Competition problem — Geometry — Maths Olympiad Prep

Library / /8 of 19

Geometry Difficulty 5.0 AIME Prove it United States

Problem:

Let $ABC$ be an acute triangle with circumcenter $O$ , incenter $I$ , orthocenter $H$ . If $OI = HI$ , what are the possible values of the angles of triangle $ABC$ ?

Solution

Solution:

Answer: this occurs if and only if some angle is $60$ degrees.

One direction is immediate; if $\angle A = 60^{\circ}$ then $BHOIC$ are cyclic since $\angle BHC = \angle BIC = \angle BOC = 120^{\circ}$ .

For the other direction, note that we have an "SSA congruence" of triangles $AIH$ and $AIO$ . Consequently, either $A$ lies on the circle $(OIH)$ or $\triangle AIH \cong \triangle AIO$ . In the latter case, $AH = AO$ , but since it's known that $AH = 2AO \cos A$ , it follows that $\cos A = \frac{1}{2} \Longrightarrow \angle A = 60^{\circ}$ .

Now it's impossible for $A, B, C \in (OIH)$ since the points $A, B, C, O$ are not concyclic. Thus some angle must be $60^{\circ}$ .

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.