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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.
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[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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0 replies
jwelsh
Jul 1, 2025
0 replies
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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
J
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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
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Polynomial NT with analytic flavour?
Aiden-1089   9
N a few seconds ago by Scilyse
Source: APMO 2025 Problem 3
Let $P(x)$ be a non-constant polynomial with integer coefficients such that $P(0) \neq 0$. Let $a_1, a_2, a_3, \dots$ be an infinite sequence of integers such that $P(i - j)$ divides $a_i-a_j$ for all distinct positive integers $i,j$. Prove that the sequence $a_1, a_2, a_3, \dots$ must be constant, that is, $a_n$ equals a constant $c$ for all positive integers $n$.
9 replies
+1 w
Aiden-1089
6 hours ago
Scilyse
a few seconds ago
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Number theory-Very hard conjucture
slimshady360   4
N 2 minutes ago by slimshady360
I'm making this conjucture on a problem i'm solving but i can't find a pattern ...
4 replies
+2 w
slimshady360
15 minutes ago
slimshady360
2 minutes ago
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Geo seems familiar?
Aiden-1089   5
N 5 minutes ago by Manteca
Source: APMO 2025 Problem 1
Let $ABC$ be an acute triangle inscribed in a circle $\Gamma$. Let $A_1$ be the orthogonal projection of $A$ onto $BC$ so that $AA_1$ is an altitude. Let $B_1$ and $C_1$ be the orthogonal projections of $A_1$ onto $AB$ and $AC$, respectively. Point $P$ is such that quadrilateral $AB_1PC_1$ is convex and has the same area as triangle $ABC$. Is it possible that $P$ strictly lies in the interior of circle $\Gamma$? Justify your answer.
5 replies
Aiden-1089
6 hours ago
Manteca
5 minutes ago
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Excircle Tangency Points Concyclic with A
tastymath75025   37
N 29 minutes ago by KevinYang2.71
Source: USA Winter TST for IMO 2019, Problem 6, by Ankan Bhattacharya
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on line $BC$ satisfying $\angle AID=90^{\circ}$. Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to $\overline{BC}$ at $A_1$. Define points $B_1$ on $\overline{CA}$ and $C_1$ on $\overline{AB}$ analogously, using the excircles opposite $B$ and $C$, respectively.

Prove that if quadrilateral $AB_1A_1C_1$ is cyclic, then $\overline{AD}$ is tangent to the circumcircle of $\triangle DB_1C_1$.

Ankan Bhattacharya
37 replies
tastymath75025
Jan 21, 2019
KevinYang2.71
29 minutes ago
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Concurrent
Omid Hatami   11
N an hour ago by Mrcuberoot
Source: Iran 2005
Suppose $H$ and $O$ are orthocenter and circumcenter of triangle $ABC$. $\omega$ is circumcircle of $ABC$. $AO$ intersects with $\omega$ at $A_1$. $A_1H$ intersects with $\omega$ at $A'$ and $A''$ is the intersection point of $\omega$ and $AH$. We define points $B',\ B'',\ C'$ and $C''$ similiarly. Prove that $A'A'',B'B''$ and $C'C''$ are concurrent in a point on the Euler line of triangle $ABC$.
11 replies
Omid Hatami
Aug 27, 2005
Mrcuberoot
an hour ago
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A legendary inequality.
Imanamiri   3
N an hour ago by erzhane
Source: Inequality Russia
Let \( x, y, z \in \mathbb{R} \) such that
\[ x^2 + y^2 + z^2 = 2. \]Prove that
\[ x + y + z \leq xyz + 2. \]
3 replies
Imanamiri
3 hours ago
erzhane
an hour ago
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the answer is 3
Aiden-1089   3
N an hour ago by ihatemath123
Source: APMO 2025 Problem 4
Let $n \geq 3$ be an integer. There are $n$ cells on a circle, and each cell is assigned either $0$ or $1$. There is a rooster on one of these cells, and it repeats the following operation:

$\bullet$ If the rooster is on a cell assigned $0$, it changes the assigned number to $1$ and moves to the next cell counterclockwise.
$\bullet$ If the rooster is on a cell assigned $1$, it changes the assigned number to $0$ and moves to the cell after the next cell counterclockwise.

Prove that the following statement holds after sufficiently many operations:
If the rooster is on a cell $C$, then the rooster would go around the circle exactly three times, stopping again at $C$. Moreover, every cell would be assigned the same number as it was assigned right before the rooster went around the circle three times.
3 replies
Aiden-1089
5 hours ago
ihatemath123
an hour ago
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Iranian TST 2019, first exam, day1, problem 2
Hamper.r   15
N 2 hours ago by megarnie
$a, a_1,a_2,\dots ,a_n$ are natural numbers. We know that for any natural number $k$ which $ak+1$ is square, at least one of $a_1k+1,\dots ,a_n k+1$ is also square.
Prove $a$ is one of $a_1,\dots ,a_n$
Proposed by Mohsen Jamali
15 replies
Hamper.r
Apr 7, 2019
megarnie
2 hours ago
J
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Divisibility Sequence
vsamc   3
N 2 hours ago by asbodke
Source: APMO 2025 Problem 5
Consider an infinite sequence $a_1,a_2, \cdots$ of positive integers such that $$100!(a_m + a_{m+1} + \cdots + a_n) \text{ is a multiple of } a_{n-m+1}a_{m+n}$$for all positive integers $m, n$ such that $m\leq n$. Prove that the sequence is either bounded or linear.
$\emph{Observation:}$ A sequence of positive integers is $\emph{bounded}$ if there exists a constant $N$ such that $a_n < N$ for all $n\in \mathbb{Z}_{>0}$. A sequence is $\emph{linear}$ if $a_n = na_1$ for all $n\in \mathbb{Z}_{>0}.$
3 replies
vsamc
5 hours ago
asbodke
2 hours ago
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The result before GMA Problem 575
mihaig   2
N 2 hours ago by mihaig
Source: Farfuridi
Given $n\geq2.$ Prove that $\sqrt{\left[\frac{n^2}4\right]}$ is the least constant $K$ such that
$$\sum_{i=1}^{n}{a_i}+K\left(\sqrt{a_1}-\sqrt{a_n}\right)^2\geq\sqrt{n\sum_{i=1}^{n}{a_i^2}}$$holds true for all $a_1\geq\cdots\geq a_n\geq0,$ with $a_1>0.$
2 replies
mihaig
Today at 4:54 AM
mihaig
2 hours ago
J
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Minimum Area for Monochromatic Triangles
steven_zhang123   1
N 3 hours ago by sarjinius
Source: 2025 Hope League Test 2 P5
Given a positive integer \( n \), find the smallest real number \( S \) such that no matter how each integer point in the coordinate plane \( xOy \) is colored with one of \( n \) different colors, there always exist three non-collinear points \( A, B, C \) of the same color such that the area of \(\triangle ABC\) is at most \( S \).
Proposed by Li Tianqin
1 reply
steven_zhang123
Jul 24, 2025
sarjinius
3 hours ago
J
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IMO ShortList 2001, combinatorics problem 7
orl   29
N 3 hours ago by eg4334
Source: IMO ShortList 2001, combinatorics problem 7
A pile of $n$ pebbles is placed in a vertical column. This configuration is modified according to the following rules. A pebble can be moved if it is at the top of a column which contains at least two more pebbles than the column immediately to its right. (If there are no pebbles to the right, think of this as a column with 0 pebbles.) At each stage, choose a pebble from among those that can be moved (if there are any) and place it at the top of the column to its right. If no pebbles can be moved, the configuration is called a final configuration. For each $n$, show that, no matter what choices are made at each stage, the final configuration obtained is unique. Describe that configuration in terms of $n$.

IMO ShortList 2001, combinatorics problem 7, alternative
29 replies
orl
Sep 30, 2004
eg4334
3 hours ago
J
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Sum is composite
anantmudgal09   8
N 4 hours ago by L13832
Source: India Practice TST 2017 D2 P2
Let $a,b,c,d$ be pairwise distinct positive integers such that $$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}$$is an integer. Prove that $a+b+c+d$ is not a prime number.
8 replies
anantmudgal09
Dec 9, 2017
L13832
4 hours ago
J
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geO 98 [convex hexagon ABCDEF with B + D + F = 360°]
Maverick   30
N 4 hours ago by Kempu33334
Source: IMO Shortlist 1998 Geometry 6
Let $ABCDEF$ be a convex hexagon such that $\angle B+\angle D+\angle F=360^{\circ }$ and \[ \frac{AB}{BC} \cdot \frac{CD}{DE} \cdot \frac{EF}{FA} = 1. \] Prove that \[ \frac{BC}{CA} \cdot \frac{AE}{EF} \cdot \frac{FD}{DB} = 1. \]
30 replies
Maverick
Oct 1, 2003
Kempu33334
4 hours ago
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J
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Combinatorics Problem
P.J   8
N May 14, 2025 by MITDragon
Source: Mexican Mathematical Olympiad Problems Book
Calculate the sum of 1 x 1000 + 2 x 999 + ... + 999 x 2 + 1000 x 1
8 replies
P.J
Dec 28, 2024
MITDragon
May 14, 2025
J
Combinatorics Problem
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Reveal topic tags
combinatorics
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G H BBookmark kLocked kLocked NReply
Source: Mexican Mathematical Olympiad Problems Book
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P.J
6 posts
P.J
#1 h  Dec 28, 2024, 10:16 AM
Y by
Calculate the sum of 1 x 1000 + 2 x 999 + ... + 999 x 2 + 1000 x 1
Z K Y
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OSAGHO
2 posts
OSAGHO
#2 h  Dec 28, 2024, 10:59 AM
Y by
Kindly correct me if i am wrong
the answer i have come up wih - 167167500
Z K Y
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Sir_Numbercrunch
176 posts
Sir_Numbercrunch
#3 h  Dec 28, 2024, 12:09 PM
Y by
you are right
Z K Y
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Just1
60 posts
Just1
#4 h  Dec 28, 2024, 4:58 PM • 1 Y
Y by P.J
Hello but I got answer 167167000.I showed each term of sum like x*(1001-x) then I got the whole sum like this: 1001*(1+2+..+1000)-(1^2+2^2+...+1000^2) so my answer is 1001*334*500 which is equal to 167167000
If I did smth wrong then please,point it polite:)
This post has been edited 1 time. Last edited by Just1, Dec 28, 2024, 5:02 PM
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Sir_Numbercrunch
176 posts
Sir_Numbercrunch
#5 h  Dec 28, 2024, 5:01 PM • 1 Y
Y by P.J
Sir_Numbercrunch wrote:
you are (almost) right
Thanks @below I didn't realize our answers differed by 500.
Z K Y
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Just1
60 posts
Just1
#6 h  Dec 28, 2024, 5:07 PM
Y by
but what is my fault?
Z K Y
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P.J
6 posts
P.J
#7 h  Dec 29, 2024, 4:58 PM
Y by
#4 & #5 Impressive
Z K Y
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Ianis
446 posts
Ianis
#8 h  Dec 29, 2024, 5:32 PM
Y by
In general the sum is\begin{align*}\sum \limits _{k=1}^nk(n+1-k) & =\sum \limits _{k=1}^nk(n+1)-\sum \limits _{k=1}^nk^2 \\ & =(n+1)\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6} \\ & =n(n+1)\left (\frac{n+1}{2}-\frac{2n+1}{6}\right ) \\ & =\frac{n(n+1)(n+2)}{6} \\ & =\binom{n+2}{3}. \end{align*}When $n=1000$ we get $167167000$.
Z K Y
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MITDragon
18 posts
MITDragon
#9 h  May 14, 2025, 12:58 PM
Y by
This can be rewritten as:
\[ \sum_{k=1}^{1000} k \cdot (1001 - k) \]
Expanding the expression:
\[ \sum_{k=1}^{1000} \left(1001k - k^2\right) = 1001 \sum_{k=1}^{1000} k - \sum_{k=1}^{1000} k^2 \]
Use the formulas for the sum of the first \(n\) natural numbers and the sum of their squares:
\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]
For \(n = 1000\):
\[ \sum_{k=1}^{1000} k = \frac{1000 \cdot 1001}{2} = 500500 \]\[ \sum_{k=1}^{1000} k^2 = \frac{1000 \cdot 1001 \cdot 2001}{6} = 333833500 \]
Therefore:
\[ 1001 \cdot 500500 - 333833500 = 501000500 - 333833500 = \boxed{167167000} \]
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