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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
concyclic
Jumbler   25
N 4 minutes 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
4 minutes ago
inequalities
pennypc123456789   4
N 6 minutes 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
6 minutes ago
Angelic tetrahedrons
v_Enhance   6
N 37 minutes ago by numbersandnumbers
Source: USA TSTST 2025/5
A tetrahedron $ABCD$ is said to be angelic if it has nonzero volume and satisfies \[ \begin{aligned} \angle BAC + \angle CAD + \angle DAB &= \angle ABC + \angle CBD + \angle DBA, \\ \angle ACB + \angle BCD + \angle DCA &= \angle ADB + \angle BDC + \angle CDA. \end{aligned} \]Across all angelic tetrahedrons, what is the maximum number of distinct lengths that could appear in the set $\{AB,AC,AD,BC,BD,CD\}$?

Karthik Vedula
6 replies
v_Enhance
Jul 1, 2025
numbersandnumbers
37 minutes ago
AOPS MO Introduce
MathMaxGreat   15
N 37 minutes ago by mongmong
$AOPS MO$

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’
15 replies
MathMaxGreat
4 hours ago
mongmong
37 minutes ago
No more topics!
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
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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.
This post has been edited 3 times. Last edited by Supercali, Jul 27, 2023, 7:50 AM
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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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