In a regular n-gon with $n>4$, side length $1$ and all diagonals drawn, can any cell have rational area?

Asked Sep 18 '22 at 07:10

Active Sep 18 '22 at 07:18

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In a regular n-gon with $n>4$, side length $1$ and all diagonals drawn, can any cell have rational area? (By "cell", I mean one of the sub-polygons, with no diagonal cutting through it.)

Here is a geogebra applet where you can see a regular polygon with $n$ sides ($3\leq n\leq50$) and all diagonals drawn.

My attempt

I know that the angles between intersecting diagonals of the n-gon are rational multiples of $\pi$, and I know that the sine or cosine of such angles are algebraic. I have tried to divide the cells into triangles (by drawing line segments between vertices) and expressing the area of the triangles as $\frac{1}{2}ab\sin{\theta}$. But I do not know how to make this approach work.

(Context: this question is a follow-up to another question about the distribution of the areas in a regular polygon with diagonals drawn.)

edited Sep 18 '22 at 07:18

asked Sep 18 '22 at 07:10

Dan

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The trig functions of rational multiples of $\pi$ are algebraic, not transcendental. E.g., $\sin(\pi/4)=\sqrt2/2$, $\sin(\pi/6)=1/2$. – Gerry Myerson Sep 18 '22 at 07:15
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@GerryMyerson Oops, I meant algebraic. Corrected. – Dan Sep 18 '22 at 07:20

In a regular *n*-gon with $n>4$, side length $1$ and all diagonals drawn, can any cell have rational area?

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In a regular n-gon with $n>4$, side length $1$ and all diagonals drawn, can any cell have rational area?