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I found this interesting fact (which I believe is non-trivial):

Let $\triangle ABC$ be a triangle. Let $\mathcal{H}_A$ be the hyperbola with foci at $B$ and $C$, passing through $A$. Define $\mathcal{H}_B$ and $\mathcal{H}_C$ similarly. Then $\mathcal{H}_A,\mathcal{H}_B,\mathcal{H}_C$ meet at exactly two points unless $\triangle ABC$ is equilateral, in which they would meet at one point (which is the orthocenter=incenter=circumcenter in that case).

I'm not very well-versed with conics, so I'm not sure how I would go about proving this. I tried using the basic definition of a hyperbola, and you get a system of three distance equations, which are quite tricky to solve. I then supposed that $X$ is some point such that $|XA-XB|=|a-b|, |XB-XC|=|b-c|$ (the intersection of two of the hyperbolas), and then tossing this on the complex plane, setting $X$ to be the origin. This didn't go so well. I don't believe any bashing approaches work, I think this problem requires some sort of projective transformation. Also, it's quite a simple conjecture, so it's likely well known, but I haven't been able to find it on the internet. All help is appreciated.

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    $\begingroup$ A related article on Arxiv $\endgroup$
    – orangeskid
    Commented Feb 7 at 11:46

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I'll say $P$ is a point on the first branch of ${\cal H}_A$ if $PB-PC=AB-AC$, while $P$ is a point on the second branch of ${\cal H}_A$ if $PB-PC=-AB+AC$, and analogous definitions for ${\cal H}_B$ and ${\cal H}_C$.

Suppose now $P$ is an intersection of the first branches of ${\cal H}_A$ and ${\cal H}_B$. We have then: $$ PB-PC=AB-AC\\ PC-PA=BC-BA\\ $$ and adding those equations we get $$ PB-PA=CB-CA, $$ that is $P$ also lies on the first branch of ${\cal H}_C$. The same goes for the point $Q$ which is the intersection of the second branches of ${\cal H}_A$ and ${\cal H}_B$.

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  • $\begingroup$ Oh damn I didn't realize it was that simple! I got caught up in all of the absolute values. Nice solution l. $\endgroup$ Commented Jul 10, 2022 at 22:22
  • $\begingroup$ @TheBestMagician Glad I was of help, but this is not the end of the story. One has to prove that ${\cal H}_A$ and ${\cal H}_B$ do intersect the right way to give those two intersections you found, which is not so obvious. $\endgroup$ Commented Jul 11, 2022 at 6:44

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