Lopsided Hexagon

Let the radius be $\sqrt{n}$ . Let the centre be $O$ and connect $OA, OB, OC, OD, OE, OF.$ We have $\angle AOB = \angle BOC = \angle COD = \alpha$ and $\angle AOF = \angle EOF = \angle DOE = \beta$ . Since the sum of the 6 angles are $360 ^\circ$ , thus $\alpha + \beta = 120 ^ \circ$ .

Now we consider the quadrilateral $CDEO$ . $\angle CDO = 90^\circ - \alpha/2$ , $\angle ODE = 90^\circ - \beta/2$ , so $\angle CDE = 180^\circ - (\alpha + \beta)/2 = 120^\circ$ .

Now we consider triangle CDE, using cosine rule, we get : $10^2 + 22^2 - 2\times 10\times 22 \times \cos120^\circ = CE^2$ , or $CE^2 = 804$ .

Now we consider triangle COE, using cosine rule again, we get : $n + n - 2n\times \cos120^\circ = CE^2$ , which simplifies to $3n = 804, n = 268$ And we are done.

[Latex edits - Calvin]

Let O be the center of the circle, and r the radius. Let G be the midpoint of AB. Let $\alpha$ be the angle AOB. ABO is an isosceles triangle with AO = BO = r and AB = 10. Then, the angle AGO is a right angle, $AOG = \alpha/2$ and by the definition of sine, we have $r\cdot sin(\alpha/2) = AG = 5$ .

The same holds for the triangles BOC and COD.

For DOE, EOF and FOA, where we let $\beta$ be the angle DOE, analog reasoning gives $r\cdot sin(\beta/2) = 11$ .

Since the sum of all angles around O is $360^\circ$ , we have $3\alpha+3\beta = 360^\circ$ , and thus $\alpha/2 = 60^\circ - \beta/2$ .

Using the formula for sine of a sum, $sin(x+y) = sin(x)cos(y)+cos(x)sin(y)$ , and that $cos(-x) = cos(x)$ and $sin(-x) = -sin(x)$ , we have $5 = r\cdot sin(\alpha/2) = r\cdot sin(60^\circ-\beta/2) =$ $= r(sin(60^\circ)cos(\beta/2)-cos(60^\circ)sin(\beta/2))$ . $sin(60^\circ)$ has the known value $\frac{\sqrt{3}}{2}$ , and $cos(60^\circ) = 1/2$ . Using that $cos(x) = \sqrt{1-sin^2(x)}$ and that $r\cdot sin(\beta/2) = 11$ from above, and rearranging, we get $5+ \frac{11}{2} = \frac{\sqrt{3}}{2} \sqrt{r^2 - 121}$ . Squaring and rearranging, we get $r^2 = 268$ . Since the problem text defined the radius as $\sqrt{n}$ , we get $n = 268$ .

10^2+22^2+10*22=804, 804/3=268

we can construct two isoscles triangles from O of bases 10 and 22

O is the center of the circle

from the given we can see that $3\theta+3\alpha$ =360

so $\theta+\alpha$ = 120 where $\alpha$ and $\theta$ are the vetrix angle of each isoscles triangle

$\Rightarrow$ $\theta$ =120- $\alpha$

therfore cos $\theta$ =cos (120- $\alpha$ ) $\Rightarrow$ $\boxed{1}$

by applying the cosine law to each triangle of base 10 and 22 we get :

$\Rightarrow$ cos $\theta$ = $\frac {2r^2-100}{2r^2}$ $\Rightarrow$ cos $\alpha$ = $\frac {2r^2-484}{2r^2}$ where r is the radius of the circle

by substituting in $\boxed{1}$ we get our answer of r= $\sqrt{268}$