Competition problem — Geometry — Maths Olympiad Prep

Library / /71 of 189

Geometry Difficulty 5.5 AIME, harder Prove it Ukraine

The problem gives us a right triangle $ABC$ with a right angle at $ACB$ . Let $W_A$ and $W_B$ be the midpoints of the smaller arcs $BC$ and $AC$ of the circumcircle of $\triangle ABC$ , and $N_A$ and $N_B$ be the midpoints of the larger arcs $BC$ and $AC$ . Let $P$ and $Q$ be the intersection points of segment $AB$ with lines $N_A W_B$ and $N_B W_A$ , respectively. Prove that $AP = BQ$ .

Fig. 7

Solution

Let $M$ be the midpoint of the hypotenuse $AB$ of triangle $ABC$ (see figure 7). It is clear that $N_A W_B W_A N_B$ is a rectangle with center $M$ . Therefore, its sides $N_A W_B$ and $N_B W_A$ are symmetric with respect to $M$ . This means that $AP = BQ$ .

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.