Hyperbolic Triangle

The vertices of a triangle lie on the hyperbola $y=\frac{1}{x}$ . Prove that its orthocentre also lies on the hyperbola.

Solution

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Let $\mathrm{A} \left( p, \frac{1}{p} \right)$ , $\mathrm{B} \left( q, \frac{1}{q} \right)$ and $\Gamma \left( r , \frac{1}{r} \right)$ . Let us denote as $\mathrm{H}$ its orthocentre. We have that:

$\lambda_{\mathrm{B}\Gamma} = \frac{\frac{1}{r} - \frac{1}{q}}{r-q} = - \frac{1}{qr}$

Hence, the slope of the altitude $\mathrm{AZ}$ is $qr$ . Similarly,

$\lambda_{\mathrm{AB}} = \frac{\frac{1}{q} - \frac{1}{p}}{q-p} = - \frac{1}{pq}$

Hence, the slope of the altitude $\Gamma \mathrm{E}$ is $pq$ . Hence,

$\left ( \varepsilon \right )_{\mathrm{B} \Gamma} : y - \frac{1}{p} = qr \left ( x - p \right )$

and

$\left ( \varepsilon \right )_{\Gamma \mathrm{E}} : y - \frac{1}{r} = pq \left ( x - r \right )$

Solving this linear system we have

$\begin{align*} \left ( \varepsilon \right )_{\mathrm{B} \Gamma} = \left ( \varepsilon \right )_{\Gamma \mathrm{E}} &\Leftrightarrow qr \left ( x- p \right ) + \frac{1}{p} = \frac{1}{r} + pq \left ( x-r \right ) \\ &\Leftrightarrow x = - \frac{1}{pqr} \end{align*}$