On a semicircle, two points are moving. What is the probability that at any given moment, the distance between the two points is not greater than the radius of the semicircle?
The sought probability depends on what quantities we use to characterize the position of the two randomly chosen points; the choice of different quantities corresponds to considering different events as equally probable.

I. solution: Let the radius of the semicircle be $r$ and denote the endpoints of the semicircular arc by $A$ and $B$ $(\widehat{A B}=r \pi)$, the two moving points by $C$ and $D$, and let $\widehat{A C}=x, \widehat{A D}=y$.

Let the random selection of points be such that we arbitrarily choose two values for $x$ and $y$ between 0 and $r \pi$.

In a Cartesian coordinate system, all possible values of $x$ and $y$ cover the square formed by the points $(0,0), (0, r \pi), (r \pi, r \pi)$, and $(r \pi, 0)$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_5d298705e0c5586f789cg-1.jpg?height=523&width=634&top_left_y=584&top_left_x=732)

Figure 1

Among these, only the points for which $|x-y| \leq \frac{r \pi}{3}$, or $x-\frac{r \pi}{3}<y<x+\frac{r \pi}{3}$, are favorable. (The shaded area in Figure 1.) Therefore, the sought probability is

$$
v_{1}=\frac{r^{2} \pi-\left(\frac{2 r \pi}{3}\right)^{2}}{r^{2} \pi^{2}}=1-\frac{4}{9}=\frac{5}{9} \approx 0.556
$$

Lajos Almási (Bp., II., Rákóczi g. III. o. t.)

Note: The present problem can be reformulated as the 1st example discussed on page 103 of Volume VI (April 1953).

$$
\text { Ágnes Roboz (Bp., VI., Varga Katalin lg. III. o. t.) }
$$

II. solution: Two points on the semicircular arc determine a chord, and each chord is uniquely determined by its midpoint, so we randomly select the midpoints of the chords, and favorable cases are those in which the chord is not longer than the radius.

First, we need to determine the geometric locus of all possible chord midpoints, i.e., the points inside the semicircle $\widehat{A B}$ for which the endpoints of the corresponding chords lie on the semicircular arc $\overrightarrow{A B}$. The geometric locus of the midpoints of chords starting from point $A$ or $B$ and lying inside the semicircle is the semicircles drawn over the radii $A O$ and $O B$. (The $1:2$ reduction of the semicircle $\widehat{A B}$ from the similarity centers $A$ and $B$.) Therefore, the points inside the region bounded by the semicircles $\widehat{A O}$, $\widehat{O B}$, and $\widehat{A B}$ represent all possible chord midpoints (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_5d298705e0c5586f789cg-1.jpg?height=437&width=721&top_left_y=2134&top_left_x=688)

The chords of length $r$ drawn in the semicircle $\widehat{A B}$ enclose an arc whose radius is clearly the distance of the chord midpoint from point $O$, i.e., (from the right triangle $A O C$) $\frac{r}{2} \sqrt{3}$. Therefore, the favorable cases for the chord midpoint are those points that lie within the annulus determined by the circles of radii $r$ and $\frac{r}{2} \sqrt{3}$ and the region bounded by the above three semicircles (the shaded area in Figure 2).

The "possible area" is thus

$$
T=\frac{r^{2} \pi}{2}-\left(\frac{r}{2}\right)^{2} \pi=\frac{r^{2} \pi}{4}
$$

In the notation of Figure 2,

$$
2 \cdot \mathrm{I}+2 \cdot \mathrm{II}+\mathrm{III}+t=\frac{r^{2} \pi}{2}
$$

The angle of the circular segment $O \widehat{C D} O$ is $180^{\circ}-2 \cdot 30^{\circ}=120^{\circ}$, and its radius is (as we have seen) $\frac{r}{2} \sqrt{3}$, and thus

$$
2 \cdot \mathrm{I}+\mathrm{III}=\frac{\left(\frac{r}{2} \sqrt{3}\right)^{2} \pi}{3}=\frac{r^{2} \pi}{4}
$$

and

$$
2 \cdot \mathrm{I}+2 \cdot \mathrm{II}=\frac{r^{2} \pi}{4}
$$

Subtracting the sum of (2) and (3) from (1),

$$
t-2 \cdot \mathbf{I}=0
$$

or

$$
t=2 \cdot \mathrm{I}=2 \cdot \frac{1}{3}\left(\frac{r^{2} \pi}{4}-\frac{3 r^{2} \sqrt{3}}{16}\right)=r^{2}\left(\frac{\pi}{6}-\frac{\sqrt{3}}{8}\right)
$$

and thus the sought probability is

$$
v_{2}=\frac{t}{T}=\frac{r^{2}\left(\frac{\pi}{6}-\frac{\sqrt{3}}{8}\right)}{r^{2} \frac{\pi}{4}}=\frac{4 \pi-3 \sqrt{3}}{6 \pi} \approx \frac{7.3702}{6 \pi} \approx 0.391
$$

Györgyi Lackner (Bp., V., Fonóip. techn. III. o. t.)