$\quad$ Inscribed and Circumscribed Circles [ Trigonometric Ratios in a Right Triangle

A regular hexagon $A B C D E K$ is inscribed in a circle of radius $3+2 \sqrt{3}$. Find the radius of the circle inscribed in triangle $B C D$.

#
The radius of the circle inscribed in a triangle is equal to the area of the triangle divided by its semiperimeter.

## Solution

Let $R$ be the radius of the given circle. In triangle $BCD$, it is known that

$$
\begin{gathered}
\angle BCD = 120^\circ, BC = CD = R, BD = R \sqrt{3} \\
S_{\triangle BCD} = \frac{1}{2} R^2 \sin 120^\circ = \frac{R^2 \sqrt{3}}{4}
\end{gathered}
$$

Then the radius of the circle inscribed in this triangle can be calculated using the formula

$$
r = \frac{2 S_{\triangle BCD}}{BC + CD + BD} = \frac{\frac{R^2 \sqrt{3}}{2}}{2R + R \sqrt{3}} = \frac{\frac{R \sqrt{3}}{2}}{2 + \sqrt{3}} = \frac{\sqrt{3}(2 \sqrt{3} + 3)}{2(2 + \sqrt{3})} = \frac{3}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-02.jpg?height=434&width=549&top_left_y=0&top_left_x=757)

## Answer

$\frac{3}{2}$.

## [Radii of inscribed, circumscribed, and exscribed circles (other).] Problem 53041 Topics: [ Ratios of linear elements of similar triangles ] Difficulty: $3+$ [ Trigonometric ratios in a right triangle ] Classes: 8,9

In triangle $ABC$, segment $MN$ with endpoints on sides $AC$ and $BC$ is parallel to the base $AB$ and touches the inscribed circle. $\angle A = 2\alpha, \angle B = 2\beta$.

Find the similarity coefficient of triangles $ABC$ and $MNC$.

## Hint

The similarity coefficient of the specified triangles is equal to the ratio of their semiperimeters.

## Solution

Let $P$ be the point of tangency of the inscribed circle with side $AC$, and $r$ be the radius of the inscribed circle. Then $AB$

$= r \operatorname{ctg} \alpha + r \operatorname{ctg} \beta, AP = r \operatorname{ctg} \alpha$

$PC = r \operatorname{ctg} \frac{1}{2} \angle C = r \operatorname{ctg}(90^\circ - \alpha - \beta) = r \operatorname{tg}(\alpha + \beta)$.

On the other hand, segment $PC$ is equal to the semiperimeter of triangle $MNC$, and the semiperimeter of triangle $ABC$ is equal to $r(\operatorname{ctg} \alpha + \operatorname{ctg} \beta + \operatorname{tg}(\alpha + \beta))$.

The similarity coefficient of triangles $ABC$ and $MNC$ is equal to the ratio of their semiperimeters, that is, $\frac{\operatorname{ctg} \alpha + \operatorname{ctg} \beta + \operatorname{tg}(\alpha + \beta)}{\operatorname{tg}(\alpha + \beta)} = \operatorname{ctg} \alpha \operatorname{ctg} \beta$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-02.jpg?height=460&width=945&top_left_y=1648&top_left_x=565)

## Answer

$\operatorname{ctg} \alpha \operatorname{ctg} \beta$.