For what values of $n$ does there exist an $n$-gon inscribed in a unit circle, with rational side lengths and irrational diagonal lengths?

Question

For what values of $n$ does there exist an $n$-gon inscribed in a unit circle, such that all the side lengths are rational, and all the diagonal lengths are irrational?

My thoughts

$n$ cannot be $3$, because a triangle has no diagonals.

$n$ can be $4$. Assuming the unit circle is $x^2+y^2=1$, the vertices could be $(1,0) \left(\frac{1}{2},\frac{\sqrt3}{2}\right) \left(-\frac{1}{2},\frac{\sqrt3}{2}\right) (-1,0)$, so the side lengths are $1,1,1,2$ and the diagonal lengths are $\sqrt3, \sqrt3$.

For $n=5$, we can inscribe in the unit circle an $n$-gon with rational side lengths (in fact, it is possible to do this for any $n\ge 3$). For example, the vertices could be $(1,0) \left(\frac{1}{2},\frac{\sqrt3}{2}\right) \left(-\frac{1}{2},\frac{\sqrt3}{2}\right) (-1,0) \left(\frac{7}{25},-\frac{24}{25}\right)$, so the side lengths are $1,1,1,\frac85,\frac65$. However, in this example the length of the diagonal between $(1,0)$ and $(-1,0)$ is rational.

Here is a visualisation of the case $n=5$. The dashed curve is a unit circle. Can the black line segments all have rational lengths while the colored line segments all have irrational lengths?

For all $n$, we can rule out the regular $n$-gon. According to Niven's Theorem, the only regular $n$-gon inscribed in the unit circle with rational side lengths is the regular $6$-gon, but in the regular $6$-gon some of the diagonal lengths are rational.

Context

This is just another question I've thought of, related to circles and rational numbers. Other questions are here and here.

Answer 1

I do not claim to be as good as Gauss, only as good as Eisenstein (for this problem), whose triples can be used to identify infinitely many solutions for $n=4$. However, this does not work for other values of $n$ and the answer is unknown even for $n=5$.

Suppose $a,b,c$ are nonnegative integers sorted according to $a<b\le c$ and satisfying $a^2+ab+b^2=c^2$. Then the reader can verify the additional relations

$(b-a)^2+(b+2a)(b-a)+(b+2a)^2=3c^2$

$(b-a)^2-(2b+a)(b-a)+(2b+a)^2=3c^2$

$(b+2a)^2-(2b+a)(b+2a)+(2b+a)^2=3c^2$

We may then define three rational quantities

$s_1=(b-a)/c, s_2=(b+2a)/c, s_3=(2b+a)/c$

and then three triangles:

• Triangle I with sides $s_1,s_2,\sqrt3$ with a $120°$ angle between the $s_1$ and $s_2$ sides

• Triangle II with sides $s_1,s_3,\sqrt3$ with a $60°$ angle between the $s_1$ and $s_3$ sides

• Triangle III with sides $s_2,s_3,\sqrt3$ with a $60°$ angle between the $s_2$ and $s_3$ sides

Given the included $60°$ or $120°$ angle opposite the $\sqrt3$ side, each triangle has unit circumradius and thus we form a quadrilateral by simply combining Triangle I with Triangle II or III along the $\sqrt3$ side (which then becomes a diagonal of the resulting quadrilateral. If Triangles I and II are combined, the result is an isosceles trapezoid with sides $s_1,s_2,s_1,s_3$ in rotational order, and both diagonals measuring $\sqrt3$. If Triangles I and III are combined, the result is an isosceles trapezoid with sides $s_2,s_1,s_2,s_3$ in rotational order, and both diagonals measuring $\sqrt3$.

For instance, we may select $(a,b,c)=(0,1,1)$. Then $s_1=s_2=1$ and we get the trapezoid with sides $1,1,1,2$ as given by the OP. If we choose the first Eisenstein triple with distinct nonzeto elements $(a,b,c)=(3,5,7)$, then we get $s_1=2/7,s_2=11/7,s_3=13/7$ giving a trapezoid with sides $2/7,11/7,2/7,13/7$ or $11/7,2/7,11/7,13/7$. In either case, the quadrilateral will remain inscribed with irrational diagonals if a pair of sides is flipped.

An attempt to extend this approach to cover $n=6$ resulted in rational diagonals and had to be abandoned.