# Three hyperbolas related to a triangle concur at two points

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.

• A related article on Arxiv Commented Feb 7 at 11:46

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 $${\cal H}_B$$.
• @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. Commented Jul 11, 2022 at 6:44