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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)!

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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
1 viewing
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
Elementary Problems Compilation
Saucepan_man02   31
N 7 minutes ago by anticodon
Could anyone send some elementary problems, which have tricky and short elegant methods to solve?

For example like this one:
Solve over reals: $$a^2 + b^2 + c^2 + d^2  -ab-bc-cd-d +2/5=0$$
31 replies
Saucepan_man02
May 26, 2025
anticodon
7 minutes ago
f(n) - n is periodic
62861   25
N 9 minutes ago by awesomeming327.
Source: IMO 2015 Shortlist, N6
Let $\mathbb{Z}_{>0}$ denote the set of positive integers. Consider a function $f: \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$. For any $m, n \in \mathbb{Z}_{>0}$ we write $f^n(m) = \underbrace{f(f(\ldots f}_{n}(m)\ldots))$. Suppose that $f$ has the following two properties:

(i) if $m, n \in \mathbb{Z}_{>0}$, then $\frac{f^n(m) - m}{n} \in \mathbb{Z}_{>0}$;
(ii) The set $\mathbb{Z}_{>0} \setminus \{f(n) \mid n\in \mathbb{Z}_{>0}\}$ is finite.

Prove that the sequence $f(1) - 1, f(2) - 2, f(3) - 3, \ldots$ is periodic.

Proposed by Ang Jie Jun, Singapore
25 replies
1 viewing
62861
Jul 7, 2016
awesomeming327.
9 minutes ago
Generic Real-valued FE
lucas3617   9
N 29 minutes ago by jasperE3
$f: \mathbb{R} -> \mathbb{R}$, find all functions where $f(2x+f(2y-x))+f(-x)+f(y)=2f(x)+f(y-2x)+f(2y)$ for all $x$,$y \in \mathbb{R}$
9 replies
1 viewing
lucas3617
Apr 25, 2025
jasperE3
29 minutes ago
Mr. Fat wandering the lattice
v_Enhance   32
N an hour ago by Ilikeminecraft
Source: ELMO 1999 Problem 2
Mr. Fat moves around on the lattice points according to the following rules: From point $(x,y)$ he may move to any of the points $(y,x)$, $(3x,-2y)$, $(-2x,3y)$, $(x+1,y+4)$ and $(x-1,y-4)$. Show that if he starts at $(0,1)$ he can never get to $(0,0)$.
32 replies
v_Enhance
Dec 29, 2012
Ilikeminecraft
an hour ago
Decreasing primes
MithsApprentice   20
N an hour ago by Ilikeminecraft
Source: USAMO 1997
Let $p_1, p_2, p_3, \ldots$ be the prime numbers listed in increasing order, and let $x_0$ be a real number between 0 and 1. For positive integer $k$, define
\[ x_k = \begin{cases} 0 & \mbox{if} \; x_{k-1} = 0, \\[.1in] {\displaystyle \left\{ \frac{p_k}{x_{k-1}} \right\}} & \mbox{if} \; x_{k-1} \neq 0, \end{cases}  \]
where $\{x\}$ denotes the fractional part of $x$. (The fractional part of $x$ is given by $x - \lfloor x \rfloor$ where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.) Find, with proof, all $x_0$ satisfying $0 < x_0 < 1$ for which the sequence $x_0, x_1, x_2, \ldots$ eventually becomes 0.
20 replies
MithsApprentice
Oct 9, 2005
Ilikeminecraft
an hour ago
Sequence of rational numbers
mojyla222   2
N an hour ago by Assassino9931
Source: Iran 2024 3rd round number theory exam P1
Given a sequence $x_1,x_2,x_3,\cdots$ of positive integers, Ali proceed the following algorythm: In the i-th step he markes all rational numbers in the interval $[0,1]$ which have denominator equal to $x_i$. Then he write down the number $a_i$ equal to the length of the smallest interval in $[0,1]$ which both two ends of that is a marked number. Find all sequences $x_1,x_2,x_3,\cdots$ with $x_5=5$ and such that for all $n\in \mathbb N$ we have
$$
a_1+a_2+\cdots+a_n= 2-\dfrac{1}{x_n}.
$$
Proposed by Mojtaba Zare
2 replies
mojyla222
Aug 27, 2024
Assassino9931
an hour ago
Midpoint in a weird configuration
Gimbrint   1
N 2 hours ago by Beelzebub
Source: Own
Let $ABC$ be an acute triangle ($AB<BC$) with circumcircle $\omega$. Point $L$ is chosen on arc $AC$, not containing $B$, so that, letting $BL$ intersect $AC$ at $S$, one has $AS<CS$. Points $D$ and $E$ lie on lines $AB$ and $BC$ respectively, such that $BELD$ is a parallelogram. Point $P$ is chosen on arc $BC$, not containing $A$, such that $\angle CBP=\angle BDE$. Line $AP$ intersects $EL$ at $X$, and line $CP$ intersects $DL$ at $Y$. Line $XY$ intersects $AB$, $BC$ and $BP$ at points $M$, $N$ and $T$ respectively.

Prove that $TN=TM$.
1 reply
Gimbrint
May 23, 2025
Beelzebub
2 hours ago
Another FE
M11100111001Y1R   2
N 2 hours ago by AblonJ
Source: Iran TST 2025 Test 2 Problem 3
Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for all $x,y>0$ we have:
$$f(f(f(xy))+x^2)=f(y)(f(x)-f(x+y))$$
2 replies
M11100111001Y1R
Today at 8:03 AM
AblonJ
2 hours ago
Shortest number theory you might've seen in your life
AlperenINAN   13
N 2 hours ago by lksb
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
13 replies
AlperenINAN
May 11, 2025
lksb
2 hours ago
n-term Sequence
MithsApprentice   15
N 2 hours ago by Ilikeminecraft
Source: USAMO 1996, Problem 4
An $n$-term sequence $(x_1, x_2, \ldots, x_n)$ in which each term is either 0 or 1 is called a binary sequence of length $n$. Let $a_n$ be the number of binary sequences of length $n$ containing no three consecutive terms equal to 0, 1, 0 in that order. Let $b_n$ be the number of binary sequences of length $n$ that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that $b_{n+1} = 2a_n$ for all positive integers $n$.
15 replies
MithsApprentice
Oct 22, 2005
Ilikeminecraft
2 hours ago
Drawing Triangles Against Your Clone
pieater314159   19
N 3 hours ago by Ilikeminecraft
Source: 2019 ELMO Shortlist C1
Elmo and Elmo's clone are playing a game. Initially, $n\geq 3$ points are given on a circle. On a player's turn, that player must draw a triangle using three unused points as vertices, without creating any crossing edges. The first player who cannot move loses. If Elmo's clone goes first and players alternate turns, who wins? (Your answer may be in terms of $n$.)

Proposed by Milan Haiman
19 replies
pieater314159
Jun 27, 2019
Ilikeminecraft
3 hours ago
Odd digit multiplication
JuanDelPan   12
N 3 hours ago by Ilikeminecraft
Source: Pan-American Girls' Mathematical Olympiad 2021, P4
Lucía multiplies some positive one-digit numbers (not necessarily distinct) and obtains a number $n$ greater than 10. Then, she multiplies all the digits of $n$ and obtains an odd number. Find all possible values of the units digit of $n$.

$\textit{Proposed by Pablo Serrano, Ecuador}$
12 replies
JuanDelPan
Oct 6, 2021
Ilikeminecraft
3 hours ago
Cup of Combinatorics
M11100111001Y1R   7
N 3 hours ago by MathematicalArceus
Source: Iran TST 2025 Test 4 Problem 2
There are \( n \) cups labeled \( 1, 2, \dots, n \), where the \( i \)-th cup has capacity \( i \) liters. In total, there are \( n \) liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.

$a)$ Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most \( \frac{4n}{3} \) steps.

$b)$ Prove that in at most \( \frac{5n}{3} \) steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
7 replies
M11100111001Y1R
May 27, 2025
MathematicalArceus
3 hours ago
Inequality
knm2608   17
N 3 hours ago by Adywastaken
Source: JBMO 2016 shortlist
If the non-negative reals $x,y,z$ satisfy $x^2+y^2+z^2=x+y+z$. Prove that
$$\displaystyle\frac{x+1}{\sqrt{x^5+x+1}}+\frac{y+1}{\sqrt{y^5+y+1}}+\frac{z+1}{\sqrt{z^5+z+1}}\geq 3.$$When does the equality occur?

Proposed by Dorlir Ahmeti, Albania
17 replies
knm2608
Jun 25, 2017
Adywastaken
3 hours ago
Iranian Geometry Olympiad (5)
MRF2017   8
N Feb 24, 2018 by uglysolutions
Source: Advanced level,P5
Do there exist six points $X_1,X_2,Y_1, Y_2,Z_1,Z_2$ in the plane such that all of the triangles $X_iY_jZ_k$ are similar for $1\leq i, j, k \leq 2$?
Proposed by Morteza Saghafian
8 replies
MRF2017
Sep 13, 2016
uglysolutions
Feb 24, 2018
Iranian Geometry Olympiad (5)
G H J
G H BBookmark kLocked kLocked NReply
Source: Advanced level,P5
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MRF2017
237 posts
#1 • 1 Y
Y by Adventure10
Do there exist six points $X_1,X_2,Y_1, Y_2,Z_1,Z_2$ in the plane such that all of the triangles $X_iY_jZ_k$ are similar for $1\leq i, j, k \leq 2$?
Proposed by Morteza Saghafian
This post has been edited 1 time. Last edited by nsato, Oct 5, 2019, 1:58 PM
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ThE-dArK-lOrD
4071 posts
#2 • 2 Y
Y by chirita.andrei, Adventure10
Bump bump :pilot:
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MRF2017
237 posts
#3 • 3 Y
Y by ThE-dArK-lOrD, Adventure10, Mango247
ThE-dArK-lOrD wrote:
Bump bump :pilot:

Hi,this problem was open before jury meeting :P :P
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Othre87
161 posts
#5 • 2 Y
Y by Adventure10, Mango247
There is a solution in the official website, check it out.
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MRF2017
237 posts
#6 • 1 Y
Y by Adventure10
Othre87 wrote:
There is a solution in the official website, check it out.

This problem was open before jury meeting ,not now.
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guangzhou-2015
778 posts
#7 • 1 Y
Y by Adventure10
The answer is EXISTS
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ThE-dArK-lOrD
4071 posts
#8 • 9 Y
Y by talkon, kk108, WizardMath, adityaguharoy, thunderz28, CyclicISLscelesTrapezoid, Adventure10, Mango247, Assassino9931
Solution by Ilya Bogdanov
The triangles are similar to $\triangle{XYZ}$ where $XY=1,XZ=t^3,YZ=t^2$ and $\angle{Z}=\angle{X}+2\angle{Y}$.

I wonder what is the motivation for this mind-blowing solution.
Attachments:
This post has been edited 1 time. Last edited by ThE-dArK-lOrD, May 28, 2017, 2:39 PM
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Ankoganit
3070 posts
#9 • 5 Y
Y by ThE-dArK-lOrD, Pluto1708, Tafi_ak, guptaamitu1, Adventure10
Here's another construction that looks somewhat more human. It was found by user deep thought on Puzzling.SE.
[asy]size(10cm);
import markers;
pair X1=(0,2),X2=(0,-2),Y1=(sqrt(7),1),Y2=(-sqrt(7),1),Z1=(sqrt(7),-1),Z2=(-sqrt(7),-1);
fill(X1--Y1--Z1--cycle^^X1--Y1--Z2--cycle,cyan);draw(X1--Y1--Z1--cycle^^X1--Y1--Z2--cycle,blue);
draw(anglemark(Y1,Z2,X1,10)^^anglemark(Z1,X1,Y1,10),blue);
draw(anglemark(X1,Y1,Z2,10,12)^^anglemark(Y1,Z1,X1,10,12),blue);
D(MP("X_1\;(0,2)",X1,N)--MP("Y_1\;(\sqrt{7},1)",Y1,NE)--MP("Z_1\;(\sqrt{7},-1)",Z1,SE)--MP("X_2\;(0,-2)",X2,S)--MP("Z_2\;(-\sqrt{7},-1)",Z2,SW)--MP("Y_2\;(-\sqrt{7},1)",Y2,NW)--X1);
label("$\triangle X_1Y_1Z_1\sim\triangle Z_2X_1Y_1$",(0,-3));
dot(X1^^X2^^Y1^^Y2^^Z1^^Z2);
[/asy]
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uglysolutions
304 posts
#10 • 3 Y
Y by MRF2017, Adventure10, Mango247
ThE-dArK-lOrD wrote:
Solution by Ilya Bogdanov
The triangles are similar to $\triangle{XYZ}$ where $XY=1,XZ=t^3,YZ=t^2$ and $\angle{Z}=\angle{X}+2\angle{Y}$.

I wonder what is the motivation for this mind-blowing solution.
I was trying to understand this as well and here's what I came up with.
We start with a generic triangle $X_1Y_2Z_2$ (though slightly annoying, I'll follow the notation of the diagram from the official solution). Now let $X_2$ be the point such that $X_1Y_2X_2Z_2$ is a parallelogram. From this we already know that $\triangle X_1Y_2Z_2 \equiv \triangle X_2Y_2Z_2$.
Next, we construct the point $Z_1$ such that $\triangle X_2Y_2Z_1 \sim \triangle Z_2Y_2X_1$ ($Z_1$ on the same side of $Z_2$ with respect to $Y_2X_2$). In other words, we consider the spiral similarity with center $Y_2$ that maps $Z_2$ to $X_2$, and define $Z_1$ as the image of $X_1$ via this transformation. Now, by definition we get that $X_2Y_2Z_1$ is similar to our original triangle $X_1Y_2Z_2$. We shall now prove that $X_1Y_2Z_1$ is similar to the original triangle as well. More precisely, we'll show that $\triangle X_1Y_2Z_1 \sim \triangle Y_2Z_2X_1$.
First notice that $\angle X_1Y_2Z_1 = \angle X_1Y_2X_2 - \angle Z_1Y_2X_2 = (180^{\circ} - \angle Y_2X_1Z_2) - \angle Z_1Y_2X_2 = \angle X_1Y_2Z_2 + \angle Y_2Z_2X_1 - \angle Z_1Y_2X_2 = \angle Y_2Z_2X_1$, where in the last step we have used that $\angle X_1Y_2Z_2 = \angle Z_1Y_2X_2$, due to the similarity. So it is enough to prove that $\frac{X_1Y_2}{Y_2Z_1} = \frac{Y_2Z_2}{Z_2X_1}$.
Let $x,y,z$ be the lengths of the sides opposite to $X_1, Y_2, Z_2$ in our original triangle. Now $Y_2X_2 = X_1Z_2 = y$, hence the ratio of the similarity $\triangle X_2Y_2Z_1 \sim \triangle Z_2Y_2X_1$ is $\frac{y}{x}$. Therefore, $Y_2Z_1 = \frac{y}{x} \cdot z$, and we have $\frac{X_1Y_2}{Y_2Z_1} = \frac{z}{\frac{yz}{x}} = \frac{x}{y} = \frac{Y_2Z_2}{Z_2X_1}$, as wanted.
So far we have that the four triangles $X_iY_2Z_k$ are similar, and this holds without any extra requirements on the initial triangle $X_1Y_2Z_2$.

If you are tired or need to go to the toilet, this is a good place to stop.

The sixth point $Y_1$ is constructed such that $Y_2X_2Y_1Z_1$ is a parallelogram. Notice this implies that $Z_1Y_1Z_2X_1$ is a parallelogram as well, because it has a pair of opposite sides that are parallel and of equal length.
Now it is easy to see that $\triangle X_1Y_2Z_1 \equiv \triangle Z_2X_2Y_1$ (respective sides are equal), and also $\triangle X_2Y_2Z_1 \equiv \triangle X_2Y_1Z_1$ (two halves of a parallelogram). So we have 6 similar triangles now.
The remaining two triangles, $X_1Y_1Z_2$ and $X_1Y_1Z_1$ are congruent (two halves of a parallelogram). So if $X_1Y_1Z_2$ happens to be similar to the initial triangle $X_1Y_2Z_2$, then all 8 triangles $X_iY_jZ_k$ are similar and we are done.
This is where the extra requirements for the initial triangle arise. Standard angle chasing yields that $\angle X_1Z_2Y_1 = \angle Y_2X_1Z_2$ if and only if $\angle X = \angle Z + 2 \angle Y$ in the starting triangle. Provided that this is the case, we further need $\frac{X_1Z_2}{Z_2Y_1} = \frac{Y_2X_1}{X_1Z_2}$, that is $\frac{y}{Z_2Y_1} = \frac{z}{y} \Rightarrow Z_2Y_1 = \frac{y^2}{z}$. Now, $Z_2Y_1 = X_1Z_1$, and because of the similarity $\triangle X_1Y_2Z_1 \sim \triangle Y_2Z_2X_1$ we have that $\frac{X_1Z_1}{X_1Y_2} = \frac{Y_2X_1}{Y_2Z_2}$, hence $X_1Z_1 = \frac{z^2}{x}$. So we need $\frac{y^2}{z} = \frac{z^2}{x}$, and if we set $x=1$ we have $y^2 = z^3$, which is the same to saying that $y=t^3$ and $z=t^2$ for some $t>0$. :)

PS. Notice the diagram from the official solution is obviously wrong, how can it be the case that $t^3 < t^2$ AND $t^3 < t^4$??
This post has been edited 1 time. Last edited by uglysolutions, Feb 24, 2018, 12:10 AM
Reason: fix typo
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