2016 IMO Problems/Problem 1

Problem

Triangle has a right angle at . Let be the point on line such that and lies between and . Point is chosen so that and is the bisector of . Point is chosen so that and is the bisector of . Let be the midpoint of . Let be the point such that is a parallelogram. Prove that and are concurrent.

Solution

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

Furthermore, $\mathrm{\angle }BFC=\mathrm{\angle }FBA+\mathrm{\angle }FAB=\mathrm{\angle }FAE=\mathrm{\angle }AFE$. Triangle $\mathrm{△}AFE$ is congruent to $\mathrm{△}FBM$, 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 $\mathrm{\angle }DAF=\mathrm{\angle }DBF=\mathrm{\angle }FXD$. Additionally, $\mathrm{\angle }MFX=\mathrm{\angle }FXD=\mathrm{\angle }FXM$ and $FE\parallel MD$ with $EF=FM=MD=DE$, making $EFMD$ a rhombus. Consequently, $\mathrm{\angle }FBD=\mathrm{\angle }MBD=\mathrm{\angle }MXF=\mathrm{\angle }DXF$ and triangle $\mathrm{△}BEM$ is congruent to $\mathrm{△}XEM$, while $\mathrm{△}MFX$ is congruent to $\mathrm{△}MBD$ which is congruent to $\mathrm{△}FEM$, and $EM=FX=BD$.

~Athmyx

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