Note that either $A C=B C$, or $A C=A B$.

Consider the first of these cases (Fig.1). Let $M$ and $H$ be the points of tangency of the inscribed circle of triangle $A B C$ with the lateral side $A C$ and the base $A B$, and $O$ be the center of the inscribed circle. Then $C H$ is the height, median, and bisector of triangle $A B C$. Denote $\angle A C H=\alpha$. In the right triangle $C O M$, it is known that

$$
O M=\frac{8}{3}, O C=C H-O H=\frac{+8}{5}-\frac{8}{3}=\frac{8 \cdot 13}{15}, \sin \alpha=\frac{O M}{O C}=\frac{\frac{8}{3}}{\frac{8 \cdot 13}{15}}=\frac{5}{13}
$$

Then

$$
\cos \alpha=\frac{12}{13}, \operatorname{tg} \alpha=\frac{5}{12}, A H=C H \operatorname{tg} \alpha=\frac{48}{5} \cdot \frac{5}{12}=4
$$

Therefore, $A B=2 A H=8$.

Consider the second case (Fig.2). Let $K, L$ and $N$ be the points of tangency of the inscribed circle of the isosceles triangle $A B C$ with the lateral sides $A B, A C$ and the base $B C$ respectively, $C H=\frac{ \pm 8}{5}$ be the height of triangle $A B C$, and $O K=O L=O N=r=\frac{8}{3}$ be the radius of its inscribed circle. Denote $B K=B N=C N=C L=a$, $A K=A L=b, \angle A B C=\angle B C A=\beta$. Then

$$
S_{\triangle A B C}=\frac{1}{2} A B \cdot C H=\frac{2 t}{5}(a+b), S_{\triangle A B C}=\frac{1}{2}(A B+A C+B C) \cdot r=\frac{8}{3}(2 a+b)
$$

From the equality $\frac{2 t}{5}(a+b)=\frac{8}{3}(2 a+b)$, we get that $a=4 b$. From the right triangle $A B N$, we find that

$$
\cos \beta=\frac{B N}{A B}=\frac{a}{a+b}=\frac{+b}{5 b}=\frac{t}{5}
$$

Then

$$
\sin \beta=\frac{3}{5}, \operatorname{ctg} \frac{\beta}{2}=\frac{1+\cos \beta}{\sin \beta}=\frac{1+\frac{4}{5}}{\frac{3}{5}}=3
$$

Finally, from the right triangle $B O K$, we find that

$$
a=B K=O K \operatorname{ctg} \beta=\frac{8}{3} \cdot 3=8
$$

Therefore,

$$
A B=a+b=a+\frac{1}{4} a=8+2=10
$$

## Answer

8 or 10.

The distance between two parallel lines is 24. On one of them lies point $C$, on the other — points $A$ and $B$, and triangle $A B C$ is isosceles and acute-angled, with its lateral side equal to 25. Find the radius of the circle inscribed in triangle $A B C$.

## Solution

Note that either $A C=B C$, or $A C=A B$.

Consider the first of these cases (Fig.1): $A C=B C=25$. Let $H$ be the point of tangency of the inscribed circle of triangle $A B C$ with the base $A B$, and $r 1$ be the radius of the circle inscribed in triangle $A B C$. Then $C H$ is the height and median of triangle $A B C$. From the right triangle $A H C$, we find that

$$
A H=\sqrt{A C^{2}-C H^{2}}=\sqrt{25^{2}-24^{2}}=\sqrt{(25-24)(25+24)}=7.
$$

Then

$$
S_{\triangle A B C}=A H \cdot C H=7 \cdot 24=168, S_{\triangle A B C}=\frac{1}{2}(A B+B C+A C) r 1=\frac{1}{2}(14+25+25) r 1=32 r 1.
$$

From the equality $32 r 1=168$, we find that $r 1=\frac{21}{4}$. Consider the second case (Fig.2): $A B=A C=25$. Let $C H$ be the height of triangle $A B C$, and the radius of the circle inscribed in triangle $A B C$ be $r 2$. Then $A H=7$, $B H=A B-A H=25-7=18$ (triangle $A B C$ is acute-angled). From the right triangle $B C H$, we find that $B C=\sqrt{B H^{2}+C H^{2}}=\sqrt{18^{2}+2 t^{2}}=6 \sqrt{3+16}=6 \cdot 5=30$,

$$
S_{\triangle A B C}=\frac{1}{2} A B \cdot C H=\frac{1}{2} \cdot 25 \cdot 24=300, S_{\triangle A B C}=\frac{1}{2}(A B+A C+B C) \cdot r 2=40 r 2.
$$

From the equality $40 r 2=300$, we get that $r 2=\frac{15}{2}$.

## Answer

$\frac{21}{4}$ or $\frac{15}{2}$.
[Supporting area. The area helps to solve the problem.] Inscribed and circumscribed circles $\quad]$

The distance between parallel lines is $\frac{\frac{18}{5}}{5}$. On one of them lies point $C$, on the other — points $A$ and $B$, and triangle $A B C$ is isosceles. It is known that the radius of the circle inscribed in triangle $A B C$ is $\frac{8}{3}$. Find $A B$.