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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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jlacosta
Apr 2, 2025
0 replies
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TST Junior Romania 2025
ant_   1
N a minute ago by wassupevery1
Source: ssmr
Consider the isosceles triangle $ABC$, with $\angle BAC > 90^\circ$, and the circle $\omega$ with center $A$ and radius $AC$. Denote by $M$ the midpoint of side $AC$. The line $BM$ intersects the circle $\omega$ for the second time in $D$. Let $E$ be a point on the circle $\omega$ such that $BE \perp AC$ and $DE \cap AC = {N}$. Show that $AN = 2AB$.
1 reply
ant_
Yesterday at 5:01 PM
wassupevery1
a minute ago
J
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Very tight inequalities
KhuongTrang   2
N 29 minutes ago by SunnyEvan
Source: own
Problem. Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that $$\color{black}{\frac{1}{35a+12b+2}+\frac{1}{35b+12c+2}+\frac{1}{35c+12a+2}\ge \frac{4}{39}.}$$$$\color{black}{\frac{1}{4a+9b+6}+\frac{1}{4b+9c+6}+\frac{1}{4c+9a+6}\le \frac{2}{9}.}$$When does equality hold?
2 replies
KhuongTrang
May 17, 2024
SunnyEvan
29 minutes ago
J
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Sum of First, Second, and Third Powers
Brut3Forc3   47
N 42 minutes ago by cubres
Source: 1973 USAMO Problem 4
Determine all roots, real or complex, of the system of simultaneous equations
\begin{align*} x+y+z &= 3, \\ x^2+y^2+z^2 &= 3, \\ x^3+y^3+z^3 &= 3.\end{align*}
47 replies
Brut3Forc3
Mar 7, 2010
cubres
42 minutes ago
J
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Modified Sum of floors
prMoLeGend42   2
N an hour ago by cubres
Find the closed form of : $\sum _{k=0}^{n-1} \left\lfloor \frac{ak+b}{n}\right \rfloor$ where $\gcd(a,n)=1$
2 replies
+1 w
prMoLeGend42
Yesterday at 9:09 AM
cubres
an hour ago
J
No more topics!
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Incenter, orthocenter and circumcenter
mitrov   4
N Sep 7, 2014 by fclvbfm934
Source: Bulgarian MO 2002 4th round day 2 problem 1
Let $I$ be the incenter of a non-equilateral triangle $ABC$ and $T_1$, $T_2$, and $T_3$ be the tangency points of the incircle with the sides $BC$, $CA$ and $AB$, respectively. Prove that the orthocenter of triangle $T_1T_2T_3$ lies on the line $OI$, where $O$ is the circumcenter of triangle $ABC$.

Proposed by Georgi Ganchev
4 replies
mitrov
Sep 7, 2014
fclvbfm934
Sep 7, 2014
J
Incenter, orthocenter and circumcenter
G H J
Reveal topic tags
geometry
incenter
circumcircle
Euler
geometry unsolved
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Source: Bulgarian MO 2002 4th round day 2 problem 1
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mitrov
28 posts
mitrov
#1 h  Sep 7, 2014, 11:29 AM • 3 Y
Y by Adventure10, Mango247, and 1 other user
Let $I$ be the incenter of a non-equilateral triangle $ABC$ and $T_1$, $T_2$, and $T_3$ be the tangency points of the incircle with the sides $BC$, $CA$ and $AB$, respectively. Prove that the orthocenter of triangle $T_1T_2T_3$ lies on the line $OI$, where $O$ is the circumcenter of triangle $ABC$.

Proposed by Georgi Ganchev
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jayme
9775 posts
jayme
#2 h  Sep 7, 2014, 11:46 AM • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,
IH' is the Euler line of T1T2T3 where H' is the orthocenter.
According to Gob theorem, O the center of the circumcircle of ABC, the tangentiel triangle of T1T2T3 lies also on IH'.

Sincerely
Jean-Louis
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jayme
9775 posts
jayme
#3 h  Sep 7, 2014, 11:50 AM • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,
for a synthetic proof, you can see

http://jl.ayme.pagesperso-orange.fr/ vol. 7 Droite de Simson de pole Fe, p. 12-16.

Sincerely
Jean-Louis
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Sardor
804 posts
Sardor
#4 h  Sep 7, 2014, 1:47 PM • 2 Y
Y by Adventure10, Mango247
See also here
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2010456&sid=3455880fe979bf79647e480a8b08cb04#p2010456
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fclvbfm934
759 posts
fclvbfm934
#5 h  Sep 7, 2014, 9:09 PM • 2 Y
Y by Adventure10, Mango247
My solution:

Let us invert about the incircle of $ABC$. Then, we clearly have $A, B, C$ going to the midpoints of $T_2T_3, T_1T_3, T_1T_2$, respectively. The circumcircle of $ABC$ therefore maps to the nine-point circle of $T_1T_2T_3$, which clearly has its center on the Euler line of $T_1T_2T_3$, namely $IH$, where $H$ is the orthocenter of $T_1T_2T_3$. Note that when we invert back, $O$ does indeed lie on $IH$, even though the nine-point center doesn't map to $O$.
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