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

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

Furthermore, $\angle BFC = \angle FBA + \angle FAB = \angle FAE = \angle AFE$. 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.

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

~Athmyx

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

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