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?


1 Answer 1


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):
120 degrees
So then we get (purple formula is $\sqrt{2^2+11^2-2\cdot2\cdot11\cdot\cos(120^\circ)}=\sqrt{147}$):
Cosine rule
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 $1\leq i\leq7,1\leq j<k\leq2i$ with $a(i,j)+a(i,k)\approx\frac{2\pi}{3}$ which gives only the found solution.

  • 1
    $\begingroup$ You can prove that $a^2+ab+b^2=3r^2$, where $a$ and $b$ are the two rod lengths. You can then show that $a$ and $b$ need to be coprime, and equal modulo $3$. That leaves only about $6$ sets of values for which you need to check they don't give a smaller $r$. $\endgroup$ Sep 3, 2018 at 11:04
  • $\begingroup$ To make the non-rigorous argument rigorous: swapping two adjacent edges leaves the vertices on the same circle by symmetry around a suitable diameter; and adjacent swaps generate the full symmetric group. $\endgroup$ Sep 3, 2018 at 13:17
  • $\begingroup$ @PeterTaylor but how do you address the part about not having two possible radii for one ordering? (In my opinion that’s the hard part to rigorise.) $\endgroup$
    – boboquack
    Sep 4, 2018 at 6:44
  • $\begingroup$ Consider the edge AB where A and B both lie on the circle. Let the centre of the circle be O and the midpoint of AB be M. Then |AM|=|BM| and |AO|=|BO|, so the triangles AMO and BMO are similar and OM is perpendicular to AB. Therefore each of the short edges subtends the same angle at O, each of the large edges subtends the same angle at O, so the combination of short edge and large edge subtends 120 degrees regardless of the overall order of the edges. Moreover, there must be some vertex where a short edge meets a large edge, at which you can apply your existing argument to fix the radius. $\endgroup$ Sep 4, 2018 at 10:19

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