[ Ratio of areas of similar triangles ]  [ Inscribed and circumscribed circles ]

In an isosceles triangle $A B C$, the angle at the base $A C$ is $\alpha$. The circle inscribed in this triangle touches the sides of the triangle at points $A 1, B 1, C 1$. Find the ratio of the area of triangle

$A 1 B 1 C 1$ to the area of triangle $A B C$.
Let the inscribed circle touch the base $AC$ of the isosceles triangle $ABC$ at point $B_1$. Then $B_1$ is the midpoint of $AC$. Denote $AB_1 = CB_1 = a$. From the right triangle $ABB_1$, we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_dc8ed1d6cdac2bf77445g-12.jpg?height=94&width=314&top_left_y=2789&top_left_x=878)

![](https://cdn.mathpix.com/cropped/2024_05_06_dc8ed1d6cdac2bf77445g-13.jpg?height=105&width=1288&top_left_y=79&top_left_x=385)

Triangle $BC_1A_1$ is similar to triangle $ABC$ with the similarity coefficient

$$
k = \frac{BC_1}{AB} = \frac{AB - AC_1}{AB} = \frac{\frac{a}{\frac{M}{M a}} - a}{\frac{a}{m 8 \alpha a}} = 1 - \cos \alpha
$$

$$
S_{\triangle B A_1 C_1} = k^2 S_{\triangle ABC} = (1 - \cos \alpha)^2 S_{\triangle ABC}
$$

Therefore,

$$
S_{\triangle A_1 B_1 C_1} = S_{\triangle ABC} - S_{\triangle B A_1 C_1} - 2 S_{\triangle A B_1 C_1} = (1 - (1 - \cos \alpha)^2 - \cos \alpha) S_{\triangle ABC} =
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_dc8ed1d6cdac2bf77445g-13.jpg?height=119&width=892&top_left_y=738&top_left_x=583)

## Answer

$\cos \alpha (1 - \cos \alpha)$.