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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}imo20161

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