Radius of circle tangent to six tangent parabolas

Question

Six congruent copies of the parabola $y = x^2$ are arranged in the plane so that each vertex is tangent to a circle, and each parabola is tangent to its two neighbors. Find the radius of the circle.

I'm honestly not sure how to start this problem. The vertices of the parabolas are spaced $60^\circ$ apart on the circumference of the circle, and the points where two parabolas are tangent are also spaced $60^\circ$ apart, $30^\circ$ from a vertex. But I'm not sure how this is useful.

I also know that the focus and directrix of each parabola are the same distance from the vertex, and the directrixes (is that the plural?) intersect $60^\circ$ apart, $30^\circ$ from a vertex. But again, I don't know how useful that is.

Normally on problems like these I'd try to draw a radius and form some right triangles but I really don't see how I can do that here.

TIA!

Answer 1

The idea is to look at a single parabola of the form $y = x^2 + r$, and determine $r$ such that the parabola is tangent to the lines $$y = \pm \sqrt{3} x.$$ This $r$ will be the radius of the inscribed circle. See the following figure, which only shows the top and bottom parabolas, but adds the lines:

To this end, we compute the derivative $$\frac{dy}{dx} = 2x$$ and equate it to $\pm \sqrt{3}$ to obtain $$x = \pm \frac{\sqrt{3}}{2}.$$ These are the $x$-values at which the parabola is tangent. So we require $$(\pm \sqrt{3}/2)^2 + r = \pm \sqrt{3}(\pm \sqrt{3}/2) = 3/2,$$ hence $$r = \frac{3}{2} - \frac{3}{4} = \frac{3}{4}.$$

Answer 2

Geometrically, consider a parabola with focus $F$, vertex $V$, and directrix through $D$ (so that $\overline{TF}\cong\overline{TT'}$, where $T'$ is the foot of the perpendicular from $T$ to the directrix). We require $T$ make $\angle TKF=30^\circ$, where $K$ is the center of the ostensible circle.

The Reflection Property tells us that $\overline{TK}$ bisects $\angle FTT'$, and we deduce that $\triangle FTT'$ is equilateral, and thus also is $\triangle FKT'$. Consequently, the directrix bisects $\overline{FK}$. Since $V$ bisects $\overline{FD}$, we conclude that $|KV| = 3|FV|$.

For the specific case of the parabola defined by $y=1x^2=\dfrac{1}{4\,|FV|}x^2$, we have $|FV|=1/4$, so that $|KV|=3/4$.

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Inspired by @JeanMarie's generalization, here's a revised argument:

For arbitrary $\theta:=\angle FKT$, and vertex-to-focus distance $a$, a little angle-chasing and the Reflection Property ultimately yield $\triangle FVM\sim\triangle MVK$, so that $$|KV|=a\cot^2\theta$$ In the context of fitting $n$ tangent $y=x^2$ parabolas together about $K$, we have $\theta=\frac{\pi}{n}$ and $a=\frac14$, so that this agrees with @JeanMarie's result. $\square$

Answer 3

This issue can be treated using complex numbers geometry.

Indeed, if we take the initial parabola with a horizontal axis of symmetry, it can be written as the locus of points

$$z=t^2+it+r, \ \ t \in \mathbb{R}\tag{1}$$

(with unknown $r$). The symmetry with respect to abscissas axis being obtained by changing $t$ into $-t$.

Therefore, as rotation by $\pi/3$ is obtained by

$$z \to \omega z, \ \ \text{with} \ \ \omega:=e^{i\pi/3}=\dfrac{1+\sqrt{3}}{2}\tag{2}$$

we can express the tangency of the initial parabola with its rotated version by saying that the following equation has a double root:

$$\omega(t^2+it+r)=(t^2\color{red}{-}it+r)\tag{3}$$

This equation can be written :

$$(\omega-1)t^2+i(\omega+1)t+(\omega-1)r=0\tag{4}$$

The double root condition is equivalent to nullify the discriminant of (4):

$$\Delta=-(\omega+1)^2-4(\omega-1)^2r=0$$

giving:

$$r=-\dfrac14\left(\dfrac{\omega+1}{\omega-1}\right)^2=-\dfrac14\left(\dfrac{e^{i\pi/3}+1}{e^{i\pi/3}-1}\right)^2=-\dfrac14\left(\dfrac{e^{i\pi/6}+e^{-i\pi/6}}{e^{i\pi/6}-e^{-i\pi/6}}\right)^2$$ $$r=-\dfrac14\left(\dfrac{\cos(\pi/6)}{i\sin(\pi/6)}\right)^2=\dfrac14\operatorname{cot}^2(\pi/6).\tag{5}$$

Extension: it happens that what has been done for $n=6$ parabola can be done for any value of $n \ge 3$ with the following immediate extension of formula (5):

$$r=\dfrac14\operatorname{cot}^2(\pi/n).\tag{5}$$

The figures represent resp. cases $n=4$ and $n=24$.

Matlab program for the figures:

   clear all;close all;hold on;axis equal off
   set(gcf,'color','w');
   n=24;
   om=exp(2*i*pi/n);
   a=0.25*cot(pi/n)^2;
   v=1+n/5;t=-v:0.01:v;
   s=t.^2+i*t+a; % initial parabola
   t=-1:0.01:1;plot(a*exp(i*pi*t),'color','r');
   for k=1:n;
       plot(s);
       s=s*om;% 2*pi/n rotation of previous parabola
   end;