Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Japan

Suppose for a triangle $ABC$ the following condition is satisfied:
A circle $O$ going through the vertices $B, C$ intersects the line segments $AB$ and $AC$ (excluding the end-points) at the points $D$ and $E$ , respectively, and $AD + AE = BC$ is satisfied.
Let $I$ be the in-center of the triangle $ABC$ and suppose that the lines $BI$ and $CI$ intersect the circle $O$ at $P, Q$ (different from $B, C$ ), respectively. Prove that the points $A, I, P, Q$ lie on the same circumference of a circle.
Here by $XY$ we mean the length of the line segment $XY$ .

Solution

From $AD + AE = BC$ it follows that we can choose a point $F$ on the side $BC$ in such a way that both $AD = CF$ and $AE = BF$ are satisfied. We have $DP = CP$ , since the angles the cords $DP$ and $CP$ subtend on the circle $O$ are equal. We also have $\angle PDA = \angle PCB = \angle PCF$ , since the points $P, C, B, D$ lie on the circumference of a same circle. These facts together with the fact $AD = FC$ imply that the triangles $PDA$ and $PCF$ are congruent. Therefore, we have

$\angle PFC = \angle PAD = \angle PAB$ , and we conclude that the points $A, R, F, B$ lie on the circumference of a same circle.

Similarly, we can conclude that the points $A, Q, F, C$ also lie on the circumference of a same circle. Consequently, we get
$\angle PAQ = \angle PAF + \angle FAQ = \angle PBF + \angle FCQ = 180^\circ - \angle CIB = 180^\circ - \angle QIP,$
from which we conclude that the points $A, I, P, Q$ lie on the circumference of a same circle.

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