IMO 2006 Shortlisted Problems

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 $$\begin{array}{rl}AM& =AD+(BD+CD-BC)\\ & =AD+AB+DC-BC\\ & =AC+AB-BC\\ & =2\cdot AL,\end{array}$$ which completes the proof.