Dodecagon in a big circle

Question

There are $12$ bars which comprise three groups of $6$, $3$ and $3$: each group has identical bars but every group has a distinct length of bars.

For example;

• Group 1 may consist of six bars each of $10$ units length,
• Group 2 may consist of three bars each of $3$ units length,
• Group 3 may consist of three bars each of $2$ units length.

By using these bars, you are forming a 12-sided convex polygon (Dodecagon) by randomly putting bars next to each other.

Interestingly, you notice that all of the vertices of this Dodecagon 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?

Hint: Hexagon in a circle try to solve this first.

Answer 1

Similarly to Hexagon in a circle I used a computer search to find the answer:

3 bars of lengths 23 and 49 and 6 bars of length 14 can be arranged around a circle of radius 49
(...well I'm pretty sure it works, the expressions from the below spoiler match up to the first 100 decimal places, but Wolfram|Alpha refuses to simplify the expression)

To prove minimality I used the same computerised technique as last time, copied here for reference:

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 $1\leq i\leq49,1\leq i,j,k\leq2i,i\neq j\neq k\neq i$ with $a(i,j)+a(i,k)+2a(i,l)\approx\frac{2\pi}{3}$ which gives only the found solution.

Note again (see previous answer for details):

The ordering of the bars is inconsequential, a circle of radius 49 can be formed at any time and this is the only radius that can be formed.