Library / South African Mathematics Olympiad Third Round /4 of 11

South African Mathematics Olympiad Third Round

Geometry Difficulty 5.0 AIME, harder Prove it South Africa

The inscribed circle of triangle $ABC$ , with centre $I$ , touches sides $BC$ , $CA$ and $AB$ at $D$ , $E$ and $F$ , respectively. Let $P$ be a point, on the same side of $FE$ as $A$ , for which $\angle PFE = \angle BCA$ and $\angle PEF = \angle ABC$ . Prove that $P$ , $I$ and $D$ lie on a straight line.

Solution

Since $\angle AEI = \angle AFI = 90^\circ$ , the points $A$ , $E$ , $F$ , $I$ lie on a circle with diameter $AI$ . Moreover, since $\angle PFE = \angle BCA$ and $\angle PEF = \angle ABC$ by our assumptions on $P$ , triangles $ABC$ and $PEF$ are similar, so $\angle EPF = \angle BAC = \angle EAF$ . $P$ was assumed to lie on the same side of $EF$ as $A$ , thus it follows that $P$ also lies on the same circle as $A$ , $E$ , $F$ and $I$ .
We also know that $BDIF$ is a cyclic quadrilateral (using the same reasoning as before, namely that $\angle BDI = \angle BFI = 90^\circ$ ), so $\angle FID + \angle FBD = 180^\circ$ .
Now we can conclude that $\angle PIF = \angle PEF = \angle ABC = \angle FBD = 180^\circ - \angle FID$ , so $\angle PIF + \angle FID = 180^\circ$ , which means that $P$ , $I$ , $D$ lie on a straight line.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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