Task B-3.3. A right-angled triangle with legs of length 6 and 8 meters is given. A wheel with a radius of 1 meter is inside the triangle and, touching one of its sides, begins to roll along the sides of the triangle (inside the triangle and always in the same direction), such that it always touches at least one side of the triangle. How much distance will the center of the wheel travel before it returns to its initial position?
## Solution.

Let the initial position of the wheel's center be at point $P$. The triangle $O P R$ which determines the trace of point $P$ is similar to triangle $B C A$, and their perimeters are in the same ratio as the corresponding sides.

![](https://cdn.mathpix.com/cropped/2024_05_30_4f4efb97ca3723b253dag-13.jpg?height=764&width=954&top_left_y=759&top_left_x=542)

The length of one leg of triangle $O P R$ is equal to the length of segment $\overline{O^{\prime} P^{\prime}}$, and $\left|O^{\prime} P^{\prime}\right|=|B C|-1-\left|B O^{\prime}\right|$. First, let's calculate $\left|B O^{\prime}\right|$.

![](https://cdn.mathpix.com/cropped/2024_05_30_4f4efb97ca3723b253dag-13.jpg?height=504&width=835&top_left_y=1770&top_left_x=623)

Point $O$ is on the angle bisector of $\beta$ because it is 1 unit away from both of its sides. Therefore, $\angle O B O^{\prime}=\frac{\beta}{2}$.

From triangle $O B O^{\prime}$, we have $\operatorname{ctg} \frac{\beta}{2}=\frac{\left|B O^{\prime}\right|}{1}=\left|B O^{\prime}\right|$.

$$
\begin{aligned}
& \operatorname{ctg} \frac{\beta}{2}=\frac{\cos \frac{\beta}{2}}{\sin \frac{\beta}{2}}=\frac{\sqrt{\frac{1+\cos \beta}{2}}}{\sqrt{\frac{1-\cos \beta}{2}}}=\frac{\sqrt{1+\cos \beta}}{\sqrt{1-\cos \beta}} \\
& \cos \beta=\frac{8}{10}=\frac{4}{5}, \quad \operatorname{ctg} \frac{\beta}{2}=\frac{\sqrt{1+\frac{4}{5}}}{\sqrt{1-\frac{4}{5}}}=\sqrt{9}=3
\end{aligned}
$$

Thus, $\left|B O^{\prime}\right|=3$ and $\left|O^{\prime} P^{\prime}\right|=8-1-3=4$.

The similarity ratio of the observed triangles is $\frac{|O P|}{|B C|}=\frac{4}{8}=\frac{1}{2}$.

We conclude that the length of the path traveled by the center of the wheel until it returns to its initial position is equal to half the perimeter of triangle $A B C$.

$$
d=\frac{10+6+8}{2}=12 \mathrm{~m}
$$