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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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Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[list][*]April 3rd (Webinar), 4pm PT/7:00pm ET, Learning with AoPS: Perspectives from a Parent, Math Camp Instructor, and University Professor
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April 9th (Webinar), 4:00pm PT/7:00pm ET, Learn about Video-based Summer Camps at the Virtual Campus
[*]April 10th (Math Jam), 4:30pm PT/7:30pm ET, 2025 MathILy and MathILy-Er Math Jam: Multibackwards Numbers
[*]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
Irrational equation
giangtruong13   1
N 3 minutes ago by Tuvshuu
Solve the equation : $$(\sqrt{x}+1)[2-(x-6)\sqrt{x-3}]=x+8$$
1 reply
+1 w
giangtruong13
40 minutes ago
Tuvshuu
3 minutes ago
IMO Shortlist 2014 N6
hajimbrak   28
N 18 minutes ago by MajesticCheese
Let $a_1 < a_2 <  \cdots <a_n$ be pairwise coprime positive integers with $a_1$ being prime and $a_1 \ge n + 2$. On the segment $I = [0, a_1 a_2  \cdots a_n ]$ of the real line, mark all integers that are divisible by at least one of the numbers $a_1 ,   \ldots , a_n$ . These points split $I$ into a number of smaller segments. Prove that the sum of the squares of the lengths of these segments is divisible by $a_1$.

Proposed by Serbia
28 replies
hajimbrak
Jul 11, 2015
MajesticCheese
18 minutes ago
3 knightlike moves is enough
sarjinius   3
N 23 minutes ago by JollyEggsBanana
Source: Philippine Mathematical Olympiad 2025 P6
An ant is on the Cartesian plane. In a single move, the ant selects a positive integer $k$, then either travels [list]
[*] $k$ units vertically (up or down) and $2k$ units horizontally (left or right); or
[*] $k$ units horizontally (left or right) and $2k$ units vertically (up or down).
[/list]
Thus, for any $k$, the ant can choose to go to one of eight possible points.
Prove that, for any integers $a$ and $b$, the ant can travel from $(0, 0)$ to $(a, b)$ using at most $3$ moves.
3 replies
sarjinius
Mar 9, 2025
JollyEggsBanana
23 minutes ago
Why is the old one deleted?
EeEeRUT   15
N 26 minutes ago by Tuvshuu
Source: EGMO 2025 P1
For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$.

Proposed by Paulius Aleknavičius, Lithuania
15 replies
1 viewing
EeEeRUT
Apr 16, 2025
Tuvshuu
26 minutes ago
Inequalities
sqing   19
N 6 hours ago by sqing
Let $ a,b,c> 0 $ and $ ab+bc+ca\leq  3abc . $ Prove that
$$ a+ b^2+c\leq a^2+ b^3+c^2 $$$$ a+ b^{11}+c\leq a^2+ b^{12}+c^2 $$
19 replies
1 viewing
sqing
Apr 22, 2025
sqing
6 hours ago
circumcenter, excenter and vertex collinear (Singapore Junior 2012)
parmenides51   6
N Today at 6:24 AM by lightsynth123
In $\vartriangle ABC$, the external bisectors of $\angle A$ and $\angle B$ meet at a point $D$. Prove that the circumcentre of $\vartriangle ABD$ and the points $C, D$ lie on the same straight line.
6 replies
parmenides51
Jul 11, 2019
lightsynth123
Today at 6:24 AM
can anyone solve this
averageguy   9
N Today at 6:18 AM by ninjaforce
Hi guys,
For some reason I can't think of a simple way to solve this problem. Is there anyway you guys can think of without trig or if it does have trig something elegant. Answer is 106 btw.
9 replies
averageguy
Dec 26, 2024
ninjaforce
Today at 6:18 AM
Inequalities
sqing   5
N Today at 4:00 AM by sqing
Let $ a,b,c $ be real numbers such that $ a^2+b^2+c^2=1. $ Prove that$$ |a-b|+|b-2c|+|c-3a|\leq 5$$$$|a-2b|+|b-3c|+|c-4a|\leq \sqrt{42}$$$$ |a-b|+|b-\frac{11}{10}c|+|c-a|\leq \frac{29}{10}$$
5 replies
sqing
Apr 22, 2025
sqing
Today at 4:00 AM
Complex Numbers Question
franklin2013   3
N Today at 3:47 AM by KSH31415
Hello everyone! This is one of my favorite complex numbers questions. Have fun!

$f(z)=z^{720}-z^{120}$. How many complex numbers $z$ are there such that $|z|=1$ and $f(z)$ is an integer.

Hint
3 replies
franklin2013
Apr 20, 2025
KSH31415
Today at 3:47 AM
Inequalities
sqing   28
N Today at 2:53 AM by sqing
Let $   a,b    $ be reals such that $  a^2+ab+b^2 =3$ . Prove that
$$ \frac{4}{ 3}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{11}{4 }$$$$ \frac{13}{ 4}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{2}{3 }$$$$ \frac{3}{ 2}\geq  \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$$$$ \frac{19}{ 6}\geq  \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2}$$Let $   a,b    $ be reals such that $  a^2-ab+b^2 =1 $ . Prove that
$$ \frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$$$$ \frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$$$$ \frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$$
28 replies
sqing
Apr 16, 2025
sqing
Today at 2:53 AM
Middle School Math <3
peace09   13
N Yesterday at 11:39 PM by Kevin2010
If $f(0)=1$ and $f(n)=\tfrac{n!}{\text{lcm}(1,2,\dots,n)}$ for each positive integer $n$, what is the value of $\tfrac{f(1)}{f(0)}+\tfrac{f(2)}{f(1)}+\dots+\tfrac{f(50)}{f(49)}$?

If you enjoyed the above problem, check out the 2024 WMC Series!
13 replies
peace09
Mar 11, 2024
Kevin2010
Yesterday at 11:39 PM
Theory of Equations
P162008   2
N Yesterday at 10:18 PM by P162008
Let $a,b,c,d$ and $e\in [-2,2]$ such that $\sum_{cyc} a = 0, \sum_{cyc} a^3 = 0, \sum_{cyc} a^5 = 10.$ Find the value of $\sum_{cyc} a^2.$
2 replies
P162008
Yesterday at 11:27 AM
P162008
Yesterday at 10:18 PM
inequalities of elements in set
toanrathay   1
N Yesterday at 7:25 PM by Lankou
Let \( m \) be a positive integer such that \( m \geq 4 \), and let the set
\[
A = \{a_1, a_2, a_3, \ldots, a_m\}
\]consist of distinct positive integers not exceeding 2025. Suppose that for every \( a, b \in A \), with \( a \ne b \), if \( a + b \leq 2025 \), then \( a + b \in A \) as well. Prove that

\[
\frac{a_1 + a_2 + a_3 + \cdots + a_m}{m} \geq 1013.
\]
1 reply
toanrathay
Yesterday at 3:33 PM
Lankou
Yesterday at 7:25 PM
2024 PUMaC Team Round, Question 14 Inquiry
A22-4   0
Yesterday at 6:39 PM
2024 PUMaC Team Round Question 14 reads as follows:

What is the largest value of $m$ for which I can find nonnegative integers $a_1, a_2, \cdots, a_m<2024$ such that for all indices $i>j$, $17$ divides $\binom{a_i}{a_j}$?
(Note: This should say "... nonnegative integers $a_1<a_2<\cdots<a_m<2024$ ...")

I interpreted this correction to mean the following:
What is the largest value of $m$ for which I there exists nonnegative integers $a_1<a_2<\cdots<a_m<2024$ such that for all indices $i>j$, $17$ divides $\binom{a_i}{a_j}$?

The official answer (https://static1.squarespace.com/static/570450471d07c094a39efaed/t/67421bd74806e80a7ab11c7d/1732385751115/PUMaC_2024_Team__Final_.pdf) is 107. However, I believe I have a construction with $108$ integers - take the set of all integers with a digit sum of $19$ in base $17$, then append $2023_{10}=700_{17}$ to the list.

I checked this with Python using the following code:
[code]def digit_sum_base(n, base):
total = 0
while n > 0:
total += n % base
n //= base
return total

target_sum = 19
base = 17
limit = 2024
qualified_numbers = [n for n in range(limit) if digit_sum_base(n, base) == target_sum]

qualified_numbers.append(2023)

from math import comb

all_divisible = True
for i in range(len(qualified_numbers)):
for j in range(i):
a, b = qualified_numbers, qualified_numbers[j]
if comb(a, b) % 17 != 0:
all_divisible = False
break
if not all_divisible:
break

print(len(qualified_numbers), all_divisible)[/code]

Am I wrong or are they wrong? Any insight would be appreciated!
0 replies
A22-4
Yesterday at 6:39 PM
0 replies
H,G,O geometry
m4thbl3nd3r   1
N Apr 5, 2025 by aidenkim119
Source: own
Let $ABC$ be a non-isosceles triangle with orthocenter $H$, circumcenter $O$. Let $M,N,P$ be midpoints of $BC,CA,AB$ and $A',B',C'$ be reflections of $M,N,P$ across $HO$. Prove that triangle $ABC,A'B'C'$ have the same centroid.
1 reply
m4thbl3nd3r
Apr 5, 2025
aidenkim119
Apr 5, 2025
H,G,O geometry
G H J
Source: own
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m4thbl3nd3r
282 posts
#1
Y by
Let $ABC$ be a non-isosceles triangle with orthocenter $H$, circumcenter $O$. Let $M,N,P$ be midpoints of $BC,CA,AB$ and $A',B',C'$ be reflections of $M,N,P$ across $HO$. Prove that triangle $ABC,A'B'C'$ have the same centroid.
Z K Y
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aidenkim119
32 posts
#2
Y by
MNO Has same centroid as ABC. It is on HO, So reflecting still keeps it same ao ok
Z K Y
N Quick Reply
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