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I ran into a need for isosceles triangles that (1) have the two equal integer side lengths $a$ (but the base $x \in \mathbb{R}$), and (2) the apex angle $\gamma$ is a rational multiple of $\pi$.
          Rational triangle
I found a few such triangles for particular $x$, but saw no general technique to solve this problem:

Given the base $x$, find all isosceles triangles over that base such that side length $a$ is an integer and $\gamma$ is a rational multiple of $\pi$. For which $x$ are there solutions?

My underlying question is for non-isosceles triangles, with the side lengths $a$ and $b$ incident to $\gamma$ both integers, but it seems already difficult for isosceles triangles.

The question arose in the design of gears :-)
          Gears4

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The base must be of the form $x = 2a\sin(k\pi/2N)$, where $a$, $k$, and $N$ are (positive) integers (giving $\gamma = k\pi/N$ as the rational multiple of $\pi$). Are you looking for a way to parlay one triple of integers for a given $x$ into others? –  Barry Cipra Jun 25 '13 at 14:36
    
Thanks, Barry, that's clear. And, Yes, I am cascading the structure, building upon itself. –  Joseph O'Rourke Jun 25 '13 at 16:24
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So are you looking for $x$ with multiple solutions? I'd pick a $k$ so that $\phi(q)=k$ has many solutions, take all the values $\sin(\frac{p}{q}\pi)$ for those $q$ and then look for pairs with rational ratio. Wouldn't solutions have to arise that way? Patterns might become obvious. –  Aaron Meyerowitz Jun 25 '13 at 22:58

2 Answers 2

up vote 3 down vote accepted

Here is an expanded take on my comment:

Imagine that we have a triangle of the sort you seek and the base angles are $\beta=\frac{p}{q}2\pi$ with the fraction in lowest terms (and $q \ge 6$). Then $x=(2cos\beta)a$

We know that $2a \cos \beta$ is an integer for $q=6.$ Otherwise it is an algebraic integer of degree $\frac{\phi(q)}{2}$ (short proof below).

So a not totally satisfactory answer to

Given a value $x$, determine if it can be a base and, if so, find all isosceles triangles over that base such that side length $a$ is an integer

is: If $x$ is rational then it must be an integer and it arises only from an equilateral triangle. Otherwise,

  • Verify that $x$ is an algebraic integer and determine the degree $d.$ (This means that you somehow know an exact value of $x.$)
  • Find the (finite list) of integers with Euler totient function $\phi(q)=2d.$ (See the comment on this below.)
  • Compute $2 cos(\frac{2p \pi}{q})$ for all cases.
  • Check if $x$ is an integral multiple of any of them (and if so, of which). Voila!

That does not immediately shed any light (that I can see) on questions such as "Is there an $x$ which arises in over $7$ ways." But it does suggest picking a value $2d$ so that there are numerous solutions to $\phi(q)=2d$ and then computing the various $cos(\frac{2p \pi}{q})$ and looking for rational ratios. Maybe some nice patterns will become clear. (What solutions do you have, if one may ask.)

The same procedure offers, in some sense, an answer to the modified question

For which $x$, algebraic of degree $d$, are there solutions?

If one wanted a huge $d$ that might not be very practical.

Further comments:

Maple has the function invphi to compute the solutions of $\phi(q)=2d$. It wasn't always perfect, but that particular bug has been fixed. See also this question .

The fact about the degree of $2 \cos \beta$ must be a standard fact involving Chebyshev polynomials, but this seems to work: Let $\zeta=e^{\frac {2\pi i}{q}}.$ Then $2\cos\beta=\zeta^p+\zeta^{-p}$ is a sum of two algebraic integers and hence itself an algebraic integer. The dimension of $\mathbb{Z}[\zeta^p]=\mathbb{Z}[\zeta]$ relative to $\mathbb{Z}$ is $\phi(q)$ but $\mathbb{Z}[\zeta^p]$ has dimension $2$ over $\mathbb{Z}[2\cos\beta] \subset \mathbb{R}$ as $t=\zeta^p$ satisfies $t^2-2\cos\beta t+1=0.$

I am not sure that you can have an $x$ in two ways in the isosceles case and feel that it might be fairly straightforward . But I have no proof. Do you have any examples? There seem to be none of degrees $4$ or $6$.

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This is very helpful, Aaron. I find it delightful that this touches upon Chebyshev polynomials and the Euler totient function! –  Joseph O'Rourke Jun 28 '13 at 11:51

I don't know if there's a good answer to this question. We have $\cos \gamma = 1 - x^2/(2a^2)$, and we need $\gamma$ to be a rational multiple of $\pi$, so want $T_n(\cos \gamma) = 1$ for some $n$, where $T_n$ is the Chebyshev polynomial. So it's equivalent to ask (given $x$) whether there are natural numbers $n$ and $a$ with $T_n( 1- x^2/(2a^2)) = 1$. For any fixed $n$, you can compute the roots and see whether there is an integer root. If one could bound the $n$ somehow for which this was possible (in terms of $x$), that might yield an algorithm. In any case, $x$ had better be algebraic.

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