Let $A$, $B$, and $C$ be three points on the curve $xy = 1$ (which is a rectangular hyperbola). Prove that the orthocenter of $\triangle ABC$ also lies on the curve $xy = 1$.
I have given up on this problem…
[Math] Prove: A triangle inscribed in a rectangular hyperbola has its orthocenter on that hyperbola
geometry
Related Solutions
Algebra isn't necessary. Instead, use geometrical definitions: in particular, the circumcentre $O$ of $ABC$ is at the intersection of the perpendicular bisectors of $AB$, $AC$, and $BC$. Do you see how to continue from there?
HINT.
From any point $A$ of the hyperbola, in the branch intersecting the parabola but external to it (see diagram), draw tangents $AD$ and $AF$ to the parabola, which intersect the other branch of the hyperbola at $B$ and $C$. Check that line $BC$ is also tangent to the parabola.
Best Answer
Let $A = (p,1/p), ~B = (q,1/q),~ C = (r, 1/r)$ be the three points of the triangle, let $H$ be the orthocenter.
Verify that the slope of $AB$ is exactly $-1/pq$, so the slope of the altitude $h_C$ is $pq$. Similarly, the slope of the altitude $h_A$ is $qr$.
Now you can calculate the position of $H$, by intersecting $h_A$ and $h_C$: We have $$A + x\cdot\begin{pmatrix}1\\qr\end{pmatrix} = H = C + y \cdot \begin{pmatrix}1\\pq\end{pmatrix}$$
for some $x$ and $y$. You can solve this linear equation system for $x$ and $y$ and get that $H = (-1/pqr, -pqr)$. This point clearly lies on your given hyperbola.