62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour

Question

In an acute triangle ABC points H and O are the orthocenter and the circumcenter correspondingly. The line HO intersects the sides AB and AC in points X and Y correspondingly, so that the point H lies on the segment OX. It turned out that XH = HO = OY. Find the angle BAC. (Oleksii Masalitin)

Accepted Answer

In the solution of this problem, we will be using the following well-known fact: in any triangle, the distance from the vertex to the orthocenter is twice larger than the distance from the circumcenter to the opposite side.

Denote by

O_{B}

and

H_{B}

the projections of

O

and

H

correspondingly onto the line

A C

, define points

O_{C}

and

H_{C}

similarly (fig. 9). From the fact above it follows that

B H = 2 O O_{B}

and

C H = 2 O O_{C}

. Also note, that

O O_{B} ∥ H H_{B}

and

H O = O Y

, and therefore

O O_{B}

is the midline of

△ Y H H_{B}

, so

H H_{B} = 2 O O_{B} = B H

. Similarly

H H_{C}

is the midline of

△ X O O_{C}

, so

H H_{C} = \frac{1}{2} O O_{C} = \frac{1}{4} C H

. As the quadrilateral

B H C_{H} C_{B}

is cyclic,

B H \cdot H H_{B} = C H \cdot H H_{C}

, and

B H^{2} = 4 H H_{C}^{2}

, so

B H = 2 H H_{C}

, and

∠ H B H_{C} = 3 0^{\circ}

. Now it's clear that

∠ B A C = 6 0^{\circ}

.