IMO Hk TST

Geometry Difficulty 7.6 National Olympiad, round 2 Prove it Hong Kong

Let $\triangle ABC$ be an acute triangle with incentre $I$ and orthocentre $H$ . $AI$ meets the circumcircle of $\triangle ABC$ again at $M$ . Suppose the length $IM$ is exactly the circumradius of $\triangle ABC$ . Show that $AH \ge AI$ .

Solution

Let $O$ be the circumcentre of $\triangle ABC$ . Since $BM = CM = IM = OM$ (where $OM$ is the circumradius of $\triangle ABC$ ), $B$ , $C$ , $I$ , $O$ are concyclic. Therefore, we have $\angle BIC = \angle BOC$ . This yields $90^\circ + \frac{A}{2} = 2A$ , and hence $A = 60^\circ$ . Now
$\angle BHC = 180^\circ - A = 120^\circ = 2A = \angle BOC,$

which implies $H$ lies on ( $BCOI$ ). So $HM = IM$ . Noting the triangle inequality $AH + HM \ge AM$ , we conclude $AH \ge AI$ as desired.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.