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

[asy] pair A = dir(130); pair B = dir(220); pair C = dir(320); draw(unitcircle, lightblue);

pair P = dir(-90); pair Q = dir(90); pair D = extension(A, P, B, C); pair O = origin; pair M = extension(B, C, O, P); pair N = 2*M-P;

draw(A--B--C--cycle, lightblue); draw(A--P--Q, lightblue); draw(A--N--D--O--A, lightblue);

draw(A--D--N--O--cycle, red);

dot(" $A$ ", A, dir(A)); dot(" $B$ ", B, dir(B)); dot(" $C$ ", C, dir(C)); dot(" $P$ ", P, dir(P)); dot(" $Q$ ", Q, dir(Q)); dot(" $D$ ", D, dir(225)); dot(" $O$ ", O, dir(315)); dot(" $M$ ", M, dir(315)); dot(" $N$ ", N, dir(315));

/* TSQ Source:

A = dir 130 B = dir 220 C = dir 320 unitcircle 0.1 lightcyan / lightblue

P = dir -90 Q = dir 90 D = extension A P B C R225 O = origin R315 M = extension B C O P R315 N = 2*M-P R315

A--B--C--cycle lightblue A--P--Q lightblue A--N--D--O--A lightblue

A--D--N--O--cycle 0.1 yellow / red [/asy]

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
1 • 2 • 3 • 4 • 5 • 6
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2017 USAJMO Problems/Problem 5

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