## Task 20/66

An ellipse with semi-axes $a$ and $b$ is to be reshaped into a circle of equal area.
The area of the ellipse is $a b \pi$, that of the desired circle is $r^{2} \pi$. Therefore, it holds that

$$
a b \pi=r^{2} \pi \quad \text { or } \quad r^{2}=a b ; \quad r=\sqrt{a b}
$$

The desired radius is thus equal to the geometric mean of the two semi-axes. The task, therefore, consists in transforming a rectangle with sides $a$ and $b$ into a square of equal area with side $r$. This is possible using the altitude theorem (or also using the theorem of Euclid). One marks off on a straight line the segments $A B=a$ and $B C=b$, draws a semicircle over $A C=a+b$, and erects a perpendicular at $B$. Its intersection with the semicircle is $D$.

The triangle $A C D$ is a right triangle at $D$ by Thales' theorem, and $B D$ is the altitude to the hypotenuse in this triangle. According to the altitude theorem, the altitude to the hypotenuse of a right triangle is the mean proportional between the two segments of the hypotenuse. Thus, it holds that

$$
B D^{2}=A B \cdot B C=a b \quad \text { or } \quad B D=\sqrt{a b}=r
$$

Thus, the problem is solved.