
G4. Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$, and let $O, H$ be the triangle's circumcenter and orthocenter respectively. Let also $A^{\prime}$ be the point where the angle bisector of angle $B A C$ meets $\Gamma$. If $A^{\prime} H=A H$, find the measure of angle $B A C$.

![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-07.jpg?height=519&width=1042&top_left_y=1756&top_left_x=541)

Figure 4: Exercise G4.


Solution. The segment $A A^{\prime}$ bisects $\angle O A H$ : if $\angle B C A=y$ (Figure 4), then $\angle B O A=$ $2 y$, and since $O A=O B$, it is $\angle O A B=\angle O B A=90^{\circ}-y$. Also since $A H \perp B C$, it is
$\angle H A C=90^{\circ}-y=\angle O A B$ and the claim follows.

Since $A . A^{\prime}$ bisects $\angle O A H$ and $A^{\prime} H=A H . O A^{\prime}=O A$, we have that the isosceles triangles $O A A^{\prime}, H A A^{\prime}$ are equal. Thus

$$
A H=O A=R
$$

where $R$ is the circumradius of triangle $A B C$.

Call $\angle A C H=a$ and recall by the law of sines that $A H=2 R^{\prime} \sin a$, where $R^{\prime}$ is the circumradius of triangle $A H C$. Then (4) implies

$$
R=2 R^{\prime} \sin a
$$

But notice that $R=R^{\prime}$ because $\frac{A C}{\sin (A H C)}=2 R^{\prime}, \frac{A C}{\sin (A B C)}=2 R$ and $\sin (A H C)=$ $\sin \left(180^{\circ}-A B C\right)=\sin (A B C)$. So (5) gives $1=2 \sin a$, and $a$ as an acuite angle can only be $30^{\circ}$. Finally, $\angle B A C=90^{\circ}-a=60^{\circ}$.

Remark. The steps in the above proof can be traced backwards making the converse also true, that is: If $\angle B A C=60^{\circ}$ then $A^{\prime} H=A H$.

- G5. Let the circles $k_{1}$ and $k_{2}$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_{1}$ and $k_{2}$ that touches them at $M$ and, $N$ respectively. If $t \perp A M$ and $M N=2 A M$, evaluate $\angle N M B$.

![](https://cdn.mathpix.com/cropped/2024_06_05_61d3145f18c90bf9f370g-08.jpg?height=540&width=1008&top_left_y=1454&top_left_x=589)

Figure 5: Exercise G5.
