Competition problem — Geometry — Maths Olympiad Prep

Library / /41 of 54

Geometry Difficulty 6.6 National olympiad Prove it China

Suppose $O$ is the circumcenter of an acute triangle $\triangle ABC$ , $P$ is a point inside $\triangle AOB$ , and $D$ , $E$ , $F$ are the projections of $P$ on three sides $BC$ , $CA$ , $AB$ of $\triangle ABC$ respectively. Prove that a parallelogram with $FE$ and $FD$ as adjacent sides lies inside $\triangle ABC$ . (posed by Leng Gangsong)

Solution

Proof As shown in the figure, we construct a parallelogram $DEFG$ with $FE$ and $FD$ as adjacent sides. To prove the proposition to be true, we need only to prove that $\angle FEG < \angle FEC$ , and $\angle FDG < \angle FDC$ . It is equivalent to proving: $\angle BFD < \angle BAC$ , and $\angle AFE < \angle ABC$ .

In fact, we construct $OH$ with $OH \perp BC$ , and $H$ is the foot of the perpendicular. From $PD \perp BC$ and $PF \perp AB$ , we know that four points $B$ , $F$ , $P$ and $D$ are concyclic. Thus $\angle BFD = \angle BPD$ . But $\angle PBD > \angle OBH$ , hence $90^\circ - \angle PBD < 90^\circ - \angle OBH$ , and that is, $\angle BPD < \angle BOH$ . Moreover, $O$ is the circumcenter of $\triangle ABC$ , so $\angle BOH = \frac{1}{2} \angle BOC = \angle BAC$ . Therefore, $\angle BFD = \angle BPD < \angle BOH = \angle BAC$ , that is, $\angle BFD < \angle BAC$ .

Similarly, we can prove that $\angle AFE < \angle ABC$ . Therefore, the proposition holds.

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.