JBMO Shortlist

Geometry Difficulty 7.2 National Olympiad, round 2 Prove it JBMO

Problem:
Let $ABC$ be an acute-angled triangle with circumcircle $\Gamma$ , and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A'$ be the point where the angle bisector of angle $BAC$ meets $\Gamma$ . If $A'H = AH$ , find the measure of angle $BAC$ .

Figure 4: Exercise G4.

Solution

Solution:
The segment $AA'$ bisects $\angle OAH$ : if $\angle BCA = y$ (Figure 4), then $\angle BOA = 2y$ , and since $OA = OB$ , it is $\angle OAB = \angle OBA = 90^{\circ} - y$ . Also since $AH \perp BC$ , it is
$\angle HAC = 90^{\circ} - y = \angle OAB$ and the claim follows.

Since $AA'$ bisects $\angle OAH$ and $A'H = AH$ , $OA' = OA$ , we have that the isosceles triangles $OAA'$ , $HAA'$ are equal. Thus
$AH = OA = R$
where $R$ is the circumradius of triangle $ABC$ .

Call $\angle ACH = a$ and recall by the law of sines that $AH = 2R' \sin a$ , where $R'$ is the circumradius of triangle $AHC$ . Then (4) implies
$R = 2R' \sin a$
But notice that $R = R'$ because $\frac{AC}{\sin(AHC)} = 2R'$ , $\frac{AC}{\sin(ABC)} = 2R$ and $\sin(AHC) = \sin\left(180^{\circ} - ABC\right) = \sin(ABC)$ . So (5) gives $1 = 2\sin a$ , and $a$ as an acute angle can only be $30^{\circ}$ . Finally, $\angle BAC = 90^{\circ} - a = 60^{\circ}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.