2017 USAJMO Problems/Problem 5

Problem

Let $O$ and $H$ be the circumcenter and the orthocenter of an acute triangle $ABC$ . Points $M$ and $D$ lie on side $BC$ such that $BM = CM$ and $\angle BAD = \angle CAD$ . Ray $MO$ intersects the circumcircle of triangle $BHC$ in point $N$ . Prove that $\angle ADO = \angle HAN$ .

Solution

THIS SOLUTION HAS NO DIAGRAM, SOMEONE WHO HAS ASYMPTOTE SKILLS PLEASE HELP. $-1$ !!! because of the diagram rule

Suppose ray $OM$ intersects the circumcircle of $BHC$ at $N'$ , and let the foot of the A-altitude of $ABC$ be $E$ . Note that $\angle BHE=90-\angle HBE=90-90+\angle C=\angle C$ . Likewise, $\angle CHE=\angle B$ . So, $\angle BHC=\angle BHE+\angle CHE=\angle B+\angle C$ . $BHCN'$ is cyclic, so $\angle BN'C=180-\angle BHC=180-\angle B-\angle C=\angle A$ . Also, $\angle BAC=\angle A$ . These two angles are on different circles and have the same measure, but they point to the same line $BC$ ! Hence, the two circles must be congruent. (This is also a well-known result)

We know, since $M$ is the midpoint of $BC$ , that $OM$ is perpendicular to $BC$ . $AH$ is also perpendicular to $BC$ , so the two lines are parallel. $AN$ is a transversal, so $\angle HAN=\angle ANO$ . We wish to prove that $\angle ANO=\angle ADO$ , which is equivalent to $AOND$ being cyclic.

Now, assume that ray $OM$ intersects the circumcircle of $ABC$ at a point $P$ . Point $P$ must be the midpoint of $\stackrel{\frown}{BC}$ . Also, since $AD$ is an angle bisector, it must also hit the circle at the point $P$ . The two circles are congruent, which implies $MN=MP\implies ND=DP\implies$ NDP is isosceles. Angle ADN is an exterior angle, so $\angle ADN=\angle DNP+\angle DPO=2\angle DPO$ . Assume WLOG that $\angle B>\angle C$ . So, $\angle DPO=\angle APO=\frac{\angle B+\angle C}{2}-\angle C=\frac{\angle B-\angle C}{2}$ . In addition, $\angle AON=\angle AOP=\angle AOB+\angle BOP=2\angle C+\angle A$ . Combining these two equations, $\angle AON+\angle ADN=\angle B-\angle C+2\angle C+\angle A=\angle A+\angle B+\angle C=180$ .

Opposite angles sum to $180$ , so quadrilateral $AOND$ is cyclic, and the condition is proved.

-william122

2017 USAJMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
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2017 USAJMO Problems/Problem 5

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