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.

2016IMOQ1.jpg

Solution

2016IMOQ1Solution.jpg

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

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

Finally AD=DB and DAF=DBF=FXD. MFX=FXD=FXM and FEMD with EF=FM=MD=DE, making EFMD a rhombus. And FBD=MBD=MXF=DXF and triangle BEM is congruent to XEM, while MFX is congruent to MBD which is congruent to 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

See Also

2016 IMO (Problems) • Resources
Preceded by
First Problem
1 2 3 4 5 6 Followed by
Problem 2
All IMO Problems and Solutions