## Task A-4.2.

We say that a semicircle is properly placed within a larger semicircle if their diameters are parallel, the endpoints of the diameter of the smaller semicircle lie on the semicircle of the larger semicircle, and the semicircle of the smaller semicircle touches the diameter of the larger semicircle.

![](https://cdn.mathpix.com/cropped/2024_05_30_a630fb9dab34a5225850g-20.jpg?height=262&width=494&top_left_y=243&top_left_x=1226)

A sequence of semicircles $K_{1}, K_{2}, K_{3}, \ldots$ is given, such that for each $n \in \mathbb{N}$, the semicircle $K_{n+1}$ is properly placed within the semicircle $K_{n}$. The region that belongs to the semicircle $K_{n}$ and does not belong to the semicircle $K_{n+1}$ is colored blue if $n$ is odd, and yellow if $n$ is even.

The radius of the semicircle $K_{1}$ is 1. Determine the total area colored blue.
## First Solution.

Let the radius of the semicircle $K_{n}$ be $r_{n}$ and its area be $P_{n}$ for $n \in \mathbb{N}$.

Let $\overline{A B}$ be the diameter of the semicircle $K_{1}$, $\overline{C D}$ be the diameter of the semicircle $K_{2}$, and let $F$ be the midpoint of the segment $\overline{C D}$. Let the point $E$ be the point of tangency between the semicircle $K_{2}$ and the line $A B$.

Then $\overline{E F}$ is the radius of the semicircle $K_{2}$, and it is perpendicular to the tangent $A B$. Since $A B$ and $C D$ are parallel, it follows that the lines $C D$ and $E F$ are also perpendicular.

Since $F$ is the midpoint of the segment $\overline{C D}$, the line $E F$ is the perpendicular bisector of that segment. The midpoint of the segment $\overline{A B}$ is the center of the semicircle on which the points $C$ and $D$ lie, and it also lies on the line $E F$. We conclude that $F$ is precisely the midpoint of the segment $\overline{A B}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a630fb9dab34a5225850g-20.jpg?height=511&width=1008&top_left_y=1378&top_left_x=455)

As shown, triangle $EFC$ is an isosceles right triangle with a right angle at vertex $F$. Its legs are of length $|F E| = |F C| = r_{2}$ and the hypotenuse is of length $|E C| = r_{1}$. From the Pythagorean theorem for this triangle, we conclude that $r_{2} = \frac{1}{\sqrt{2}} r_{1}$.

Similarly, $r_{n+1} = \frac{1}{\sqrt{2}} r_{n}$, or

$$
r_{n} = \left(\frac{1}{\sqrt{2}}\right)^{n-1}, \quad \text{for every } n \in \mathbb{N}
$$

Next, we calculate that

$$
P_{n} = \frac{1}{2} \cdot r_{n}^{2} \cdot \pi = \frac{1}{2} \cdot \left(\frac{1}{2}\right)^{n-1} \cdot \pi = \frac{\pi}{2^{n}}, \quad \text{for every } n \in \mathbb{N}
$$

The area covered by the blue color can be calculated by first adding the area of the blue $P_{1}$, then subtracting the area of the yellow $P_{2}$, then adding the area of the blue $P_{3}$, then subtracting the area of the yellow $P_{4}$, and so on. Thus, we seek

$$
P = P_{1} - P_{2} + P_{3} - P_{4} + \cdots
$$

Substituting the expressions for $P_{n}$, we get

$$
\begin{aligned}
P & = \frac{\pi}{2} - \frac{\pi}{4} + \frac{\pi}{8} - \frac{\pi}{16} + \cdots \\
& = \frac{\pi}{2} \left(1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots\right) \\
& = \frac{\pi}{2} \cdot \frac{1}{1 + \frac{1}{2}} = \frac{\pi}{2} \cdot \frac{2}{3} = \frac{\pi}{3}
\end{aligned}
$$

In the second-to-last step, we used the formula for the sum of a geometric series:

$$
\sum_{n=0}^{+\infty} q^{n} = \frac{1}{1 - q}, \quad \text{for } q = -\frac{1}{2} \in (-1, 1)
$$