# Triangle on a Rectangular Hyperbola

### Solution

In a suitably selected coordinate system, a rectangular hyperbola may be represented by the equation $\displaystyle y=\frac{1}{x}.$ Let points $A,B,C$ have abscissas $a,b,c,$ respectively. Then, for example, the slope $m_{AB}$ of $AB$ is

$\displaystyle m_{AB}=\frac{\displaystyle \frac{1}{b}-\frac{1}{a}}{\displaystyle b-a}=-\frac{1}{ab}.$

Let $\displaystyle H\left(h,\frac{1}{h}\right)$ be the point where the altitude from $C$ meets the hyperbola. The slope of $CH$ is

$\displaystyle m_{CH}=\frac{\displaystyle \frac{1}{h}-\frac{1}{c}}{h-c}=-\frac{1}{ch},$

and the orthogonality of $AB$ and $CH$ gives $\displaystyle \frac{1}{abch^2}=-1$ so that $\displaystyle h=-\frac{1}{abc}.$ The symmetry of this expression in $a,b,c$ shows that the remaining two altitudes would intersect the hyperbola in the same point $H,$ which is therefore the orthocenter of $\Delta ABC.$ Its coordinates are $\displaystyle \left(-\frac{1}{abc},-abc\right).$

### Acknowledgment

This is problem 15 from the Canadian Crux Mathematicorum (n 4, 1974). The problem was proposed by H. Dworechak and solved by Léo Sauvé.

Francisco Javier Garcia Capitan has kindly pointed out that the statement above is known as Brianchon-Poncelet as it was published as Theorem 1 in their paper Géométrie des courbes. Recherches sur la détermination d'une hyperbole équilatère, au moyen de quatre conditions données, Annales de Mathématiquess pures et appliquves, tome 11 (1820-1821), p. 205-220.