The broken wheel

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

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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.

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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.

Answer 4

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.