It is known that the distance from the center of the circumscribed circle to the side $AB$ of triangle $ABC$ is equal to half the radius of this circle. Find the height of triangle $ABC$, dropped to the side $AB$, if it is less than $\sqrt{\frac{3}{2}}$, and the other two sides of the triangle are equal to 2 and 3.
If vertex $C$ and the center of the circumscribed circle of the given triangle $A B C$ are located on the same side of the line $A B$, then $\angle A C B=60^{\circ}$, and if - on opposite sides, then $\angle A C B=120^{\circ}$

## Solution

Let $O$ be the center of the circle with radius $R$ circumscribed around triangle $A B C, C H-$ the height of the triangle, $O M=\frac{1}{2}$ $R$ - the perpendicular dropped from the center of the circle to the side $A B, A C=2, B C=3$.

In the right triangle $O M B$, the leg $O M$ is half the hypotenuse $O B=R$. Therefore, $\angle O B M=30^{\circ}$. Then $\angle$ $A O B=120^{\circ}$.

If points $O$ and $C$ are located on the same side of the line $A B$ (Fig.1), then $\angle A C B=\frac{1}{2} \angle A O B=60^{\circ}$ (the inscribed angle is equal to half the corresponding central angle).

By the cosine theorem

$$
A B=\sqrt{C A^{2}+C B^{2}-2 \cdot C A \cdot C B \cos 60^{\circ}}=\sqrt{4+9-6}=\sqrt{7} .
$$

Since

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} \cdot A B \cdot C H=\frac{1}{2} \cdot \sqrt{7} \cdot C H \text { and } S_{\Delta \mathrm{ABC}}=\frac{1}{2} \cdot C A \cdot C B \cdot \sin 60^{\circ}=\frac{1}{2} \cdot 2 \cdot 3 \cdot \frac{\sqrt{3}}{2}
$$

then

$$
\frac{1}{2} \cdot \sqrt{7} \cdot C H=\frac{1}{2} \cdot 2 \cdot 3 \cdot \frac{\sqrt{3}}{2}
$$

From this, we find that $C H=\frac{3 \sqrt{3}}{\sqrt{7}}$.

Since

$$
C H=\frac{3 \sqrt{3}}{\sqrt{7}}>\sqrt{\frac{3}{2}} \Leftarrow 3 \sqrt{2}>\sqrt{7} \Leftarrow 18>7
$$

the found value of the height does not satisfy the condition of the problem.

Let points $O$ and $C$ be located on opposite sides of the line $A B$ (Fig.2). Then

$$
\begin{gathered}
\angle A C B=180^{\circ}-\frac{1}{2} \angle A O B=180^{\circ}-60^{\circ}=120^{\circ}, \\
A B=\sqrt{C A^{2}+C B^{2}-2 \cdot C A \cdot C B \cos 120^{\circ}}=\sqrt{4+9+6}=\sqrt{19} \\
S_{\triangle \mathrm{ABC}}=\frac{1}{2} \cdot \sqrt{19} \cdot C \text { and and } S_{\triangle \mathrm{ABC}}=\frac{1}{2} \cdot C A \cdot C B \cdot \sin 120^{\circ}=\frac{1}{2} \cdot 2 \cdot 3 \cdot \frac{\sqrt{3}}{2} \\
\frac{1}{2} \cdot \sqrt{19} \cdot C H=\frac{1}{2} \cdot 2 \cdot 3 \cdot \frac{\sqrt{3}}{2}
\end{gathered}
$$

From this, we find that $C H=\frac{3 \sqrt{3}}{\sqrt{19}}$. Since in this case

$$
C H=\frac{3 \sqrt{3}}{\sqrt{19}}<\sqrt{\frac{3}{2}} \Leftarrow 3 \sqrt{2}<\sqrt{19} \Leftarrow 18<19
$$

## Answer

$3 \sqrt{\frac{3}{19}}$.