IMO Shortlist, 2006

Geometry Difficulty 8.7 Shortlist Prove it IMO

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^{\circ}$ in such a way that $BD = BA$ . The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$ , respectively. Let $J$ be the incentre of triangle $BCD$ . Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.
(Russia)

Solution

Denote by $P$ the common point of $AJ$ and $KL$ . Let the parallel to $KL$ through $J$ meet $AC$ at $M$ . Then $P$ is the midpoint of $AJ$ if and only if $AM = 2 \cdot AL$ , which we are about to show.

Denoting $\angle BAC = 2\alpha$ , the equalities $BA = BD$ and $AK = AL$ imply $\angle ADB = 2\alpha$ and $\angle ALK = 90^{\circ} - \alpha$ . Since $DJ$ bisects $\angle BDC$ , we obtain $\angle CDJ = \frac{1}{2} \cdot (180^{\circ} - \angle ADB) = 90^{\circ} - \alpha$ . Also $\angle DMJ = \angle ALK = 90^{\circ} - \alpha$ since $JM \parallel KL$ . It follows that $JD = JM$ .

Let the incircle of triangle $BCD$ touch its side $CD$ at $T$ . Then $JT \perp CD$ , meaning that $JT$ is the altitude to the base $DM$ of the isosceles triangle $DMJ$ . It now follows that $DT = MT$ , and we have
$DM = 2 \cdot DT = BD + CD - BC.$
Therefore
$\begin{aligned} AM & = AD + (BD + CD - BC) \\ & = AD + AB + DC - BC \\ & = AC + AB - BC \\ & = 2 \cdot AL, \end{aligned}$
which completes the proof.

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