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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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0 replies
jlacosta
Apr 2, 2025
0 replies
J
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f.e with finite number of f(t)=-t
jjkim0336   2
N 32 minutes ago by jasperE3
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
2 replies
jjkim0336
3 hours ago
jasperE3
32 minutes ago
J
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inequalities
Cobedangiu   1
N an hour ago by pooh123
Source: UCT
Let $a,b,c>0$ and $a+b+c=3$ Prove that:
$\sum \dfrac{\sqrt{a^2+a+1}}{2a^2+10a+9}\ge \dfrac{\sqrt{3}}{7}$
1 reply
Cobedangiu
Yesterday at 4:14 PM
pooh123
an hour ago
J
H
Solve $\sin(17x)+\sin(13x)=\sin(7x)$
Speed2001   0
an hour ago
How to solve the equation:
$$ \sin(17x)+\sin(13x)=\sin(7x),\;0<x<24^{\circ} $$Approach: I'm trying to factor $\sin(18x)$ to get $x=10^{\circ}$.

Any hint would be appreciated.
0 replies
Speed2001
an hour ago
0 replies
J
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Symmedian line
April   91
N an hour ago by BS2012
Source: All Russian Olympiad - Problem 9.2, 10.2
Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.
91 replies
April
May 10, 2009
BS2012
an hour ago
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No more topics!
H
J
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collinear points out of 3 circumcirlcles
parmenides51   1
N Dec 1, 2017 by GGPiku
Source: Czech-Polish-Slovak Match 2017 day 2 P1
Let ${ABC}$ be a triangle. Line l is parallel to ${BC}$ and it respectively intersects side ${AB}$ at point ${D}$, side ${AC}$ at point ${E}$, and the circumcircle of the triangle ${ABC}$ at points ${F}$ and ${G}$, where points ${F,D,E,G}$ lie in this order on l. The circumcircles of triangles ${FEB}$ and ${DGC}$ intersect at points ${P}$ and ${Q}$. Prove that points ${A, P,Q}$ are collinear.

(Slovakia)
1 reply
parmenides51
Sep 28, 2017
GGPiku
Dec 1, 2017
J
collinear points out of 3 circumcirlcles
G H J
Reveal topic tags
geometry
collinear
L
G H BBookmark kLocked kLocked NReply
Source: Czech-Polish-Slovak Match 2017 day 2 P1
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parmenides51
30650 posts
parmenides51
#1 h  Sep 28, 2017, 3:29 PM • 2 Y
Y by Adventure10, Mango247
Let ${ABC}$ be a triangle. Line l is parallel to ${BC}$ and it respectively intersects side ${AB}$ at point ${D}$, side ${AC}$ at point ${E}$, and the circumcircle of the triangle ${ABC}$ at points ${F}$ and ${G}$, where points ${F,D,E,G}$ lie in this order on l. The circumcircles of triangles ${FEB}$ and ${DGC}$ intersect at points ${P}$ and ${Q}$. Prove that points ${A, P,Q}$ are collinear.

(Slovakia)
This post has been edited 1 time. Last edited by parmenides51, Sep 28, 2017, 3:29 PM
Reason: country proposer added
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GGPiku
402 posts
GGPiku
#2 h  Dec 1, 2017, 9:17 AM • 1 Y
Y by Adventure10
Let the circle $FEB$ intersect $AC,AB$ in $H,I$ and the circle $DGC$ intersect $AB,AC$ in $J,K$. Easy angle chasing leads to $\angle BFE=\angle BHC=\angle BJC=\angle DGC$. So $B,J,H,C$ are concyclic, so $D,J,H,E$ are concyclic. This implies $AJ*AD=AH*AE$, so $A$ is on the radical axis, and the conclusion follows.
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