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## IMO 2016 – Problem 1

IMO 2016, Problem 1. 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 such that ${DA=DC}$ and ${AC}$ is the bisector of ${\angle DAB}$. Point ${E}$ is chosen such 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 (where ${AM || EX}$ and ${AE || MX}$). Prove that the lines ${BD, FX}$ and ${ME}$ are concurrent.

Solution: When trying to see what’s going on, by drawing the figure associated to this problem you can easily get lost in the details… Let’s just stop after drawing points ${A}$ and ${D}$. If drawn properly one immediately notices that ${D}$ seems to be the circumcenter of ${ABC}$… Let ${D'}$ be the circumcenter of ${ABC}$. Since ${AF=FB}$ and ${AD=DB}$ it follows that ${\angle DAF = \angle DBF = \angle DCA}$ so ${BCDF}$ is cyclic. If we denote the common measure of ${\angle DAF, \angle CAB, \angle FBD, \angle DBF, \angle DCF}$ by ${x}$ we find that ${\angle BDC = \angle BFC = 2x}$ and since ${\angle DBC = \angle DCB = 90^\circ -x}$ we see that indeed ${\Delta DBC}$ is isosceles and thus ${DA=DB=DC}$.

We go on and look at point ${E}$ which seems to lie on ${BF}$. How do we prove this? First we note that ${M}$, the midpoint of ${CF}$ is the circumcenter of ${DFBC}$ and thus ${MF=MD=MC=MB}$.

Furthermore, ${\angle MDC = \angle MCD = \angle EAD = \angle EDA = x}$ and ${AD = DC}$. Therefore triangles ${\Delta EAD}$ and ${\Delta MDC}$ are congruent. As a consequence ${AE=ED = MD = MF}$. The properties of ${E}$ imply immediately that ${ED || FM}$ so ${EDMF}$ is a parallelogram, and moreover a rhombus, since ${DE=DM}$.

Now we ca finish by noting that ${ME}$ is an axis of symmetry for the rhombus ${EFMD}$ and triangles ${DMB}$ and ${FMX}$ are congruent. The symmetry of the figure implies that the lines ${BD, FX}$ and ${ME}$ are concurrent. ${\square}$