Given a rectangle inscribed in a circle, we extend it to an octagon, four of whose vertices coincide with the vertices of the rectangle, and the other four are the intersection points of the perpendicular bisectors of the rectangle's sides and the circle. How should we choose the rectangle so that the ratio of the area of the octagon to the area of the rectangle is minimized?
Let's choose the radius of the circle to be one unit, and let the lengths of the sides of the rectangle $B D F H$ be $2 a$ and $2 b (0 < a, b < 2)$. The area of the rectangle is then $4 a b$. If $O$ is the center of the circle, then the area of the octagon is the sum of the areas of the kites $O F G H$, $O H A B$, $O B C D$, and $O D E F$.

$1988-05-213-1$. eps

The area of a kite is half the product of its diagonals, so the areas of the above kites are respectively $\frac{1}{2} \cdot 2 b \cdot 1 ; \frac{1}{2} \cdot 2 a \cdot 1 ; \frac{1}{2} \cdot 2 b \cdot 1$; $\frac{1}{2} \cdot 2 a \cdot 1$, and thus the area of the octagon is $2 a + 2 b$. The ratio of the areas of the two figures is therefore

$$
\frac{2 a + 2 b}{4 a b} = \frac{a + b}{2 a b}
$$

Since $a$ and $b$ are positive, the expression $\frac{a + b}{2 a b}$ is minimized when its reciprocal, $\frac{2 a b}{a + b}$, is maximized. The latter is precisely the harmonic mean of the numbers $a$ and $b$, which cannot be greater than their quadratic mean, $\sqrt{\frac{a^2 + b^2}{2}}$, and is equal to it only when $a = b$. Since in our case, by the Pythagorean theorem, $a^2 + b^2 = 1^2 = 1$,

$$
\frac{2 a b}{a + b} \leqq \sqrt{\frac{a^2 + b^2}{2}} = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}
$$

Thus, $\frac{2 a b}{a + b}$ is maximized when $a = b$, and its value is then $\frac{\sqrt{2}}{2}$, so the ratio of the areas of the octagon and the rectangle is minimized when $a = b$, i.e., when the rectangle is a square (and the octagon is regular), and the minimum value is $\sqrt{2}$.

Remark. The reader can easily prove the inequality sequence for the harmonic, geometric, arithmetic, and quadratic means of positive numbers $a$ and $b$:

$$
\frac{2 a b}{a + b} \leqq \sqrt{a b} \leqq \frac{a + b}{2} \leqq \sqrt{\frac{a^2 + b^2}{2}}
$$

Equality holds in all cases only when $a = b$.