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In Yahoo Answers, here, Rita the dog defined a pointy triangle, (more or less) as having three properties.

  1. The lengths of two sides are rational and greater than 1.

  2. The length of the third side is 1/n for some integer n.

  3. The area of the triangle is 1/n.

    We say that a pointy triangle likes n if its area is 1/n.

Prove that pointy triangles like an infinite number of n's.

As Pauley Morph, I posted the same question in Yahoo Answers, here, and got no response. I also posted the right triangle lemma, here, and got one tepid answer.

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  • $\begingroup$ Why all the negative votes? $\endgroup$ May 12, 2015 at 4:04
  • $\begingroup$ I think it's interesting that your question got downvoted but your answer got upvoted. $\endgroup$ May 12, 2015 at 4:40
  • $\begingroup$ @AlfredYerger: I checked and SE said I could answer my own questions. So I did. I spent (well, actually, enjoyed) a lot of time working on this problem and tried several approaches that led me nowhere. I was staring at a table of solutions and decided to try $\alpha \beta = u/v$. When the Pells equations popped up, I got that rush that I always get at at such Eurika moments. It would be nice if the down votes were accompanied by comments. $\endgroup$ May 12, 2015 at 4:51
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    $\begingroup$ Yeah it's a perfectly fine thing to do. I'm just intrigued by the fact that people like the answer, but not the question. It's unusual. $\endgroup$ May 12, 2015 at 4:52
  • $\begingroup$ The two right triangles in the picture below are $(5, 12, 13)$ and $(7, 24, 25)$ right triangles scaled down to $(5/6, 2, 13/6)$ and $(7/12, 2, 25/12)$. Corresponding to $\alpha = 3/2$ and $\beta = 4/3$.In this case, $(\alpha - \dfrac{1}{\alpha}) - (\beta - \dfrac{1}{\beta}) = 1/4$. $\endgroup$ May 12, 2015 at 5:05

1 Answer 1

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parameterizing a pointy triangle

Clearly the altitude at the base with length $1/n$ is $2$. It can be shown, if any right triangle has one side of length 2, then there is a real number $\gamma > 1$ such that the other side has a lenght of $\gamma - \dfrac{1}{\gamma}$ and a hypotenuse of length $\gamma + \dfrac{1}{\gamma}$. Also the lengths are all rational if and only if $\gamma$ is rational. Using the image above, we must find rational numbers $\alpha$ and $\beta$ such that \begin{equation} (\alpha - \dfrac{1}{\alpha}) - (\beta - \dfrac{1}{\beta}) = \dfrac 1 n \end{equation} for some integer n.

We suppose that there are positive, relatively prime, odd integers u > v such that $\alpha \beta = \dfrac{u}{v}$. It is perfectly fine for one of u or v to be even, but the equations have to be changed slightly. Since I don't need to find all pointy triangles, I won't add that complication to the solution. Replacing $n$ with $u v k$, $\beta$ with $\dfrac{u}{v \alpha}$, and simplifying, we wind up with the equation

\begin{equation} v(u+v)k\alpha^2 - \alpha - u(u+v)k = 0 \end{equation}

The positive solution is

\begin{align} \alpha = \dfrac{\sqrt{1 + 4uv(u+v)^2k^2}+1}{2v(u+v)k}\cr \beta = \dfrac{\sqrt{1 + 4uv(u+v)^2k^2}-1}{2v(u+v)k} \end{align}

Clearly $\alpha$ and $\beta$ will exists and be rational if there exists integers $N_k$, that depend on $k$, such that $N_k^2 = 1 + 4uv(u+v)^2k^2$. But these are Pell's equations and such equations are guaranteed to have an infinite number of solutions.

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  • $\begingroup$ Writing $\alpha = a/b$ and $\beta = c/d$ (in lowest terms) we see that a, b, c, and d solve this problem $\endgroup$ May 12, 2015 at 0:45
  • $\begingroup$ Note also that $\alpha - \beta = \dfrac{1}{v(u+v)k}$ and $\dfrac{1}{\alpha} - \dfrac{1}{\beta} = \dfrac{1}{u(u+v)k}$ So $(\alpha - \dfrac{1}{\alpha}) - (\beta - \dfrac{1}{\beta}) = \dfrac{1}{u v k}$ becomes $\dfrac{1}{u(u+v)k} + \dfrac{1}{v(u+v)k} = \dfrac{1}{u v k}$ which solves this $\endgroup$ May 12, 2015 at 16:40

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