# Justify this is locus is indeed a parabola

This problem consists of a geometric construction and proving that the locus generated is a parabola. Here's the construction:

1. Trace a circumference with radius $r$ and center $A$.

2. From any point $C$ on $\bigcirc A$, trace the line $\overleftrightarrow{AC}$. Let $D$ be a point on this line such that $C$ is the midpoint of $\overline{AD}$.

3. Let $E$ be a point on $\bigcirc A$. Trace the line $\overleftrightarrow{ED}$.

4. Let $E^\prime$ be reflection of $E$ in $\overleftrightarrow{AC}$. Trace the line $\overleftrightarrow{AE^\prime}$.

5. Let $F$ the intersection of $\overleftrightarrow{AE^\prime}$ with $\overleftrightarrow{ED}$.

Prove the locus described by $F$, as the point $E$ moves through the circumference, is a parabola.

It might seem simple, but this problem has been giving me some troubles trying to find the key idea.

Let $G$ be the foot of the perpendicular from $F$ to the perpendicular bisector of $\overline{AD}$. When $\overleftrightarrow{E^\prime A}$ is not parallel to that bisector (that is, for $\overleftrightarrow{E^\prime A} \not\perp \overline{AC}$), let $P$ be the point where the two lines meet.

Then,

$$\frac{|\overline{AF}|}{|\overline{AE}|} \;\underbrace{\phantom{\huge|}\quad=\quad\phantom{\huge|}}_{\triangle FAE\sim\triangle FPD}\; \frac{|\overline{PF}|}{|\overline{PD}|} \;\underbrace{\phantom{\huge|}\quad=\quad\phantom{\huge|}}_{\triangle APD \text{ is isos.}}\; \frac{|\overline{PF}|}{|\overline{PA}|} \;\underbrace{\phantom{\huge|}\quad=\quad\phantom{\huge|}}_{\triangle PFG \sim \triangle PAC} \; \frac{|\overline{FG}|}{|\overline{AC}|}$$

But, $$\overline{AE} \cong \overline{AC}$$

So, $$\overline{AF} \cong \overline{FG}$$

Thus,

$F$ is equidistant from $A$ and the perpendicular bisector of $\overline{AD}$. (Note that this is trivially true when $\overleftrightarrow{E^\prime A}\perp\overline{AD}$, as well, since then $F$ and $E$ coincide as a vertex of square $\square ACGF$.) Consequently, its locus, by definition, is the parabola with the latter elements as focus and directrix. $\square$

• Amazing as always! – Hari Shankar Sep 30 '17 at 7:40

Let $A(0,0)$ be the centre and C be $(a,0)$. So D is $(2a,0)$ Note that for C and its diametrically opposite point, the images are $C$. For any other point on the circumference, the typical point is $E(a\cos \theta, a\sin \theta)$ and its image $E'(a\cos \theta, -a\sin \theta)$

So $AE' \equiv y+x \tan \theta =0$ and $DE \equiv \frac{y}{x-2a} = \frac{\sin \theta}{2-\cos \theta}$.

Hence, $F = AE' \cap DE = \left(-\frac{a}{1-\cos \theta}, \frac{a \sin \theta}{1-\cos \theta} \right).$

Its easily seen that $F$ satisfies $y^2 =-4a(x-a)$ which is a parabola with focus $(0,0)$ and vertex $(a,0)$. $C(a,0)$ belongs to the parabola, which completes the proof

I would be interested in seeing a pure geometric solution though

• Two geometric solutions have appeared. – DanielWainfleet Sep 30 '17 at 7:10

Let $d$ be the line through $A$ orthogonal to $AC$. Let $\mathrm{H}$ be the harmonic homology of center $D$ and axis $d$, let $\Gamma$ be the polarity associated to the circle.

Let consider the polarity $\Psi=\mathrm{H}\circ\Gamma\circ\mathrm{H}$. Then the conic associated to $\Psi$ is a parabola because $\mathrm{H}$ maps the tangent $t$ to the circle at $C$ to the line at infinity.

Finally, if $V$ denote the point at infinity of the line $EE'$ and $M=EE'\cap AC$ then the group $EE'MV$ is harmonic, because $M$ is the midpoint of $EE'$. The projection of center $A$ sends the group $EE'MV$ onto the group $EFDN$ where $N=ED\cap d$. Thus $EFDN$ is harmonic, hence $F=\mathrm{H}(E)$. This proves that the parabola associated to $\Psi$ is given by the geometric construction described in OP.