IMO 2018 P1

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M Ahsan Al Mahir
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Joined: Wed Aug 10, 2016 1:29 am

IMO 2018 P1

Unread post by M Ahsan Al Mahir » Wed Jan 09, 2019 11:44 pm

Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

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Tasnood
Posts: 73
Joined: Tue Jan 06, 2015 1:46 pm

Re: IMO 2018 P1

Unread post by Tasnood » Thu Jan 10, 2019 9:46 am

Let's use our contradiction! ;)
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Assume, $FG$ and $DE$ are not parallel. So let $G'$ be on the circle such that $FG' \parallel DE$ and $E'$ on $AC$ such that $G'E'=G'C$. Let $FG'$ intersects $AB$ and $AC$ at $X$ and $Y$ respectively.
We have these property: $FD=FB, GE=GC$
$\blacktriangleright$ $\angle FDX=\angle FBD=\angle FBA=\angle FG'A=\angle AG'Y$
Again, $DE \parallel XY$ yields $\angle AYX=\angle AXY \Rightarrow \angle AYG'=\angle AXF$
$\blacktriangleright \triangle AYG' \sim \triangle DXF..........(1)$
Similarly we can prove: $\angle G'E'Y= \angle G'CA= \angle AFX, \angle E'YG'=\angle DXF$
$\blacktriangleright \triangle E'YG' \sim \triangle FXA..........(2)$
From $(1)$, $\frac{AG'}{DF}=\frac{AY}{FX}=\frac{YG'}{DX}$
From $(2)$, $\frac{E'G'}{AF}=\frac{E'Y}{FX}=frac{G'Y}{AX}$
From the ratio, $DX \times AY=FX \times YG'=E'Y \times AX=E'Y \times AY$
$\blacktriangleright DX=E'Y$ We know, $AD=AE$ and so, $AX=AY$ for, $DE \parallel XY$
$\blacktriangleright AX-DX=AY-E'Y \Rightarrow AD=AE'$ But, $AD=AE$
So, $AE=AE'$ yields $\boxed {E=E'}$
:P

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