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. enter image description here

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.


  • 1
    $\begingroup$ Hint:Let the radius be $r$ then we have to find $r$ such that the pair of tangents to $y=x^2$ from the point$(-r,0)$ make an angle of $60^{\circ}$ which is easy coordinate geometry $\endgroup$ Commented Mar 2, 2021 at 16:22

3 Answers 3


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:

enter image description here

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 $$r = \frac{3}{2} - \frac{3}{4} = \frac{3}{4}.$$

  • $\begingroup$ You may be interested by the solution I just gave using complex numbers geometry. $\endgroup$
    – Jean Marie
    Commented Mar 10, 2021 at 10:11

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.

enter image description here

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 $|KV|=3/4$.

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

enter image description here

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. $\square$

  • $\begingroup$ I have written a generalization of the given issue with any number of parabolas. $\endgroup$
    – Jean Marie
    Commented Mar 11, 2021 at 17:39
  • $\begingroup$ @JeanMarie: Indeed! ... In my figure, taking $\theta:=\angle FKT$ to be arbitrary, a little angle chasing shows that $\square KFTT'$ is a rhombus (but with a non-congruent diagonal $\overline{FT'}$), and that $\angle FT'D=\theta$. Then $$|DF|=|FT'|\sin\theta=2|KF|\sin^2\theta$$ so that $$\frac{|KV|}{|VF|}=\frac{|KF|-\frac12|DF|}{\frac12|DF|}=\frac{\cos^2\theta}{\sin^2\theta}\quad\to\quad |KV|=|VF|\cot^2\theta=\frac14\cot^2\theta$$ and we agree. :) $\endgroup$
    – Blue
    Commented Mar 11, 2021 at 18:23
  • $\begingroup$ Thank you for giving a "geometric flavor" to this $\dfrac14 \operatorname{cot}^2\theta$... $\endgroup$
    – Jean Marie
    Commented Mar 11, 2021 at 18:28
  • $\begingroup$ @JeanMarie: Happy to help. :) ... BTW: Here's a more-direct geometric calculation: Taking $M$ the midpoint of $\overline{FT'}$ (and noting that $\overline{MV}$, as a midpoint segment of $\triangle FT'D$, is parallel to $\overline{T'D}$) we have $$|FV|=|MV|\tan\theta=|KV|\tan^2\theta$$ $\endgroup$
    – Blue
    Commented Mar 11, 2021 at 18:40
  • $\begingroup$ Indeed much more direct. $\endgroup$
    – Jean Marie
    Commented Mar 11, 2021 at 18:42

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:


This equation can be written :


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



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

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):


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

enter image description here

enter image description here

Matlab program for the figures:

   clear all;close all;hold on;axis equal off
   s=t.^2+i*t+a; % initial parabola
   for k=1:n;
       s=s*om;% 2*pi/n rotation of previous parabola

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