, 2014

Solution:
Answer: $\frac{2 \sqrt{3}-1}{3}$

We first cut the regular hexagon $\mathcal{H}$ by segments connecting its center to each vertex into six different equilateral triangles with side lengths $1$. Therefore, each point inside $\mathcal{H}$ is contained in some equilateral triangle. We first see that for each point inside an equilateral triangle, the radius of the largest circle with center $P$ which is contained in $\mathcal{H}$ equals the shortest distance from $P$ to the nearest side of the hexagon, which is also a side of the triangle in which it is contained.

Consider that the height of each triangle is $\frac{\sqrt{3}}{2}$. Therefore, the region inside the triangle containing all points with distance more than $\frac{1}{2}$ to the side of the hexagon is an equilateral triangle with a height of $\frac{\sqrt{3}-1}{2}$. Consequently, the area inside the triangle containing all points with distance less than $\frac{1}{2}$ to the side of the hexagon has area $\frac{\sqrt{3}}{4}\left(1-\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)^{2}\right)=\frac{\sqrt{3}}{4} \cdot\left(\frac{2 \sqrt{3}-1}{3}\right)$. This is of the ratio $\frac{2 \sqrt{3}-1}{3}$ to the area of the triangle, which is $\frac{\sqrt{3}}{4}$.

Since all triangles are identical and the point $P$ is picked uniformly within $\mathcal{H}$, the probability that the radius of the largest circle with center $P$ which is contained in $\mathcal{H}$ is less than $\frac{1}{2}$ is $\frac{2 \sqrt{3}-1}{3}$, as desired.