Stay ahead of learning milestones! Enroll in a class over the summer!

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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
f(f(x)+y) = x+f(f(y))
NicoN9   1
N a few seconds ago by InterLoop
Source: own, well this is my first problem I've ever write
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that\[
f(f(x)+y) = x+f(f(y))
\]for all $x, y\in \mathbb{R}$.
1 reply
NicoN9
8 minutes ago
InterLoop
a few seconds ago
China TST 1986 4k circle markers
orl   3
N 9 minutes ago by TUAN2k8
Source: China TST 1986, problem 8
Mark $4 \cdot k$ points in a circle and number them arbitrarily with numbers from $1$ to $4 \cdot k$. The chords cannot share common endpoints, also, the endpoints of these chords should be among the $4 \cdot k$ points.

i. Prove that $2 \cdot k$ pairwisely non-intersecting chords can be drawn for each of whom its endpoints differ in at most $3 \cdot k - 1$.
ii. Prove that the $3 \cdot k - 1$ cannot be improved.
3 replies
orl
May 16, 2005
TUAN2k8
9 minutes ago
China South East Mathematical Olympiad 2021 Grade11 P8
Henry_2001   2
N 18 minutes ago by parkjungmin
A sequence $\{z_n\}$ satisfies that for any positive integer $i,$ $z_i\in\{0,1,\cdots,9\}$ and $z_i\equiv i-1 \pmod {10}.$ Suppose there is $2021$ non-negative reals $x_1,x_2,\cdots,x_{2021}$ such that for $k=1,2,\cdots,2021,$ $$\sum_{i=1}^kx_i\geq\sum_{i=1}^kz_i,\sum_{i=1}^kx_i\leq\sum_{i=1}^kz_i+\sum_{j=1}^{10}\dfrac{10-j}{50}z_{k+j}.$$Determine the least possible value of $\sum_{i=1}^{2021}x_i^2.$
2 replies
Henry_2001
Aug 8, 2021
parkjungmin
18 minutes ago
Killer NT that nobody solved (also my hardest NT ever created)
mshtand1   10
N 42 minutes ago by mshtand1
Source: Ukraine IMO 2025 TST P8
A positive integer number \( a \) is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence \( \{b_k\}_{k=1}^{\infty} \), where
\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]
Proposed by Arsenii Nikolaev and Mykhailo Shtandenko
10 replies
mshtand1
Apr 19, 2025
mshtand1
42 minutes ago
2009 PUMaC Algebra B2: polynomial of nested roots
PUMACuploader   2
N 2 hours ago by P162008
Let $p(x)$ be the polynomial with leading coefficent 1 and rational coefficents, such that \[p\left(\sqrt{3 + \sqrt{3 + \sqrt{3 + \ldots}}}\right) = 0,\] and with the least degree among all such polynomials. Find $p(5)$.
2 replies
PUMACuploader
Aug 31, 2011
P162008
2 hours ago
hmmt quadratic power of a prime
martianrunner   2
N 4 hours ago by lgx57
I was practicing problems and came across one as such:

"Find all integers $x$ such that $2x^2 + x-6$ is a positive integral power of a prime positive integer."

I mean after factoring I don't really know where to go...

A hint would be appreciated, and if you want to solve it, please hide your solutions!

Thanks :)
2 replies
martianrunner
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lgx57
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n! \prod f(i)^{-1.5} is divisible by 4 but not 8 2024 TMC AIME Mock #10
parmenides51   2
N 5 hours ago by Amkan2022
Let $f(k)$ for positive integral values of $k$ be the largest perfect square that divides $k$. Find the number of positive integers $n$ less than $1024$ such that $$n! \cdot  \prod^n_{i=1}f(i)^{-1.5}$$is divisible by $4$ but not $8$.
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parmenides51
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parmenides51   7
N 5 hours ago by persamaankuadrat
Let $P_n$ be the product of all positive integers less than $n$ not divisible by $5$. Let $N$ be the remainder when $P_{3749}$ is divided by $15625 = 5^6$. If $N =\overline{ABCDE}_{10}$, find $\overline{ABC} + \overline{DE}$.
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parmenides51
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persamaankuadrat
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ReticulatedPython   4
N Today at 3:28 AM by soryn
Let $OABC$ be a tetrahedron such that $\angle{AOB}=\angle{AOC}=\angle{BOC}=90^\circ.$ A sphere of radius $r$ is circumscribed about tetrahedron $OABC.$ Given that $OA=a$, $OB=b$, and $OC=c$, prove that $$r^2+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9\sqrt[3]{4}}{4}$$with equality at $a=b=c=\sqrt[3]{2}.$
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ReticulatedPython
Apr 21, 2025
soryn
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girishpimoli   1
N Today at 3:26 AM by lpieleanu
A fair coin is repeatedly tossed. If the probability that the first time head is tossed , twice in a row , is on the $9$ th and $10$ th toss, is
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girishpimoli
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lpieleanu
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BrUMO 2025 Team Round Problem 7
lpieleanu   1
N Today at 2:45 AM by enaschair
Digits $1$ through $9$ are placed on a $3 \times 3$ square such that all rows and columns sum to the same value. Please note that diagonals do not need to sum to the same value. How many ways can this be done?
1 reply
lpieleanu
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enaschair
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A Hard Function Problem
Saucepan_man02   1
N Today at 2:05 AM by Saucepan_man02
If $f(x)=\frac{(\sqrt{5}+1)x-1}{(10-2 \sqrt{5})x+(\sqrt{5}+1)}$, find the value of$\underbrace{f(f(....f(\sqrt{5})..)}_{61 \text{times}}$.
1 reply
Saucepan_man02
Today at 1:39 AM
Saucepan_man02
Today at 2:05 AM
Geometry Basic
AlexCenteno2007   2
N Today at 1:44 AM by AlexCenteno2007
Let $ABC$ be an isosceles triangle such that $AC=BC$. Let $P$ be a dot on the $AC$ side.
The tangent to the circumcircle of $ABP$ at point $P$ intersects the circumcircle of $BCP$ at $D$. Prove that CD$ \parallel$AB
2 replies
AlexCenteno2007
Today at 12:11 AM
AlexCenteno2007
Today at 1:44 AM
Basic geometry
AlexCenteno2007   1
N Today at 1:37 AM by imbadatmath1233
Let $ABC$ be an equilateral triangle.M and N are the midpoints of $AB$ and $BC$ respectively. Externally to the triangle $ABC$ an isosceles right triangle APC is constructed, with the angle $APC =90°$.If point $I$ is the intersection of $AN$ and $MP$, show that $CI$ is a bisector of the angle $ACM$
1 reply
AlexCenteno2007
Today at 12:21 AM
imbadatmath1233
Today at 1:37 AM
<PAC=2<CPA
Igor   6
N Jun 1, 2005 by Ghang Hwan
Source: French TST 2005 pb 5.
Let $ABC$ be a triangle such that $BC=AC+\frac{1}{2}AB$. Let $P$ be a point of $AB$ such that $AP=3PB$.

Show that $\widehat{PAC} = 2 \widehat{CPA}.$
6 replies
Igor
May 27, 2005
Ghang Hwan
Jun 1, 2005
<PAC=2<CPA
G H J
G H BBookmark kLocked kLocked NReply
Source: French TST 2005 pb 5.
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Igor
137 posts
#1 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Let $ABC$ be a triangle such that $BC=AC+\frac{1}{2}AB$. Let $P$ be a point of $AB$ such that $AP=3PB$.

Show that $\widehat{PAC} = 2 \widehat{CPA}.$
This post has been edited 1 time. Last edited by Igor, May 27, 2005, 12:56 PM
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Igor
137 posts
#2 • 2 Y
Y by Adventure10, Mango247
Main idea (I'll hide it if you don't want to see it) :
Click to reveal hidden text
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Anto
213 posts
#3 • 2 Y
Y by Adventure10, Mango247
If i am not mistaken, this problem appered in MediteraneanMC 2003.
Where the official solution was very ungly solution using only the rule of sine and cosine.
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Igor
137 posts
#4 • 2 Y
Y by Adventure10 and 1 other user
Ok, so here is my solution:

On $AB$, take $I$ such that $IC=AC$. It sufficient to prove that $PI=IC$. But $PI=PA-AI=\frac{3}{4}AB-AI$
Hence :
$IC=AC \Longleftrightarrow \frac{3}{4}AB-AC=AI\Longleftrightarrow \frac{3AB}{4AC}-1=\frac{AI}{AC}$.
$IC=AC \Longleftrightarrow \frac{3AB}{4AC}-1=2.\textrm{cos}(\widehat{PAC})$ by the sine law.
$IC=AC \Longleftrightarrow \frac{3AB}{4AC}-1=\frac{AB^2+AC^2-BC^2}{AB.AC}$ by the cosine law.
$IC=AC \Longleftrightarrow 3AB^2-4AC.AB=4AB^2+4AC^2-4BC^2$.
$IC=AC \Longleftrightarrow BC^2=(AC+\frac{1}{2}AB)^2$, and we are done.
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Sailor
256 posts
#5 • 1 Y
Y by Adventure10
From Steward's theorem:
$CP^2=b^2+\frac{3bc}{4}$. (1)

But we shall prove that: $CP=2b\cos(\frac{A}{2})$, or $CP^2=4b^2\frac{p(p-a)}{bc}$, since:
$p=b+\frac{3c}{4}$ ,$p-a=\frac{c}{4}$ and using (1) we get the result.
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xiaozg
9 posts
#6 • 2 Y
Y by Adventure10, Mango247
it's Stewart's theorem.
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Ghang Hwan
40 posts
#7 • 2 Y
Y by Adventure10, Mango247
So cool :lol: :lol:

Here is my solution which use only circles.

Let Q On AC with AB=AQ and opposite with C.................1

Let R ON BC with BC* CR=AC*CQ................................2

and easy to get BP*BA=BR*B C (by length condition).........3

0.5*<BAC = <AQB (by ........1)

=<ARC (by .........2)

=<APC (by...........3)

and WE CAN SAY "Q.E.D." :D :D :D
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