Competition problem — Geometry — Maths Olympiad Prep

Library / /3 of 10

Geometry Difficulty 7.9 National olympiad, round 2 Prove it Netherlands

Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude from $A$ . The circle with centre $A$ passing through $D$ intersects the circumcircle of triangle $ABC$ in $X$ and $Y$ , in such a way that the order of the points on this circumcircle is: $A$ , $X$ , $B$ , $C$ , $Y$ . Show that $\angle BXD = \angle CYD$ .

Solution

As the radius $AD$ is perpendicular to $BC$ , the line $BC$ is tangent to the circumcircle of $\triangle DXY$ . By the inscribed angle theorem (tangent case), we have $\angle XDB = \angle XYD$ . Moreover, the quadrilateral $BCYX$ is cyclic, so $\angle CBX + \angle XYC = 180^\circ$ . By the sum of angles in $\triangle BDX$ , we have $\angle BXD = 180^\circ - \angle DBX - \angle XDB = (180^\circ - \angle CBX) - \angle XDB = \angle XYC - \angle XYD$ . As $\angle XYC - \angle XYD = \angle DYC$ , we obtain $\angle BXD = \angle DYC$ . $\square$

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.