## Task B-4.6.

In an isosceles triangle with a base length of $14 \mathrm{~cm}$ and legs of length $25 \mathrm{~cm}$, a circle is inscribed. Above it, another circle is inscribed which touches both legs and the first circle, above this one a third circle is inscribed, and so on, continuing the process of inscribing circles. If all the inscribed circles are painted with a certain color, what part of the triangle's area will remain unpainted?
## Second Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8cadc17e7912dc7108beg-22.jpg?height=834&width=554&top_left_y=314&top_left_x=662)

Using the Pythagorean theorem, we calculate the length of the height to the base.

$$
v=|\overline{C P}|=\sqrt{25^{2}-7^{2}}=24 \mathrm{~cm}
$$

The triangles $A P C, S_{1} D_{1} C, S_{2} D_{2} C, S_{3} D_{3} C, \ldots$ are mutually similar (all angles are congruent). Let $\left|D_{1} S_{1}\right|=\left|P S_{1}\right|=r_{1}$. From the similarity of triangles $A P C$ and $S_{1} D_{1} C$, we get

$$
\begin{aligned}
& \frac{r_{1}}{24-r_{1}}=\frac{7}{25} \\
& r_{1}=\frac{21}{4} \mathrm{~cm}
\end{aligned}
$$

Let $\left|D_{2} S_{2}\right|=r_{2}$. From the similar triangles $A P C$ and $S_{2} D_{2} C$, we get

$$
\begin{aligned}
& \frac{r_{2}}{24-2 r_{2}-r_{2}}=\frac{7}{25} \\
& r_{2}=\frac{189}{64} \mathrm{~cm}
\end{aligned}
$$

Similarly, we obtain the radius of the third circle $r_{3}=\frac{1701}{1024} \mathrm{~cm}$.

The areas of the circles are $r_{1}^{2} \pi, r_{2}^{2} \pi, r_{3}^{2} \pi, \ldots$

They form a geometric sequence with the common ratio $q=\frac{r_{2}^{2}}{r_{1}^{2}}=\frac{r_{3}^{2}}{r_{3}^{2}}$.

Let's check that these ratios are equal:

$$
\begin{aligned}
& \frac{\left(\frac{189}{64}\right)^{2}}{\left(\frac{21}{4}\right)^{2}}=\left(\frac{189 \cdot 4}{21 \cdot 64}\right)^{2}=\left(\frac{9}{16}\right)^{2} \\
& \frac{\left(\frac{1701}{1024}\right)^{2}}{\left(\frac{189}{64}\right)^{2}}=\left(\frac{1701 \cdot 64}{1024 \cdot 189}\right)^{2}=\left(\frac{9}{16}\right)^{2}
\end{aligned}
$$

Calculate the difference between the area of the triangle and the area of all the circles.

$$
P=\frac{14 \cdot 24}{2}-\frac{\left(\frac{21}{4}\right)^{2} \pi}{1-\left(\frac{9}{16}\right)^{2}}=168-\frac{441 \pi}{16 \cdot\left(1-\left(\frac{9}{16}\right)^{2}\right)}=168-\frac{441 \pi}{16 \cdot \frac{175}{256}}=168-\frac{1008}{25} \pi \mathrm{cm}^{2} .2 \text { points }
$$