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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
Given k>=2 integers. Find polynomials P,Q with real coefficients satisfying 1/2
kyotaro   1
N 2 minutes ago by Tintarn

Let $k>=2$ be an integer. Find polynomials $P,Q$ with real coefficients satisfying $1/2^k+1/3^k+...+1/n^k=P(n)/Q(n)$
for all $n\ge1$.
1 reply
kyotaro
Jul 19, 2025
Tintarn
2 minutes ago
Geometry
LuxusN   0
2 minutes ago
Let triangle $ABC$ be inscribed in circle $(O)$. Let $J$ be a point on the angle bisector of $\angle BAC$. Draw a circle centered at $J$ tangent to $BC$ at $D$. Draw the two common external tangents to $(J;JD)$ and $(O)$, which intersect line $BC$ at $P$ and $Q$. Prove that $AP$ and $AQ$ are isogonal conjugates with respect to $\angle BAC$.
0 replies
LuxusN
2 minutes ago
0 replies
orang NT
KevinYang2.71   32
N 12 minutes ago by ioannism45
Source: ISL 2024 N1
Find all positive integers $n$ with the following property: for all positive divisors $d$ of $n$, we have $d+1\mid n$ or $d+1$ is prime.
32 replies
KevinYang2.71
Jul 16, 2025
ioannism45
12 minutes ago
sumdigits
teomihai   3
N 15 minutes ago by CuriousMathBoy72
Find sum digits all number 1 to 2001.
3 replies
teomihai
4 hours ago
CuriousMathBoy72
15 minutes ago
intersection of lines through midpoints are met at perpendicular bisector
azzam2912   4
N 15 minutes ago by historypasser-by
Given a triangle $\triangle ABC$. The incircle of triangle $\triangle ABC$ is tangent to the sides $BC, CA,$ and $AB$ at points $D, E,$ and $F$, respectively. Let $X$ and $Y$ be the midpoints of $BD$ and $DC$, respectively. Let $P$ and $Q$ be the midpoints of $BF$ and $CE$, respectively. Prove that $XP$ and $YQ$ intersect on the perpendicular bisector of side $BC$.

notes. : does anyone knows from which source this question comes from?
4 replies
azzam2912
Today at 1:00 AM
historypasser-by
15 minutes ago
2024 International Math Olympiad Number Theory Shortlist, Problem 2
brainfertilzer   15
N 22 minutes ago by cursed_tangent1434
Source: 2024 ISL N2
Determine all finite, nonempty sets $\mathcal{S}$ of positive integers such that for every $a,b\in\mathcal{S}$ there exists $c\in\mathcal{S}$ with $a\mid b+2c$.
15 replies
brainfertilzer
Jul 16, 2025
cursed_tangent1434
22 minutes ago
what accursed word contains "ZAX"???
Scilyse   19
N 26 minutes ago by cursed_tangent1434
Source: 2024 IMOSL G5
Let $ABC$ be a triangle with incentre $I$, and let $\Omega$ be the circumcircle of triangle $BIC$. Let $K$ be a point in the interior of segment $BC$ such that $\angle BAK < \angle KAC$. The angle bisector of $\angle BKA$ intersects $\Omega$ at points $W$ and $X$ such that $A$ and $W$ lie on the same side of $BC$, and the angle bisector of $\angle CKA$ intersects $\Omega$ at points $Y$ and $Z$ such that $A$ and $Y$ lie on the same side of $BC$.

Prove that $\angle WAY = \angle ZAX$.

Proposed by David Brodsky, Uzbekistan
19 replies
Scilyse
Jul 16, 2025
cursed_tangent1434
26 minutes ago
2025 Greece IMO TST P3
brainfertilzer   18
N 29 minutes ago by cursed_tangent1434
Source: 2024 ISL G3
Let $ABCDE$ be a convex pentagon and let $M$ be the midpoint of $AB$. Suppose that segment $AB$ is tangent to the circumcircle of triangle $CME$ at $M$ and that $D$ lies on the circumircles of $AME$ and $BMC$. Lines $AD$ and $ME$ interesect at $K$, and lines $BD$ and $MC$ intersect at $L$. Points $P$ and $Q$ lie on line $EC$ so that $\angle PDC = \angle EDQ = \angle ADB$.

Prove that lines $KP, LQ,$ and $MD$ are concurrent.
18 replies
brainfertilzer
Jul 16, 2025
cursed_tangent1434
29 minutes ago
quadrilateral geo with length conditions
OronSH   20
N 31 minutes ago by cursed_tangent1434
Source: IMO Shortlist 2024 G1
Let $ABCD$ be a cyclic quadrilateral such that $AC<BD<AD$ and $\angle DBA<90^\circ$. Point $E$ lies on the line through $D$ parallel to $AB$ such that $E$ and $C$ lie on opposite sides of line $AD$, and $AC=DE$. Point $F$ lies on the line through $A$ parallel to $CD$ such that $F$ and $C$ lie on opposite sides of line $AD$, and $BD=AF$.

Prove that the perpendicular bisectors of segments $BC$ and $EF$ intersect on the circumcircle of $ABCD$.

Proposed by Mykhailo Shtandenko, Ukraine
20 replies
OronSH
Jul 16, 2025
cursed_tangent1434
31 minutes ago
partitioning boards
KevinYang2.71   8
N 33 minutes ago by cursed_tangent1434
Source: ISL 2024 C2
Let $n$ be a positive integer. The integers $1,\,2,\,3,\,\ldots,\,n^2$ are to be written in the cells of an $n\times n$ board such that each integer is written in exactly one cell and each cell contains exactly one integer. For every integer $d$ with $d\mid n$, the $d$-division of the board is the division of the board into $(n/d)^2$ nonoverlapping sub-boards, each of size $d\times d$, such that each cell is contained in exactly one $d\times d$ sub-board.

We say that $n$ is a cool number if the integers can be written on the $n\times n$ board such that, for each integer $d$ with $d\mid n$ and $1<d<n$, in the $d$-division of the board, the sum of the integers written in each $d\times d$ sub-board is not a multiple of $d$.

Determine all even cool numbers.

Proposed by Melek Güngör, Türkiye
8 replies
KevinYang2.71
Jul 16, 2025
cursed_tangent1434
33 minutes ago
Iran TST P10
TheBarioBario   2
N 37 minutes ago by megarnie
Source: Iranian TST problem 10
We call an infinite set $S\subseteq\mathbb{N}$ good if for all parwise different integers $a,b,c\in S$, all positive divisors of $\frac{a^c-b^c}{a-b}$ are in $S$. for all positive integers $n>1$, prove that there exists a good set $S$ such that $n \not \in S$.

Proposed by Seyed Reza Hosseini Dolatabadi
2 replies
TheBarioBario
Apr 2, 2022
megarnie
37 minutes ago
Ellipse and Incircle
Eeightqx   4
N 41 minutes ago by IsinAHEDRON
Source: 2025 China South East Mathematical Olympiad Grade11 P4
A convex quadrilateral $ABCD$ satisfies that $AB+AD=CB+CD$. The incircle of $\triangle ABD$ touches $AB$ at $M$, the incircle of $\triangle BCD$ touches $BC$ at $N$. Show that the intersection of $AN$ and $CM$ lies on the another inner common tangent of two incircles.
4 replies
Eeightqx
Today at 6:12 AM
IsinAHEDRON
41 minutes ago
Power polynomial
asbodke   7
N an hour ago by Roots_Of_Moksha
Source: 2025 ELMO Shortlist N5
Let $n$ and $k$ be positive integers. Let $P$ be a polynomial with degree at most $k$ and real coefficients such that for each integer $0 \leq i \leq k$, we have $P(i) = n^i$. For which pairs $(n,k)$ does there exist an integer $m$ such that $P(m)=n^{m-1}$?

Grant Blitz
7 replies
asbodke
Jun 30, 2025
Roots_Of_Moksha
an hour ago
Another Tau Sequence
Eeightqx   1
N an hour ago by hung9A
Source: 2025 China South East Mathematical Olympiad Grade11 P1
Given an even number $m\ge6$.

Say a positive integer sequence $\{a_n\}_{n=1}^{m}$ satisfies property $\tau$, if
(i) $a_1=1$;
(ii) $\forall 1\le k\le m-1$, $a_{k+1}-a_k\in\{a_i\mid 1\le i\le k\}$;
(iii) $a_2,\, a_4,\,\cdots,\, a_m$ is an geometric sequence with common quotient $2$.

(1) Write a $\tau$ sequence when $m=22$;
(2) Find the total number of the sequences satisfy $\tau$ when given $m$.
1 reply
Eeightqx
Today at 7:39 AM
hung9A
an hour 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
1264 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.
This post has been edited 3 times. Last edited by Supercali, Jul 27, 2023, 7:50 AM
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i3435
1353 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
285 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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