In an acute-angled triangle $A B C$, the bisector $A D$ bisects the segment $O H$, where $O$ is the center of the circumscribed circle, and $H$ is the intersection point of the altitudes. It is known that $A C=2, A D=\sqrt{3}+\sqrt{2}$ - 1. Find the radius of the circumscribed circle around triangle $A B C$.

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Let $A K$ be the height of triangle $A B C$, $P$ and $M$ be the midpoints of sides $A C$ and $B C$ respectively, $Q$ be the midpoint of $O H$, and $R$ be the desired radius.

Then $\angle A B C=\frac{1}{2} \angle A O C=\angle A O P$ (the inscribed angle is equal to half of the corresponding central angle). In the right triangles $A K B$ and $A P O$, the two acute angles are respectively equal, so $\angle B A K=\angle O A C$. Since $A D$ is the angle bisector of $\angle B A C$, then $\angle H A D=\angle O A D$. Thus, the median $A Q$ of triangle $O A H$ is its angle bisector. Therefore, triangle $O A H$ is isosceles and $A H=O A=R$.

Since $M P$ is the midline of triangle $A B C$, then $M P \| A B$. Moreover, $O M \| A H$ and $O P \| B H$. Therefore, triangle $M O P$ is similar to triangle $A H B$, with a similarity ratio of $\frac{1}{2}$. Consequently, $O M=\frac{1}{2} A H=\frac{1}{2} R$.

In the right triangle $O M C$, the leg $O M=\frac{1}{2} R$ is half the hypotenuse $O C=R$. Therefore, $\angle M O C=60^{\circ}$. Then $\angle B O C=2 \angle M O C=120^{\circ}$. Consequently, $\angle B A C=\frac{1}{2} \angle B O C=60^{\circ}$ (the inscribed angle is equal to half of the corresponding central angle), and $\angle D A C=30^{\circ}$.

By the cosine theorem,

$$
C D^{2}=A C^{2}+A D^{2}-2 \cdot A C \cdot A D \cdot \cos 30^{\circ}=4+(\sqrt{3}+\sqrt{2}-1)^{2}-2 \cdot 2 \cdot(\sqrt{3}+\sqrt{2}-1) \cdot \frac{\sqrt{3}}{2}=4-2 \sqrt{2}
$$

By the sine theorem,

$$
\frac{C D}{\sin \angle C A D}=\frac{A C}{\sin \angle A D C}
$$

from which

$$
\sin \angle A D C=\frac{A C \sin 30^{\circ}}{C D}=\frac{2 \cdot \frac{1}{2}}{\sqrt{4-2 \sqrt{2}}}=\frac{1}{\sqrt{4-2 \sqrt{2}}}=\frac{\sqrt{2+\sqrt{2}}}{2}=\sin 67.5^{\circ}
$$

Notice that $A C<A D$, since

$$
2<\sqrt{3}+\sqrt{2}-1 \Leftarrow 3<\sqrt{3}+\sqrt{2} \Leftarrow 9<(\sqrt{3}+\sqrt{2})^{2}=5+2 \sqrt{6} \Leftarrow 4<2 \sqrt{6} \Leftarrow 4<6
$$

Thus, the angle $\angle A D C$ in triangle $A D C$ lies opposite the non-largest side $A C$. Therefore, $\angle A D C<90^{\circ}$. Then $\angle A D C=67.5^{\circ}$.

Since $\angle A D C$ is the exterior angle of triangle $A B D$, then

$$
\angle A B C=\angle A D C-\angle B A D=67.5^{\circ}-30^{\circ}=37.5^{\circ} .
$$

Therefore,

$$
R=\frac{A C}{2 \sin \angle A B C}=\frac{2}{2 \sin 37.5^{\circ}}=\frac{1}{\sin 37.5^{\circ}}=\frac{1}{\sin \left(60^{\circ}-\frac{1}{2} \cdot 45^{\circ}\right)}=\frac{\sqrt{6}-\sqrt{2}+2}{\sqrt{2+\sqrt{2}}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e1837d72990d2467a29dg-52.jpg?height=623&width=828&top_left_y=506&top_left_x=615)

## Answer

Prove that $\angle B A H=\angle O A C$, triangle $A O H$ is isosceles, the distance from the center of the circle to side $B C$ is half of $A H$, and angle $\angle A D C$ is acute. Then apply the cosine and sine theorems to triangle $A D C$.