<s>[INST] Find circumradius of an octagon inscribable into a circle with side lengths $1,1,1,1,3\sqrt2,3\sqrt2,3\sqrt2$ and $3\sqrt2$ An octagon has side lengths $1,1,1,1,3\sqrt2,3\sqrt2,3\sqrt2$ and $3\sqrt2.$ What should be the length of its circumradius?
I tried solving it using elementary geometry, but that was of no use. I tried my hand at complex number geometry as well, but that didn't work. 
Can someone provide a formal answer (considering that I am just in my 11th grade)? Is there an elegant approach to this question using complex numbers?
 [/INST] The circumradius does not depend on the order of the sides.  To show this, draw the isosceles triangle from the center to the vertices.  The angle subtended by each of the $1$ sides is the same, as is the angle subtended by each of the $3\sqrt 2$ sides.  These angles must sum to $\frac \pi 2$.  Draw the octagon with the sides alternating.  
If $\theta$ is the angle subtended by a side of $1$ and $r$ is the radius of the circle we have $\sin \frac \theta 2 = \frac 1{2r}$.  The angle subtended by a side of $1$ and a side of $3\sqrt 2$ must add to $\frac \pi 2$ by symmetry, so the angle subtended by a side of $3\sqrt 2$ is $\frac \pi 2-\theta$.  That gives $\sin \left(\frac \pi 4-\frac \theta 2\right)=\frac 3{r\sqrt 2 }$
$$\sin \left(\frac \pi 4-\frac \theta 2\right)=\frac 3{r\sqrt 2 }\\
\sin \frac \pi 4 \cos \frac \theta 2-\cos \frac \pi 4 \sin \frac\theta 2 =\frac 3{r\sqrt 2 }\\
\frac {\sqrt 2}2\cos \frac \theta 2-\frac {\sqrt 2}{4r}=\frac 3{r \sqrt 2}\\
\cos \frac \theta 2-\frac 1{2r}=\frac 3r\\
\cos \frac \theta 2=\frac 7{2r}\\
\left(\frac 7{2r}\right)^2+\left(\frac 1{2r}\right)^2=1\\50=4r^2\\
r=\frac 5{\sqrt 2}$$
Done by hand, checked with Alpha (click exact form on the final result).
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