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k a August Highlights and 2025 AoPS Online Class Information
jwelsh   0
Friday at 2:14 PM
CONGRATULATIONS to all the competitors at this year’s International Mathematical Olympiad (IMO)! The US Team took second place with 5 gold medals and 1 silver - we are proud to say that each member of the 2025 IMO team has participated in an AoPS WOOT (Worldwide Online Olympiad Training) class!

"As a parent, I'm deeply grateful to AoPS. Tiger has taken very few math courses outside of AoPS, except for a local Math Circle that doesn't focus on Olympiad math. AoPS has been one of the most important resources in his journey. Without AoPS, Tiger wouldn't be where he is today — especially considering he's grown up in a family with no STEM background at all."
— Doreen Dai, parent of IMO US Team Member Tiger Zhang

Interested to learn more about our WOOT programs? Check out the course page here or join a Free Scheduled Info Session. Early bird pricing ends August 19th!:
CodeWOOT Code Jam - Monday, August 11th
ChemWOOT Chemistry Jam - Wednesday, August 13th
PhysicsWOOT Physics Jam - Thursday, August 14th
MathWOOT Math Jam - Friday, August 15th

There is still time to enroll in our last wave of summer camps that start in August at the Virtual Campus, our video-based platform, for math and language arts! From Math Beasts Camp 6 (Prealgebra Prep) to AMC 10/12 Prep, you can find an informative 2-week camp before school starts. Plus, our math camps don’t have homework and cover cool enrichment topics like graph theory. Our language arts courses will build the foundation for next year’s challenges, such as Language Arts Triathlon for levels 5-6 and Academic Essay Writing for high school students.

Lastly, Fall is right around the corner! 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. We’ve opened new Academy locations in San Mateo, CA, Pasadena, CA, Saratoga, CA, Johns Creek, GA, Northbrook, IL, and Upper West Side (NYC), New York.

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.
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0 replies
jwelsh
Friday at 2:14 PM
0 replies
How to solve this?
Mr._Calculator   0
11 minutes ago
In the diagram below, $ABC$ is a triangle, where the angle bisectors of $\angle ABC$ and $\angle ACB$ intersect at point $P$. Let points $D$ and $E$ lie on sides $AB$ and $AC$, respectively, such that $DE$ passes through point $P$ and $\angle AED = \angle ABP$. If the areas of triangles $BDP$, $CEP$ and $BPC$ are equal to $18 \text{ cm}^2$, $36 \text{ cm}^2$ and $57 \text{ cm}^2$, respectively, then what is the area, in $\text{cm}^2$, of $ABC$?
0 replies
Mr._Calculator
11 minutes ago
0 replies
Number Theory
JetFire008   1
N 21 minutes ago by blug
Source: Elementary Number Theory by David M. Burton
Modify Euclid's proof that there are infinitely many primes by assuming the existence of a largest prime $p$ and using the integer $N=p!+1$ to arrive at a contradiction.
1 reply
JetFire008
an hour ago
blug
21 minutes ago
Partial products from p-2 numbers
JustPostNorthKoreaTST   2
N 22 minutes ago by navid
Source: 2015 North Korea Mathematical Olympiad P2
Let $p$ be an odd prime and $a_1,a_2,\ldots,a_{p-2}$ be positive integers (not necessarily different), such that for any $k \in \{1,2,\ldots,p-2\}$, we have $p \nmid a_k$ and $p \nmid (a_k^k-1)$. Show that one can pick some numbers from $a_1,a_2,\ldots,a_{p-2}$ such that their product is $\equiv 2 \pmod p$.
2 replies
JustPostNorthKoreaTST
3 hours ago
navid
22 minutes ago
Bonza functions
KevinYang2.71   72
N 24 minutes ago by pie854
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
72 replies
KevinYang2.71
Jul 15, 2025
pie854
24 minutes ago
No more topics!
Difficult combinatorics problem
shactal   7
N May 19, 2025 by shactal
Can someone help me with this problem? Let $n\in \mathbb N^*$. We call a distribution the act of distributing the integers from $1$
to $n^2$ represented by tokens to players $A_1$ to $A_n$ so that they all have the same number of tokens in their urns.
We say that $A_i$ beats $A_j$ when, when $A_i$ and $A_j$ each draw a token from their urn, $A_i$ has a strictly greater chance of drawing a larger number than $A_j$. We then denote $A_i>A_j$. A distribution is said to be chicken-fox-viper when $A_1>A_2>\ldots>A_n>A_1$ What is $R(n)$
, the number of chicken-fox-viper distributions?
7 replies
shactal
May 18, 2025
shactal
May 19, 2025
Difficult combinatorics problem
G H J
G H BBookmark kLocked kLocked NReply
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shactal
14 posts
#1
Y by
Can someone help me with this problem? Let $n\in \mathbb N^*$. We call a distribution the act of distributing the integers from $1$
to $n^2$ represented by tokens to players $A_1$ to $A_n$ so that they all have the same number of tokens in their urns.
We say that $A_i$ beats $A_j$ when, when $A_i$ and $A_j$ each draw a token from their urn, $A_i$ has a strictly greater chance of drawing a larger number than $A_j$. We then denote $A_i>A_j$. A distribution is said to be chicken-fox-viper when $A_1>A_2>\ldots>A_n>A_1$ What is $R(n)$
, the number of chicken-fox-viper distributions?
Z K Y
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shactal
14 posts
#2
Y by
Someone?
Z K Y
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aaravdodhia
2705 posts
#3
Y by
Isn't it $0$?

Note that $A_i$ beats $A_j$ when $E(d_i-d_j)>0$, where $E$ represents expected value and $d_i, d_j$ are the draws of players $i$ and $j$. That happens when $E(d_i) - E(d_j)>0$, or the sum of $A_i$'s collection is greater than the sum of $A_j$'s collection. In the problem, the sum of $A_1$'s collection must be greater than the sum of everybody else's, contradicting $A_n > A_1$.

This logic is due to the distribution being given prior to the players drawing and comparing their numbers. But if all distributions were considered at once, any pair $(i,j)$ would satisfy $A_i > A_j$ with equal probability $\tfrac12\left(\text{probability }E(d_i-d_j)\ne0\right)$, so the player's expected draws would all be the same. Hence there cannot be an order to $A_1\dots A_n$.

If this problem is from another source, I'd suggest reading their explanation. :)
Z K Y
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shactal
14 posts
#4
Y by
Well, the thing is I don't have the solution and I would like to know the method to solve this type of problems
Z K Y
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Ash_the_Bash07
1336 posts
#5
Y by
ok$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$
Z K Y
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shactal
14 posts
#6
Y by
But I don't think the answer is $0$, because I already found some examples where the condition is satisfied
Z K Y
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shactal
14 posts
#7
Y by
Here is an example that satisfies the condition: Player $A$ has numbers $\{2,4,9\}$, player B has $\{1,6,8\}$ and player $C$ has $\{3,5,7\}$
Z K Y
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shactal
14 posts
#8
Y by
If I can show that the events "$A$ wins against $B$" and "$B$ wins against $C$" are independent, then the problem is trivial. But how to prove this?
Z K Y
N Quick Reply
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