Competition problem — Geometry — Maths Olympiad Prep

Library / /6 of 21

Geometry Difficulty 6.4 National Olympiad Prove it South Korea

Let $ABC$ be an isosceles triangle with $AC = BC$ . Let $D$ be a point on a line $BA$ such that $A$ lies between $B$ and $D$ . Let $O_1$ be the circumcircle of triangle $DAC$ . $O_1$ meets $BC$ at point $E$ . Let $F$ be the point on the line $BC$ such that $FD$ is tangent to circle $O_1$ , and let $O_2$ be the circumcircle of triangle $DBF$ . Two circles $O_1, O_2$ meet at point $G$ ( $G \neq D$ ). Let $O$ be the circumcenter of triangle $BEG$ . Prove that the line $FG$ is tangent to circle $O$ if and only if $DG$ is perpendicular to $FO$ .

Solution

We first show that both $DB$ and $DE$ are tangent to circle $O$ . Since $DFBG$ is concyclic, we have $\angle FDG = \angle GBE$ . Since $FD$ is tangent to $O_1$ , we have $\angle DEG = \angle FDG$ . Hence $\angle GBE = \angle DEG$ , which means that $DE$ is tangent to $O$ . On the other hand, since $ACED$ is concyclic, $\angle BAC = \angle CED$ . Since $ABC$ is an isosceles triangle, $\angle BAC = \angle ABC = \angle DBE$ . Thus we have $\angle DBE = \angle DEB$ , that is, the triangle $DBE$ is an isosceles triangle with $DB = DE$ , which means that $DB$ is also tangent to $O$ .

Since $DB$ and $DE$ are tangent to $O$ , the line $FE$ is the polar of the pole $D$ with respect to $O$ . By La Hire theorem, $D$ lies on the polar of $F$ . Hence $FG$ is tangent to circle $O$ if and only if $DG$ is the polar of $F$ , which is equivalent to the fact that $DG$ is perpendicular to $FO$ .

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.