## 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 uncolored?
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8cadc17e7912dc7108beg-21.jpg?height=862&width=576&top_left_y=300&top_left_x=680)

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

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

The radius of the largest inscribed circle is calculated from \( P=r_{1} s \),

$$
r_{1}=\frac{14 \cdot 24}{25+25+14}=\frac{21}{4}
$$

Consider the similar triangles \( A B C, A_{1} B_{1} C_{1}, A_{2} B_{2} C_{2} \ldots \)

The similarity coefficient of the triangles is calculated from the ratio of the heights

$$
k=\frac{24-2 r_{1}}{24}=\frac{9}{16}
$$

The same applies to the other triangles.

If the triangles are similar with a similarity coefficient \( k \), then the areas of the inscribed circles are proportional to \( k^{2}=\frac{81}{256} \).

We calculate the sum of the areas of all the circles

$$
P=P_{1}+k^{2} P_{1}+k^{4} P_{1}+\ldots=\frac{P_{1}}{1-k^{2}}=\frac{\frac{441}{16}}{\frac{1-81}{256}} \pi=\frac{1008}{25} \pi
$$

The unshaded area is

$$
\frac{14 \cdot 24}{2}-\frac{1008}{25} \pi=168-\frac{1008}{25} \pi \text{ cm}^{2}
$$