logical deduction - Hexagon in a circle

Similar and Hint to: Dodecagon in a big circle

There are $6$ bars which comprise two groups of $3$, $3$: each group has identical bars but every group has a distinct length of bars.

For example;

• Group 1 may consist of three bars each of $10$ units length,


• Group 2 may consist of three bars each of $3$ units length.

By using these bars, you are forming a 6-sided convex polygon (Hexagon) by randomly putting bars next to each other.

Interestingly, you notice that all of the vertices of this Hexagon are on a circle with an integer-valued radius and more interestingly all bars have integer-valued lengths as well.

In this case,

What is the minimum value for the radius that satisfies the above condition?

Answer

Well I noticed that:

If you have 3 bars of length 2 and 3 bars of length 11 and place them alternatingly, they can form a circle of radius 7.

This is because:

Each group of two bars subtends 120 degrees, so we have the following diagram (arrows mean implication, click image for full diagram):

So then we get (purple formula is $\sqrt{2^2+11^2-2\cdot2\cdot11\cdot\cos(120^\circ)}=\sqrt{147}$):

And easy trigonometry gives us that the radius is $\frac{1}{\sqrt{3}}$ times this, i.e. $\sqrt{49}=7$

I have no proof this is minimal. See below for computer proof.

Actually, I do note that:

The order of the 6 pieces doesn't matter to the radius of the circle, and it is still possible to make a cyclic hexagon with any order.

since (non-rigorous argument):

Consider rearranging the arc lengths: clearly one can make a circle of the same size with any arrangement of the arcs. But treating the vertices of the n-gon like hinges and cutting a hole at one vertex, if we shrink the circle the end of the chain will slide too far to meet the start of the chain perfectly, and conversely if we enlarge the circle the end of the chain won't meet the start of the chain at all.

Proof of minimality by computer - sketch (I have run the program this specifies but it was in Python IDLE so I don't have access to it currently):

By the cosine rule, a secant of length $s$ in a circle with radius $r$ subtends an angle of $a(r,s)=\cos^{-1}(1-\frac{s^2}{2r^2})$. So we want integers $1\leq i\leq7,1\leq j