A circle is circumscribed around a 12-sided convex polygon, with 6 of its sides each being $\sqrt{2}$ units long, and the other sides each being $\sqrt{24}$ units long. What is the radius of the circle?
Let the central angle corresponding to the sides of length $\sqrt{2}$ be denoted by $\alpha$, and the one corresponding to the sides of length $\sqrt{24}$ by $\beta$. Since the polygon is convex, when traversing its perimeter in one direction, we circle the center of the circle exactly once, thus

$$
6 \alpha+6 \beta=360^{\circ}, \quad \alpha+\beta=60^{\circ}
$$

There must be a vertex of the polygon where a side of length $\sqrt{2}$ and a side of length $\sqrt{24}$ meet. Let one of these be denoted by $B$, and let the adjacent vertices be $A$ for the side of length $\sqrt{2}$, and $C$ for the side of length $\sqrt{24}$. The center of the circle is denoted by $O$. The sum of the angles at $A$, $B$, and $C$ in the quadrilateral $ABCO$ is $360^{\circ}-60^{\circ}=300^{\circ}$, and since the triangles $ABO$ and $BCO$ are isosceles,

$$
\angle OAB = \angle ABO, \quad \angle OBC = \angle BCO,
$$

thus the angle at $B$ in the quadrilateral is equal to the sum of the angles at $A$ and $C$:

$$
\angle ABC = 150^{\circ}.
$$

In general, let $AB = c$, $BC = a$, and $\angle ABC = 150^{\circ}$ in a triangle $ABC$. Denote the projection of $A$ onto the line $BC$ by $D$, and the reflection of $A$ across $D$ by $E$. The triangle $ABD$ is right-angled, and

$$
\angle ABD = 180^{\circ} - 150^{\circ} = 30^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_93863387c6fe0ba048fbg-1.jpg?height=596&width=1762&top_left_y=1025&top_left_x=168)

Therefore, $ABE$ is an equilateral triangle, and

$$
AD = c / 2, \quad BD = c \frac{\sqrt{3}}{2}
$$

In the right-angled triangle $ACD$, by the Pythagorean theorem,

$$
AC^2 = \left(a + c \frac{\sqrt{3}}{2}\right)^2 + \left(\frac{c}{2}\right)^2 = a^2 + c^2 + ac \sqrt{3}
$$

In our case, $a = \sqrt{24}$ and $c = \sqrt{2}$, so $AC^2 = 24 + 2 + \sqrt{24 \cdot 2 \cdot 3} = 38$, and $AC = \sqrt{38} = 6.164$.

Remark. We could have obtained (2) from (1) using the cosine rule as well. Our solution essentially described the proof of the theorem, taking advantage of the simplifications provided by the case $\angle ABC = 150^{\circ}$.