## Task B-2.8.

Two circles touch each other internally at point $F$. The diameter of one circle is $8 \mathrm{~cm}$, and the diameter of the other is half as large. From the boundary point $T$ of the diameter $\overline{T F}$ of the larger circle, we construct a tangent to the smaller circle. If point $E$ is the intersection $(E \neq T)$ of the tangent and the larger circle, and $S_{1}$ is the center of the smaller circle, determine the perimeter of triangle $T S_{1} E$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_c3f5968a3447b3c0b4b2g-10.jpg?height=854&width=1144&top_left_y=657&top_left_x=469)

Sketch...

(1 point)

It is true that $\angle T D S_{1}=90^{\circ}$ (tangent) and $\angle T E F=90^{\circ}$ (angle over the diameter of the circle), so triangles $T D S_{1}$ and $T E F$ are similar.

(2 points)

It is true that $\left|T S_{1}\right|=6 \mathrm{~cm}, |T F|=8 \mathrm{~cm}$, and $\left|S_{1} D\right|=2 \mathrm{~cm}$. Due to the similarity of the triangles, $\left|T S_{1}\right|:|T F|=\left|S_{1} D\right|:|E F|$, so $6: 8=2:|E F|$, i.e., $|E F|=\frac{8}{3}$.

(1 point)

Now, $|T E|=\sqrt{|T F|^{2}-|E F|^{2}}=\sqrt{64-\frac{64}{9}}=\frac{16 \sqrt{2}}{3}$.

By the Pythagorean theorem,

$$
|T D|=\sqrt{\left|T S_{1}\right|^{2}-\left|S_{1} D\right|^{2}}=\sqrt{36-4}=4 \sqrt{2}
$$

Now,

$$
\left|S_{1} E\right|=\sqrt{\left|S_{1} D\right|^{2}+|D E|^{2}}=\sqrt{4+\left(\frac{16 \sqrt{2}}{3}-4 \sqrt{2}\right)^{2}}=\sqrt{\frac{68}{9}}=\frac{2 \sqrt{17}}{3}
$$

The perimeter of the triangle is $o=6+\frac{2 \sqrt{17}}{3}+\frac{16 \sqrt{2}}{3} \mathrm{~cm}$.

(2 points)

## SCHOOL/CITY COMPETITION IN MATHEMATICS

## 3rd grade - high school - B category

February 4, 2010.

IF A STUDENT HAS A DIFFERENT METHOD OF SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO EVALUATE THE METHOD AND SCORE IT APPROPRIATELY.