Competition problem — Geometry — Maths Olympiad Prep

Library / /37 of 87

Geometry Difficulty 5.5 AIME, harder Prove it Austria

Let $k_1$ and $k_2$ be internally tangent circles with common point $X$ . Let $P$ be a point lying neither on one of the two circles nor on the line through the two centers. Let $N_1$ be the point on $k_1$ closest to $P$ and $F_1$ be the point on $k_1$ that is farthest from $P$ . Analogously, let $N_2$ be the point on $k_2$ closest to $P$ and $F_2$ be the point on $k_2$ that is farthest from $P$ .
Prove that $\angle N_1XN_2 = \angle F_1XF_2$ .
(Robert Geretschläger)

Solution

The line segment $N_1F_1$ is a diameter of $k_1$ passing through $P$ . Similarly, $N_2F_2$ is a diameter of $k_2$ passing through $P$ .
Due to Thales's theorem, we have $\angle N_1XF_1 = 90^\circ$ and $\angle N_2XF_2 = 90^\circ$ .
Let $\angle N_2XF_1 = \alpha$ , we obtain
$\angle N_1XN_2 = 90^\circ - \alpha \quad \text{and} \quad \angle F_1XF_2 = 90^\circ - \alpha,$
which proves the equality of the angles.

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 reproduced verbatim; metadata (topic, difficulty) added by this project.