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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 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
Prove that IMO is isosceles
YLG_123   4
N 16 minutes 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
16 minutes ago
Geometric mean of squares a knight's move away
Pompombojam   0
20 minutes 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
20 minutes ago
0 replies
Circumcircle of ADM
v_Enhance   71
N 21 minutes 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
21 minutes ago
Three operations make any number
awesomeming327.   2
N 43 minutes 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
awesomeming327.
4 hours ago
happymoose666
43 minutes ago
Inspired by RMO 2006
sqing   3
N an hour ago by Marrelia
Source: Own
Let $ a,b >0  . $ Prove that
$$  \frac {a^{2}+1}{b+k}+\frac { b^{2}+1}{ka+1}+\frac {2}{a+kb}  \geq \frac {6}{k+1}  $$Where $k\geq 0.03 $
$$  \frac {a^{2}+1}{b+1}+\frac { b^{2}+1}{a+1}+\frac {2}{a+b}  \geq 3  $$
3 replies
sqing
Saturday at 3:24 PM
Marrelia
an hour ago
Inspired by 2025 Beijing
sqing   10
N 2 hours ago by ytChen
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
10 replies
sqing
Saturday at 4:56 PM
ytChen
2 hours ago
IMO 2017 Problem 4
Amir Hossein   116
N 4 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
4 hours ago
A sharp one with 3 var
mihaig   10
N 4 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
4 hours ago
Another right angled triangle
ariopro1387   1
N 4 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
4 hours ago
four points lie on a circle
pohoatza   78
N 5 hours ago 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
5 hours ago
JBMO TST Bosnia and Herzegovina 2023 P4
FishkoBiH   2
N 5 hours ago by Stear14
Source: JBMO TST Bosnia and Herzegovina 2023 P4
Let $n$ be a positive integer. A board with a format $n*n$ is divided in $n*n$ equal squares.Determine all integers $n$3 such that the board can be covered in $2*1$ (or $1*2$) pieces so that there is exactly one empty square in each row and each column.
2 replies
FishkoBiH
Yesterday at 1:38 PM
Stear14
5 hours ago
Does there exist 2011 numbers?
cyshine   8
N 5 hours ago by TheBaiano
Source: Brazil MO, Problem 4
Do there exist $2011$ positive integers $a_1 < a_2 < \ldots < a_{2011}$ such that $\gcd(a_i,a_j) = a_j - a_i$ for any $i$, $j$ such that $1 \le i < j \le 2011$?
8 replies
cyshine
Oct 20, 2011
TheBaiano
5 hours ago
D1036 : Composition of polynomials
Dattier   1
N 5 hours ago by Dattier
Source: les dattes à Dattier
Find all $A \in \mathbb Q[x]$ with $\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$ and $\deg(A)>1$.
1 reply
Dattier
Saturday at 1:52 PM
Dattier
5 hours ago
number sequence contains every large number
mathematics2003   3
N 5 hours ago by sttsmet
Source: 2021ChinaTST test3 day1 P2
Given distinct positive integer $ a_1,a_2,…,a_{2020} $. For $ n \ge 2021 $, $a_n$ is the smallest number different from $a_1,a_2,…,a_{n-1}$ which doesn't divide $a_{n-2020}...a_{n-2}a_{n-1}$. Proof that every number large enough appears in the sequence.
3 replies
mathematics2003
Apr 13, 2021
sttsmet
5 hours ago
Points and triangles in the plane
pablock   6
N Feb 23, 2025 by awesomeming327.
Source: Brazilian Mathematical Olympiad 2018 - Q6
Consider $4n$ points in the plane, with no three points collinear. Using these points as vertices, we form $\binom{4n}{3}$ triangles. Show that there exists a point $X$ of the plane that belongs to the interior of at least $2n^3$ of these triangles.
6 replies
pablock
Nov 16, 2018
awesomeming327.
Feb 23, 2025
Points and triangles in the plane
G H J
Source: Brazilian Mathematical Olympiad 2018 - Q6
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pablock
168 posts
#1 • 10 Y
Y by UK2019Project, anantmudgal09, rcorreaa, rkm0959, lminsl, nguyendangkhoa17112003, Aarth, kiyoras_2001, Adventure10, Mango247
Consider $4n$ points in the plane, with no three points collinear. Using these points as vertices, we form $\binom{4n}{3}$ triangles. Show that there exists a point $X$ of the plane that belongs to the interior of at least $2n^3$ of these triangles.
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MellowMelon
5850 posts
#2 • 13 Y
Y by rkm0959, stroller, DanDumitrescu, ZeusDM, nguyendangkhoa17112003, Aarth, kiyoras_2001, lminsl, yayups, edfearay123, wateringanddrowned, CyclicISLscelesTrapezoid, Adventure10
Let $S$ be the set of $4n$ points.

1. Let $T$ be the set of all lines through the origin. For each $\ell \in T$, let $f(\ell)$ be a line parallel to $\ell$ that does not pass through any point in $S$ and has $2n$ points of $S$ on each side. Such an $f(\ell)$ exists for each $\ell$ since we can move the line continuously in the plane and have the number of points on each side of the line vary continuously in the discrete sense.

2. Fix an $\ell \in T$. Let $m$ be a variable line in $T$ that starts as $\ell$ and rotates clockwise continuously until it becomes $\ell$ again. At any time, $f(\ell)$, $f(m)$ split the plane into four sections, and the number of points in each section varies continuously in the discrete sense. Initially the number of points in each section is $0, 2n, 0, 2n$, and at the end it is $2n, 0, 2n, 0$. This means there exists some line $m'$ such that $f(\ell)$ and $f(m')$ split the plane into four sections each with exactly $n$ points.

3. Choose $X$ to be the intersection of $f(\ell)$ and $f(m')$. Consider each of the $n^4$ ways to choose one point from each of the four sections formed by $f(\ell)$ and $f(m')$. These four points form four triangles. Exactly two of the four triangles contain $X$. This method of counting will obtain each triangle exactly $n$ times. This gives a total of $n^4 \cdot 2 / n = 2n^3$ triangles containing $X$.
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Aarth
137 posts
#3 • 2 Y
Y by Adventure10, Mango247
MellowMelon wrote:

2. Fix an $\ell \in T$. Let $m$ be a variable line in $T$ that starts as $\ell$ and rotates clockwise continuously until it becomes $\ell$ again. At any time, $f(\ell)$, $f(m)$ split the plane into four sections, and the number of points in each section varies continuously in the discrete sense. Initially the number of points in each section is $0, 2n, 0, 2n$, and at the end it is $2n, 0, 2n, 0$. This means there exists some line $m'$ such that $f(\ell)$ and $f(m')$ split the plane into four sections each with exactly $n$ points.

Why does the number of points in each section vary continuously?
This post has been edited 1 time. Last edited by Aarth, Nov 26, 2019, 3:51 PM
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Letteer
56 posts
#4 • 2 Y
Y by wateringanddrowned, Adventure10
Solution. part
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Eyed
1065 posts
#5
Y by
I claim there exists two lines, $\ell_{1}, \ell_{2}$, such that the $4$ regions divided by $\ell_1, \ell_2$ each contains $n$ points. Define $S$ as the set of the $4n$ points.

Let $\ell_{1}$ be an arbitrary balancing line, so each side of $\ell_{1}$ contains $2n$ points. Then, consider a point $P\in \ell_{1}$, and consider the line $\ell$ going through $P$ that will have $n$ points on its left side, and $n$ points on the right side of $\ell_{1}$. If we put $P$ at the very beginning of line $\ell_{1}$, then the amount of points on the left side of $\ell_{1}$ and the left side of $\ell$ is $2n$. However, moving $P$ to the very end of line $\ell_{1}$ gives $0$ points on the lefts ide of $\ell_{1}, \ell$. Thus, since by moving $P$, at each step we change at most $1$ in the number of points on the left side of $\ell, \ell_1$, this means at some point, there are $n$ points on the left side of $\ell, \ell_{1}$. Letting $\ell_2 = \ell$ gives the desired result.

Now, if $O = \ell_1\cap \ell_2$, we can choose $O$ such that it does not lie on any of the lines joining $2$ points in $S$. Consider the four regions $A,B,C,D$, each with $n$ points, and in clockwise order (so $A,C$ are opposite, same with $B,D$). Now, for each $n^2$ pairs of points in $A,C$, if $E\in A, F\in C$, then either for any point $X\in B$, we have $O\in \triangle EFX$, or for any point in $Y\in D$ we have $O\in \triangle EFY$ (this is because either $O$ lies above or below line $EF$. Therefore, there are $n^3$ triangles that contain $O$ with two vertices in $A,C$. Similarly, there are $n^3$ triangles that contain $O$ with two vertices in $B,D$, so we conclude that there are at least $2n^3$ triangles that $O$ is a part of.
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Leo.Euler
577 posts
#6
Y by
Let $S$ be the set of $4n$ points in the plane.

Claim: There exist lines $\ell_1$ and $\ell_2$ that divide the plane into $4$ regions with $n$ points contained in each region and no points on either line.
Proof. Consider all of the lines through any two points of $S$, and take $\ell$ to be a line parallel to none of these lines. Realize that if we first take it to be arbitrarily far away from the points in $S$, and then translate it in the direction closer to the points, then the number of points on the side that initially has no points starts with $0$ and increases by exactly $1$ whenever it increases. Since it eventually reaches $4n$ points, we can move it to contain exactly $2n$ points on both sides. Call this line $\ell_1$.

Now, color the points on the left side of $\ell_1$ blue and the points on the right side of the line red. Let $P_0$ be a point of minimal distance to $\ell_1$. Then slightly rotate $\ell_1$ so that no points other than $P_0$ lie on the line or switch sides, and $P_0$ lies through the rotated line. Consider the following windmill process. First, rotate $P_0$ clockwise until it passes through another point of different color, $P_1$. Then for the $i$th iteration of the process, when $\ell_1$ is $\overline{P_iP_{i+1}}$, rotate it clockwise until it hits a new point $P_{i+2}$ with color different from $P_{i+1}$, and then rotate $\ell_1$ about $P_{i+1}$ so that it becomes $\overline{P_{i+1}P_{i+2}}$. Eventually, we can get the line to be in a position in which it passes through a red point and a blue point, dividing the plane into quadrants with $(n-1, n, n-1, n)$ points in them. Let this line intersect the original $\ell_1$ at point $X$. Then rotate the line by a sufficiently small angular displacement about $X$ so that it passes through no point in $S$. Call the resulting line $\ell_2$. Then $\ell_1$ and $\ell_2$ divide the plane into quadrants with $n$ points in each quadrant, as claimed
:yoda:

Consider the two lines as in the above claim, and let their intersection be $X$. Color the points in each quarter-plane red, blue, green, and yellow, in the clockwise order. I claim that the size $|T|$ of the set $T$ of triangles with all vertices of distinct colors containing the point $X$ in their interior is exactly $2n^3$. Let $Q$ be the set of quadrilaterals with all vertices having distinct colors. Then for each quadrilateral $q \in Q$, draw its diagonals and consider the $4$ subtriangles it is split into. Since $X$ lies in exactly one of those triangles, there are exactly $2$ triangles whose vertices are a subset of those of $Q$ that contain $X$ in its interior. Thus, $n|T|=2|Q|=2n^4$, so $|T|=2n^3$ as desired.
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awesomeming327.
1735 posts
#7
Y by
Another way to think about the problem after finding the first balancing line (assumed to be y=0) is to take a variable shear causing the median of the x-values on either side of the line to shift towards each other continuous, until we have n points in each quadrant.

Since isometries and shears are affine transforms, which preserve area ratios, the statement “X lies in ABC” is preserved, so we can make any translation, rotation, or shear to the original configuration that we want.
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