Parabolic Eye

Call the black parabola $P_1$ , the blue parabola $P_2$ , and the circle $C$ .

These curves have the following equations:

$\begin{aligned} P_1&: \quad y=x^2 \\ P_2&: \quad y=a-bx^2 \\ C&: \quad x^2+(y-c)^2=r^2 \end{aligned}$

where $r$ is the radius of $C$ and its centre (which, by symmetry, is on the $y$ -axis) is $(0,c)$ .

$P_1$ and $P_2$ intersect when $x^2=a-bx^2$ , that is when $x=\pm \sqrt{\frac{a}{b+1}}$

The gradient of $P_1$ is $2x$ ; the gradient of $P_2$ is $-2bx$ . Since they are orthogonal, the product of their gradients at their intersections is $-1$ ; by symmetry we can just look at the positive intersection to get

$\left( 2\sqrt{\frac{a}{b+1}} \right) \cdot \left( -2b \sqrt{\frac{a}{b+1}} \right)=-1$

which after tidying up becomes

$b=\frac{1}{4a-1}$

which means the positive intersection point is at $x=\frac12 \sqrt{4a-1}$ .

Now let's look at $C$ . It touches $P_2$ at the point $(0,a)$ ; so we have

$c=a-r$

(where we've used the geometry to discard the case $c=a+r$ .)

It is also tangent to $P_1$ . The intersections of $C$ and $P_1$ are found by plugging $y=x^2$ into the equation for $C$ ; that is at the intersection points

$y+(y-c)^2=r^2$

Expanding, this is

$y^2+(1-2c)y+c^2-r^2=0$

Since $C$ is tangent to $P_1$ , we need this quadratic to have exactly one real root; so its discriminant must be zero, ie

$(1-2c)^2=4\left(c^2-r^2\right)$

Tidying again, this gives

$1-4c+4r^2=0$

Putting these equations together, we can express $a$ in terms of $r$ :

$a=\left(r+\frac12 \right)^2$

Finally, let $A$ be the area between the two parabolas:

$\begin{aligned} A&=2\int_{0}^{\frac12 \sqrt{4a-1}} a-\frac{x^2}{4a-1}-x^2 dx \\ &=\frac{2a}{3} \sqrt{4a-1} \end{aligned}$

The fraction we wish to maximise is $\frac{\pi r^2}{A}$ .

Since we're after the diameter of the circle, it makes sense to write everything in terms of $r$ ; the area fraction is then

$f(r)=\frac{3\pi r^2 }{(2r+1)^2 \sqrt{r(r+1)}}$

To find the maximum, set $f'(r)=0$ ; this gives (after some messy differentiation)

$f'(r) = -\frac{3 \pi r^2 (4 r^2 - 3)}{2 (r (1 + r))^{3/2} (1 + 2 r)^3} = 0$

which, when we discount $r \leq 0$ becomes the much nicer $4r^2-3=0$ .

The question asks for the square of the diameter; this is just $4r^2=\boxed3$ . The fraction of the area between the two parabolas occupied by the circle is around $74.5\%$ .

Let $A(0,a)$ be the vertex of $y=a - bx^2$ , $B$ and $B'$ be the points the two parabolas intersect, $C(0,y_0)$ the center of the circle and $r$ its radiux, and $P(x,y)$ the point in the positive quadrant where the circle touches $y=x^2$ .

The point $P(x,y)$ satisfies both $\begin{cases} x^2 + (y-y_0)^2 = r^2 \\ y = x^2 \end{cases} \implies y + (y-y_0)^2 = r^2$ . The gradients of the circle and parabola at $P$ are the same. For the circle $\dfrac {dy}{dx} + 2(y-y_0)\dfrac {dy}{dx} = 0$ . Since $\dfrac {dy}{dx} = 2x \ne 0$ , $\implies y-y_0 = - \dfrac 12$ and $y = \blue{y_0} - \dfrac 12 = \blue{a - r} - \dfrac 12$ , as $y_0 = a - r$ . Then we have:

$\begin{aligned} y + (y-y_0)^2 & = r^2 \\ a - r - \frac 12 + \frac 14 & = r^2 \\ \implies a & = r^2 + r + \frac 14 = \left(r + \frac 12\right)^2 \end{aligned}$

When the two parabolas intersect, we have $a - bx^2 = x^2 \implies x_B = \pm \sqrt{\dfrac a{b+1}}$ . Since the parabolas intersect orthogonally, the product of their gradients is $-1$ . That is $(-2bx_B) (2x_B) = -1$ or:

$\begin{aligned} \left(-2b\sqrt{\frac a{b+1}}\right) \left(2\sqrt{\frac a{b+1}}\right) & = -1 \\ 4ab & = b+1 \\ \implies b & = \frac 1{4a-1} \end{aligned}$

Note that the area enclosed by a parabola is (\frac 23) the area of rectangle inscribing it. Therefore the area of the area enclosed by the two parabolas is directly proportional to $a\sqrt{\dfrac a{b+1}} = a \sqrt{\dfrac a{4ab}} = a \sqrt{a-\dfrac 14}$ . Since the area of the circle is directly proportional to $r^2$ , we need to find $r$ when $\dfrac {r^2}{a \sqrt{a-\frac 14}}$ is maximum or $\dfrac {a \sqrt{a-\frac 14}}{r^2}$ is minimum.

$\begin{aligned} f(r) & = \frac {a \sqrt{a-\frac 14}}{r^2} = \frac {\left(r+\frac 12\right)^2\sqrt{r(r+1}}{r^2} = \left(r+1+\frac 1{4r}\right)\sqrt{1+\frac 1r} \\ \frac {df(r)}{dr} & = \left(1-\frac 1{4r^2}\right)\sqrt{1+\frac 1r} - \frac {r+1+\frac 1{4r}}{2r^2\sqrt{1+\frac 1r}} \end{aligned}$

Equating $\dfrac {df(r)}{dr}$ to zero.

$\begin{aligned} 2r^2\left(1-\frac 1{4r^2}\right)\left(1+\frac 1r\right) & = r+1+\frac 1{4r} \\ 8r^3 + 4r^2 - 6r - 3 & = 0 \\ (2r-\sqrt 3)(2r+\sqrt 3)(2r+1) & = 0 & \small \blue{\text{Since }r > 0} \\ \implies r & = \frac {\sqrt 3}2 \\ 2r & = \sqrt 3 \end{aligned}$

Therefore, $p=\boxed 3$ .