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

*

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


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


*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.
 A: 
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.
