Library / /17 of 27

, 2008

Geometry Difficulty 5.9 AIME, harder Prove it India

Let $ABC$ be a non-isosceles triangle, and let $\Gamma$ be its in-circle. Let $D$ , $E$ , $F$ be the points of contact of $\Gamma$ with the sides $BC$ , $CA$ , $AB$ respectively. Suppose $FD$ , $DE$ , $EF$ intersect $CA$ , $AB$ , $BC$ in $U$ , $V$ , $W$ respectively. If $L$ , $M$ , $N$ are respectively the mid-points of $DW$ , $EU$ , $FV$ prove that $L$ , $M$ , $N$ are collinear.

Solution

Draw a line through $N$ which is parallel to $ED$ . This passes through the midpoints $P$ of $DF$ and $Q$ of $FE$ . Similarly the line through $M$ parallel to $FD$ passes through the mid-point $Q$ of $FE$ and $R$ of $ED$ ; the line through $L$ parallel to $FE$ passes through the mid-point $P$ of $FD$ and $R$ of $ED$ . We thus obtain the medial triangle $PRQ$ of $ABC$ . Since $AF = AE$ , the line $AQ$ is also the bisector of $\angle A$ . Similarly $BP$ bisects $\angle B$ and $CR$ bisects $\angle C$ .

Thus $AQ$ , $BP$ , $CR$ concur at $I$ , the in-centre of $ABC$ . Now Desargues' theorem is applicable to the triangles $ABC$ and $QPR$ . It follows that $PR \cap BC$ , $RQ \cap CA$ and $QP \cap AB$ are collinear. Thus $L$ , $M$ and $N$ are collinear.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.