Competition problem — Geometry — Maths Olympiad Prep

Library / /36 of 128

Geometry Difficulty 5.1 AIME, harder Prove it Philippines

Problem:

Let $B H$ be the altitude from the vertex $B$ to the side $A C$ of an acute-angled triangle $A B C$ . Let $D$ and $E$ be the midpoints of $A B$ and $A C$ , respectively, and $F$ the reflection of $H$ across the line segment $E D$ . Prove that the line $B F$ passes through the circumcenter of $\triangle A B C$ .

Solutions — 2

Solution 1

Solution:

Let $O$ be the circumcenter of $\triangle A B C$ . Since $\angle O B A = 90^{\circ} - \angle C$ , it suffices to show that $\angle F B A = 90^{\circ} - \angle C$ .

Note that $A D = B D = D H$ and $D H = D F$ . Therefore, quadrilateral $A H F B$ is cyclic (with circumcenter $D$ ), and so $\angle F B A = \angle F H E = 90^{\circ} - \angle D E H$ . Since $D E$ is parallel to $B C$ , $\angle D E H = \angle C$ , and $\angle F B A = 90^{\circ} - \angle C$ .

Solution 2

Solution:

As before, denote by $O$ the circumcenter of $\triangle A B C$ . Then the quadrilateral $A D O E$ is cyclic. Also, we know that $A D = H D = D B$ , therefore,
$\angle A = \angle D H A = 180^{\circ} - \angle D H E = 180^{\circ} - \angle D F E$
Therefore, $A D F E$ is cyclic. Since $A D F O E$ is cyclic, $D F O E$ is also cyclic, and
$\angle C = \angle D E A = \angle D E F = \angle D O F$
On the other hand, $\angle C = \angle D O B$ , so $\angle D O F = \angle D O B$ , therefore $B, F$ , and $O$ are collinear.

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.