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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
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non-perfect square is non-quadratic residue mod some p
SpecialBeing2017   3
N 11 minutes ago by ilovemath0402
If $n$ is not a perfect square, then there exists an odd prime $p$ s.t. $n$ is a quadratic non-residue mod $p$.
3 replies
SpecialBeing2017
Apr 14, 2023
ilovemath0402
11 minutes ago
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Circles tangent at orthocenter
Achillys   62
N 23 minutes ago by Rayvhs
Source: APMO 2018 P1
Let $H$ be the orthocenter of the triangle $ABC$. Let $M$ and $N$ be the midpoints of the sides $AB$ and $AC$, respectively. Assume that $H$ lies inside the quadrilateral $BMNC$ and that the circumcircles of triangles $BMH$ and $CNH$ are tangent to each other. The line through $H$ parallel to $BC$ intersects the circumcircles of the triangles $BMH$ and $CNH$ in the points $K$ and $L$, respectively. Let $F$ be the intersection point of $MK$ and $NL$ and let $J$ be the incenter of triangle $MHN$. Prove that $F J = F A$.
62 replies
Achillys
Jun 24, 2018
Rayvhs
23 minutes ago
J
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Unsymmetric FE
Lahmacuncu   1
N 35 minutes ago by ja.
Source: Own
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that satisfies $f(x^2+xy+y)+f(x^2y)+f(xy^2)=2f(xy)+f(x)+f(y)$ for all real $(x,y)$
1 reply
Lahmacuncu
2 hours ago
ja.
35 minutes ago
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find angle
TBazar   3
N an hour ago by TBazar
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
3 replies
TBazar
6 hours ago
TBazar
an hour ago
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No more topics!
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A coincidence about triangles with common incenter
flower417477   7
N May 4, 2025 by flower417477
$\triangle ABC,\triangle ADE$ have the same incenter $I$.Prove that $BCDE$ is concyclic iff $BC,DE,AI$ is concurrent
7 replies
flower417477
Apr 30, 2025
flower417477
May 4, 2025
J
A coincidence about triangles with common incenter
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Reveal topic tags
incenter
geometry
Concyclic
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flower417477
374 posts
flower417477
#1 h  Apr 30, 2025, 2:08 PM • 1 Y
Y by PikaPika999
$\triangle ABC,\triangle ADE$ have the same incenter $I$.Prove that $BCDE$ is concyclic iff $BC,DE,AI$ is concurrent
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flower417477
374 posts
flower417477
#2 h  Apr 30, 2025, 2:12 PM • 1 Y
Y by PikaPika999
It is adapted from 2011 USATSTP7,but I think it's harder?
This post has been edited 1 time. Last edited by flower417477, Apr 30, 2025, 2:18 PM
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flower417477
374 posts
flower417477
#3 h  May 2, 2025, 12:17 PM • 1 Y
Y by PikaPika999
I should probably bump this?
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mashumaro
41 posts
mashumaro
#4 h  May 3, 2025, 12:02 PM
Y by
Let $AI\cap BC = K$ and $A$-excenter of $\triangle ABC$ be $I_A$.

Claim 1. $BC$, $DE$, $AI$ concurrent $\Rightarrow$ $BCDE$ concyclic
Proof

Claim 2. $BCDE$ concyclic $\nRightarrow$ $BC$, $DE$, $AI$ concurrent
See diagram below.
Attachments:
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flower417477
374 posts
flower417477
#5 h  May 3, 2025, 1:27 PM
Y by
Sorry for my mistake.
But thanks for your proof!
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mashumaro
41 posts
mashumaro
#6 h  May 3, 2025, 1:35 PM • 1 Y
Y by flower417477
flower417477 wrote:
Sorry for my mistake.
But thanks for your proof!

You could fix is as such:
If $BCDE$ is cyclic, then $BC \cap DE$ either lies on $AI$ or the line perpendicular to $AI$ at $I$.
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flower417477
374 posts
flower417477
#7 h  May 4, 2025, 12:12 PM
Y by
mashumaro wrote:
flower417477 wrote:
Sorry for my mistake.
But thanks for your proof!

You could fix is as such:
If $BCDE$ is cyclic, then $BC \cap DE$ either lies on $AI$ or the line perpendicular to $AI$ at $I$.

Can you give me the proof of this proposition?Thanks!
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flower417477
374 posts
flower417477
#8 h  May 4, 2025, 12:43 PM
Y by
Sorry,it's simple by radical center...
My fool
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