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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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
Miklos Schweitzer 1971_7
ehsan2004   1
N 4 minutes ago by pi_quadrat_sechstel
Let $ n \geq 2$ be an integer, let $ S$ be a set of $ n$ elements, and let $ A_i , \; 1\leq i \leq m$, be distinct subsets of $ S$ of size at least $ 2$ such that \[ A_i \cap A_j \not= \emptyset, A_i \cap A_k \not= \emptyset, A_j \cap A_k \not= \emptyset, \;\textrm{imply}\ \;A_i \cap A_j \cap A_k \not= \emptyset \ .\] Show that $ m \leq 2^{n-1}-1$.

P. Erdos
1 reply
ehsan2004
Oct 29, 2008
pi_quadrat_sechstel
4 minutes ago
Functional equation with a twist (it's number theory)
Davdav1232   0
13 minutes ago
Source: Israel TST 8 2025 p2
Prove that for all primes \( p \) such that \( p \equiv 3 \pmod{4} \) or \( p \equiv 5 \pmod{8} \), there exist integers
\[
1 \leq a_1 < a_2 < \cdots < a_{(p-1)/2} < p
\]such that
\[
\prod_{\substack{1 \leq i < j \leq (p-1)/2}} (a_i + a_j)^2 \equiv 1 \pmod{p}.
\]
0 replies
Davdav1232
13 minutes ago
0 replies
Grid combi with T-tetrominos
Davdav1232   0
16 minutes ago
Source: Israel TST 8 2025 p1
Let \( f(N) \) denote the maximum number of \( T \)-tetrominoes that can be placed on an \( N \times N \) board such that each \( T \)-tetromino covers at least one cell that is not covered by any other \( T \)-tetromino.

Find the smallest real number \( c \) such that
\[
f(N) \leq cN^2
\]for all positive integers \( N \).
0 replies
1 viewing
Davdav1232
16 minutes ago
0 replies
forced vertices in graphs
Davdav1232   0
18 minutes ago
Source: Israel TST 7 2025 p2
Let \( G \) be a graph colored using \( k \) colors. We say that a vertex is forced if it has neighbors in all the other \( k - 1 \) colors.

Prove that for any \( 2024 \)-regular graph \( G \), there exists a coloring using \( 2025 \) colors such that at least \( 1013 \) of the colors have a forced vertex of that color.

Note: The graph coloring must be valid, this means no \( 2 \) vertices of the same color may be adjacent.
0 replies
Davdav1232
18 minutes ago
0 replies
Can this sequence be bounded?
darij grinberg   70
N 19 minutes ago by ezpotd
Source: German pre-TST 2005, problem 4, ISL 2004, algebra problem 2
Let $a_0$, $a_1$, $a_2$, ... be an infinite sequence of real numbers satisfying the equation $a_n=\left|a_{n+1}-a_{n+2}\right|$ for all $n\geq 0$, where $a_0$ and $a_1$ are two different positive reals.

Can this sequence $a_0$, $a_1$, $a_2$, ... be bounded?

Proposed by Mihai Bălună, Romania
70 replies
darij grinberg
Jan 19, 2005
ezpotd
19 minutes ago
weird conditions in geo
Davdav1232   0
21 minutes ago
Source: Israel TST 7 2025 p1
Let \( \triangle ABC \) be an isosceles triangle with \( AB = AC \). Let \( D \) be a point on \( AC \). Let \( L \) be a point inside the triangle such that \( \angle CLD = 90^\circ \) and
\[
CL \cdot BD = BL \cdot CD.
\]Prove that the circumcenter of triangle \( \triangle BDL \) lies on line \( AB \).
0 replies
Davdav1232
21 minutes ago
0 replies
find angle
TBazar   4
N 31 minutes ago by vanstraelen
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
4 replies
TBazar
Today at 6:57 AM
vanstraelen
31 minutes ago
Polys with int coefficients
adihaya   4
N 41 minutes ago by sangsidhya
Source: 2012 INMO (India National Olympiad), Problem #3
Define a sequence $<f_0 (x), f_1 (x), f_2 (x), \dots>$ of functions by $$f_0 (x) = 1$$$$f_1(x)=x$$$$(f_n(x))^2 - 1 = f_{n+1}(x) f_{n-1}(x)$$for $n \ge 1$. Prove that each $f_n (x)$ is a polynomial with integer coefficients.
4 replies
adihaya
Mar 30, 2016
sangsidhya
41 minutes ago
Italian WinterCamps test07 Problem4
mattilgale   89
N an hour ago by cj13609517288
Source: ISL 2006, G3, VAIMO 2007/5
Let $ ABCDE$ be a convex pentagon such that
\[ \angle BAC = \angle CAD = \angle DAE\qquad \text{and}\qquad \angle ABC = \angle ACD = \angle ADE.
\]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$.

Proposed by Zuming Feng, USA
89 replies
mattilgale
Jan 29, 2007
cj13609517288
an hour ago
Simple triangle geometry [a fixed point]
darij grinberg   49
N an hour ago by cj13609517288
Source: German TST 2004, IMO ShortList 2003, geometry problem 2
Three distinct points $A$, $B$, and $C$ are fixed on a line in this order. Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$. Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$. Suppose $\Gamma$ meets the segment $PB$ at $Q$. Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$.
49 replies
darij grinberg
May 18, 2004
cj13609517288
an hour ago
Kosovo MO 2010 Problem 5
Com10atorics   19
N an hour ago by CM1910
Source: Kosovo MO 2010 Problem 5
Let $x,y$ be positive real numbers such that $x+y=1$. Prove that
$\left(1+\frac {1}{x}\right)\left(1+\frac {1}{y}\right)\geq 9$.
19 replies
Com10atorics
Jun 7, 2021
CM1910
an hour ago
Hard combi
EeEApO   1
N an hour ago by EeEApO
In a quiz competition, there are a total of $100 $questions, each with $4$ answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?

Now, suppose my spouse and I move into a new home. Every year, we have twins. Starting at the age of $16$, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
1 reply
EeEApO
3 hours ago
EeEApO
an hour ago
Problem on symmetric polynomial
ayan_mathematics_king   5
N an hour ago by bjump
If $a^3+b^3+c^3=(a+b+c)^3$, prove that $a^5+b^5+c^5=(a+b+c)^5$ where $a,b,c \in \mathbb{R}$
5 replies
ayan_mathematics_king
Jul 28, 2019
bjump
an hour ago
Simple inequality
sqing   57
N 2 hours ago by Sh309had
Source: Shortlist BMO 2018, A1
Let $a, b, c $ be positive real numbers such that $abc = \frac {2} {3}. $ Prove that:

$$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant  \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$$
57 replies
sqing
May 3, 2019
Sh309had
2 hours ago
trolling geometry problem
iStud   4
N Apr 22, 2025 by GreenTea2593
Source: Monthly Contest KTOM April P3 Essay
Given a cyclic quadrilateral $ABCD$ with $BC<AD$ and $CD<AB$. Lines $BC$ and $AD$ intersect at $X$, and lines $CD$ and $AB$ intersect at $Y$. Let $E,F,G,H$ be the midpoints of sides $AB,BC,CD,DA$, respectively. Let $S$ and $T$ be points on segment $EG$ and $FH$, respectively, so that $XS$ is the angle bisector of $\angle{DXA}$ and $YT$ is the angle bisector of $\angle{DYA}$. Prove that $TS$ is parallel to $BD$ if and only if $AC$ divides $ABCD$ into two triangles with equal area.
4 replies
iStud
Apr 21, 2025
GreenTea2593
Apr 22, 2025
trolling geometry problem
G H J
Source: Monthly Contest KTOM April P3 Essay
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iStud
268 posts
#1
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Given a cyclic quadrilateral $ABCD$ with $BC<AD$ and $CD<AB$. Lines $BC$ and $AD$ intersect at $X$, and lines $CD$ and $AB$ intersect at $Y$. Let $E,F,G,H$ be the midpoints of sides $AB,BC,CD,DA$, respectively. Let $S$ and $T$ be points on segment $EG$ and $FH$, respectively, so that $XS$ is the angle bisector of $\angle{DXA}$ and $YT$ is the angle bisector of $\angle{DYA}$. Prove that $TS$ is parallel to $BD$ if and only if $AC$ divides $ABCD$ into two triangles with equal area.
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iStud
268 posts
#2
Y by
i can prove the $\Longleftarrow$ direction but not pretty sure with the $\Longrightarrow$ direction, can anyone help me?
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mathuz
1524 posts
#3
Y by
Note that $EFGH$ is a parallelogram and $FG\parallel EH \parallel BD$. Therefore, $TS\parallel BD$ iff $ES:SG = HT:TF$.

Assume now $TS\parallel BD$. Then $ES:SG = HT:TF$.
We have $\triangle XCD \sim \triangle XAB$ and $\triangle YBC\sim \triangle YDA$. This implies $XS$ is the angle bisector of $\angle EXG$ and $YT$ is the angle bisector of $\angle HYF$. Therefore, we have \[ \frac{EX}{GX} = \frac{ES}{SG} = \frac{HT}{TF} = \frac{HY}{FY}. \]Again, by the similarity this means \[ \frac{AB}{CD} = \frac{DA}{BC} \quad \Rightarrow \quad AB\cdot BC = CD\cdot DA, \]which essentially means the areas of $ABC$ and $ADC$ are equal.

This gives $\Longrightarrow $ direction.
This post has been edited 1 time. Last edited by mathuz, Apr 22, 2025, 1:35 AM
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iStud
268 posts
#4
Y by
mathuz wrote:
Note that $EFGH$ is a parallelogram and $FG\parallel EH \parallel BD$. Therefore, $TS\parallel BD$ iff $ES:SG = HT:TF$.

Assume now $TS\parallel BD$. Then $ES:SG = HT:TF$.
We have $\triangle XCD \sim \triangle XAB$ and $\triangle YBC\sim \triangle YDA$. This implies $XS$ is the angle bisector of $\angle EXG$ and $YT$ is the angle bisector of $\angle HYF$. Therefore, we have \[ \frac{EX}{GX} = \frac{ES}{SG} = \frac{HT}{TF} = \frac{HY}{FY}. \]Again, by the similarity this means \[ \frac{AB}{CD} = \frac{DA}{BC} \quad \Rightarrow \quad AB\cdot BC = CD\cdot DA, \]which essentially means the areas of $ABC$ and $ADC$ are equal.

This gives $\Longrightarrow $ direction.

demn, all of my Menelause bash worth nothing in front of this dude's solution
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GreenTea2593
236 posts
#5
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Oops, I took this from 2016 Japan MO Finals Problem 2
This post has been edited 1 time. Last edited by GreenTea2593, Apr 22, 2025, 3:12 AM
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