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.

  • $\begingroup$ When you write "by randomly putting", do you mean that for any random order of the bars the vertices can all fit on the circle? And also, can the bars intersect (as they do for example on a pentagram)? $\endgroup$
    – Penguino
    Sep 2, 2018 at 22:31
  • $\begingroup$ Just to clarify, do all bars within the same group have the same length? Sorry, the wording is confusing me a little bit. $\endgroup$
    – hexomino
    Sep 2, 2018 at 23:19
  • $\begingroup$ @hexomino yes they have $\endgroup$
    – Oray
    Sep 3, 2018 at 3:05
  • $\begingroup$ @Penguino order does not matter. nar cannot intersect or polygon is convex $\endgroup$
    – Oray
    Sep 3, 2018 at 3:07

1 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.


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