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?
 A: 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).
A: You can arrange the sides arbitrarily. Therefore let the $1$s and $3\sqrt{2}$s follow alternatively. This means that you have a quarter circle with two chords of length $1$ and $3\sqrt{2}$. If $\alpha$ and $\beta$ are the half angles belonging to these chords you have
$$r\sin\alpha={1\over2},\qquad r\sin\beta={3\over\sqrt{2}}\ .$$
From  $\alpha+\beta={\pi\over4}$ you get 
$$\sin\alpha={1\over\sqrt{2}}(\cos\beta-\sin\beta)\ .$$
These facts should allow you to compute $r$.
A: Consider an octagon, four of whose sides have length $a$ and four of whose sides have length $3a\sqrt2,$ and suppose it is inscribed in a circle whose diameter is $1.$ When we have found $a,$ we will then divide all lengths by $a$ to conclude that the diameter that you seek is $1/a.$
First I will work more generally, taking the sides of an octagon inscribed in a circle of diameter $1$ to be $a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8.$
Each side of length $a_i$ for $i=1,\ldots,8$ divide the circle into two arcs: one that joins the endpoints $p,q$ of that side without passing through the endpoints of any of the other sides, and the other arc. Let $v$ by any point on that other arc. Let $\alpha_i$ be the measure of the angle $\angle pvq.$ A theorem of elementary geometry says this angle is the same regardless of which point is chosen as $v.$ Then trigonometry tells us that $\sin\alpha_i = a_i$ so that $\cos\alpha_i= \pm \sqrt{1-a_i^2},$ and we must have $\text{“}+\text{''}$ rather than $\text{“}-\text{''}$ in this case because with four "long" sides, none of the sides can have the corresponding angle larger than a right angle.
Thus in four cases we have $\cos\alpha_i= \sqrt{1-a^2}$ and in the other four cases $\cos\alpha_i = \sqrt{1-18a^2}.$
\begin{align}
& a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 + a_6^2 + a_7^2 + a_8^2 \\[10pt]
= {} & \phantom{{}+{}} 2\big(a_1 a_2\big) \big(\cos\alpha_3\cos\alpha_4 \cos\alpha_5\cos\alpha_6\cos\alpha_7\cos\alpha_8\big) \\
& {} + \text{$27$ other terms since $\tbinom 8 2 = 28$} \\[10pt]
& {} - 4\big(a_1 a_2 a_3 a_4\big) \big( \cos\alpha_5\cos\alpha_6\cos\alpha_7\cos\alpha_8\big) \\
& {}\qquad \text{similarly followed by $69$ other terms since $\tbinom 8 4 = 70$)} \\[10pt]
& {} + 6\big( a_1 a_2 a_3 a_4 a_5 a_6 \big) \big( \cos\alpha_7 \cos\alpha_8 \big) \\
& {} \qquad \text{(and then $27$ other such terms)} \\[10pt]
& {} - 8\big( a_1 a_2 a_3 a_4 a_5 a_6 a_7 a_8\big) \\
& {} \qquad \text{(just one term here since $\tbinom 8 8 =1$)}.
\end{align}
Among the $28$ terms with coefficient $2,$ there are


*

*$6$ terms with $a_i a_j=a^2,$

*$16$ with $a_i a_j = 3a^2\sqrt2,$ and

*$6$ with $a_i a_j = 18a^2.$
And so on.
Pursue this to an algebraic equation in satisfied by $a.$
Doubtless this method is far less efficient than some others, so I will skip some details. It's the first thing that comes to mind for me only because I have worked with trigonometric identities of this kind.
How do we prove this identity? Use the law of sines and the law of cosines. And this is among the aforementioned parts of the argument in which I will skip details.
A: 
Given the octagon, we have $\angle AOC = 90^\circ$, which leads to $\alpha+ \beta = 135^\circ$. Apply the cosine rule to the triangle $ABC$,
$$AC = \sqrt2 r = \sqrt{1^2 + (3\sqrt2)^2 - 2\cdot 1 (3\sqrt2)\cos135^\circ}=5$$
which yields,
$$r = \frac5{\sqrt2}$$
