# A simple conjecture (and a question) about three parabolas related to any triangle

Given any triangle, we can build three parabolas, each with focus on one vertex and with directrix the opposing side, as illustrated here:

My first conjecture, likely trivial, is that, given any triangle,

The three parabolas never intersect, but they are tangent to one another in at most three points.

For instance, in case of an equilateral triangle, it seems that the three parabolas "touch" each other in three points $$E,F,G$$

However, it is not obvious to me whether the equilateral triangle is the only case in which we can find three tangential points $$E,F,G$$.

The question is, then:

In which conditions (on the initial triangle) can we find three, two, one or no tangential points?

I apologize in case the question is trivial. But thank you very much for your hints, comments, suggestions!

• Let the triangle be $ABC$. Let $a,b,c$ be the lines $BC, CA, AB$. For each point $P$ and each line $g$, let $Pg$ be the distance from $P$ to $g$. If a point $P$ lies on the parabola with focus $B$ and on the parabola with focus $C$ at the same time, then $PB = Pb$ and $PC = Pc$. But $PB \geq Pc$ (since $c \in B$), with equality if and only if $PB \perp c$. Similarly, $PC \geq Pb$, with equality if and only if $PC \perp b$. Summing up these two inequalities, we get $PB + PC \geq Pc + Pb$, which must become an equality because $PB = Pb$ and $PC = Pc$. Thus, ... – darij grinberg Nov 6 '18 at 22:20
• ... we must have $PB \perp c$ and $PC \perp b$. This uniquely determines $P$ as the point where the perpendicular to $b$ through $C$ meets the perpendicular to $c$ through $B$. (Equivalently, this point is the antipode of $A$ on the circumcircle of $\triangle ABC$.) But of course, just letting $P$ be this point doesn't actually guarantee that $PB = Pb$ and $PC = Pc$. Argue that this happens precisely when $AB = AC$. Conclude. – darij grinberg Nov 6 '18 at 22:22
• The triangle $(0,0),(1,0),(1,1)$ has parabolae that intersect in only one place: $(0,1)$. – Jam Nov 6 '18 at 22:23
• Thanks for your comments! So only the equilateral triangle yields to three tangential points... – user559615 Nov 6 '18 at 22:25
• Fun fact: The parabola with focus $A$ is tangent to the perpendicular bisectors of $\overline{AB}$ and $\overline{AC}$, and also the angle bisectors (both interior and exterior) at $B$ and $C$. – Blue Nov 7 '18 at 14:02

Let $$\ell$$ be the perpendicular bisector of $$\overline{BC}$$, and let $$B^\prime$$ be the point where $$\ell$$ meets the perpendicular at $$B$$ to $$\overline{AB}$$. Then $$B^\prime$$ is equidistant from $$C$$ and $$\overleftrightarrow{AB}$$, so $$B^\prime$$ is on the $$C$$-focus parabola. Since $$\ell$$ bisects $$\angle BB^\prime C$$, an aspect of the parabolic reflection property implies that $$\ell$$ is tangent to the parabola.
Similarly, $$\ell$$ is tangent to the $$B$$-focus parabola (say, at $$C^\prime$$), so $$\ell$$ separates the two parabolas, making the only possibility of their meeting when $$B^\prime$$ and $$C^\prime$$ coincide. This happens if and only if $$\triangle ABC$$ is isosceles with base $$\overline{BC}$$. $$\square$$
From this, we see that the three parabolas admit no tangential points iff the triangle is scalene, three tangential points iff the triangle is equilateral, and exactly one tangential point iff the triangle is non-equilaterally isosceles. There are never exactly two tangential points. $$\square$$