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sqing 5
N
by GeoMorocco
Source: Own
Let 
Prove that
Let 
Prove that
Let 
Prove that
Prove that
Let 
Prove that
Let 
Prove that
5 replies
one cyclic formed by two cyclic
CrazyInMath 37
N
by G81928128
Source: EGMO 2025/3
Let 
be an acute triangle. Points 
, and 
lie on a line in this order and satisfy 
. Let 
and 
be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
be an acute triangle. Points 
, and 
lie on a line in this order and satisfy 
. Let 
and 
be the midpoints of 
and 
, respectively. Suppose triangle 
is acute, and let 
be its orthocentre. Points 
and 
lie on lines 
and 
, respectively, such that 
and are concyclic and pairwise different, and 
and are concyclic and pairwise different. Prove that 
and are concyclic.
37 replies
Iran second round 2025-q1
mohsen 0
Find all positive integers n>2 such that sum of n and any of its prime divisors is a perfect square.
0 replies
FE inequality from Iran
mojyla222 1
N
by bin_sherlo
Source: Iran 2025 second round P5
Find all functions 
such that for all 
such that for all 
1 reply
such that 
.
Advanced topics in Inequalities
va2010 10
N
by Novmath
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
10 replies
Geometry Problem
Itoz 2
N
by Itoz
Source: Own
Given 
. Let the perpendicular line from 
to 
meets 
at points 
, respectively, and the foot from 
to 
is 
. 
intersects line 
at 
, 
intersects line at 
, and lines 
intersect at .
Prove that
is tangent to 
.
. Let the perpendicular line from 
to 
meets 
at points 
, respectively, and the foot from 
to 
is 
. 
intersects line 
at 
, 
intersects line at 
, and lines 
intersect at .Prove that
is tangent to 
.
2 replies
1 viewing
Why is the old one deleted?
EeEeRUT 11
N
by Mathgloggers
Source: EGMO 2025 P1
For a positive integer , let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
Here
is the largest positive integer that divides both 
and 
. Integers and are coprime if 
.
Proposed by Paulius Aleknavičius, Lithuania
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
Here
is the largest positive integer that divides both 
and 
. Integers and are coprime if 
.Proposed by Paulius Aleknavičius, Lithuania
11 replies
Congruence related perimeter
egxa 2
N
by LoloChen
Source: All Russian 2025 9.8 and 10.8
On the sides of triangle 
, points 
are chosen such that when going around the triangle, the points occur in the order 
. It is given that
![\[
AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2.
\]](//latex.artofproblemsolving.com/4/c/f/4cf9e7bc009736836bcd10c1d8ba46bd3a91cd7b.png)
Prove that the perimeters of the triangles formed by the triplets 
and 
are equal.
, points 
are chosen such that when going around the triangle, the points occur in the order 
. It is given that![\[
AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2.
\]](http://latex.artofproblemsolving.com/4/c/f/4cf9e7bc009736836bcd10c1d8ba46bd3a91cd7b.png)
Prove that the perimeters of the triangles formed by the triplets 
and 
are equal.
2 replies
number theory
Levieee 7
N
by g0USinsane777
Idk where it went wrong, marks was deducted for this solution
Show that for a fixed pair of distinct positive integers
and 
, there cannot exist infinitely many 
such that
![\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]](//latex.artofproblemsolving.com/5/4/d/54d20546de5736770a9929ac86163e3b06553b7d.png)
Let
![\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]](//latex.artofproblemsolving.com/6/6/8/668d53723d307556536736fda0e4a19be848834f.png)
Then,
![\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]](//latex.artofproblemsolving.com/1/9/f/19f2627a432354a757d3c9e8ba42e96f83f46726.png)
So:
![\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]](//latex.artofproblemsolving.com/0/7/3/073602693604bbf3cb1f5c1c2e5051e8994a5e78.png)
Therefore,
![\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]](//latex.artofproblemsolving.com/1/1/4/11441fab3ba768fee7a0eef4dbf68124d6637aec.png)
Let
![\[
(n + a)(n + b) = k^2.
\]](//latex.artofproblemsolving.com/0/e/c/0ec5e7bac2aebe9ec507727ed21f7ca028655222.png)
Assume 
. Then we have:
![\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]](//latex.artofproblemsolving.com/7/5/f/75fc52b6662df3a4c0cb2f40f49c4b3351c90a4d.png)
or it could also be that 
.
Without loss of generality, we take the first case:
![\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]](//latex.artofproblemsolving.com/1/f/5/1f5936241f7bebcf15c68fed5267bbfad2d40c2b.png)
Thus,
![\[
k_1 k_2 = \frac{n + b}{n + a}.
\]](//latex.artofproblemsolving.com/d/4/c/d4c1958e3554b58c3def282070f849842cc10912.png)
Since
, we have:
![\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]](//latex.artofproblemsolving.com/3/0/6/306a6a00c44d4e4db1b36c7e588f684f28fdbf44.png)
For infinitely many
, 
must be an integer, which is not possible.
Therefore, there cannot be infinitely many such.
Show that for a fixed pair of distinct positive integers
and 
, there cannot exist infinitely many 
such that![\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/5/4/d/54d20546de5736770a9929ac86163e3b06553b7d.png)
Let
![\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/6/6/8/668d53723d307556536736fda0e4a19be848834f.png)
Then,
![\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/1/9/f/19f2627a432354a757d3c9e8ba42e96f83f46726.png)
So:![\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/0/7/3/073602693604bbf3cb1f5c1c2e5051e8994a5e78.png)
Therefore,
![\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/1/1/4/11441fab3ba768fee7a0eef4dbf68124d6637aec.png)
Let
![\[
(n + a)(n + b) = k^2.
\]](http://latex.artofproblemsolving.com/0/e/c/0ec5e7bac2aebe9ec507727ed21f7ca028655222.png)
Assume 
. Then we have:![\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]](http://latex.artofproblemsolving.com/7/5/f/75fc52b6662df3a4c0cb2f40f49c4b3351c90a4d.png)
or it could also be that 
.Without loss of generality, we take the first case:
![\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]](http://latex.artofproblemsolving.com/1/f/5/1f5936241f7bebcf15c68fed5267bbfad2d40c2b.png)
Thus,
![\[
k_1 k_2 = \frac{n + b}{n + a}.
\]](http://latex.artofproblemsolving.com/d/4/c/d4c1958e3554b58c3def282070f849842cc10912.png)
Since
, we have:![\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]](http://latex.artofproblemsolving.com/3/0/6/306a6a00c44d4e4db1b36c7e588f684f28fdbf44.png)
For infinitely many
, 
must be an integer, which is not possible.Therefore, there cannot be infinitely many such
.
7 replies
and 
.
