Solutions of Triangles

Consider half of the octagon with sides $3,3,4,4$ . Let the radius of the circumcircle be $r$ . Then the half octagon is formed by two $r$ - $3$ - $r$ isosceles triangles and two $r$ - $4$ - $r$ isosceles triangles. Let the angles of the $r$ - $3$ - $r$ and $r$ - $4$ - $r$ isosceles triangles at the center of the circumcircle be $\alpha$ and $\beta$ respectively. Then, we have:

$2\alpha + 2\beta = \pi \quad \Rightarrow \alpha + \beta = \frac {\pi}{2} \quad \Rightarrow \beta = \frac {\pi}{2} - \alpha$

By Cosine Rule:

$\begin {cases} \space\space 9 &= r^2 + r^2 - 2r^2\cos {\alpha} & = 2 r^2 (1 - \cos{\alpha} )\\ 16 &= 2 r^2 (1 - \cos{\beta} ) & = 2 r^2 (1 - \sin{\alpha} ) \end {cases}$

We note that $\cos{\beta} = \cos{(\frac {\pi}{2} - \alpha)} = \sin{\alpha}$ .

Dividing the two equations, we have:

$\dfrac {9}{16} = \dfrac {1 - \cos{\alpha} }{1 - \sin{\alpha} } \quad \Rightarrow 9(1 - \sin{\alpha}) = 16(1 - \cos{\alpha})$

$\Rightarrow - 9\sin{\alpha} = 7 - 16\cos{\alpha}$

Squaring both sides:

$81\sin^2{\alpha} = 49 - 224\cos{\alpha} + 256\cos^2{\alpha}$

$\Rightarrow 81(1- \cos^2{\alpha} ) = 49 - 224\cos{\alpha} + 256\cos^2{\alpha}$

$\Rightarrow 337\cos^2{\alpha} - 224\cos{\alpha} + 49 = 0$

$\Rightarrow \cos{\alpha} = 0.78556399$ or $-0.120875563$

As $\alpha < \frac {\pi}{2} \quad \Rightarrow \cos{\alpha} = 0.78556399$

Substituting $\cos{\alpha} = 0.78556399$ in $9=2 r^2 (1 - \cos{\alpha} )$ , we get $r = \boxed {4.581}$ .

We can rearrange the side lengths such that $AB = CD = EF = GH = 3$ and $BC = DE = FG = HA = 4$ without changing the radius of the circumcircle.

Let $O$ be the center of the circumcircle and $R$ denote the length of its radius.

Since there is $4$ -fold rotational symmetry about $O$ , we have $\angle AOC = 90^{\circ}$ .

Using the Pythagorean Theorem on $\Delta OAC$, we get $AC = R\sqrt{2}$ .

Since $\angle ABC$ cuts the circumcircle into arcs of $90^{\circ}$ and $270^{\circ}$ , $\angle ABC = 135^{\circ}$ .

Now, apply the Law of Cosines to $\Delta ABC$ to get:

$AC^2 = AB^2+BC^2-2 \cdot AB \cdot BC \cdot \cos \angle ABC$

$2R^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cdot \cos 135^{\circ}$

$R = \sqrt{\dfrac{25+12\sqrt{2}}{2}} \approx 4.581$

The different sides of the quadrilateral act as chords of the circle. We already know that the chords of same length subtend the same angle at the center of the circle. Let the chord of length 3 subtend an angle x, and the chord of length 4 subtend an angle y. Since sum of angles subtended by all 8 chords will be 360, we instantly have x+y=90 degrees. Let the radius of the circle be r. now, 2 r Sin(0.5 x)=3 and 2 r Sin(0.5 y)=4. substitute y=90 degree - x in the above equation. Compute Cos(x/2) using Sin(x/2). We obtain a quadratic equation in r whose solution is r = sqrt((25+12sqrt(2))/2)=4.58