Let $ k$ be a circle and $ k_{1},k_{2},k_{3},k_{4}$ four smaller circles with their centres $ O_{1},O_{2},O_{3},O_{4}$ respectively, on $ k$. For $ i = 1,2,3,4$ and $ k_{5}= k_{1}$ the circles $ k_{i}$ and $ k_{i+1}$ meet at $ A_{i}$ and $ B_{i}$ such that $ A_{i}$ lies on $ k$. The points $ O_{1},A_{1},O_{2},A_{2},O_{3},A_{3},O_{4},A_{4}$ lie in that order on $ k$ and are pairwise different. Prove that $ B_{1}B_{2}B_{3}B_{4}$ is a rectangle.
Problem
Source: MEMO Individual Competition, Question 3
Tags: geometry, rectangle, geometry proposed
weiquan
30.09.2007 14:51
It is true regardless of the order of $ O_i$,$ A_i$ on $ k$ (directed angles were used in what I hope is a correct way) $ \angle B_1 A_1 A_2$ $ = \angle B_1 A_1 O_2 - \angle A_2 A_1 O_2$ $ = \left( 90^{\circ} - \frac {1}{2} \angle A_1 O_2 B_1 \right) - \angle O_2 A_2 A_1$ $ = 90^{\circ} - \angle A_1 A_2 B_1 - \angle O_2 A_2 A_1$ $ = 90^{\circ} - \angle O_2 A_2 O_1$ Also, $ \angle A_4 A_1 B_1$ $ = 90^{\circ} - \angle O_2 A_4 O_1$ $ = 90^{\circ} - \angle O_2 A_2 O_1$ $ = \angle B_1 A_1 A_2$ Thus, $ \angle B_1 A_1 A_2 = \frac {1}{2} \angle A_4 A_1 A_2$ Similarly, $ \angle A_2 A_3 B_3 = \frac {1}{2} \angle A_2 A_3 A_4$ Therefore, $ \angle B_1 B_2 B_3$ $ = \angle B_1 B_2 A_2 + \angle A_2 B_2 B_3$ $ = \angle B_1 A_1 A_2 + \angle A_2 A_3 B_3$ $ = \frac {1}{2} \angle A_4 A_1 A_2 + \frac {1}{2} \angle A_2 A_3 A_4$ $ = 90^{\circ}$ $ (QED)$
weiquan
07.10.2007 03:48
weiquan wrote: $ = 90^{\circ} - \angle A_1 A_2 B_1 - \angle O_2 A_2 A_1$ $ = 90^{\circ} - \angle O_2 A_2 O_1$ I just realised I forgot to justify that $ O_1 B_1 A_2$ are collinear, which I proved it first and conveniently assumed that it was given $ = 90^{\circ} - \angle A_1 A_2 B_1 - \angle O_2 A_2 A_1$ $ = 90^{\circ} - \angle A_1 O_2 O_1 - \angle O_2 A_2 A_1$ $ = 90^{\circ} - \angle A_1 A_2 O_1 - \angle O_2 A_2 A_1$ $ = 90^{\circ} - \angle O_2 A_2 O_1$