Putnam 1989 
Topic: Probability Calculus Geometry

Problem: A dart, thrown at random, hits a square target. Assuming that any two parts of the target of equal area are equally likely to be hit, find the probability that the point hit is nearer to the center than to any edge. Express your answer in the form $\displaystyle{\frac{a\sqrt{b} + c}{d}}$, where $a,\,b,\,c,\,d$ are integers.

Solution: Consider a $2 \times 2$ dartboard centered at the origin with vertices at $(1, 1)$, $(-1, 1)$, $(-1, -1)$, and $(1, -1)$. A point $(x, y)$ in the square is closer to the center of the board than the top edge of the board if and only if \n\[ \n\sqrt{x^2 + y^2} \leq 1 - y, \n\] or equivalently, \n\[ \ny \leq \frac{1 - x^2}{2}. \n\] A similar parabolic region arises for each of the other three sides, and therefore the region of points closer to the center than any edge has the shape of the shaded region in the following figure.\n\nLet $A$ denote the area of this region in the first quadrant bounded by $y = \frac{(1 - x^2)}{2}$ and $y = x$. Then, the probability we desire is \n\[ \n\frac{8 \cdot \text{Area } A}{4} = 2 \cdot \text{Area } A = 2 \int_0^{\sqrt{2}-1} \left( \frac{1 - x^2}{2} - x \right) dx = \frac{4\sqrt{2} - 5}{3}. \n\] Thus, the final answer is $\boxed{\frac{4\sqrt{2} - 5}{3}}$.