IMC 2009 · Day 2 · P6

We can choose a coordinate system of the space such that the line $\ell$ is the $z$-axis and the point $P$ is $(d, 0, 0)$. The distance from the point $(x, y, z)$ to $\ell$ is $\sqrt{x^2 + y^2}$, while the distance from $P$ to $X$ is $|PX| = \sqrt{(x - d)^2 + y^2 + z^2}$. Square everything to get rid of the square roots. The condition can be reformulated as follows: the square of the distance from $\ell$ to $X$ is at least $4 |PX|^2$. $$x^2 + y^2 \ge 4 ((x - d)^2 + y^2 + z^2)$$ $$0 \ge 3x^2 - 8dx + 4d^2 + 3y^2 + 4z^2$$ $$\frac{4}{3} d^2 \ge 3 \left( x - \frac{4}{3} d \right)^2 + 3y^2 + 4z^2$$ A translation by $\frac{4}{3} d$ in the $x$-direction does not change the volume, so we get $$\frac{4}{3} d^2 \ge 3 x_1^2 + 3y^2 + 4z^2$$ $$1 \ge \left( \frac{3 x_1}{2d} \right)^2 + \left( \frac{3 y}{2d} \right)^2 + \left( \frac{\sqrt{3} z}{d} \right)^2,$$ where $x_1 = x - \frac{4}{3} d$. This equation defines a solid ellipsoid in canonical form. To compute its volume, perform a linear transformation: we divide $x_1$ and $y$ by $\frac{2d}{3}$ and $z$ by $\frac{d}{\sqrt{3}}$. This changes the volume by the factor $\left( \frac{2d}{3} \right)^2 \frac{d}{\sqrt{3}} = \frac{4d^3}{9\sqrt{3}}$ and turns the ellipsoid into the unit ball of volume $\frac{4\pi}{3}$. So before the transformation the volume was $\frac{4d^3}{9\sqrt{3}} \cdot \frac{4\pi}{3} = \frac{16 \pi d^3}{27 \sqrt{3}}$.