Prove that a conic section through the vertices of a triangle and its orthocentre is a rectangular hyperbola.

Question

I can see that a rectangular hyperbola through the vertices of a triangle $ABC$ passes through the orthocentre $H$. But how do you prove the converse that a circumhyperbola through $H$ is rectangular?

Any proofs or hints would be appreciated! Thanks

Answer 1

I'll prove this result:

Given $\triangle ABC$ with orthocenter $H$, let $P$ be any distinct point on the triangle's circumcircle. These five points determine a rectangular hyperbola (whose center happens to be on the triangle's "nine-point circle").

Coordinatize with $$A := r(1,0) \qquad B := r (\cos 2C,\sin 2C) \qquad C := r (\cos(-2B),\sin(-2B))$$ where $r$ is the circumradius (and $A$, $B$, $C$ also serve to name the angles at each vertex), so that $$H = r(1 - 2 \cos A \cos(B - C), 2 \cos A \sin(B - C))$$ We can take $P := r (\cos2\theta, \sin2\theta)$.

The unique conic through these points has this equation: $$\left| \begin{array}{cccccc} x^2 & x y & y^2 & x & y & 1 \\ A_x^2 & A_x A_y & A_y^2 & A_x & A_y & 1 \\ B_x^2 & B_x B_y & B_y^2 & B_x & B_y & 1 \\ C_x^2 & C_x C_y & C_y^2 & C_x & C_y & 1 \\ H_x^2 & H_x H_y & H_y^2 & H_x & H_y & 1 \\ P_x^2 & P_x P_y & P_y^2 & P_x & P_y & 1 \\ \end{array} \right| = 0$$

This expands and reduces (with the help of Mathematica) to $$\begin{align} 0 &= x^2 \cos(B - C - \theta) - y^2 \cos(B - C - \theta) - 2 x y \sin(B - C - \theta) \\ &- 4 x r \cos B \cos C \cos\theta \\ &- 2 y r (\sin C \cos(B + \theta) - \cos C \sin(B - \theta))\\ &+ r^2 (2 \cos(B+C) \cos\theta + \cos(B-C+\theta) ) \end{align} \tag{1}$$

That the sum of $x^2$ and $y^2$ coefficients vanishes indicates that the conic is specifically a rectangular hyperbola. $\square$

Now, it is known that nine-point circle is the dilation of the circumcircle with respect to the orthocenter, with scale factor $1/2$. Thus, to show that the center of the hyperbola (call it $Q$) lies on the nine-point circle, it suffices to show that $Q$ is the midpoint of $H$ and some point on the circumcircle; in fact, as suggested by the animation, that "some point" is $P$ itself.

We can find $Q$ by differentiating $(1)$ with respect to $x$ and with respect to $y$, and then solving the resulting linear system for $Q$'s $xy$-coordinates. The reader can check that they're equivalent to those of the midpoint of $P$ and $H$:

$$Q = r\,\left( \cos^2\theta - \cos A \cos(B-C), \sin\theta \cos\theta + \cos A \sin(B-C)\right) =\frac12(P+H) \tag2$$