## Task B-2.3.

The lengths of the sides of the hexagon $A B C D E F$ are in order $|A B|=b,|B C|=|C D|=a,|D E|=b$, $|E F|=|F A|=a, a \neq b$. A circle with radius 8 and center at point $O$ is circumscribed around the hexagon. If $\cos \varangle E F O=\frac{\sqrt{2}}{4}$, calculate the perimeter and area of this hexagon.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_8b541dbeede02a7e62f7g-07.jpg?height=677&width=740&top_left_y=415&top_left_x=658)

The following triangles are congruent:

$$
\begin{aligned}
& \triangle O A B \cong \triangle O E D \quad(\mathrm{~S}-\mathrm{S}-\mathrm{S}) \\
& \triangle O B C \cong \triangle O C D \cong \triangle O E F \cong \triangle O F A \quad \text { (S-S-S) }
\end{aligned}
$$

Therefore, we have the equality of the following angles: $\varangle D O E=\varangle A O B=x, \varangle B O C=\varangle C O D=$ $\varangle E O F=\varangle F O A=y$.

From this, it follows that $2 x+4 y=360^{\circ}$, that is, $x+2 y=180^{\circ}$. We conclude that points $C, O$, and $F$ are collinear, and that quadrilaterals $A B C F$ and $C D E F$ are inscribed isosceles trapezoids. Let $h=|E N|$ be the length of the height from $E$ to $C F$.

Then we have

$$
\cos \varangle E F O=\frac{\sqrt{2}}{4}=\frac{8-\frac{b}{2}}{a} \Rightarrow b=16-\frac{a \sqrt{2}}{2} .
$$

By applying the Pythagorean theorem to triangles $E N O$ and $E N F$, we get

$$
8^{2}-\left(\frac{b}{2}\right)^{2}=a^{2}-\left(8-\frac{b}{2}\right)^{2} \Rightarrow a^{2}+8 b-128=0
$$

The last equation together with (1) gives

$$
a^{2}+8\left(16-\frac{a \sqrt{2}}{2}\right)^{2}-128=0, \text { or } a^{2}-4 a \sqrt{2}=0
$$

The only possible solution is $a=4 \sqrt{2}$. Then $b=16-\frac{4 \sqrt{2} \sqrt{2}}{2}=12$. The height $h=\sqrt{r^{2}-6^{2}}=$ $\sqrt{28}=2 \sqrt{7}$.

The perimeter of the hexagon is, therefore, $o=4 a+2 b=16 \sqrt{2}+24$.

The area of the hexagon is twice the area of trapezoid $A B C F: P=2 \cdot \frac{12+16}{2} \cdot 2 \sqrt{7}=56 \sqrt{7}$.