Problem 1409

Solution:

Answer: $\frac{2 \pi}{3}+\frac{\sqrt{3}}{4}$

Solution 1. An arbitrary point $\left(x_{0}, y_{0}\right)$ is contained in $S$ if and only if there exists some $(x, y)$ on the curve $\left(x, x^{2}-\frac{3}{4}\right)$ such that $\left(x-x_{0}\right)^{2}+\left(y-y_{0}\right)^{2}<y^{2}$, since the radius of the circle is at most the distance from $(x, y)$ to the $x$-axis. Some manipulation yields $x^{2}-2 y_{0}\left(x^{2}-\frac{3}{4}\right)-2 x x_{0}+x_{0}^{2}+y_{0}^{2}<0$.

Observe that $\left(x_{0}, y_{0}\right) \in S$ if and only if the optimal choice for $x$ that minimizes the expression satisfies the inequality. The minimum is achieved for $x=\frac{x_{0}}{1-2 y_{0}}$. After substituting and simplifying, we obtain $y_{0}\left(\frac{-x_{0}^{2}}{1-2 y_{0}}+x_{0}^{2}+y_{0}^{2}+\frac{3}{2} y_{0}\right)<0$. Since $y_{0}<0$ and $1-2 y_{0}>0$, we find that we need $-2 x_{0}^{2}-2 y_{0}^{2}+\frac{3}{2}-2 y_{0}>0 \Longleftrightarrow 1>x_{0}^{2}+\left(y_{0}+\frac{1}{2}\right)^{2}$.

$S$ is therefore the intersection of the lower half-plane and a circle centered at $\left(0,-\frac{1}{2}\right)$ of radius 1. This is a circle of sector angle $4 \pi / 3$ and an isosceles triangle with vertex angle $2 \pi / 3$. The sum of these areas is $\frac{2 \pi}{3}+\frac{\sqrt{3}}{4}$.

Solution 2. Let $O=\left(0,-\frac{1}{2}\right)$ and $\ell=\{y=-1\}$ be the focus and directrix of the given parabola. Let $\ell^{\prime}$ denote the $x$-axis. Note that a point $P^{\prime}$ is in $S$ iff there exists a point $P$ on the parabola in the lower half-plane for which $d\left(P, P^{\prime}\right)<d\left(P, \ell^{\prime}\right)$. However, for all such $P, d\left(P, \ell^{\prime}\right)=1-d(P, \ell)=1-d(P, O)$, which means that $P^{\prime}$ is in $S$ iff there exists a $P$ on the parabola for which $d\left(P^{\prime}, P\right)+d(P, O)<1$. It is not hard to see that this is precisely the intersection of the unit circle centered at $O$ and the lower half-plane, so now we can proceed as in Solution 1.