<s>[INST] Find maximum lattice-point triangle with orthocenter at origin I was wondering if someone could help me understand this problem:
Let $S$ be the set of points $(x,y)$ in the Cartesian plane for which $\left(\tfrac{x}{x^2+y^2},\tfrac{y}{x^2+y^2}\right)$ is a lattice point. Find the largest possible area of a triangle with vertices in $S$ and orthocenter at the origin.
So I assume that means find all $(x,y)\in\mathbb{R}:\left(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}\right)\in\mathbb{Z}\backslash\{0,0\}$ correct?  If that's the case, then it's easy to identify these points as solutions to the equations
$$
\begin{align}
\frac{x}{x^2+y^2}=n\\
\frac{y}{x^2+y^2}=m
\end{align}
$$
with $(n,m)\in\mathbb{Z}$.  This turns out to be the intersections of four (infinitely) concentric circles shown in the plot below with the dot in $S$ and so the matter then becomes finding the largest triangle with such vertices and orthocenter at the origin.
Is this the correct way of interpreting this problem?  I get $3/10$ as the answer if so but that's only by brute-force checking and I at present do not have a rigorous proof.
Thanks guys.
:
 [/INST] $$\frac{x}{x^2+y^2}=n \qquad \frac{y}{x^2+y^2}=m$$
$$x^2 = n^2(x^2+y^2)^2 \qquad y^2 = m^2(x^2+y^2)^2$$
$$(x^2+y^2)=(m^2+n^2)(x^2+y^2)^2$$
$$(m^2+n^2)(x^2+y^2) = 1$$
$$x^2+y^2 = \dfrac{1}{m^2+n^2}$$
So $(x,y)$ can be any point on the circle with center at the origin and radius of $\dfrac{1}{\sqrt{m^2+n^2}}$
With $m,n \in \mathbb Z$, the largest possible radius is $\dfrac{1}{2}$.
From this answer the largest triangle will be any equilateral triangle inscribed on the circle $x^2+y^2=\dfrac 12$.  Which, as you commented, has an area of $\dfrac 38 \sqrt 3$
</s>