Plan ahead for the next school year. Schedule your class today!

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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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Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

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0 replies
jwelsh
Jul 1, 2025
0 replies
sumdigits
teomihai   2
N 3 minutes ago by teomihai
Find sum digits all number 1 to 2001.
2 replies
teomihai
2 hours ago
teomihai
3 minutes ago
2020 IGO Advanced P5
Gaussian_cyber   17
N 4 minutes ago by bin_sherlo
Source: 7th Iranian Geometry Olympiad (Advanced) P5
Consider an acute-angled triangle $\triangle ABC$ ($AC>AB$) with its orthocenter $H$ and circumcircle $\Gamma$.Points $M$,$P$ are midpoints of $BC$ and $AH$ respectively.The line $\overline{AM}$ meets $\Gamma$ again at $X$ and point $N$ lies on the line $\overline{BC}$ so that $\overline{NX}$ is tangent to $\Gamma$.
Points $J$ and $K$ lie on the circle with diameter $MP$ such that $\angle AJP=\angle HNM$ ($B$ and $J$ lie one the same side of $\overline{AH}$) and circle $\omega_1$, passing through $K,H$, and $J$, and circle $\omega_2$ passing through $K,M$, and $N$, are externally tangent to each other. Prove that the common external tangents of $\omega_1$ and $\omega_2$ meet on the line $\overline{NH}$.
Proposed by Alireza Dadgarnia
17 replies
Gaussian_cyber
Nov 4, 2020
bin_sherlo
4 minutes ago
About APMO 2025
AlexanderWangUSA   5
N 9 minutes ago by megarnie
When will the APMO 2025 problems be posted?
5 replies
AlexanderWangUSA
Yesterday at 8:55 PM
megarnie
9 minutes ago
locus of medicenter, equilateral triangle
p.lazarov06   4
N 18 minutes ago by Siddharthmaybe
Source: Sharygin Finals 2022 8.3
A circle $\omega$ and a point $P$ not lying on it are given. Let $ABC$ be an arbitrary equilateral triangle inscribed into $\omega$ and $A', B', C'$ be the projections of $P$ to $BC, CA, AB$. Find the locus of centroids of triangles $A' B'C'$.
4 replies
p.lazarov06
Oct 3, 2022
Siddharthmaybe
18 minutes ago
No more topics!
Computing functions
BBNoDollar   8
N May 24, 2025 by wh0nix
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
8 replies
BBNoDollar
May 18, 2025
wh0nix
May 24, 2025
Computing functions
G H J
G H BBookmark kLocked kLocked NReply
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BBNoDollar
15 posts
#1
Y by
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )
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alinazarboland
172 posts
#2
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Here's a sketch of a method which solves every single mobius tranform problem I saw.
Let $z_1,z_2$ be the two complex roots of $f(z)=z$. Then, since a mobius transform is just a combination of shifting,scaling,rotating, and inversion, for any complex number $z$ we have:
$$(z , \infty ; z_1,z_2) = (f(z) , \frac{a}{c} ; z_1,z_2)$$If you write this $n$ times you'd get:
$$k^n .\frac{z-z_1}{z-z_2} = \frac{f_n - z_1}{f_n - z_2}$$Where $k = \frac{a/c - z_1}{a/c - z_2}$.Now let $f_n(x) = \frac{x}{1 + nx}$ for some $n$. One can easily get $k^n=1$(by comparing the coefficient of $x$ in the respective polynomial identity) and so $x_1=x_2$(comparing $x^2$s).
Now, $x_1=x_2$ means we have a double root for $f(x)=x$ and delta=0 so $(d-a)^2+4bc=0$. Combining with the fact that $x_1,x_2$ are fix points of every $f_k$ , we'll get $(n-1)^2+0=0$ and $n=1$
This post has been edited 2 times. Last edited by alinazarboland, May 18, 2025, 7:42 PM
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alinazarboland
172 posts
#3
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Here are two old problems one from $2012$ IMC and one from Iranian Olympiad which are trivial with this method
https://artofproblemsolving.com/community/c7h491145p2754513
https://artofproblemsolving.com/community/c6h368215p2026678
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BBNoDollar
15 posts
#4
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alinazarboland wrote:
Here's a sketch of a method which solves every single mobius tranform problem I saw.
Let $z_1,z_2$ be the two complex roots of $f(z)=z$. Then, since a mobius transform is just a combination of shifting,scaling,rotating, and inversion, for any complex number $z$ we have:
$$(z , \infty ; z_1,z_2) = (f(z) , \frac{a}{c} ; z_1,z_2)$$If you write this $n$ times you'd get:
$$k^n .\frac{z-z_1}{z-z_2} = \frac{f_n - z_1}{f_n - z_2}$$Where $k = \frac{a/c - z_1}{a/c - z_2}$.Now let $f_n(x) = \frac{x}{1 + nx}$ for some $n$. One can easily get $k^n=1$(by comparing the coefficient of $x$ in the respective polynomial identity) and so $x_1=x_2$(comparing $x^2$s).
Now, $x_1=x_2$ means we have a double root for $f(x)=x$ and delta=0 so $(d-a)^2+4bc=0$. Combining with the fact that $x_1,x_2$ are fix points of every $f_k$ , we'll get $(n-1)^2+0=0$ and $n=1$

Thank you very much, i appreciate this solution ! I can understand it, but i need a 9th grade solution. I solved the ''reciprocal'' implication by induction, now i need to demonstrate the ''direct'' one. Can you or anyone help me ?
This post has been edited 1 time. Last edited by BBNoDollar, May 18, 2025, 10:15 PM
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ICE_CNME_4
22 posts
#5
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Bumping this
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ICE_CNME_4
22 posts
#6
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Bump. Bump
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BBNoDollar
15 posts
#7
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BUMPING for 9th grade solution
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ICE_CNME_4
22 posts
#8
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Someone for this?
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wh0nix
27 posts
#9
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Hint
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