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Problem 2118

National Olympiad second round; IMO P1/P4

Geometry Difficulty 7.7 Prove it Team Selection Test for IMO · Turkey · 2011

Let $I$ be the incenter and $AD$ be a diameter of the circumcircle of a triangle $ABC$ . If the point $E$ on the ray $BA$ and the point $F$ on the ray $CA$ satisfy the condition
$BE = CF = \frac{AB + BC + CA}{2},$
show that the lines $EF$ and $DI$ are perpendicular.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $a = BC$ , $b = CA$ , $c = AB$ , and $u = (a + b + c)/2$ . Let $P$ be the point where the line drawn parallel to $AB$ through $I$ meets $BD$ . Similarly define $Q$ . Then $IP = u - b = AF$ , $IQ = u - c = AE$ and $\angle PIQ = \angle BAC = \angle FAE$ . Hence the triangles $PIQ$ and $FAE$ are congruent.

Let $P'$ be the point on the ray $IQ$ with $IP' = IP$ and $Q'$ be the point on the ray $IP$ with $IQ' = IQ$ . Then $P'IQ'$ and $FAE$ are congruent, and since two pairs of their sides are parallel, so is the third: $FE \parallel Q'P'$ .

As $I, P, D, Q$ are concyclic, $\angle DIP + \angle IQ'P' = \angle DIP + \angle IQP = \angle DIP + \angle IDP = 90^\circ$ . Hence $ID$ is perpendicular to $P'Q'$ and consequently to $EF$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.