# The broken wheel

In a regular polygon, we can connect six of the vertices to form a convex hexagon.

I construct a hexagon by picking six vertices out of a regular $$n$$-sided polygon. In this hexagon, if you connect the $$3$$ pairs of opposite vertices, the resulting lines meet at a common point. Furthermore, none of the hexagon's sides are of the same length.

What’s the minimal $$n$$ side lengths such that I can construct this hexagon?

• Corresponds means that they all match with a regular polygon. – Display maths Jul 6 '20 at 21:42
• I didn't understand the question at first but I think I do now. I wonder if it would be better to introduce the regular polygon first, and then consider the hexagon as formed by a subset of the vertices? I think this may be a cleaner presentation and you wouldn't need to specify that the hexagon is cyclic because it is a corollary. – hexomino Jul 6 '20 at 23:31
• I think I have the answer - would be great if I could actually answer the question :) – subrunner Jul 8 '20 at 13:11
• Can you guys please open the question. I edited it. – Display maths Jul 8 '20 at 22:04
• Reopened. @subrunner post away :-) – Rand al'Thor Jul 9 '20 at 7:33

First let's establish some trigonometric identities

$$\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\,\,,\,\, \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}\,\,, \,\,\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$, $$\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\,\,, \,\,\cos\left(\frac{\pi}{12}\right) = \frac{1+\sqrt{3}}{2\sqrt{2}}\,\,, \,\,\cos\left(\frac{5\pi}{12}\right) = \frac{\sqrt{3}-1}{2\sqrt{2}}$$ From this we find that $$\left(1-\cos\left(\frac{3\pi}{4}\right)\right)\left(1-\cos\left(\frac{\pi}{12}\right)\right)\left(1-\cos\left(\frac{\pi}{3}\right)\right) = \left(\frac{\sqrt{2}+1}{\sqrt{2}}\right)\left(\frac{2\sqrt{2}-1-\sqrt{3}}{2\sqrt{2}}\right)\left(\frac{1}{2}\right)$$ $$=\frac{4-\sqrt{2}-\sqrt{6}+2\sqrt{2}-1-\sqrt{3}}{8} = \frac{3+\sqrt{2}-\sqrt{3}-\sqrt{6}}{8}$$ Also $$\left(1-\cos\left(\frac{\pi}{4}\right)\right)\left(1-\cos\left(\frac{5\pi}{12}\right)\right)\left(1-\cos\left(\frac{\pi}{6}\right)\right) = \left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)\left(\frac{2\sqrt{2}+1-\sqrt{3}}{2\sqrt{2}}\right)\left(\frac{2-\sqrt{3}}{2}\right)$$ $$= \frac{3-\sqrt{2}+\sqrt{3}-\sqrt{6}}{8}\left(2-\sqrt{3}\right) = \frac{6-2\sqrt{2}+2\sqrt{3}-2\sqrt{6}-3\sqrt{3}+\sqrt{6}-3+3\sqrt{2}}{8}$$ $$= \frac{3+\sqrt{2}-\sqrt{3}-\sqrt{6}}{8}$$ Altogether, this means that $$\left(1-\cos\left(\frac{3\pi}{4}\right)\right)\left(1-\cos\left(\frac{\pi}{12}\right)\right)\left(1-\cos\left(\frac{\pi}{3}\right)\right) = \left(1-\cos\left(\frac{\pi}{4}\right)\right)\left(1-\cos\left(\frac{5\pi}{12}\right)\right)\left(1-\cos\left(\frac{\pi}{6}\right)\right)$$ Multiplying across by $$8$$ and then taking the square root of both sides this may be rewritten as $$\sqrt{2-2\cos\left(9.\frac{2\pi}{24}\right)}\sqrt{2-2\cos\left(\frac{2\pi}{24}\right)}\sqrt{2-2\cos\left(4.\frac{2\pi}{24}\right)} = \sqrt{2-2\cos\left(3.\frac{2\pi}{24}\right)}\sqrt{2-2\cos\left(5.\frac{2\pi}{24}\right)}\sqrt{2-2\cos\left(2.\frac{2\pi}{24}\right)}$$

Why is this relevant?

As stated in Helen's answer, the lengths of the edges of a cyclic polygon correspond to the angles between vertices and the circumcentre. In particular, if this angle is $$\theta$$ and the circumradius is $$1$$, the length of the corresponding edge is $$\sqrt{2-2\cos \theta}$$.

Furthermore, if the vertices coincide with the vertices of a regular polygon with $$n$$ sides and there are $$m$$ edges between the vertices then the angle at the circumcentre is $$m.\frac{2\pi}{n}$$.

Lastly, if $$ABCDEF$$ is a cyclic hexagon, then its three main diagonals are concurrent iff $$|AB|.|CD|.|EF| = |BC|.|DE|.|FA|$$ A nice proof of this fact is provided on Maths Stack Exchange here: https://math.stackexchange.com/a/360120/314970

Putting all of that together means that

If we consider a regular icositetragon ($$24$$-gon) and join the vertices such that the number of edges between $$AB$$, $$BC$$, $$CD$$, $$DE$$, $$EF$$, $$FA$$ is $$9, 3, 1, 5, 4, 2$$ respectively, then the main diagonals of the hexagon $$ABCDEF$$ will be concurrent. Wikipedia has a nice picture of an icositetragon which I've played around with to check this fact:

Conclusion

Thus we have shown that the minimal $$n \leq 24$$. Helen has convincingly argued that the minimal $$n \geq 21$$. Furthermore, we can rule out the cases $$21$$ and $$23$$ because for any regular polygon with an odd number of sides no three diagonals are concurrent (considering the full set of diagonals here). This fact is noted here which they cite as a result originally due to a mathematician called Heineken.

This means that the only case to rule out is $$n=22$$ (an icosidigon). Wikipedia also has a nice picture of this shape with which we can play around. It is easy to determine that the number of edges between consecutive vertices of our cyclic hexagon must be $$1,2,3,4,5,7$$ in some order so there isn't a lot to play around with here. The closest you can seem to get is the following:

But these lines can be found not to be concurrent using the previous formula.
Overall, it looks like $$n=24$$ is the minimum possible.

• Fixing the circumradius of the shape at 1 is such a clever idea, I tried to have a look at edge lengths as well but without fixing the radius I got massive expressions that I had no clue how to compress. Very nice answer! +1 from me – Helen Jul 17 '20 at 19:55

Answer (if "hexagon with different side lengths" means "at least 1 side has a different length" and not "no side has the same length as another one"):

8 (regular octagon)

Reason:

Octagon works because it is symmetric in regards to the octagon's diagonals (meaning: the octagon's diagonals all meet in 1 spot). If you inscribe the hexagon by leaving out 2 opposing corners, the diagonals of the hexagon are the octagon diagonals and thus meet in 1 spot:

Why not Heptagon? The only way to inscribe a hexagon in a heptagon is to leave out 1 corner. The 1-4, 2-5, 3-6 lines are the opposing diagonals of the hexagon. However, the heptagon requires that those lines never meet in 1 spot (due to the rotational symmetry of the heptagon and the lines in regards to the heptagon center, they can only meet in 1 spot if that spot is the heptagon center. However, the heptagon center does not lie on any of the diagonals.). Since they cannot meet in 1 spot, a heptagon is not a solution.

• I assume the different lengths means all 6 sides must be different. But OP has not clarified that one, so have a +1! – justhalf Jul 9 '20 at 9:50
• The hexagon has different side lengths though. – Display maths Jul 9 '20 at 23:41
• It means that all six sides must be different. – Display maths Jul 10 '20 at 1:04

Very low non-spoiler text content as most of the text in the answer gives hints to the possible solution. Spoilers are grouped so that opening one spoiler gives another "step" towards the solution.

Lemma: The lengths of the edges of a cyclic polygon correspond to the angles between vertices and the circumcentre (which I'll call circumangles for lack of a better name).

This is because the angles directly determine the rotational distance of points around the circumference, which directly determines the positional distance of vertices in the polygon.

Creating a regular $$n$$-sided polygon means that

All circumangles must be $$2\pi\over n$$.
This means in order to construct a hexagon from a bigger polygon, we must choose $$6$$ unique (positive) multiples of the circumangle, which is the same as picking $$6$$ unique natural numbers.

In order to get the lowest possible $$n$$,

We must minimize the sum of the natural numbers we pick.
Trivially, this means that the numbers we pick are $$1,2,3,4,5,6$$.
Therefore, the smallest possible polygon encompassing the special case of hexagon must be a $$21$$-gon.

Note: This answer is a completely new version describing the conditions necessary to create a hexagon like the one described in the question (as well as proving the minimum number of necessary edges). The old answer was a proof that it was impossible, but it failed to consider if the point of concurrence was not the circumcentre of the hexagon.

The old answer has been removed from the post, as it was wrong and would make this post very long, but can be seen in the edit history.

Major credit to hexomino who provided a very intuitive counterexample in the comments to the original answer.

• Yes the main argument is based on that premise but it seemed to me that it would be impossible to create a hexagon with 6 different side lengths and concurrent diagonals. – Helen Jul 11 '20 at 17:51
• See this example: i.stack.imgur.com/RCcfj.png I couldn't find an example online so I created it myself. I think it should be fairly clear that all side lengths are different and even how to put them in order of size. The construction here works by creating the diagonals first. – hexomino Jul 11 '20 at 18:06
• Ohhh of course they can all intersect somewhere else, thank you for that example – Helen Jul 11 '20 at 18:23
• The actual answer is pretty close to 21 sides. – Display maths Jul 11 '20 at 23:00
• Somehow I completely forgot the concurrent diagonals, I'll be back tomorrow with another (hopefully the final) update – Helen Jul 12 '20 at 0:25

Let's draw a circle with radius R=5 units and mark on the circumference the six vertices of a regular hexagon. Let's draw a chord AB with length equal to R. From the point B measure a chord equal to two-fifths of R, cutting the circle at point C, and draw the straight line BC. From point A measure a chord equal to three-fifths of R, cutting the circle at point F, and draw the line AF. From the point C draw a chord CD, where D is the fourth vertex of the hexagon. Draw the diagonals AD and CF. From point B draw a straight line BE which passes through the intersection of the two diagonals and cuts the circle at point E. Draw the chords EF and DE. The hexagon ABCDEF is the one you seek. All lengths have been measured precisely and all are different from each other.

• The hexagon must have different side lengths. – Display maths Jul 10 '20 at 22:29
• In your answer, there are four sides that are congruent to each other and two sides that are congruent to each other. All 6 sides must be different to each other. – Display maths Jul 10 '20 at 22:32
• Τhis was not clear from the way you posted the question. Are you now saying that each side of the hexagon has to have a different length? – Vassilis Parassidis Jul 11 '20 at 0:31
• Yes thats exactly what I mean – Display maths Jul 11 '20 at 1:12