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IMO 2016 Problem 1

Posted: Sun Aug 07, 2016 12:57 pm
by Phlembac Adib Hasan
Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.

Re: IMO 2016 Problem 1

Posted: Sun Jan 01, 2017 4:00 am
by ahmedittihad
We will show the result by the** radical axis theorem**. So, we will show that $BFDX, BMDE, FEMX$ is concyclic.

$ \angle ADF = \angle ACB$ by **spiral similarity**.

$\angle FDC = 180 - \angle ADE - \angle CDX = 180 - 2\angle ADE - \angle ADF = 180 - \angle BFC - \angle ACB = 90 $.
So, $BCDF$ is a cyclic quadrilateral and $M$ is the center of this circle. So, $\angle AMD = 2\angle ACD = \angle MAE$. Along with the fact that $ AC \parallel DE$ we get that $AMDE$ is an isosceles trapizium. Meaning also that it is cyclic.

Now, $ \angle BAD + \angle BMD = 2\angle BAC + 2\angle ACD + 2\angle BCF = 2( 2\angle BAC + \angle BCF) = 2(\angle BFC + \angle BCF) = 180 $. So,$ABMD$ is cyclic. Also, $ \angle ABD = \angle ABF + \angle FBD = \angle ABF + \angle ACD = 2\angle BAC$. So, $ \angle ABD + \angle AED = 2\angle BAC +180 - 2\angle DAE = 180$.
So we get, $ABMDE$ is cyclic.

Now, we prove that $BDFX$ is cyclic. $\angle AMD = \angle EAM = \angle EXM = \angle MDX$. So, $\triangle MDX$ is isosceles. So, $X$ lies on the circle $BFDC$.

At last, we need to prove that $FEXM$ is cyclic. See that. $A$ is the center of similitude that sends $BD$ to $FE$. So, $ \angle ABD = \angle AFE = 2\angle BAC$.
We already have that $\angle MXE = 2\angle BAC$. And the result follows.

Re: IMO 2016 Problem 1

Posted: Mon Apr 17, 2017 1:56 am
by Thanic Nur Samin
Let $\angle FAB=x$.

Since $\triangle AFB \sim \triangle ADC$, $\triangle AFD \sim \triangle ABC$ [sprial similarity]

Thus $\angle AFD = 90^{\circ}+x$, and so $FD\perp AB$. Since $FB=FA$, $FD$ is the perpendicular biscector of $AB$. Now, $\angle DBF = x$. Since $\angle AED=180^{\circ}-2x$, $ABDE$ is cyclic and since $BF$ is the angle biscector of $\angle ABD$ and $E$ is the midpoint of arc $AD$ not containing $B$, we get that $B,F,E$ are collinear.

$\angle AMB=2\angle FCB=180^{\circ}-4x=\angle AEB$, and consequently $AEMB$ is cyclic. Note that since $FA=FB$ and $FA.FM=FB.FE$, $FE=FM$ and $AM=BE$ and $BM=AE$.

Now, since $CD||AB$ we get $FD\perp CD$ and so $BCDF$ cyclic with center $M$, so $MF=MD$. And $AEMB$ is an isosceles trapezoid, so $ME||AB$ implying $FD\perp ME$.

Now, $MB=AE=MX$ and $EB=AM=EX$.

So $BD$ and $XF$ are reflections of each other wrt $ME$ and hence they are collinear.

Re: IMO 2016 Problem 1

Posted: Tue Sep 18, 2018 11:52 am
by Olympiadsuccess
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