A round extendable table has a diameter of 1 meter. When extended, a $1 \mathrm{~m} \times 0.5 \mathrm{~m}$ rectangle fits between the two semicircular parts of the tabletop. Does the tabletop then have two points that are more than $150 \mathrm{~cm}$ apart?
Solution. When the table is in its extended state, let's complete the semicircles to form the circles $k$ and $\ell$, with centers $K$ and $L$.

The line $KL$ intersects the circles $k$ and $\ell$ at points $A$ and $B$ respectively. Enlarge the circle $k$ from point $A$ by a factor of $\frac{3}{2}$. The image of this transformation consists of points $X'$, which are obtained as follows: for any point $X$ on circle $k$, measure the distance $AX' = \frac{3}{2} AX$ on the ray $AX$. Under this transformation, the image of $K$ on the ray $AK$ is the point $K' = O$ such that $AO = \frac{3}{2} AK = \frac{3}{2} \cdot 0.5 = 0.75$; thus, $O$ is the midpoint of $AB$. By the converse of the parallel chords theorem, $X'O$ is parallel to $XK$, and $X'O = \frac{3}{2} XK = 0.75$. Therefore, $X'$ lies on the circle $n$ with center $O$ and radius 0.75. Since $X$ is an internal point of the segment $AX'$, every point $X$ on circle $k$ lies on the disk of circle $n$. Thus, the disk of circle $n$ contains the circle $k$. Similarly, it can be shown that enlarging circle $\ell$ from point $B$ by a factor of $\frac{3}{2}$ also results in circle $n$. Therefore, the disk of circle $n$ contains both circles $k$ and $\ell$; hence, the points $P, Q, R, S$ are also on this disk, so circle $n$ covers the extended table. Since any two points on the table are points on the disk of circle $n$, their distance is at most the diameter of circle $n$, which is $2 - 0.5 = 1.5 \, \text{m}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_32b82df25d05a12294b6g-1.jpg?height=461&width=501&top_left_y=778&top_left_x=801)

Remarks. 1. From the description in the solution, it can also be easily seen that the image of a circle under a central enlargement is a complete circle; however, this was not necessary for the solution.

2. However, we did use the fact that the disk of a circle is convex, i.e., if $T$ and $U$ are on the disk of a circle with center $O$ and radius $r$, then any point $Y$ on the segment $TU$ is also on the disk. In other words: if $OT \leq r$ and $OU \leq r$, then $OY \leq r$. To prove this, let $O_1$ be the foot of the perpendicular from $O$ to $TU$. One of the segments $O_1T$ and $O_1U$ contains $Y$; if, for example, it is $O_1T$, then by the Pythagorean theorem,

$$
OY = \sqrt{OO_1^2 + YO_1^2} \leq \sqrt{OO_1^2 + TO_1^2} = OT \leq r
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_32b82df25d05a12294b6g-1.jpg?height=214&width=331&top_left_y=1731&top_left_x=889)

3. Mátyás Aradi (Budapest, Németh László Gymnasium, 9th grade) made an interesting observation about the problem: If the rectangle is fitted to the semicircles as shown in the figure, then the distance between points $A$ and $B$ is $2 \, \text{m}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_32b82df25d05a12294b6g-1.jpg?height=211&width=442&top_left_y=2125&top_left_x=836)