Home » Uncategorized » Hyperbolic Triangle

Hyperbolic Triangle

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


We are working on the following figure

Rendered by QuickLaTeX.com

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,

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


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

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*}

and finally y=-pqr. So, \mathrm{H} \left( - \frac{1}{pqr} , -pqr \right). This proves the claim.

Read more

1 Comment

Leave a comment

Donate to Tolaso Network