Art of Problem Solving
AoPS Online AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy AoPS Academy
Small live classes for advanced math
and language arts learners in grades 1-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Happy Memorial Day! Please note that AoPS Online is closed May 24-26th.
combinatorics
B
No tags match your search
M
G
Topic
First Poster
Last Poster
H
Inspired by 2025 Beijing
sqing   11
N 38 minutes ago by sqing
Source: Own
Let $ a,b,c,d >0 $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
11 replies
sqing
Saturday at 4:56 PM
sqing
38 minutes ago
J
H
A functional equation
super1978   1
N an hour ago by CheerfulZebra68
Source: Somewhere
Find all functions $f: \mathbb R \to \mathbb R$ such that:$$ f(f(x)(y+f(y)))=xf(y)+f(xy) $$for all $x,y \in \mathbb R$
1 reply
super1978
an hour ago
CheerfulZebra68
an hour ago
J
H
Prove that IMO is isosceles
YLG_123   4
N 2 hours ago by Blackbeam999
Source: 2024 Brazil Ibero TST P2
Let \( ABC \) be an acute-angled scalene triangle with circumcenter \( O \). Denote by \( M \), \( N \), and \( P \) the midpoints of sides \( BC \), \( CA \), and \( AB \), respectively. Let \( \omega \) be the circle passing through \( A \) and tangent to \( OM \) at \( O \). The circle \( \omega \) intersects \( AB \) and \( AC \) at points \( E \) and \( F \), respectively (where \( E \) and \( F \) are distinct from \( A \)). Let \( I \) be the midpoint of segment \( EF \), and let \( K \) be the intersection of lines \( EF \) and \( NP \). Prove that \( AO = 2IK \) and that triangle \( IMO \) is isosceles.
4 replies
YLG_123
Oct 12, 2024
Blackbeam999
2 hours ago
J
H
Geometric mean of squares a knight's move away
Pompombojam   0
2 hours ago
Source: Problem Solving Tactics Methods of Proof Q27
Each square of an $8 \times 8$ chessboard has a real number written in it in such a way that each number is equal to the geometric mean of all the numbers a knight's move away from it.

Is it true that all of the numbers must be equal?
0 replies
Pompombojam
2 hours ago
0 replies
J
H
Circumcircle of ADM
v_Enhance   71
N 2 hours ago by judokid
Source: USA TSTST 2012, Problem 7
Triangle $ABC$ is inscribed in circle $\Omega$. The interior angle bisector of angle $A$ intersects side $BC$ and $\Omega$ at $D$ and $L$ (other than $A$), respectively. Let $M$ be the midpoint of side $BC$. The circumcircle of triangle $ADM$ intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. Let $N$ be the midpoint of segment $PQ$, and let $H$ be the foot of the perpendicular from $L$ to line $ND$. Prove that line $ML$ is tangent to the circumcircle of triangle $HMN$.
71 replies
v_Enhance
Jul 19, 2012
judokid
2 hours ago
J
H
Three operations make any number
awesomeming327.   2
N 3 hours ago by happymoose666
Source: own
The number $3$ is written on the board. Anna, Boris, and Charlie can do the following actions: Anna can replace the number with its floor. Boris can replace any integer number with its factorial. Charlie can replace any nonnegative number with its square root. Prove that the three can work together to make any positive integer in finitely many steps.
2 replies
1 viewing
awesomeming327.
6 hours ago
happymoose666
3 hours ago
J
H
IMO 2017 Problem 4
Amir Hossein   116
N 6 hours ago by cj13609517288
Source: IMO 2017, Day 2, P4
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Proposed by Charles Leytem, Luxembourg
116 replies
Amir Hossein
Jul 19, 2017
cj13609517288
6 hours ago
J
H
A sharp one with 3 var
mihaig   10
N 6 hours ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$ab+bc+ca+abc\geq4.$$
10 replies
mihaig
May 13, 2025
mihaig
6 hours ago
J
H
Another right angled triangle
ariopro1387   1
N 6 hours ago by lolsamo
Source: Iran Team selection test 2025 - P7
Let $ABC$ be a right angled triangle with $\angle A=90$. Point $M$ is the midpoint of side $BC$ And $P$ be an arbitrary point on $AM$. The reflection of $BP$ over $AB$ intersects lines $AC$ and $AM$ at $T$ and $Q$, respectively. The circumcircles of $BPQ$ and $ABC$ intersect again at $F$. Prove that the center of the circumcircle of $CFT$ lies on $BQ$.
1 reply
ariopro1387
Yesterday at 4:13 PM
lolsamo
6 hours ago
J
H
four points lie on a circle
pohoatza   78
N Yesterday at 8:27 PM by ezpotd
Source: IMO Shortlist 2006, Geometry 2, AIMO 2007, TST 1, P2
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.

Proposed by Vyacheslev Yasinskiy, Ukraine
78 replies
pohoatza
Jun 28, 2007
ezpotd
Yesterday at 8:27 PM
J
H
J
H
Outcome related combinatorics problem
egxa   1
N Apr 29, 2025 by iliya8788
Source: All Russian 2025 10.7
A competition consists of $25$ sports, each awarding one gold medal to a winner. $25$ athletes participate, each in all $25$ sports. There are also $25$ experts, each of whom must predict the number of gold medals each athlete will win. In each prediction, the medal counts must be non-negative integers summing to $25$. An expert is called competent if they correctly guess the number of gold medals for at least one athlete. What is the maximum number \( k \) such that the experts can make their predictions so that at least \( k \) of them are guaranteed to be competent regardless of the outcome?
1 reply
egxa
Apr 18, 2025
iliya8788
Apr 29, 2025
J
Outcome related combinatorics problem
G H J
Reveal topic tags
pigeonhole principle
combinatorics
L
G H BBookmark kLocked kLocked NReply
Source: All Russian 2025 10.7
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
egxa
211 posts
egxa
#1 h  Apr 18, 2025, 5:12 PM
Y by
A competition consists of $25$ sports, each awarding one gold medal to a winner. $25$ athletes participate, each in all $25$ sports. There are also $25$ experts, each of whom must predict the number of gold medals each athlete will win. In each prediction, the medal counts must be non-negative integers summing to $25$. An expert is called competent if they correctly guess the number of gold medals for at least one athlete. What is the maximum number \( k \) such that the experts can make their predictions so that at least \( k \) of them are guaranteed to be competent regardless of the outcome?
This post has been edited 1 time. Last edited by egxa, Apr 18, 2025, 5:21 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
iliya8788
8 posts
iliya8788
#2 h  Apr 29, 2025, 10:46 AM
Y by
We claim that the maximum $k$ is equal to $24$.
We correspond each athlete to a number from $1$ to $25$.
Define Podium as the $n$-tuple $(a_{1},a_{2},...,a_{25})$ where $a_{i}$ is the amount of medals won by athlete $i$.
First notice that none of the experts can predict the $n$-tuple $(1,1,...,1)$ because then if the podium is equal to $(25,0,0,...,0)$ then the most possible amount of judges predicting correctly is equal to $24$. From the previous statement it follows that one of the judges predicts a $n$-tuple which has a $0$. So we proceed with the following algorithm. Pick one of the prediction $n$-tuples and convert it with the following algorithm to a new $n$-tuple: turn anything that isn't equal to $0$ to $0$ and then spread the 25 medals among the zero's.
Now if the podium $n$-tuple is the same as this $n$-tuple then at least one of the experts isn't competent and so maximum $k$ is equal to $24$.
Now we will prove that no matter what the podium looks like we can achieve $k=24$.
So we propose the following predictions: $(24,1,0,0,...,0),(24,0,1,0,0,...,0),(24,0,0,1,0,0,...0),...,(24,0,0,...,1),(1,1,...,1,1,1)$.
Define the count of zeros of the podium as $C$.
Now 3 cases might happen:
$C>2$: then it's easy to notice that every expert except the last one is competent.
$C=2$: an expert isn't competent if and only if both zeros are in the same place as the $1$ and $24$. notice that no $2$ of the predictions have these 2 numbers in the same spot so worst case possible $24$ of the experts are competent.
$C=1$: notice that the podium $n$-tuple must have $23$, $1$'s and a single $2$ and a single $0$. an expert isn't competent if and only if one of the following 2 cases occur:
First case: the $1$ in the prediction is in the same place as the $0$ of the podium.
Second case: the $24$ in the prediction is in the same place as the $0$ of the podium and the $1$ in the prediction is in the same place as the $2$ of the podium.
notice that the position of the $1$ differs among the predictions so for a fixed podium only one expert can be not competent because of the first case. The same applies for the second case as well. Since the union of the positions of $1$ and $24$ differs among the predictions there can't be 2 experts that are incompetent because of the second case. Also both cases can't make experts incompetent simultaneously since in case of both cases happening the $1$ and $24$ must basically switch places but in each of the first $24$ predictions $24$ is not in the first spot and all of the $1$'s are in the first spot. So only one of these $24$ experts can be incompetent and the last expert is competent so at least $24$ of the experts are competent. $\blacksquare$
This post has been edited 5 times. Last edited by iliya8788, Apr 29, 2025, 12:11 PM
Z K Y
N Quick Reply
G
H
=
a