8,9
[ Area of a circle, sector, and segment ]

In the acute angles of a right triangle, two equal touching circles are inscribed. The sum of the areas of these circles is equal to the area of the circle inscribed in the triangle. Find the angles of the triangle.
Let circles of radius $r$ with centers $O_1$ and $O_2$ be inscribed in the acute angles $A$ and $B$ of the right triangle $ABC$ (Fig.1), and the first circle touches the hypotenuse $AB$ at point $N$, the leg $AC$ at point $M$, while the second circle touches the hypotenuse at point $K$, the leg $BC$ at point $L$. Denote $\angle BAC = \alpha$. From the right triangles $AO_1N$ and $BO_2K$, we find that

$$
\begin{gathered}
AN = \frac{O_1N}{\lg \angle NAO_1} = \frac{\varphi}{\lg \frac{\kappa}{2}} \\
KB = \frac{O_2K}{\lg \angle KBO_2} = \frac{r}{\lg \left(45^3 - \frac{\alpha}{2}\right)} = \frac{\frac{\varphi}{1 - \operatorname{tg} \frac{\Gamma}{2}}}{1 + \frac{\alpha}{2} \frac{r}{2}} = \frac{r\left(1 + \lg \frac{\alpha}{2}\right)}{1 - \lg \frac{\alpha}{2}}
\end{gathered}
$$

Then

$$
\begin{aligned}
AB = AN + NK + KB = & \frac{\varphi}{\lg \frac{\gamma}{2}} + 2r + \frac{\varphi\left(1 + \lg \frac{\alpha}{2}\right)}{1 - \lg \frac{\alpha}{2}} = r\left(\frac{1}{\lg \frac{\kappa}{2}} + 2 + \frac{1 + \lg \frac{\alpha}{2}}{1 - \lg \frac{\alpha}{2}}\right) = \\
& = \frac{\varphi\left(1 + 2 \lg \frac{\alpha}{2} - \lg^2 \frac{\alpha}{2}\right)}{\lg \frac{\gamma}{2}\left(1 - \lg \frac{\gamma}{2}\right)}
\end{aligned}
$$

Let the circle of radius $R$ with center $O$, inscribed in triangle $ABC$, touch the leg $AC$ at point $P$ (Fig.2). Then

![](https://cdn.mathpix.com/cropped/2024_05_06_a98533106cc029c284d7g-39.jpg?height=123&width=1083&top_left_y=1505&top_left_x=499)

and since

$$
\begin{gathered}
AB = \frac{AC}{\cos \angle BAC} = \frac{AC}{\cos \frac{\alpha}{2}} = \frac{\frac{AC}{1 - \operatorname{tg}^2 \frac{\alpha}{2}}}{1 + \operatorname{tg}^2 \frac{\alpha}{2}} = \frac{AC\left(1 + \operatorname{tg}^2 \frac{\alpha}{2}\right)}{1 - \operatorname{tg}^2 \frac{\alpha}{2}} = \\
= \frac{R\left(1 + \lg \frac{\alpha}{2}\right)}{\lg \frac{\alpha}{2}} \cdot \frac{1 + \operatorname{tg}^2 \frac{\alpha}{2}}{1 - \lg^2 \frac{\alpha}{2}} = \frac{R\left(1 + \operatorname{tg}^2 \frac{\alpha}{2}\right)}{\lg \frac{\alpha}{2}\left(1 - \lg^2 \frac{\alpha}{2}\right)}
\end{gathered}
$$

then

$$
\frac{r\left(1 + 2 \lg \frac{\alpha}{2} - \lg^2 \frac{\alpha}{2}\right)}{\lg \left(1 - \lg^2 \frac{\alpha}{2}\right)} = \frac{r\left(1 + \operatorname{tg}^2 \frac{\alpha}{2}\right)}{\lg \left(\frac{\sin \alpha}{2}\left(1 - \lg \frac{\alpha}{2}\right)\right)}
$$

By the problem's condition, $2 \pi r^2 = \pi R^2$, so $R = r \sqrt{2}$, hence,

![](https://cdn.mathpix.com/cropped/2024_05_06_a98533106cc029c284d7g-39.jpg?height=112&width=958&top_left_y=2288&top_left_x=550)

$$
\begin{aligned}
& (\sqrt{2} + 1) \operatorname{tg}^2 \frac{\alpha}{2} - 2 \operatorname{tg} \frac{\alpha}{2} + \sqrt{2} - 1 = 0, \operatorname{tg} \frac{\alpha}{2} = \sqrt{2} - 1 \\
& \operatorname{tg} \alpha = \frac{2 \lg \frac{\alpha}{2}}{1 - \lg^2 \frac{\alpha}{2}} = \frac{2(\sqrt{2} - 1)}{1 - (\sqrt{2} - 1)^2} = \frac{2(\sqrt{2} - 1)}{2(\sqrt{2} - 1)} = 1
\end{aligned}
$$

Therefore, $\alpha = 45^\circ$.

| 111534 Topics: [ | Tangent Circles  Heron's Formula |
| :---: | :---: |

On a segment and two unequal parts of lengths $2a$ and $2b$, semicircles are constructed on the same side of the segment. Find the radius of the circle that is tangent to the three constructed semicircles.

## Solution

Let $AB = 2a + 2b$ be the diameter of the larger semicircle with center $O$, $AC = 2a$ be the diameter of the semicircle with center $O_1$, $BC = 2b$ be the diameter of the semicircle with center $O_2$, and $x$ be the radius of the circle with center $O_3$ that is tangent to the three given semicircles. The line of centers of two tangent circles passes through their point of tangency, so

$$
\begin{gathered}
OO_3 = a + b - x, O_1O_3 = a + x, O_2O_3 = b + x \\
OO_1 = OA - O_1A = a + b - a = b, OO_2 = OB - O_2B = a + b - b = a
\end{gathered}
$$

Let $p_1$ and $p_2$ be the semiperimeters of triangles $OO_1O_3$ and $OO_2O_3$. Then

$$
p_1 = \frac{(a + x) + (a + b - x) + b}{2} = a + b, p_2 = \frac{(b + x) + (a + b - x) + a}{2} = a + b
$$

By Heron's formula,

$$
S_{\triangle O O_1 O_3} = \sqrt{(a + b)(b - x) x a}, S_{\triangle O O_2 O_3} = \sqrt{(a + b)(a - x) x b}
$$

so

$$
\frac{S_{\Delta O O_1 O_3}}{S_{\Delta O O_2 O_3}} = \frac{\sqrt{(a + b)(b - x) x a}}{\sqrt{(a + b)(a - x) x b}} = \frac{\sqrt{(b - x) a}}{\sqrt{(a - x) b}}.
$$

On the other hand, the triangles $OO_1O_3$ and $OO_2O_3$ have a common height drawn from vertex $O_3$, so

$$
\frac{S_{\Delta O O_1 O_3}}{S_{\Delta O O_2 O_3}} = \frac{OO_1}{OO_2} = \frac{b}{a}
$$

From the equation

$$
\frac{\sqrt{(b - x) a}}{\sqrt{(a - x) b}} = \frac{b}{a}
$$

we find that

$$
x = \frac{\frac{ab^3 - a^3b}{b^3 - a^3}}{= \frac{ab(a + b)}{a^2 + ab + b^2}}
$$

## Answer