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I found a geometry question posted today very interesting and would like to clear a doubt regarding a parabola if it can be inscribed (as described below).

Setup : Two equilateral triangles $\triangle ABC$, $\triangle CDE$ stand on $\overline{BD}$ as shown. $AD, BE$ intersect at $F$. Show that $FC^2 = AF \cdot EF$

enter image description here

If tangents at two points $A,E$ on the parabola meet at $C$, $F$ being the focus of parabola, then $FC^2=AF \cdot EF$ holds. Now if I try to inscribe a parabola in the region of $\angle ACE$ such that it is tangent to $AC$ at $A$ and $CE$ at $E$, does the condition $FC^2=AF \cdot EF$ uniquely identifies $F$ as focus of such a parabola, or are there more than one parabola with different focus also possible?

I thought that five conditions should uniquely determine any conic. Those known conditions here are points $A,E$, tangents at $A,E$ and the property that $FA, FC, FE$ are in geometric progression. But I am not sure if this direction of last condition is sufficient since perhaps we can find $F'$ somewhere else for which similar triangles $AF'C, CF'E$ can be made, and $F'$ will become the focus of second such parabola.

To be precise, can somebody suggest a proof for the converse of the property that if $FC^2 = AF \cdot FE$, where $AC, EC$ are tangents to the parabola, then $F$ must be the focus of such parabola.

Here is one such parabola with focus $F$ and meeting above conditions.

enter image description here

Bonus for those interested in a fun challenge in this setup, not related to the problem above :

Let $BE$ cut $AC$ in $P$ and $AD$ cut $CE$ in $Q$. Given $AF = 4, FC = 6$, calculate with proof, $BP + DQ$. Answer : $19$

MyMolecules
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  • Nice challenge. There's a "one-liner" explanation using the congruent triangles and $FC^2$ identity. $\quad$ You might want to restructure the question so that it's focused on the doubt, instead of the context. – Calvin Lin Jun 27 '23 at 20:53
  • @CalvinLin The tricky part is proving $PCQ$ is equilateral. Indeed I want the converse proof, I will highlight so. – MyMolecules Jun 27 '23 at 20:58
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    Not that tricky if you know the congruent triangles map. That $60^\circ$ rotation about $C$ takes lines $BE$ to $AD$ and lines (not as points) $AC$ to $CE$, hence by considering the intersections of these lines, take $P$ to $Q$. Thus $PC=CQ$ (and hence $PCQ$ is equilateral). – Calvin Lin Jun 27 '23 at 21:05
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    @CalvinLin Oh now I see it, benefits of rotation! – MyMolecules Jun 27 '23 at 21:19

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Unfortunately the converse is false: there are infinitely many points $F$ such that $FC^2=AF\cdot FE$, but only one of them is the focus of the parabola tangent to lines $AC$, $EC$ at points $A$, $E$ respectively.

To see why, choose any point $P$ on $AC$ and construct $Q$ on $EC$ such that $$ CQ:QE=AP:PC. $$ We can then construct the circle of Apollonius through $P$ whose points have all the same ratio of distances from $A$ and $C$, and also the circle of Apollonius through $Q$ whose points have all the same ratio of distances from $C$ and $E$. If $F$ is an intersection of those circles, then: $$ AF:FC=AP:PC=CQ:QE=FC:FE $$ and consequently $FC^2=AF\cdot FE$.

Varying the position of $P$ we can thus construct many points having that property. But only one of them is the focus $G$ of the parabola tangent to lines $AC$, $EC$ at points $A$, $E$, which is uniquely determined.

enter image description here

EDIT.

We can construct the focus $G$ of the parabola as follows. Draw a line $r$ through $C$ and through the midpoint of $AE$: this line is parallel to the axis by a property of the parabola.

Draw two lines $s$, $t$ parallel to $r$ and passing through $A$, $E$ respectively. By the reflective property of the parabola, the reflection of $s$ about the tangent at $A$ passes through the focus, and the same is true for the reflection of $t$ about the tangent at $E$. Hence focus $G$ is the intersection of those reflected lines.

enter image description here

Intelligenti pauca
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  • Thank you! So the tangent parabola is unique but to determine its focus $G$, the ratio $AG : GC$ should be known - did I understand it correct? (Because we are varying $P$, we are essentially varying the ratio $AF:FC$) – MyMolecules Jun 28 '23 at 08:42
  • @MyMolecules I added a construction for focus $G$. – Intelligenti pauca Jun 28 '23 at 13:21
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    N.B. The "five constraints" rule holds for a generic conic. Knowing it is a parabola reduces the number to four. – Intelligenti pauca Jun 28 '23 at 14:02
  • Amazing! Thank you for nice insights : a) "$h^2=ab$ reduces number of constraints to $4$ b) Reflecting two diameters in respective tangents gives focal radii; their intersection gives focus! – MyMolecules Jun 28 '23 at 17:02