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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.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[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
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
AOPS MO Introduce
MathMaxGreat   19
N 12 minutes ago by Jackson0423
$AOPS MO$

Problems: post it as a private message to me or @jerryZYang, please post it in $LATEX$ and have answers

6 Problems for two rounds, easier than $IMO$

If you want to do the problems or be interested, reply ’+1’
Want to post a problem reply’+2’ and message me
Want to be in the problem selection committee, reply’+3’
19 replies
MathMaxGreat
Today at 1:04 AM
Jackson0423
12 minutes ago
IMO Shortlist 2010 - Problem G1
Amir Hossein   138
N 15 minutes ago by mahyar_ais
Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proposed by Christopher Bradley, United Kingdom
138 replies
Amir Hossein
Jul 17, 2011
mahyar_ais
15 minutes ago
Two lengths are equal
62861   32
N 24 minutes ago by OronSH
Source: IMO 2015 Shortlist, G5
Let $ABC$ be a triangle with $CA \neq CB$. Let $D$, $F$, and $G$ be the midpoints of the sides $AB$, $AC$, and $BC$ respectively. A circle $\Gamma$ passing through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$, respectively. The points $H'$ and $I'$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H'I'$ meets $CD$ and $FG$ at $Q$ and $M$, respectively. The line $CM$ meets $\Gamma$ again at $P$. Prove that $CQ = QP$.

Proposed by El Salvador
32 replies
62861
Jul 7, 2016
OronSH
24 minutes ago
Number Theory Marahon
Jupiterballs   15
N 26 minutes ago by ItzsleepyXD
Let's start a number theory marathon
Rules:-
just don't post >2 problems before a solution and be friendly :)

I'll start
P1
15 replies
Jupiterballs
Jun 23, 2025
ItzsleepyXD
26 minutes ago
bounded or all n?
X.Allaberdiyev   2
N 35 minutes ago by young_desi
Source: IMSC Problem 4
Determine all $n$ such that it is possible to find a convex $n$-gon which can be tiled with triangles having angles $15^\circ$, $75^\circ$ and $90^\circ$.
2 replies
X.Allaberdiyev
Jul 7, 2025
young_desi
35 minutes ago
Parallel lines (extension of previous problem)
RANDOM__USER   0
an hour ago
Source: Own
Let \(D\) be an arbitrary point on the side \(BC\) in a triangle \(\triangle{ABC}\). Let \(E\) and \(F\) be the intersection of the lines parallel to \(AC\) and \(AB\) through \(D\) with \(AB\) and \(AC\). Let \(G\) be the intersection of \((AFE)\) with \((ABC)\). Let \(M\) be the midpoint of \(BC\) and \(X\) the intersection of \(AM\) with \((ABC)\). Let \(H\) be the intersection of \((XMG)\) with \(BC\). Prove that \(EF\) is parallel to \(AH\).

IMAGE

Note: This is another property of a configuration I posted before where one needed to prove that \(X, D\) and \(G\) are collinear. There are surprisingly many properties in the configuration posted earlier :P
0 replies
RANDOM__USER
an hour ago
0 replies
CGMO8: How many k such that 2012 divides nCr(2012,k)
v_Enhance   18
N an hour ago by SomeonecoolLovesMaths
Source: 2012 China Girl's Mathematical Olympiad
Find the number of integers $k$ in the set $\{0, 1, 2, \dots, 2012\}$ such that $\binom{2012}{k}$ is a multiple of $2012$.
18 replies
1 viewing
v_Enhance
Aug 13, 2012
SomeonecoolLovesMaths
an hour ago
AOPS MO discussions
MathMaxGreat   1
N an hour ago by MathMaxGreat
MathMaxGreat
3 hours ago
MathMaxGreat
an hour ago
IMO Genre Predictions
ohiorizzler1434   112
N an hour ago by compoly2010
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
112 replies
ohiorizzler1434
May 3, 2025
compoly2010
an hour ago
same vibes
X.Allaberdiyev   44
N 2 hours ago by logic101
Source: JBMO 2025 Problem 3
Let $ABC$ be a right-angled triangle with $\angle A = 90º$, let $D$ be the foot of the altitude from $A$ to $BC$, and let $E$ be the midpoint of $DC$. The circumcircle of $ABD$ intersects $AE$ again at point $F$. Let $X$ be the intersection of the lines $AB$ and $DF$. Prove that $XD = XC$.

Proposed by Dren Neziri, Albania
44 replies
1 viewing
X.Allaberdiyev
Jun 26, 2025
logic101
2 hours ago
sum(sqrt(x))<product(sqrt(x))+product(sqrt(x+1))
Marius_Avion_De_Vanatoare   3
N 2 hours ago by aecksteinul
Source: Moldova TST 2025 P9
Prove that for all positive real numbers \( x, y, z \), the inequality
\[
\sqrt{x} + \sqrt{y} + \sqrt{z} - \sqrt{xyz} \le \sqrt{(x+1)(y+1)(z+1)}
\]holds.
3 replies
Marius_Avion_De_Vanatoare
Jun 14, 2025
aecksteinul
2 hours ago
Circumcenter on angle bisector
DottedCaculator   8
N 2 hours ago by Korean_fish_Kaohsiung
Source: 2025 ELMO Shortlist G6
Let $ABC$ be a triangle with incenter $I$, and let the midpoint of $BC$ be $M$. Ray $MI$ intersects $AB$ and $AC$ at $P$ and $Q$. If $N$ is the midpoint of major arc $BC$, prove that the circumcenter of triangle $NPQ$ lies on $AI$.

Rohan Bodke
8 replies
DottedCaculator
Jun 30, 2025
Korean_fish_Kaohsiung
2 hours ago
concyclic
Jumbler   25
N 3 hours ago by Learning11
Source: Chinese Western Mathematical Olympiad 2006, Problem 6
$AB$ is a diameter of the circle $O$, the point $C$ lies on the line $AB$ produced. A line passing though $C$ intersects with the circle $O$ at the point $D$ and $E$. $OF$ is a diameter of circumcircle $O_{1}$ of $\triangle BOD$. Join $CF$ and produce, cutting the circle $O_{1}$ at $G$. Prove that points $O,A,E,G$ are concyclic.
25 replies
Jumbler
Nov 7, 2006
Learning11
3 hours ago
inequalities
pennypc123456789   4
N 3 hours ago by SunnyEvan
If $a,b,c$ are positive real numbers, then
$$
\frac{a + b}{a + 7b + c} + \dfrac{b + c}{b + 7c + a}+\dfrac{c + a}{c + 7a + b} \geq \dfrac{2}{3}$$
we can generalize this problem
4 replies
pennypc123456789
Apr 17, 2025
SunnyEvan
3 hours ago
Decimal functions in binary
Pranav1056   3
N May 23, 2025 by ihategeo_1969
Source: India TST 2023 Day 3 P1
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $f(x) + y$ and $f(y) + x$ have the same number of $1$'s in their binary representations, for any $x,y \in \mathbb{N}$.
3 replies
Pranav1056
Jul 9, 2023
ihategeo_1969
May 23, 2025
Decimal functions in binary
G H J
G H BBookmark kLocked kLocked NReply
Source: India TST 2023 Day 3 P1
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Pranav1056
35 posts
#1 • 4 Y
Y by GeoKing, CahitArf, Supercali, Siddharth03
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that $f(x) + y$ and $f(y) + x$ have the same number of $1$'s in their binary representations, for any $x,y \in \mathbb{N}$.
This post has been edited 2 times. Last edited by Pranav1056, Jul 9, 2023, 6:22 AM
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Supercali
1263 posts
#2 • 2 Y
Y by Om245, thepassionatepotato
A story about this problem: It was originally meant to be D3 P3 (basically the hardest problem in the TSTs), but a few days before the test, some of us found an easier solution while trying. Hence the problem had to be demoted to D3 P1. D4 P3 at that time, which was a very hard geo, was shifted to D3 P3 (I think it was more suitable for that position anyway), and we had to use the shortlist for D4 P3. Anyway, I think this is a very cute problem.

Here is the solution that we found:

For $n\in\mathbb N$, let $d(n)$ denote the number of $1$'s in the binary representation of $n$. Let $P(x,y)$ denote the statement that $f(x)+y$ and $f(y)+x$ have the same number of $1$'s in their binary representation.

Claim 1: For any $y,n \in \mathbb{N}$ with $2^n>f(y)$, $f(2^n-f(y))+y$ is a power of two.
Proof: $P(2^n-f(y),y)$ gives us that $2^n$ and $f(2^n-f(y))+y$ have the same number of $1$'s, and the former has exactly one $1$, so $f(2^n-f(y))+y$ has exactly one $1$, from which the claim follows. $\blacksquare$

Claim 2: $f(y+2^k)-f(y)$ is a power of two for any $k \geq 0$ and $y \geq 2^k$.
Proof: Choose an $n$ such that $n>1000+\log_2(10+|f(y+2^k)-f(y)|)$. By Claim 1, $f(2^n-f(y))=2^t-y >0$ for some $t$ with $t \geq k+1$. Therefore $P(2^n-f(y),y+2^k)$ gives $$d(2^n-f(y)+f(y+2^k))=d(2^t+2^k)=2$$since $t \geq k+1$. If $f(y)=f(y+2^k)$, then LHS is $d(2^n)=1$, contradiction! If $f(y)>f(y+2^k)$, and $f(y)-f(y+2^k)$ has $m<\log_2(10+|f(y+2^k)-f(y)|)$ digits, then $$d(2^n-f(y)+f(y+2^k)) \geq n-m-1 \geq 999>2$$since $2^n-f(y)+f(y+2^k)$ starts with at least $n-m-1$ ones, contradiction! Therefore $f(y+2^k)>f(y)$, and since $n$ is bigger that the number of digits in $f(y+2^k)-f(y)$, there is no carry-over, so
$$2=d(2^n-f(y)+f(y+2^k))=1+d(f(y+2^k)-f(y))$$which gives us $f(y+2^k)-f(y)$ is a power of $2$, as required. $\blacksquare$


Claim 2 gives us $f(y+1)-f(y)=2^{t(y)}$ for some $t(y)$, for all $y$. But for $y \geq 2$, $f(y+2)-f(y)$ is also a power of two $\implies$ $2^{t(y)}+2^{t(y+1)}$ is a power of two, which is only possible if $t(y)=t(y+1)$ for all $y \geq 2$. Therefore $f(y+1)-f(y)$ is a constant power of two for all $y \geq 2$, say $2^k$. This gives us $f(y)=2^ky+c$ for some constant $c$, for all $y \geq 2$. Putting this in Claim 1, we get
$$2^{k+n}-(2^{2k}-1)y-(2^k-1)c$$is a power of two for any $y \geq 2$ and any sufficiently large $n$. This is only possible if, for all $y \geq 2$,
$$(2^{2k}-1)y+(2^k-1)c=0$$$$\iff (2^k-1)((2^k+1)y+c)=0$$which can only hold for all $y \geq 2$ if $2^k=1$, i.e., $f(y)=y+c$ for all $y \geq 2$. But Claim 1 for $y=1$ and large $n$ gives
$$2^n-f(1)+1+c$$is a power of two for all sufficiently large $n$, which is only possible if $f(1)=1+c$. Therefore the only solutions are
$$\boxed{f(x)=x+c \ \ \forall x \in \mathbb{N}}$$where $c$ is a non-negative integer.
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i3435
1351 posts
#3 • 2 Y
Y by GeoKing, Om245
$P(2^n-f(x),x)$ means $x=2^k-f(2^n-f(x))$ for some $k$, for all $x,n$. $P(2^a+2^k-f(x),x)$ means that $f(2^a+2^k-f(x))+x$ is of the form $2^c+2^d$, where $c\neq d$ when $a\neq k$ and $c=d$ when $a=k$. Replacing $x$ in the previous equation with $2^n-f(x)$, $f(2^a+x)+2^n-f(x)$ either has one or two ones in its binary representation. If you make $n$ large, we get that $f(2^a+x)-f(x)$ is a power of two for all $a,x$. In the same manner as the previous post, you can get $f(x)=2^kx+c$ for some $k,c$. $P(x,2^kx)$ means $2^{k+1}x+c$ and $(2^{2k}+1)x+c$ have the same number of $1$'s in their binary representation. If $x$ is a large power of $2$, then the second one will have one more $1$ than the first one unless $k=0$. Thus $k=0$ and $f(x)=x+c$, which works.
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ihategeo_1969
273 posts
#4
Y by
Let $P(x,y)$ denote the assertion. Call $2$ such numbers quirky.

$P(2^n-f(y),y)$ gives us $f(2^n-f(y))+y=2^{g(n,y)}$ for any $n>f(y)$ where $g: \mathbb{N}^2 \to \mathbb N$ is a function. See that $g(n,y)$ is unbounded.

Claim: $f(x+2^\ell)-f(x)$ is a power of $2$ for any $\ell \ge 0$ and $x \ge 2^\ell$.
Proof: $P(x+2^\ell,2^n-f(x))$ gives us that $f(x+2^\ell)+2^n-f(x)$ and $2^{g(n,x)}+2^\ell$ are quirky.

Now $2^{g(n,x)}>2^\ell$ so $f(x+2^\ell)+2^n-f(x)=2^{g_1(n,x,\ell)}+2^{g_2(n,x,\ell)}$ where $g_1$, $g_2: \mathbb{N}^3 \to \mathbb Z _{\ge 0}$ and $g_1(n,x,\ell) \neq g_2(n,x,\ell)$. Fix $x$ and $\ell$ and we will abuse some notation by letting $g_i(n,x,\ell)=g_i(n)$ because I am lazy. So we have \begin{align*}
& 2^{g_1(n)}+2^{g_2(n)}-2^n \text{ is constant} \\
\implies & 2^{g_1(n)}+2^{g_2(n)}+2^m=2^{g_1(m)}+2^{g_2(m)}+2^n
\end{align*}Say $m>ng_1(n)g_2(n)$ and $g_1(n)$, $g_2(n)$, $m$ are all distinct and so is $g_1(m)$, $g_2(m)$. If $n=g_2(m)$ then LHS have $3$ $1$'s in their binary representation but RHS has atmost $2$.

Now as $n \neq m$ so $n \in \{g_1(n),g_2(n)\}$ and hence we get that $2^{g_1(n)}+2^{g_2(n)}-2^n$ is a power of $2$ and so we are done. $\square$

Choose $\ell=0$ and $1$ and easily get that $f(x+1)-f(x)$ is a constant power of $2$ for $x \ge 2$. By a bit case bash we get that that must be $1$. And similarly we get that $f(2)-f(1)=1$ as well.

Hence the only solution is $\boxed{f(x) \equiv x+c \text{ } \forall \text{ }x \in \mathbb{N}}$ where $c \ge 0$; which obviously works.
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