Competition problem — Geometry — Maths Olympiad Prep

Library / /40 of 50

Geometry Difficulty 6.0 AIME, harder Prove it Belarus

Given the parallelogram $ABCD$ . The circle $S_1$ passes through the vertex $C$ and touches the sides $BA$ and $AD$ at points $P_1$ and $Q_1$ respectively. The circle $S_2$ passes through the vertex $B$ and touches the sides $DC$ and $AD$ at points $P_2$ and $Q_2$ respectively. Let $d_1, d_2$ be the distances from $C$ and $B$ to the lines $P_1Q_1$ and $P_2Q_2$ respectively.
Find all possible values of the ratio $d_1 : d_2$ .

Solution

Answer : 1.
We prove that $d_1 = d_2$ . Let $\angle A = \angle C = \alpha$ , $\angle B = \angle D = -\alpha$ . Choose points $F$ and $T$ on $P_1Q_1$ so that $\angle P_1FC = \angle Q_1TC = \angle D$ . Then $CF = CT$ . Further, $\triangle P_1BC \sim \triangle Q_1TC$ , thus we have $\frac{BC}{TC} = \frac{PC}{QC}$ . Similarly, $\triangle Q_1DC \sim \triangle P_1FC$ , so $\frac{DC}{FC} = \frac{QC}{PC}$ . Multiplying these equalities we have $BC \cdot DC = TC^2$ . Note that $d_1$ is the length of the altitude in $\triangle FCT$ , so $d_1 = TC \sin \angle FTC = TC \sin \alpha$ , or

$d_1 = \sqrt{BC \cdot DC} \sin \alpha.$ In the same way we obtain that $d_2 = \sqrt{BC \cdot DC} \sin \alpha = d_1$ , as required.

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.