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BdMO Online Forum • View topic - IMO 2015 - Problem 4

IMO 2015 - Problem 4

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IMO 2015 - Problem 4

Post Number:#1  Unread postby Nirjhor » Wed Jul 15, 2015 1:49 am

Triangle $ABC$ has circumcircle $\Omega$ and circumcenter $O$. A circle $\Gamma$ with center $A$ intersects the segment $BC$ at points $D$ and $E$, such that $B$, $D$, $E$, and $C$ are all different and lie on line $BC$ in this order. Let $F$ and $G$ be the points of intersection of $\Gamma$ and $\Omega$, such that $A$, $F$, $B$, $C$, and $G$ lie on $\Omega$ in this order. Let $K$ be the second point of intersection of the circumcircle of triangle $BDF$ and the segment $AB$. Let $L$ be the second point of intersection of the circumcircle of triangle $CGE$ and the segment $CA$.

Suppose that the lines $FK$ and $GL$ are different and intersect at the point $X$. Prove that $X$ lies on the line $AO$.

Proposed by Greece.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


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Re: IMO 2015 - Problem 4

Post Number:#2  Unread postby tanmoy » Tue Oct 06, 2015 10:42 pm

$\text{My Solution}$:
Because $GA=FA$ and $GO=FO$, it is enough to show that $XG=XF$.
Let $FG\cap \odot BFD=T,FG\cap \odot CEG=S$ then $AG=AF \Longrightarrow \measuredangle GCA=\measuredangle ABF\Longrightarrow \measuredangle KBF=\measuredangle GCL=\measuredangle GEL................(1)$
Note that $\measuredangle TBD=\measuredangle DFT=\measuredangle GEC\Longrightarrow GE\parallel BT........(2)$.
Also $\measuredangle SEC=\pi-\angle FGC=\measuredangle FBC\Longrightarrow SE\parallel FB..........(3)$.
From $(2)$ and $(3)$,$\measuredangle TBF=\measuredangle GES..................(4)$.
From $(1)$ and $(4)$,we get $\measuredangle KBT=\measuredangle LCS\Longrightarrow \measuredangle SGL=\measuredangle TFK\Longrightarrow XG=XF$. :D
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