3. Given triangle $A B C$. The lengths of its sides are known: $A B=B C=80, A C=96$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ddf1918985530a395e14g-02.jpg?height=720&width=634&top_left_y=651&top_left_x=797)

The circle $Q_{1}$ is inscribed in triangle $A B C$. The circle $Q_{2}$ is tangent to $Q_{1}$ and the sides $A B$ and $B C$. The circle $Q_{3}$ is tangent to $Q_{2}$ and also the sides $A B$ and $B C$. Find the radius of circle $Q_{3}$.
Answer: 1.5.

Solution. Let $r_{i}$ be the radius of the circle $Q_{i}(i=1,2,3)$. It is easy to find $r_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ddf1918985530a395e14g-02.jpg?height=756&width=640&top_left_y=1701&top_left_x=797)

Let $B D$ be the height. From triangle $A B D$, we find its length:

$$
B D=\sqrt{A B^{2}-A D^{2}}=\sqrt{80^{2}-48^{2}}=64
$$

Thus,

$$
r_{1}=\frac{S_{A B C}}{A B+A D}=24
$$

The height of triangle $A_{1} B C_{1}$, drawn from vertex $B$, is

$$
B D-2 r_{1}=64-48=16
$$

We draw the common internal tangents to the circles (see figure). Triangle $A_{1} B C_{1}$ is similar to triangle $A B C$. The similarity coefficient is the ratio of the heights: $k=\frac{16}{64}=\frac{1}{4}$. Therefore, $r_{2}=k r_{1}=6$. Triangle $A_{2} B C_{2}$ is similar to triangle $A_{1} B C_{1}$ with the same similarity coefficient $k$. Therefore, $r_{3}=k r_{2}=1.5$.

Evaluation. Full solution: 13 points. If only the radius of the 1st circle is found, 4 points.