Six parabolas are arranged around a circle. Each parabola is congruent to the graph of y = x^2/4. Also, each parabola is tangent to the circle, and its two neighboring parabolas. Find the radius of the circle.
Let the radius of the circle be $r$. Then the horizontal distance from the center of the circle to any vertex of one of the parabolas is $a = 2r$. Let the parabolas be named $P_1$, $P_2$, $P_3$, $P_4$, $P_5$, $P_6$ in clockwise order around the circle. Let $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$ be the points where each of the parabolas is tangent to the circle, and let $H_1$, $H_2$, $H_3$, $H_4$, $H_5$, $H_6$ be the corresponding points where they are tangent to their neighbors.
[asy]
void axes(real x0, real x1, real y0, real y1)
{
draw((x0,0)--(x1,0),EndArrow);
draw((0,y0)--(0,y1),EndArrow);
label("$x$",(x1,0),E);
label("$y$",(0,y1),N);
for (int i=floor(x0)+1; i<x1; ++i)
draw((i,.1)--(i,-.1));
for (int i=floor(y0)+1; i<y1; ++i)
draw((.1,i)--(-.1,i));
}
path[] yh(real a, real b, real h, real k, real x0, real x1, bool upper=true, bool lower=true, pen color=black)
{
real f(real x) { return a/(b*b)*(x-h)*(x-h)+k; }
path[] arr;
if (upper) { arr.push( graph(f, x0, x1), color ); }
if (lower) { arr.push( graph(-f, x0, x1), color ); }
return arr;
}
path[] y(real a, real b, real h, real k, real x0, real x1, bool upper=true, bool lower=true, pen color=black)
{
real f(real x) { return a*x*x/b/b + k; }
path[] arr;
if (upper) { arr.push( graph(f, x0, x1), color ); }
if (lower) { arr.push( graph(-f, x0, x1), color ); }
return arr;
}
void tangent(pair A, real a, real b, real h, real k, bool parabola=true, pen color=black)
{
real x = A.x;
real y = A.y;
real m = -a*h/b/b;
real n = a*h*h/b/b+k;
draw((x-1.5,y-m*1.5-n)--(x+1.5,y+m*1.5+n), color, dashed);
if (parabola) {
draw(shift((x,y))*y(a,b,h,k,-2*abs(x-h)+x,2*abs(x-h)+x), color+dashed);
}
dot(A);
}
size(8cm);
axes(-5,5,-1,9);
real r = 3;
pair O = (0,0);
draw(Circle(O,r), red);
picture[] pics;
for (int i=0; i<6; ++i)
pics.push(new picture);
for (int i=0; i<6; ++i)
{
pair A = r*dir(60*i);
draw(pics[i], y(1,4,0,0,-5,5), gray(0.7));
tangent(pics[i], A, 1, 4, 0, 0);
tangent(pics[i], A, 1, 4, 0, 0, false);
tangent(pics[i], A, -1, 4, 0, 0);
tangent(pics[i], A, -1, 4, 0, 0, false);
tangent(pics[i], A, 1, 4, 0, 0, false);
tangent(pics[i], A, -1, 4, 0, 0, false);
clip(pics[i],arc(O,r,60*(i-0.5),60*(i+0.5))--cycle);
add(pics[i]);
}
[/asy]
By symmetry, $P_1$ and $P_4$ must share the same axis of symmetry, which is a vertical line passing through the center of the circle. Let this line have equation $x=c$. Let parabola $P_1$ have equation
\[y=\frac{1}{4}(x-d)^2+e.\]Since it is tangent to the circle at $A_1$ and to $P_2$ and $P_6$ at $H_1$ and $H_6$, respectively, we have
\begin{align*}
e &= r, \\
\frac{1}{4}(d-c)^2+e &= r, \\
\frac{1}{4}(d-c+a)^2+e &= \frac{1}{4}(d-c-a)^2+r, \\
\frac{1}{4}(d-c-a)^2+e &= \frac{1}{4}(d-c+a)^2+r.
\end{align*}Subtracting the first two equations yields
\[\frac{1}{4}(d-c-a)^2-\frac{1}{4}(d-c+a)^2=a.\]The two sides of this equation differ by $2a$, so canceling the $a$ gives
\[d-c = -a,\]which means that $d=c-a$.
Let parabola $P_2$ have equation
\[y=\frac{1}{4}(x-f)^2+g.\]Then we have $g=r$ and
\[\frac{1}{4}(f-d-a)^2+g = r,\]and
\[\frac{1}{4}(f-c)^2+g = \frac{1}{4}(f-d-2a)^2+r.\]Substituting $d=c-a$ and canceling the $r$ yields
\[\frac{1}{4}(a)^2+g = \frac{1}{4}(f-c+a)^2+g,\]which gives $a=0$. Therefore, we have $d=c$, $H_1=H_6$, and
\begin{align*}
\frac{1}{4}(d-c)^2+r&=r, \\
\frac{1}{4}(d-c+a)^2+r&=\frac{1}{4}(d-c-a)^2+r,
\end{align*}which simplifies to $a^2=(d-c)^2/3$. Thus, $2a=2r/3$, so $r=\boxed{9/2}$.