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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
[EGMO2] Domino tilings
socrates   22
N an hour ago by lpieleanu
Source: EGMO 2015, Problem 2
A domino is a $2 \times 1$ or $1 \times  2$ tile. Determine in how many ways exactly $n^2$ dominoes can be placed without overlapping on a $2n \times  2n$ chessboard so that every $2 \times  2$ square contains at least two uncovered unit squares which lie in the same row or column.
22 replies
socrates
Apr 16, 2015
lpieleanu
an hour ago
IMO online scoreboard
Shayan-TayefehIR   97
N an hour ago by PokemonMaster2012
Is there still an active link for IMO's online scoreboard?, I guess the scoring process is not over yet and that old link doesn't work...
97 replies
Shayan-TayefehIR
Today at 5:31 AM
PokemonMaster2012
an hour ago
2025 German Mathematical Olympiad \\[3pt] Problem 4
unicornfly   11
N an hour ago by Shan3t
there is another solution ?

\textbf{Problem.}
Let \( a, b, c > 0 \). Prove that
\[
\boxed{
\frac{a^5}{b^2} + \frac{b}{c} + \frac{c^3}{a^2} > 2a
}
\]
\textbf{Solution.}

We apply the AM–GM inequality to the following decomposition of the left-hand side:

\[
\frac{a^5}{b^2} + \frac{b}{c} + \frac{c^3}{a^2}
=
3 \cdot \frac{a^5}{3b^2}
+ 6 \cdot \frac{b}{6c}
+ 2 \cdot \frac{c^3}{2a^2}.
\]
This gives a total of 11 positive terms, so by the AM–GM inequality:

\[
\frac{a^5}{b^2} + \frac{b}{c} + \frac{c^3}{a^2}
\ge
11 \cdot
\sqrt[11]{\left( \frac{a^5}{3b^2} \right)^3
\cdot \left( \frac{b}{6c} \right)^6
\cdot \left( \frac{c^3}{2a^2} \right)^2}.
\]
Let us simplify the product inside the root:

\[
\left( \frac{a^5}{3b^2} \right)^3
= \frac{a^{15}}{3^3 b^6}, \qquad
\left( \frac{b}{6c} \right)^6
= \frac{b^6}{6^6 c^6}, \qquad
\left( \frac{c^3}{2a^2} \right)^2
= \frac{c^6}{2^2 a^4}.
\]
Multiplying them gives:

\[
\frac{a^{15}}{3^3 b^6} \cdot
\frac{b^6}{6^6 c^6} \cdot
\frac{c^6}{2^2 a^4}
=
\frac{a^{15 - 4} \cdot b^6 \cdot c^6}
{3^3 \cdot 6^6 \cdot 2^2 \cdot b^6 \cdot c^6}
=
\frac{a^{11}}{3^3 \cdot 2^2 \cdot 6^6}.
\]
Now, we use the estimate:
\[
3^3 \cdot 2^2 \cdot 6^6
= 27 \cdot 4 \cdot (2 \cdot 3)^6
= 27 \cdot 4 \cdot 2^6 \cdot 3^6
= 2^8 \cdot 3^9.
\]
Hence,
\[
\frac{a^5}{b^2} + \frac{b}{c} + \frac{c^3}{a^2}
\ge
11 \cdot \sqrt[11]{\frac{a^{11}}{2^8 \cdot 3^9}}
= \frac{11a}{\sqrt[11]{2^8 \cdot 3^9}}.
\]
Finally, we observe that
\[
\sqrt[11]{2^8 \cdot 3^9}
< \frac{11}{2} \quad \Rightarrow \quad
\frac{11a}{\sqrt[11]{2^8 \cdot 3^9}} > 2a.
\]
\textbf{Conclusion:}

\[
\boxed{
\frac{a^5}{b^2} + \frac{b}{c} + \frac{c^3}{a^2} > 2a
}
\qquad \text{for all } a,b,c>0.
\]
11 replies
unicornfly
Today at 1:48 AM
Shan3t
an hour ago
Nice and easy FE on R+
sttsmet   23
N an hour ago by Maths_VC
Source: EMC 2024 Problem 4, Seniors
Find all functions $ f: \mathbb{R}^{+} \to \mathbb{R}^{+}$ such that $f(x+yf(x)) = xf(1+y)$
for all x, y positive reals.
23 replies
sttsmet
Dec 23, 2024
Maths_VC
an hour ago
Tangents inducing isogonals
nikolapavlovic   58
N an hour ago by hectorleo123
Source: Serbian MO 2017 6
Let $k$ be the circumcircle of $\triangle ABC$ and let $k_a$ be A-excircle .Let the two common tangents of $k,k_a$ cut $BC$ in $P,Q$.Prove that $\measuredangle PAB=\measuredangle CAQ$.
58 replies
nikolapavlovic
Apr 2, 2017
hectorleo123
an hour ago
geometry is mAJorly BACk
EpicBird08   11
N an hour ago by hectorleo123
Source: ISL 2024/G6
Let $ABC$ be an acute triangle with $AB < AC$, and let $\Gamma$ be the circumcircle of $ABC$. Points $X$ and $Y$ lie on $\Gamma$ so that $XY$ and $BC$ meet on the external angle bisector of $\angle BAC$. Suppose that the tangents to $\Gamma$ at $x$ and $Y$ intersect at a point $T$ on the same side of $BC$ as $A$, and that $TX$ and $TY$ intersect $BC$ at $U$ and $V$, respectively. Let $J$ be the centre of the excircle of triangle $TUV$ opposite the vertex $T$.

Prove that $AJ$ bisects $\angle BAC$.
11 replies
EpicBird08
Jul 16, 2025
hectorleo123
an hour ago
Lots of perpendiculars; compute HQ/HR
MellowMelon   56
N 2 hours ago by Kempu33334
Source: USA TST 2011 P1
In an acute scalene triangle $ABC$, points $D,E,F$ lie on sides $BC, CA, AB$, respectively, such that $AD \perp BC, BE \perp CA, CF \perp AB$. Altitudes $AD, BE, CF$ meet at orthocenter $H$. Points $P$ and $Q$ lie on segment $EF$ such that $AP \perp EF$ and $HQ \perp EF$. Lines $DP$ and $QH$ intersect at point $R$. Compute $HQ/HR$.

Proposed by Zuming Feng
56 replies
MellowMelon
Jul 26, 2011
Kempu33334
2 hours ago
Find all real functions withf(x^2 + yf(z)) = xf(x) + zf(y)
Rushil   33
N 2 hours ago by Fly_into_the_sky
Source: INMO 2005 Problem 6
Find all functions $f : \mathbb{R} \longrightarrow \mathbb{R}$ such that \[ f(x^2 + yf(z)) = xf(x) + zf(y) , \] for all $x, y, z \in \mathbb{R}$.
33 replies
Rushil
Aug 23, 2005
Fly_into_the_sky
2 hours ago
Hard tiling problem
Anto0110   3
N 2 hours ago by Anto0110
Source: Belarus I think
A pointed corner is defined as the union of the unit square \([0,1] \times [0,1]\) with the two right isosceles triangles
with vertices \((0,1), (1,1), (0,2)\) and \((1,0), (2,1), (1,1)\).
Prove that you cannot tile a rectangle with pointed corners.
3 replies
Anto0110
Jul 16, 2025
Anto0110
2 hours ago
Minimal of xy subjected to a constraint!
persamaankuadrat   8
N 2 hours ago by iniffur
Let $x,y$ be positive real numbers such that

$$x+y^{2}+x^{3} = 1481$$
Find the minimal value of $xy$
8 replies
persamaankuadrat
Yesterday at 1:44 PM
iniffur
2 hours ago
Club Truncator
4everwise   12
N 2 hours ago by neeyakkid23
Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac{1}{3}$. The probability that Club Truncator will finish the season with more wins than losses is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
12 replies
4everwise
Dec 28, 2006
neeyakkid23
2 hours ago
IMO 2025 facts and figures
oolite   0
2 hours ago
Source: https://www.imo-official.org
Note: the official statistics are here.

Medal boundaries

One property you'd hope for in a good IMO is that there are relatively few contestants at the borderline between medals.

For each of the past several years, we calculate the percentage number of contestants having the highest silver/bronze/no-medal score (unlucky) or the lowest gold/silver/bronze score (lucky). In the following chart, that number is shown in red text. The coloured bars show the relative proportions of none:bronze:silver:gold medals, with red hatching over the "danger zone" of lucky and unlucky scores: chart.This year, an unremarkable 19.7% of students found themselves in the danger zone, but it was a happy IMO for most of those on the borderline; just look at how many borderline golds were awarded!

Question difficulty and "impact"

The following scatterplot shows impact (defined here) vs. mean score for all problems from 2000 to the present, with this year's problems plotted in yellow (day 1) and cyan (day 2). Problems with higher impact are those, like Turbo the snail from 2024, that have a randomising effect on candidates' total scores:scatterplot.There was no Turbo-like grade-wrecker problem this year.

The following table shows the mean score per problem, and overall. The rank is relative to other IMOs since 2000, with rank #1 being the hardest example of that problem and #26 being the easiest:table.Problems 2, 3, 4 and 5 were uncharacteristically easy (or high-scoring) this year. Just look at the contrast between P3 (easiest this millennium, relative to other P3s) and P6 (third hardest)!

Rare scores

Here, we're considering single-day scores from IMO 1979 to the present, i.e. the "modern era" during which each day was worth $3\times 7$ marks. All bar 42 of the $512$ possible ordered triples have been seen in that period.

From this year's IMO, here are the single-day scores that, until this year, had not been seen for a very long time:lost & found.Alternatively, sorting by the number of prior sightings of the score, we get:low counts.No prizes this year for unearthing any hitherto-unobserved scores, and the devilish 6,6,6 remains reassuringly absent from the scoreboards.
0 replies
oolite
2 hours ago
0 replies
Right Triangle and Circles
4everwise   11
N 3 hours ago by neeyakkid23
Let $\triangle{PQR}$ be a right triangle with $PQ=90$, $PR=120$, and $QR=150$. Let $C_{1}$ be the inscribed circle. Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$, such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$. Construct $\overline{UV}$ with $U$ on $\overline{PQ}$ and $V$ on $\overline{QR}$ such that $\overline{UV}$ is perpendicular to $\overline{PQ}$ and tangent to $C_{1}$. Let $C_{2}$ be the inscribed circle of $\triangle{RST}$ and $C_{3}$ the inscribed circle of $\triangle{QUV}$. The distance between the centers of $C_{2}$ and $C_{3}$ can be written as $\sqrt{10n}$. What is $n$?
11 replies
4everwise
Dec 28, 2006
neeyakkid23
3 hours ago
Trivial Fe
Tofa7a._.36   3
N 4 hours ago by ulis7s
Find all functions $f : \mathbb R \rightarrow \mathbb R$ such that:
$$f(f(2x+y)) + f(x) = 2x + f(x+y)$$For all real numbers $x,y \in \mathbb{R}$.
3 replies
Tofa7a._.36
Jul 15, 2025
ulis7s
4 hours ago
Combinatorics
AlexCenteno2007   2
N May 11, 2025 by Royal_mhyasd
Adrian and Bertrand take turns as follows: Adrian starts with a pile of ($n\geq 3$) stones. On their turn, each player must divide a pile. The player who can make all piles have at most 2 stones wins. Depending on n, determine which player has a winning strategy.
2 replies
AlexCenteno2007
May 9, 2025
Royal_mhyasd
May 11, 2025
Combinatorics
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AlexCenteno2007
166 posts
#1
Y by
Adrian and Bertrand take turns as follows: Adrian starts with a pile of ($n\geq 3$) stones. On their turn, each player must divide a pile. The player who can make all piles have at most 2 stones wins. Depending on n, determine which player has a winning strategy.
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AlexCenteno2007
166 posts
#2
Y by
Could you help me?
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Royal_mhyasd
115 posts
#3 • 1 Y
Y by AlexCenteno2007
Nice one. There's quite a few solutions as well but I'll just leave the one I thought of
We're going to prove that Adrian wins for n=3 and n=even and Bertrand wins for n=odd, n>3.
n=3 is obvious - A wins by dividing it into a pile of 1 and a pile of 2.
For n>=4, we're gonna use induction. (our induction is that A wins when n=3 and n=even and B wins when n=odd, n>3)
If n=even, A can simply divide his pile into a pile of 1 and a pile of n-1. n-1 is odd, so the player who is about to make a move (player B) is gonna lose, so A wins.
If n=odd, A is gonna divide the pile into two piles of a and b stones, a<b, a+b = n so either a is odd and b is even or a is even and b is odd.
If b-a isn't 3 :
Bertrand is gonna divide the current piles into 3 piles of : b-a, a, a stones. We define "the first zone" as any move that can be traced back to the pile of b-a stones and "the second zone" as any love that can be traced back to the piles of a stones. If A makes a move in the second zone, B will simply mirror him in the other sub-zone (the one that can be traced back to the other pile of a stones). Therefore, it's pretty obvious that B will make the last move in the second zone, seeing as he can always mirror A. b-a is odd because a and b arent equal mod 2, so if A makes a move in the first zone then B can simply apply his winning strategy for b-a (it's odd and it isn't 3 so the second player wins), ensuring that he doesn't lose. Also, as long as the second zone still hasn't been reduced to piles of at most 2 stones, moves can still be made in the first zone even if B already technically won since there will still be piles of 2 stones. But it's impossible for A to make the last move in the first zone since that would mean that an odd number of total moves were made, so there's an even number of piles of 1 and no other piles, so b-a is even which isn't true. Therefore, B makes the last move in the first zone. So B wins.
If b-a = 3:
If the odd number out of a and b isn't 3 :
Then B can simply split the 2 piles into 2 piles of x/2, x/2 and y where x is the even number and y is the odd number. He can apply the same strategy as before since y isn't 3.
If the odd number out of a and b is 3:
Then we have n=9, b=6, a=3.
In this case, B will divide the pile of 6 into two piles of 3. This forces A to divide one of the piles of 3 into 2 piles of 1 and 2 so we will have
2 3 3 (the pile of 1 doesn't matter)
B divides the pile of 2 into 2 piles of 1,so we have
3 3
A divides one of the piles of 3 into 2 piles of 1 and 2,and then B will do the same for the other pile of 3 so B wins, which concludes our proof.
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