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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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[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 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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jlacosta
Apr 2, 2025
0 replies
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Hard Inequality Problem
Omerking   1
N 10 minutes ago by lpieleanu
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$ is given where $a,b,c$ are positive reals. Prove that:
$$\frac{1}{\sqrt{a^3+1}}+\frac{1}{\sqrt{b^3+1}}+\frac{1}{\sqrt{c^3+1}} \le \frac{3}{\sqrt{2}}$$
1 reply
1 viewing
Omerking
Today at 3:51 PM
lpieleanu
10 minutes ago
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USAMO 2000 Problem 5
MithsApprentice   22
N 10 minutes ago by Maximilian113
Let $A_1A_2A_3$ be a triangle and let $\omega_1$ be a circle in its plane passing through $A_1$ and $A_2.$ Suppose there exist circles $\omega_2, \omega_3, \dots, \omega_7$ such that for $k = 2, 3, \dots, 7,$ $\omega_k$ is externally tangent to $\omega_{k-1}$ and passes through $A_k$ and $A_{k+1},$ where $A_{n+3} = A_{n}$ for all $n \ge 1$. Prove that $\omega_7 = \omega_1.$
22 replies
MithsApprentice
Oct 1, 2005
Maximilian113
10 minutes ago
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f.e with finite number of f(t)=-t
jjkim0336   0
17 minutes ago
Source: own
f:R->R
f(xf(y)+y)=yf(x)+f(f(y)) and there are finite number of t such that f(t)= - t
0 replies
jjkim0336
17 minutes ago
0 replies
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Common external tangents of two circles
a1267ab   55
N 18 minutes ago by awesomeming327.
Source: USA Winter TST for IMO 2020, Problem 2, by Merlijn Staps
Two circles $\Gamma_1$ and $\Gamma_2$ have common external tangents $\ell_1$ and $\ell_2$ meeting at $T$. Suppose $\ell_1$ touches $\Gamma_1$ at $A$ and $\ell_2$ touches $\Gamma_2$ at $B$. A circle $\Omega$ through $A$ and $B$ intersects $\Gamma_1$ again at $C$ and $\Gamma_2$ again at $D$, such that quadrilateral $ABCD$ is convex.

Suppose lines $AC$ and $BD$ meet at point $X$, while lines $AD$ and $BC$ meet at point $Y$. Show that $T$, $X$, $Y$ are collinear.

Merlijn Staps
55 replies
a1267ab
Dec 16, 2019
awesomeming327.
18 minutes ago
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Equilateral triangle and incenter
crazyfehmy   2
N Jun 1, 2013 by mathuz
Source: Turkey JBMO Team Selection Test 2013, P1
Let $D$ be a point on the side $BC$ of an equilateral triangle $ABC$ where $D$ is different than the vertices. Let $I$ be the excenter of the triangle $ABD$ opposite to the side $AB$ and $J$ be the excenter of the triangle $ACD$ opposite to the side $AC$. Let $E$ be the second intersection point of the circumcircles of triangles $AIB$ and $AJC$. Prove that $A$ is the incenter of the triangle $IEJ$.
2 replies
crazyfehmy
May 31, 2013
mathuz
Jun 1, 2013
J
Equilateral triangle and incenter
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Reveal topic tags
geometry
incenter
circumcircle
geometry proposed
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Source: Turkey JBMO Team Selection Test 2013, P1
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crazyfehmy
1345 posts
crazyfehmy
#1 h  May 31, 2013, 7:13 PM • 1 Y
Y by Adventure10
Let $D$ be a point on the side $BC$ of an equilateral triangle $ABC$ where $D$ is different than the vertices. Let $I$ be the excenter of the triangle $ABD$ opposite to the side $AB$ and $J$ be the excenter of the triangle $ACD$ opposite to the side $AC$. Let $E$ be the second intersection point of the circumcircles of triangles $AIB$ and $AJC$. Prove that $A$ is the incenter of the triangle $IEJ$.
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ThirdTimeLucky
402 posts
ThirdTimeLucky
#2 h  May 31, 2013, 7:42 PM • 3 Y
Y by Adventure10, Mango247, Nurefsan17
$ \angle IEA= \angle JEA $. $ \angle IEJ=120^{\circ} $. $ \angle IAJ=150^{\circ} $. $ \implies A $ is the incentre of $ \triangle IEJ $.
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mathuz
1520 posts
mathuz
#3 h  Jun 1, 2013, 10:02 AM • 3 Y
Y by Adventure10, Mango247, Nurefsan17
$ \angle IEA= \angle JEA=60^\circ $ and $A$ lies on the bisector of the $ \angle IEJ$.
Let $ \angle EJA=x $, $ \angle EIA=y $ and $ \angle AJI=z $, $ \angle AIJ=t. $
We have that $x+y=30^\circ $, because $ \angle BEC=90^\circ .$
Since Cheva's theorem , $x=z$ and $y=t$.
So, $A$ is incenter of the triangle $IEJ.$ :wink:
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