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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

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  • $\begingroup$ I know that it is true as I have found it quoted in various places, but I was looking for a proof! $\endgroup$
    – JamesDixon
    Sep 8, 2017 at 11:18
  • $\begingroup$ There was a deleted answer. Anyway this answer might be of help, if only to leave a comment there. $\endgroup$ Sep 8, 2017 at 11:27
  • $\begingroup$ Oh. Why was the answer deleted - was it wrong? The other question is the converse of what I am trying to prove, which I have seen before. But thanks all the same. $\endgroup$
    – JamesDixon
    Sep 8, 2017 at 11:39
  • $\begingroup$ I'm considering a full proof. In the meantime, if you want quick assurance that "some" conic through the four points is a rectangular hyperbola, consider a fifth point where one of the altitudes meets one of the sides. The conic determined by those five points is the union of the extended side and the extended altitude; that's a rectangular hyperbola, albeit a degenerate one. :) $\endgroup$
    – Blue
    Sep 8, 2017 at 11:55

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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").

enter image description here

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 $P := r (\cos\theta, \sin\theta)$.

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 $$\begin{align} 0 &= x^2 \cos(B - C - \theta) - y^2 \cos(B - C - \theta) - 2 x y \sin(B - C - \theta) \\ &- 4 x r \cos B \cos C \cos\theta \\ &- 2 y r (\sin C \cos(B + \theta) - \cos C \sin(B - \theta))\\ &+ r^2 (2 \cos(B+C) \cos\theta + \cos(B-C+\theta) ) \end{align}$$

The discriminant of this equation is $$(2\sin(\cdots))^2 - 4(\cos(\cdots))(-\cos(\cdots)) = 4(\sin^2(\cdots)+\cos^2(\cdots)) = 4$$ which, being positive, indicates that the represented conic is a hyperbola. That the sum of $x^2$ and $y^2$ coefficients vanishes indicates that the conic is specifically a rectangular hyperbola. (See Wikipedia's "Discriminant" subsection of its "Conic section" entry.) $\square$

Showing that the center of the conic lies on the nine-point circle is left as an exercise to the reader. (The Penguin entry referenced by @Jan-MagnusØkland's comment to the question affirms this fact.) I'll note, however, that this property was leveraged in the conception of this proof: It is known that the nine-point circle is the dilation, with scale factor $1/2$, of the circumcircle with respect to the orthocenter; consequently, for any center, $K$, chosen on the nine-point circle, the reflection of $H$ in $K$ would lie on the circumcircle and on the corresponding hyperbola. We took our $P$ to be that reflection of $H$.

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  • $\begingroup$ What is $\cos{A}$ in the description of $H$? $\endgroup$ Sep 11, 2017 at 9:16
  • $\begingroup$ Thanks Blue - it's quite a 'messy' proof, but it works. I have actually found a proof using the projective version of Pascal's theorem since I posted this. $\endgroup$
    – JamesDixon
    Sep 11, 2017 at 9:33
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    $\begingroup$ Jan - $cos A$ is the cosine of the angle at $A$ in the triangle $\endgroup$
    – JamesDixon
    Sep 11, 2017 at 9:34
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    $\begingroup$ @Jan-MagnusØkland: I double-checked my Mathematica calculations. The formula for $H$ appears to be correct. $\endgroup$
    – Blue
    Sep 11, 2017 at 10:13
  • $\begingroup$ @JamesDixon: I'm very interested in your projective proof; please post it. (As for mine ... This is actually a far cleaner approach than the "messy" version I mentioned in a previous comment. :) $\endgroup$
    – Blue
    Sep 11, 2017 at 10:19

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