Plan ahead for the next school year. Schedule your class today!

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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.

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0 replies
jwelsh
Jul 1, 2025
0 replies
Bonza functions
KevinYang2.71   67
N 5 minutes ago by PokemonMaster2012
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
67 replies
KevinYang2.71
Jul 15, 2025
PokemonMaster2012
5 minutes ago
Orochenter, parallel lines, circumcircle, concyclic
geometry6   6
N 7 minutes ago by Shan3t
Source: MOP 2006
Let $H$ be the orthocenter of $\triangle ABC$. Let $D, E, F$ lie on the circumcircle of $\triangle ABC$ such that $AD \parallel BE \parallel CF$. Let $S, T, U$ respectively denote the reflections of $D, E, F$ across $BC, CA, AB$. Prove that points $S, T, U, H$ are concyclic.
6 replies
geometry6
Sep 7, 2020
Shan3t
7 minutes ago
interesting problem
Giahuytls2326   1
N 15 minutes ago by Giahuytls2326
Source: my teacher
Find all pairs of positive integers \((m, n)\) with \(m, n > 1\) such that \(m \mid a^n - 1\) for every \(a \in \{1, 2, \ldots, n\}\).
1 reply
Giahuytls2326
an hour ago
Giahuytls2326
15 minutes ago
Function with primes
dangerousliri   44
N 17 minutes ago by math-olympiad-clown
Source: BMO 2019, Problem 1
Let $\mathbb{P}$ be the set of all prime numbers. Find all functions $f:\mathbb{P}\rightarrow\mathbb{P}$ such that:
$$f(p)^{f(q)}+q^p=f(q)^{f(p)}+p^q$$holds for all $p,q\in\mathbb{P}$.

Proposed by Dorlir Ahmeti, Albania
44 replies
dangerousliri
May 2, 2019
math-olympiad-clown
17 minutes ago
Question about Fubini integration in Euclidean spaces
euclides05   0
5 hours ago
Fubini's theorem in Lebesgue integration uses the product measure. My question is, how can this be applied in Euclidean spaces, where if $d= d_1+d_2$, the lebesgue measure of $R^d$ is the completion of the product measures
of $R^{d_1}$, $R^{d_2}$
and not the actual product measure?
I would be very pleased if someone provided a thorough answer cause i'm studying convolution stuff where it's used heavily and i haven't been able to find a stict explanation for it.
0 replies
euclides05
5 hours ago
0 replies
On the number of isomorphism classes of subgroups and normal subgroups
Sivin   3
N Today at 4:02 AM by Sivin
Let $(G,\circ)$ be a group. Let $\mathcal{I}(G)$ denote the number of subgroups of $G$ up to isomorphism (i.e. the number of isomorphism classes of subgroups of $G$) and let $\mathcal{N}(G)$ denote the number of normal subgroups of $G$ (not necessarily up to isomorphism). Find all groups $G$ with $\mathcal{I}(G)=\mathcal{N}(G).$

For example, the dihedral group of degree $10$:
$$D_{10}=\langle s,t\mid s^{10},t^{2},stst\rangle$$has $7$ isomorphism classes of subgroups, namely $C_1,C_2,K,C_5,C_{10},D_5,D_{10},$ so $\mathcal{I}(D_{10})=7.$ It also has $7$ normal subgroups (see https://groupprops.subwiki.org/wiki/Dihedral_group:D20#Subgroups). Hence $\mathcal{N}(D_{10})=7.$ It follows that $D_{10}$ is a group with $\mathcal{I}(G)=\mathcal{N}(G).$
3 replies
Sivin
Aug 17, 2024
Sivin
Today at 4:02 AM
Frankl Theorem?
EthanWYX2009   0
Today at 2:55 AM
Source: 2024 February 谜之竞赛-3
Let \( n \) be a positive integer, \( f(n) \) denote the maximum possible size of a family \(\mathcal{F}\) of subsets of \(\{1, 2, \cdots, n\}\) such that for any two subsets \( X, Y \in \mathcal{F} \), \( |X \cap Y| \) is a perfect square.

Show that $2^{\lfloor \sqrt{n} \rfloor} \leq f(n) \leq 100^{\sqrt{n} \ln n}.$

Proposed by Qianhong Cheng, Guangdong Experimental High School
0 replies
EthanWYX2009
Today at 2:55 AM
0 replies
Studying for Putnam
chessboy_123   0
Yesterday at 9:18 PM
Hello everyone,
I'm going to be a rising freshman in college and I want to start studying for the Putnam. In HS I only ever made AIME with a highest score of 8. Are there any books or other resources to help me bridge the gap between AIME and Putnam?

Thanks!
0 replies
chessboy_123
Yesterday at 9:18 PM
0 replies
On f(x) and f([x])
ajovanovic   3
N Yesterday at 8:39 PM by ajovanovic
For which $f:\mathbb{R}\to\mathbb{R}$ does $f(x)$ ~$f([x])$ hold?

Note: I have proved this for concave, unbounded, monotone increasing $f$. Is this a necessary condition?
3 replies
ajovanovic
Yesterday at 8:18 PM
ajovanovic
Yesterday at 8:39 PM
archemedian property
We2592   1
N Yesterday at 5:16 PM by Etkan
Q) Let $I_n := (n,\infty)$ then find the $\bigcap_{n\in \mathbb{N}} I_n$ ?
Q) Let $I_n := [n,\infty)$ then find the $\bigcap_{n\in \mathbb{N}} I_n$ ?
and also find the supremum and infimum of this ?

now how to approach it with help of archemedian property help
1 reply
We2592
Yesterday at 2:50 PM
Etkan
Yesterday at 5:16 PM
f(z) = z^n!
fxandi   5
N Yesterday at 2:53 PM by fxandi
Let $D \subseteq \mathbb{C}$ be a region and the function $f : D \to \mathbb{C}$ defined by $f(z) = \displaystyle\sum_{n = 0}^\infty z^{n!}$ is well-defined and analytic in $D.$
If $\{z \in \mathbb{C} : |z| < 1\} \subseteq D,$ show that $D = \{z \in \mathbb{C} : |z| < 1\}.$
5 replies
fxandi
Jul 24, 2025
fxandi
Yesterday at 2:53 PM
Inequality with sum of sines
TheIntegral   1
N Yesterday at 9:17 AM by smartvong
Source: UM Mathematical Olympiad 2025 (B3)
Given $a_1 = \sqrt{3}$, $a_{n + 1} = \frac{3\pi}{2}\sin\left(\frac{1}{a_n}\right)$ for all integers $n \ge 1$, show that for all integers $n \ge 2$,
$$\sin a_1 + \sin a_2 + \cdots + \sin a_n \le 0.98n.$$
1 reply
TheIntegral
Jun 15, 2025
smartvong
Yesterday at 9:17 AM
2022 Putnam B6
giginori   11
N Yesterday at 8:03 AM by Ritwin
Find all continuous functions $f:\mathbb{R}^+\rightarrow \mathbb{R}^+$ such that $$f(xf(y))+f(yf(x))=1+f(x+y)$$for all $x, y>0.$
11 replies
giginori
Dec 4, 2022
Ritwin
Yesterday at 8:03 AM
Analytic Number Theory
EthanWYX2009   1
N Yesterday at 6:10 AM by paxtonw
Source: 2024 Jan 谜之竞赛-7
For positive integer \( n \), define \(\lambda(n)\) as the smallest positive integer satisfying the following property: for any integer \( a \) coprime with \( n \), we have \( a^{\lambda(n)} \equiv 1 \pmod{n} \).

Given an integer \( m \geq \lambda(n) \left( 1 + \ln \frac{n}{\lambda(n)} \right) \), and integers \( a_1, a_2, \cdots, a_m \) all coprime with \( n \), prove that there exists a non-empty subset \( I \) of \(\{1, 2, \cdots, m\}\) such that
\[\prod_{i \in I} a_i \equiv 1 \pmod{n}.\]Proposed by Zhenqian Peng from High School Affiliated to Renmin University of China
1 reply
EthanWYX2009
Jul 23, 2025
paxtonw
Yesterday at 6:10 AM
Reflections and midpoints in triangle
TUAN2k8   2
N May 24, 2025 by Double07
Source: Own
Given an triangle $ABC$ and a line $\ell$ in the plane.Let $A_1,B_1,C_1$ be reflections of $A,B,C$ across the line $\ell$, respectively.Let $D,E,F$ be the midpoints of $B_1C_1,C_1A_1,A_1B_1$, respectively.Let $A_2,B_2,C_2$ be the reflections of $A,B,C$ across $D,E,F$, respectively.Prove that the points $A_2,B_2,C_2$ lie on a line parallel to $\ell$.
2 replies
TUAN2k8
May 24, 2025
Double07
May 24, 2025
Reflections and midpoints in triangle
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TUAN2k8
60 posts
#1 • 1 Y
Y by Phuc62794
Given an triangle $ABC$ and a line $\ell$ in the plane.Let $A_1,B_1,C_1$ be reflections of $A,B,C$ across the line $\ell$, respectively.Let $D,E,F$ be the midpoints of $B_1C_1,C_1A_1,A_1B_1$, respectively.Let $A_2,B_2,C_2$ be the reflections of $A,B,C$ across $D,E,F$, respectively.Prove that the points $A_2,B_2,C_2$ lie on a line parallel to $\ell$.
This post has been edited 1 time. Last edited by TUAN2k8, May 24, 2025, 1:44 PM
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Funcshun840
64 posts
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It suffices to prove that $B_2 C_2$ is parallel to $l$, which we will do by proving $A A_1$ is its perpendicular bisector. Note that $A_1 B_2 = B C_1 = C B_1 = A_1 C_2$ by parallelograms. Now reflect $A$ over $E$, $F$ to get $K$, $L$. Then $K$ lies on $B B_1$ and $L$ on $C C_1$, additionally, $KL$ is parallel to $EF$, which is parallel to $B_1 C_1$, so $KBCL$ is an isosceles trapezoid. This implies $A B_2 = BL = CK = A C_2$, hence we are done.
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Double07
95 posts
#3
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Here is a complex bash:

Every line is defined by the equation $\ell: \overline{z}=\lambda z+\nu$, where $\lambda=\dfrac{\overline{p_1}-\overline{p_2}}{p_1-p_2}$ and $\nu=\dfrac{p_1\overline{p_2}-\overline{p_1}p_2}{p_1-p_2}$ and $P_1, P_2\in\ell$.

Notice that $|\lambda|=1$.

Then the reflection of a point $P$ in $\ell$ is $q=\overline{\lambda p+\nu}$.

So $a_1=\dfrac{\overline{\lambda}}{a}+\overline{\nu}$, $b_1=\dfrac{\overline{\lambda}}{b}+\overline{\nu}$ and $c_1=\dfrac{\overline{\lambda}}{c}+\overline{\nu}$.

Since $B_1A_2C_1A$ is a parallelogram, we have $a_2=b_1+c_1-a=\dfrac{\overline{\lambda}(b+c)}{bc}+2\overline{\nu}-a$.

Similarly $b_2=\dfrac{\overline{\lambda}(a+c)}{ac}+2\overline{\nu}-b$.

Notice that it suffices to prove that $A_2B_2\parallel \ell$, which is equivallent to proving that $\dfrac{\overline{a_2}-\overline{b_2}}{a_2-b_2}=\lambda$.

$a_2-b_2=\overline{\lambda}\cdot\dfrac{a-b}{ab}-(a-b)=(a-b)\left(\dfrac{\overline{\lambda}}{ab}-1\right)$.

So $\dfrac{\overline{a_2}-\overline{b_2}}{a_2-b_2}=-\dfrac{\lambda ab-1}{\overline{\lambda}-ab}=\lambda$, since $|\lambda|=1$.

So $A_2B_2\parallel \ell$, so we're done.
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