Library / /35 of 120

, 2012

Geometry Difficulty 4.9 AIME Prove it Saudi Arabia

Let $ABC$ be a triangle. Point $D$ lies on side $BC$ . Let $O$ , $O_1$ , and $O_2$ be the circumcenters of triangle $ABC$ , $ABD$ , and $ACD$ , respectively. Prove that circumcircles of triangles $BOO_1$ and $COO_2$ meet on line $BC$ .

Solution

Let $P$ be on $BC$ such that $OP$ is parallel to $AD$ . If $O$ is on $BC$ , then $P = O$ and the result is clear. We claim that the circumcircles of $BOO_1$ and $COO_2$ both pass through $P$ . One of the angles $\widehat{ADB}$ and $\widehat{ADC}$ is not acute. Without loss of generality, assume that $\widehat{ADB} \ge 90^\circ$ . Then $O_1$ does not lie in the interior of triangle $ADB$ . Note that
$\widehat{OO_1B} = \frac{\widehat{AO_1B}}{2} = 180^\circ - \widehat{ADB} = 180^\circ - \widehat{OPB},$
implying that $BO_1OP$ is cyclic.

Similarly, we can show that $CO_2OP$ is cyclic. Therefore, the circumcircles of $BOO_1$ and $COO_2$ pass through $P$ .

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.