Two disjoint circles $k_1$ and $k_2$ with centers $O_1$ and $O_2$ respectively lie on the same side of a line $p$ and touch the line at $A_1$ and $A_2$ respectively. The segment $O_1O_2$ intersects $k_1$ at $B_1$ and $k_2$ at $B_2$. Prove that $A_1B_1\perp A_2B_2$.
Problem
Source: Slovenia 1997 4th Grade P3
Tags: geometry
AKS_9_54_61
03.05.2021 14:08
Let $\angle A_1O_1O_2 = \theta$, then $\angle O_1O_2A_2 = \pi -\theta$. Also $\angle O_1B_1A_1 = \frac{\pi-\angle A_1O_1B_1}{2} = \frac{\pi-\theta}{2}$ .Similarly $\angle B_2O_2A_2 = \frac{\theta}{2}$. Hence $\angle O_1B_1A_1 + \angle B_2O_2A_2 = \frac{\pi}{2} \implies A_1B_1 \perp A_2B_2$
CT17
03.05.2021 15:14
$\angle B_1A_1A_2 = \frac{1}{2}\overarc{A1B1} = \frac{1}{2}\angle A_1O_1B_1$. Similarly, $\angle B_2A_2A_1 = \frac{1}{2}\angle A_2O_2B_2$ and $\angle B_1A_1A_2 + \angle B_2A_2A_1 = \frac{1}{2}(\angle A_1O_1B_1 + \angle A_2O_2B_2) = 90^\circ$ from quadrilateral $O_1A_1A_2O_2$, which solves the problem.
jayme
03.05.2021 15:21
Dear Mathlinkers, 1. B3 the second point of intersection of O2B2 and k2 2. A1B1 //A2B3 and we are done... Sincerely Jean-Louis