 ISL N3
Determine all sequences $a_1, a_2, \dots$ of positive integers such that for any pair of positive integers $m\leqslant n$, the arithmetic and geometric means
\[ \frac{a_m + a_{m+1} + \cdots + a_n}{n-m+1}\quad\text{and}\quad (a_ma_{m+1}\cdots a_n)^{\frac{1}{n-m+1}}\]are both integers.
10 replies
brainfertilzer
Today at 3:00 AM
vsamc
2 hours ago
Fixed and variable points
BR1F1SZ   2
N Jun 4, 2025 by hukilau17
Source: 2025 Francophone MO Seniors P3
Let $\omega$ be a circle with center $O$. Let $B$ and $C$ be two fixed points on the circle $\omega$ and let $A$ be a variable point on $\omega$. We denote by $X$ the intersection point of lines $OB$ and $AC$, assuming $X \neq O$. Let $\gamma$ be the circumcircle of triangle $\triangle AOX$. Let $Y$ be the second intersection point of $\gamma$ with $\omega$. The tangent to $\gamma$ at $Y$ intersects $\omega$ at $I$. The line $OI$ intersects $\omega$ at $J$. The perpendicular bisector of segment $OY$ intersects line $YI$ at $T$, and line $AJ$ intersects $\gamma$ at $P$. We denote by $Z$ the second intersection point of the circumcircle of triangle $\triangle PYT$ with $\omega$. Prove that, as point $A$ varies, points $Y$ and $Z$ remain fixed.
2 replies
BR1F1SZ
May 11, 2025
hukilau17
Jun 4, 2025
Fixed and variable points
G H J
G H BBookmark kLocked kLocked NReply
Source: 2025 Francophone MO Seniors P3
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BR1F1SZ
592 posts
#1
Y by
Let $\omega$ be a circle with center $O$. Let $B$ and $C$ be two fixed points on the circle $\omega$ and let $A$ be a variable point on $\omega$. We denote by $X$ the intersection point of lines $OB$ and $AC$, assuming $X \neq O$. Let $\gamma$ be the circumcircle of triangle $\triangle AOX$. Let $Y$ be the second intersection point of $\gamma$ with $\omega$. The tangent to $\gamma$ at $Y$ intersects $\omega$ at $I$. The line $OI$ intersects $\omega$ at $J$. The perpendicular bisector of segment $OY$ intersects line $YI$ at $T$, and line $AJ$ intersects $\gamma$ at $P$. We denote by $Z$ the second intersection point of the circumcircle of triangle $\triangle PYT$ with $\omega$. Prove that, as point $A$ varies, points $Y$ and $Z$ remain fixed.
Z K Y
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Lam040208
15 posts
#2
Y by
BR1F1SZ wrote:
Let $\omega$ be a circle with center $O$. Let $B$ and $C$ be two fixed points on the circle $\omega$ and let $A$ be a variable point on $\omega$. We denote by $X$ the intersection point of lines $OB$ and $AC$, assuming $X \neq O$. Let $\gamma$ be the circumcircle of triangle $\triangle AOX$. Let $Y$ be the second intersection point of $\gamma$ with $\omega$. The tangent to $\gamma$ at $Y$ intersects $\omega$ at $I$. The line $OI$ intersects $\omega$ at $J$. The perpendicular bisector of segment $OY$ intersects line $YI$ at $T$, and line $AJ$ intersects $\gamma$ at $P$. We denote by $Z$ the second intersection point of the circumcircle of triangle $\triangle PYT$ with $\omega$. Prove that, as point $A$ varies, points $Y$ and $Z$ remain fixed.
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hukilau17
294 posts
#3
Y by
Quick complex bash. Identify $\omega$ with the unit circle so that
$$o=0$$$$|a|=|b|=|c|=1$$$$x=\frac{ac(b-b) - b(-b)(a+c)}{ac-b(-b)} = \frac{b^2(a+c)}{b^2+ac}$$Let $G$ be the center of $\gamma$ so that
$$g = \frac{ax(\overline{a}-\overline{x})}{\overline{a}x-a\overline{x}} = \frac{ax\left(\frac1a-\frac{x}{b^2}\right)}{\frac{x}a-\frac{ax}{b^2}} = \frac{a(ax-b^2)}{a^2-b^2} = \frac{\frac{a^2b^2(a+c)}{b^2+ac}-ab^2}{a^2-b^2} = \frac{ab^2}{b^2+ac}$$Then
$$y = \frac{\overline{a}g}{\overline{g}} = \frac{\frac{b^2}{b^2+ac}}{\frac{c}{b^2+ac}} = \frac{b^2}c$$Note that the coordinate of $Y$ does not depend on $a$. We also have (letting $k$ denote the coordinate of $I$)
$$k = -\frac{y-g}{y\overline{g}-1} = -\frac{\frac{b^2}c-\frac{ab^2}{b^2+ac}}{\frac{b^2}{b^2+ac}-1} = \frac{b^4}{ac^2}$$$$j = -k = -\frac{b^4}{ac^2}$$$$p = j + g - aj\overline{g} = -\frac{b^4}{ac^2} + \frac{ab^2}{b^2+ac} + \frac{b^4}{c(b^2+ac)} = \frac{b^2(a^2c^2-b^4)}{ac^2(b^2+ac)} = \frac{b^2(ac-b^2)}{ac^2}$$Now we find the coordinate of $T$. Since $T$ lies on the perpendicular bisector of $OY$, we have
$$|t-o| = |t-y|$$$$t\overline{t} = \left(t-\frac{b^2}c\right)\left(\overline{t}-\frac{c}{b^2}\right) = t\overline{t} - \frac{ct}{b^2} - \frac{b^2\overline{t}}c + 1$$$$b^4\overline{t} + c^2t = b^2c \implies \overline{t} = \frac{c(b^2-ct)}{b^4}$$Since $T$ lies on line $YI$, we have
$$\overline{t} = \frac{y+k-t}{yk} = \frac{\frac{b^2}c+\frac{b^4}{ac^2}-t}{\frac{b^6}{ac^3}} = \frac{c(ab^2c+b^4-ac^2t)}{b^6}$$Intersecting these gives
$$\frac{c(b^2-ct)}{b^4} = \frac{c(ab^2c+b^4-ac^2t)}{b^6}$$$$b^4-b^2ct = ab^2c+b^4-ac^2t$$$$t = \frac{ab^2}{ac-b^2}$$Let $U$ denote the circumcenter of $\triangle PYT$. We have the vectors
$$p' = p-y = -\frac{b^4}{ac^2}$$$$t' = t-y = \frac{b^4}{c(ac-b^2)}$$Then if $u' = u-y$, we have
\begin{align*}
u' &= \frac{p't'(\overline{p'} - \overline{t'})}{\overline{p'}t' - p'\overline{t'}} \\
&= \frac{-\frac{b^8}{ac^3(ac-b^2)}\left[-\frac{ac^2}{b^4} + \frac{ac^2}{b^2(ac-b^2)}\right]}{-\frac{ac}{ac-b^2} - \frac{b^2}{ac-b^2}} \\
&= \frac{\frac{b^6}{c(ac-b^2)^2} - \frac{b^4}{c(ac-b^2)}}{\frac{ac}{ac-b^2} + \frac{b^2}{ac-b^2}} \\
&= \frac{b^6 - b^4(ac-b^2)}{c(ac-b^2)(ac+b^2)} \\
&= \frac{b^4(2b^2-ac)}{c(a^2c^2-b^4)}
\end{align*}and then
$$u = u' + y = \frac{b^2(a^2c^2-ab^2c+b^4)}{c(a^2c^2-b^4)}$$$$z = \frac{\overline{y}u}{\overline{u}} = \frac{\frac{a^2c^2-ab^2c+b^4}{a^2c^2-b^4}}{\frac{c(b^4-ab^2c+a^2c^2)}{b^2(b^4-a^2c^2)}} = -\frac{b^2}c$$So the coordinate of $Z$ does not depend on $a$ either. $\blacksquare$
Z K Y
N Quick Reply
G
H
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a