Wikipedia:Reference desk/Archives/Mathematics/2022 May 15

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May 15

Areas of the unit regular n-gons.

Let the set S defined where s(n) = the area of the regular n+2 gon of side 1. (So s(1) is the area of the regular triangle of side 1, s(2) is the area of the square of side 1, etc.) are s(1) and s(4) (regular triangle and hexagon) the only entries whose ratio is a whole number? are s(1) and s(4) the only entries whose ratio is a rational number?Naraht (talk) 01:09, 15 May 2022 (UTC)

Not a solution, but a simpler, equivalent formulation. Define, for integer $n\geq 3,$ $h_{n}=\tan {\tfrac {\pi }{n}}.$ Then $h_{3}:h_{6}=$ ${\sqrt {3}}:{\tfrac {1}{3}}{\sqrt {3}}=$ $3:1.$ Are there any other values of $p,q,3\leq p<q,$ for which the ratio $h_{p}:h_{q}$ is rational?  --Lambiam 06:38, 15 May 2022 (UTC)

Lambian. Not sure why that is equivalent. The triangle and hexagon are in a ration of 1:6, and the tan values are in a ratio of 1:3.Naraht (talk) 19:26, 16 May 2022 (UTC)

Denoting the area of the regular $n$-gon with unit side length as $A_{n}$ – so your $s(i)=A_{i+2}$ – the following relation between $A_{n}$ and $h_{n}$ holds: $A_{n}={\tfrac {1}{4}}nh_{n}^{{-}1}.$ So $A_{p}:A_{q}=ph_{q}:qh_{p}.$ If one ratio is rational, so is the other.  --Lambiam 20:16, 16 May 2022 (UTC)

I think finding if ${\frac {h_{k_{1}}}{h_{k_{2}}}}={\frac {tan({\frac {\pi }{k_{1}}})}{tan({\frac {\pi }{k_{2}}})}}\in \mathbb {Q}$ can be shown to be equivalent to finding if ${\frac {sin({\frac {\pi }{k_{1}}}+{\frac {\pi }{k_{2}}})}{sin({\frac {\pi }{k_{1}}}-{\frac {\pi }{k_{2}}})}}\in \mathbb {Q}$ since:

${\frac {tan({\frac {\pi }{k_{1}}})}{tan({\frac {\pi }{k_{2}}})}}={\frac {sin({\frac {\pi }{k_{1}}})cos({\frac {\pi }{k_{2}}})}{sin({\frac {\pi }{k_{2}}})cos({\frac {\pi }{k_{1}}})}}={\frac {(e^{\frac {i\pi }{k_{1}}}-e^{-{\frac {i\pi }{k_{1}}}})(e^{\frac {i\pi }{k_{2}}}+e^{-{\frac {i\pi }{k_{2}}}})}{(e^{\frac {i\pi }{k_{2}}}-e^{-{\frac {i\pi }{k_{2}}}})(e^{\frac {i\pi }{k_{1}}}+e^{\frac {i\pi }{k_{1}}})}}={\frac {e^{{\frac {i\pi }{k_{1}}}+{\frac {i\pi }{k_{2}}}}+e^{{\frac {i\pi }{k_{1}}}-{\frac {i\pi }{k_{2}}}}-e^{-{\frac {i\pi }{k_{1}}}+{\frac {i\pi }{k_{2}}}}-e^{-{\frac {i\pi }{k_{1}}}-{\frac {i\pi }{k_{2}}}}}{e^{{\frac {i\pi }{k_{1}}}+{\frac {i\pi }{k_{2}}}}-e^{{\frac {i\pi }{k_{1}}}-{\frac {i\pi }{k_{2}}}}+e^{-{\frac {i\pi }{k_{1}}}+{\frac {i\pi }{k_{2}}}}-e^{-{\frac {i\pi }{k_{1}}}-{\frac {i\pi }{k_{2}}}}}}\in \mathbb {Q}$ if and only if ${\frac {2e^{{\frac {i\pi }{k_{1}}}-{\frac {i\pi }{k_{2}}}}-2e^{-{\frac {i\pi }{k_{1}}}+{\frac {i\pi }{k_{2}}}}}{e^{{\frac {i\pi }{k_{1}}}+{\frac {i\pi }{k_{2}}}}-e^{{\frac {i\pi }{k_{1}}}-{\frac {i\pi }{k_{2}}}}+e^{-{\frac {i\pi }{k_{1}}}+{\frac {i\pi }{k_{2}}}}-e^{-{\frac {i\pi }{k_{1}}}-{\frac {i\pi }{k_{2}}}}}}\in \mathbb {Q}$ if and only if (assuming that $k_{1}\neq k_{2}$ and thus the numerator is nonzero) ${\frac {e^{{\frac {i\pi }{k_{1}}}+{\frac {i\pi }{k_{2}}}}-e^{{\frac {i\pi }{k_{1}}}-{\frac {i\pi }{k_{2}}}}+e^{-{\frac {i\pi }{k_{1}}}+{\frac {i\pi }{k_{2}}}}-e^{-{\frac {i\pi }{k_{1}}}-{\frac {i\pi }{k_{2}}}}}{2e^{{\frac {i\pi }{k_{1}}}-{\frac {i\pi }{k_{2}}}}-2e^{-{\frac {i\pi }{k_{1}}}+{\frac {i\pi }{k_{2}}}}}}\in \mathbb {Q}$ if and only if ${\frac {2e^{{\frac {i\pi }{k_{1}}}+{\frac {i\pi }{k_{2}}}}-2e^{-{\frac {i\pi }{k_{1}}}-{\frac {i\pi }{k_{2}}}}}{2e^{{\frac {i\pi }{k_{1}}}-{\frac {i\pi }{k_{2}}}}-2e^{-{\frac {i\pi }{k_{1}}}+{\frac {i\pi }{k_{2}}}}}}={\frac {sin({\frac {\pi }{k_{1}}}+{\frac {\pi }{k_{2}}})}{sin({\frac {\pi }{k_{1}}}-{\frac {\pi }{k_{2}}})}}\in \mathbb {Q}$.

Similarly, if we have two rational numbers $a,b\in \mathbb {Q}$ for which we know ${\frac {sin(a\pi )}{sin(b\pi )}}\in \mathbb {Q}$, then we can obtain $k_{1}$ and $k_{2}$ (which may or may not be integers) through $k_{1}={\frac {2}{a+b}}$ and $k_{2}={\frac {2}{a-b}}$. So if we can find all rational numbers $a,b\in \mathbb {Q}$ for which the ratio ${\frac {sin(a\pi )}{sin(b\pi )}}\in \mathbb {Q}$, then finding all such $a,b$ for which the ratios ${\frac {2}{a+b}},{\frac {2}{a-b}}$ are integers greater than or equal to $3$ should solve the problem.

GalacticShoe (talk) 17:53, 17 May 2022 (UTC)

Okay, so according to a paper by Arno Berger, if $r_{1},r_{2}\in \mathbb {Q}$ with neither $r_{1}-r_{2}$ nor $r_{1}+r_{2}$ being an integer, then $1,cos(\pi r_{1}),cos(\pi r_{2})$ are $\mathbb {Q}$-independent if and only if $N(r_{1}),N(r_{2})\geq 4$ and $(N(r_{1}),N(r_{2}))\neq (5,5)$, where $N(r)$ is the lowest possible positive denominator of $r$ expressed as a ratio of integers. Since $sin(x)=cos({\frac {\pi }{2}}-x)$, we can write ${\frac {sin({\frac {\pi }{k_{1}}}+{\frac {\pi }{k_{2}}})}{sin({\frac {\pi }{k_{1}}}-{\frac {\pi }{k_{2}}})}}$ as ${\frac {cos({\frac {\pi }{2}}-{\frac {\pi }{k_{1}}}-{\frac {\pi }{k_{2}}})}{cos({\frac {\pi }{2}}-{\frac {\pi }{k_{1}}}+{\frac {\pi }{k_{2}}})}}$. Let $r_{1}={\frac {1}{2}}-{\frac {1}{k_{1}}}-{\frac {1}{k_{2}}},r_{2}={\frac {1}{2}}-{\frac {1}{k_{1}}}+{\frac {1}{k_{2}}}$. $r_{1}+r_{2}=1-{\frac {2}{k_{1}}}$ is always noninteger for $k_{1}\geq 3$. $r_{1}-r_{2}=-{\frac {2}{k_{2}}}$ is likewise always noninteger for $k_{2}\geq 3$. So for $k_{1},k_{2}\geq 3$, we have that $1,cos(\pi r_{1}),cos(\pi r_{2})$ are $\mathbb {Q}$-independent if and only if $N(r_{1}),N(r_{2})\geq 4$ and $(N(r_{1}),N(r_{2}))\neq (5,5)$. Now ${\frac {cos({\frac {\pi }{2}}-{\frac {\pi }{k_{1}}}-{\frac {\pi }{k_{2}}})}{cos({\frac {\pi }{2}}-{\frac {\pi }{k_{1}}}+{\frac {\pi }{k_{2}}})}}$ being rational implies that ${\frac {cos({\frac {\pi }{2}}-{\frac {\pi }{k_{1}}}-{\frac {\pi }{k_{2}}})}{cos({\frac {\pi }{2}}-{\frac {\pi }{k_{1}}}+{\frac {\pi }{k_{2}}})}}={\frac {p}{q}}$ for some integers $p,q$ where $q\neq 0$ and so $q*cos({\frac {\pi }{2}}-{\frac {\pi }{k_{1}}}-{\frac {\pi }{k_{2}}})-p*cos({\frac {\pi }{2}}-{\frac {\pi }{k_{1}}}+{\frac {\pi }{k_{2}}})=0$ makes $1,cos(\pi r_{1}),cos(\pi r_{2})$ not $\mathbb {Q}$-independent, so by contrapositive and equivalence we see that $N(r_{1}),N(r_{2})\geq 4$ and $(N(r_{1}),N(r_{2}))\neq (5,5)$ implies that ${\frac {cos({\frac {\pi }{2}}-{\frac {\pi }{k_{1}}}-{\frac {\pi }{k_{2}}})}{cos({\frac {\pi }{2}}-{\frac {\pi }{k_{1}}}+{\frac {\pi }{k_{2}}})}}={\frac {sin({\frac {\pi }{k_{1}}}+{\frac {\pi }{k_{2}}})}{sin({\frac {\pi }{k_{1}}}-{\frac {\pi }{k_{2}}})}}$ is irrational. So we can reduce the problem to finding values of $N(r_{1}),N(r_{2})$ for different $k_{1},k_{2}\geq 3$.

GalacticShoe (talk) 07:34, 18 May 2022 (UTC)

This contains the formula for a regular n-gon, whereby A = 1/2 a * p, where a is the length of the apothem and p is the perimeter. A polygon with unit sides will always have a rational perimeter (it is always equal to n units of length). The question of a rational area then breaks down to finding polygons with a rational apothem, which itself reduces to finding pythagorean triples, which are right triangles with rational side lengths. I have no idea if there are any further than the hexagon and the equilateral triangle that work, but if you are going to find them, you could work backwards from known pythagorean triple triangles, and see if you can construct a regular n-gon from them. Not sure if that will lead anywhere, but it's another way to conceptualize the problem; remember that all regular n-gons can be constructed from n-isosceles triangles, and any such triangle can be split into two right triangles by the apothem. The problem gets reduced to playing around with triangles and finding ones that fit your strictures. --Jayron32 12:40, 17 May 2022 (UTC)

Actually, I'm not even sure it needs to be a pythagorean triple; you only need two rational legs on your right triangle; the hypotenuse can be irrational for all we care; it doesn't enter into the calculation of the area. It doesn't mean there are any such polygons, but it does expand our search space in terms of finding the proper right triangles (which again, may be a fruitless exercise, it's just one way I thought of to play with the geometry.) --Jayron32 12:45, 17 May 2022 (UTC)

The apothem article gives a formula of $a={\frac {s}{2}}\tan {\frac {\pi (n-2)}{2n}}.$ That means that we need to find values of n for which the tan bit of that is rational. Another way to work this. --Jayron32 12:50, 17 May 2022 (UTC)

Finding rational values of trig functions is related to Niven's theorem, and my brain is getting tired of working on this now, but I suspect that may help someway. --Jayron32 12:53, 17 May 2022 (UTC)

Never mind. I misread the OP's post. The OP is looking for ratios of areas of two polygons that come out rational. That is workable with the above stuff, but I went off on a different tangent. You'd still use the apothem/area formula, but what you're looking for is situations where the ratio of two different tangent functions comes out to be rational, not the tangent function itself. --Jayron32 13:14, 17 May 2022 (UTC)

What you are looking for then is where ${\frac {\tan \ a}{\tan \ b}}$ is rational, and where a and b are both equal to ${\frac {\pi (n-2)}{2n}}$ for different values of n. --Jayron32 13:16, 17 May 2022 (UTC)

That is equivalent to what I wrote, since $\tan {\frac {\pi (n-2)}{2n}}=h_{n}^{{-}1},$ where $h_{n}$ is as defined above.  --Lambiam 18:25, 17 May 2022 (UTC)

Yes. I just did a lot more "thinking out loud" to get there. Sorry to imply you were wrong. I never said that, and if my ramblings have that impression, that is entirely my fault, and I am deeply sorry for doing so. --Jayron32 12:19, 18 May 2022 (UTC)