AIME 2010 II · 第 2 题

Solution

Any point outside the square with side length $\frac{1}{3}$ that has the same center and orientation as the unit square and inside the square with side length $\frac{3}{5}$ that has the same center and orientation as the unit square has $\frac{1}{5}\le d(P)\le\frac{1}{3}$.

Since the area of the unit square is $1$, the probability of a point $P$ with $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is the area of the shaded region, which is the difference of the area of two squares.

$\left(\frac{3}{5}\right)^2-\left(\frac{1}{3}\right)^2=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}$

Thus, the answer is $56 + 225 = \boxed{281}.$

Solution 2

First, let's figure out $d(P) \geq \frac{1}{3}$ which is

$$\left(\frac{3}{5}\right)^2=\frac{9}{25}.$$ Then, $d(P) \geq \frac{1}{5}$ is a square inside $d(P) \geq \frac{1}{3}$, so

$$\left(\frac{1}{3}\right)^2=\frac{1}{9}.$$ Therefore, the probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is

$$\frac{9}{25}-\frac{1}{9}=\frac{56}{225}$$ So, the answer is $56+225=\boxed{281}$

Solution 3

First, lets assume that point $P$ is closest to a side $S$ of the square. If it is $\frac{1}{5}$ far from $S$, then it should be at least $\frac{1}{5}$ from both the adjacent sides of $S$ in the square. This leaves a segment of $1 - 2 \cdot \frac{1}{5} = \frac{3}{5}$. If the distance from $P$ to $S$ is $\frac{1}{3}$, then notice the length of the side-ways segment for $P$ is $1 - 2 \cdot \frac{1}{3} = \frac{1}{3}$. Notice that as the distance from $P$ to $S$ increases, the possible points for the side-ways decreases. This produces a trapezoid with parallel sides $\frac{3}{5}$ and $\frac{1}{3}$ with height $\frac{1}{3} - \frac{1}{5} = \frac{2}{15}$. This trapezoid has area (or probability for one side) $\frac{1}{2} \cdot \left(\frac{1}{3}+\frac{3}{5}\right)\cdot \frac{2}{15} = \frac{14}{225}$. Since the square has $4$ sides, we multiply by $4$. Hence, the probability is $\frac{56}{225}$. The answer is $\boxed{281}$. ~Saucepan_man02