Let $ABC$ be an acutangle scalene triangle with $\angle BAC = 60^{\circ}$ and orthocenter $H$. Let $\omega_b$ be the circumference passing through $H$ and tangent to $AB$ at $B$, and $\omega_c$ the circumference passing through $H$ and tangent to $AC$ at $C$. Prove that $\omega_b$ and $\omega_c$ only have $H$ as common point. Prove that the line passing through $H$ and the circumcenter $O$ of triangle $ABC$ is a common tangent to $\omega_b$ and $\omega_c$. Note: The orthocenter of a triangle is the intersection point of the three altitudes, whereas the circumcenter of a triangle is the center of the circumference passing through it's three vertices.
Problem
Source: Mexico National Olympiad 2021Problem 4
Tags: geometry, tangent circles, orthocenter, Euler Line, circumcircle
MP8148
11.11.2021 02:14
Let $O_B$, $O_C$ denote the centers of $\omega_b$, $\omega_c$. We have $O_BB = O_BH$ and $\angle BO_BH = 2\angle ABH = 2(90^\circ - \angle A) = 60^\circ$, so $BHO_B$ is equilateral, so $\overline{HO_B}$ bisects $\angle BHC$ because $\angle BHC = 180^\circ - \angle A = 120^\circ$. Similarly $O_C$ also lies on this bisector, so $\omega_b$ and $\omega_c$ are tangent at $H$.
WLOG $AB < AC$. Then $\angle BOC = 2\angle A = 120^\circ = \angle BHC$, so $BHOC$ is cyclic. Thus \[\angle OHO_C = \angle OHC + \angle CHO_C = \angle OBC + 60^\circ = (90^\circ - \angle A) + 60^\circ = 90^\circ,\]so $\overline{OH}$ is tangent to $\omega_c$. By the first part of the problem, it is tangent to $\omega_b$ as well.
ike.chen
18.11.2021 05:01
Sketch: $\bullet$ Let $BB \cap AH = P$ and $CC \cap AH = Q$. Basic angle chasing implies $\omega_b = (BHP)$ and $\omega_c = (CHQ)$. $\bullet$ It's easy to see that $$\angle HBP + \angle HCQ = (\angle A + [90^{\circ} - \angle C])$$$$+ (\angle A + [90^{\circ} - \angle B]) = 180^{\circ}$$follows from $\angle A = 60^{\circ}$. This observation yields the first tangency condition. $\bullet$ The angle condition also implies $BHOC$ is cyclic. Hence, we have $$\angle CHO = \angle CBO = \angle OCB = \angle ACH = \angle CQH$$which proves the second tangency condition. $\blacksquare$ Remark: Finding a third point on $\omega_b$ and $\omega_c$ is imperative for angle chasing. As a result, we are motivated to characterize the second intersection between some secant and $\omega_b, \omega_c$. It would also be adventageous to make these second intersections symmetric wrt $B$ and $C$, i.e. on $AH$.