Given any triangle $\triangle ABC$, we build the hyperbole with foci in $A$ and $B$ and passing through $C$.

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Similarly, we can build other two hyperboles, one with foci in $A$ and $C$ and passing through $B$ (red), and one with foci in $B$ and $C$ and passing through $A$ (green).

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The first part of my conjecture is that the three hyperboles always intersect in two points $D$ and $E$.

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Moreover, the ellipse with foci in these two points $D$ and $E$, and passing through one of the three vertices of the triangle $\triangle ABC$, pass also through the other two vertices.

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These are probably obvious results. However, is there an elementary proof for these conjectures?

Thanks for your help! Sorry in case this is too trivial.

EDIT: You might be interested also in this other post.

Second part: Let $a=BC$, $b = CA$ and $c =AB$.

We have $$EB-EC = DC-DB = c-b$$ $$\color{red}{EB-EA = DA-DB = a-b}$$ $$DA-DC = FC-FA =c-a$$

If $$AD+AE = 2d$$ then since $$BD+BE = (DA+b-a)+(EA+a-b)=2d$$

so $B$ also lies on this ellipse. The same is true for $C$.

Try this:

Let E be a point of intersection of black and red hyperbole. Then

black hyperbole: $EA-EB=AC-CB$

red hyperbole: $EA-EC=AB-BC$

Subtracting, you get

$$EC-EB=AC-AB$$ therefore green hyperbole also passes through E.

Do the same for point D, saying that D is a point of intersection of red and green hyperbole, and you will get that black hyperbole also passes through D.

  • Thanks for the elegant and compact approach! A comment: How can we know that we can find always the two points $D$ and $E$? – Andrea Prunotto Oct 17 at 4:23

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