2016 IMO Problems/Problem 1

Problem

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.

Solution

The Problem shows that $$\angle DAC = \angle DCA = \angle CAD$$, it follows that $$AB \parallel CD$$. Extend $$DC$$ to intersect $$AB$$ at $$G$$, we get $$\angle GFA = \angle GFB = \angle CFD$$. Making triangles $$\triangle CDF$$ and $$\triangle AGF$$ similar. Also, $$\angle FDC = \angle FGA = 90^\circ$$ and $$\angle FBC = 90^\circ$$, which points $$D$$, $$C$$, $$B$$, and $$F$$ are concyclic.

And $$\angle BFC = \angle FBA + \angle FAB = \angle FAE = \angle AFE$$. Triangle $$\triangle AFE$$ is congruent to $$\triangle FBM$$, and $$AE = EF = FM = MB$$. Let $$MX = EA = MF$$, then points $$B$$, $$C$$, $$D$$, $$F$$, and $$X$$ are concyclic.

Finally $$AD = DB$$ and $$\angle DAF = \angle DBF = \angle FXD$$. $$\angle MFX = \angle FXD = \angle FXM$$ and $$FE \parallel MD$$ with $$EF = FM = MD = DE$$, making $$EFMD$$ a rhombus. And $$\angle FBD = \angle MBD = \angle MXF = \angle DXF$$ and triangle $$\triangle BEM$$ is congruent to $$\triangle XEM$$, while $$\triangle MFX$$ is congruent to $$\triangle MBD$$ which is congruent to $$\triangle FEM$$, so $$EM = FX = BD$$.

~Athmyx

Solution 2

Let $\angle FBA = \angle FAB = \angle FAD = \angle FCD = \angle DAE = \angle ADE = \alpha$. And WLOG, $MF = 1$. Hence, $CF = 2$,

$\implies$ $BF = CF.cos(2\alpha) = 2.cos(2\alpha) = FA$,

$\implies$ $DA = \frac{AC}{2cos(\alpha)} = \frac{CF+FA}{2cos(\alpha)} = \frac{2+2cos(2\alpha)}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{cos(\alpha)}$ and

$\implies$ $DE = AE = \frac{DA}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{2.(cos(\alpha))^2} = 1$.

So $MX = DE = 1$ which means $B$, $C$, $X$ and $F$ are concyclic. We know that $DE || MC$ and $DE = 1 = MC$, so we conclude $MCDE$ is parallelogram. So $\angle AME = \alpha$. That means $MDEA$ is isosceles trapezoid. Hence, $MD = EA = 1$. By basic angle chasing,

$\angle MBF = \angle MFB = 2\alpha$ and $\angle MXD = \angle MDX = 2\alpha$ and we have seen that $MB = MF = MD = MX$, so $BFDX$ is isosceles trapezoid. And we know that $ME$ bisects $\angle FMD$, so $ME$ is the symmetrical axis of $BFDX$.

$B$ and $X$, $D$ and $E$ are symmetrical respect to $ME$. Hence, the symmetry of $BD$ with respect to $ME$ is $FX$. And we are done $\blacksquare$.

~EgeSaribas