1976 AHSME Problems/Problem 8

In order for the coordinates to have absolute values less than $4$, the points must lie in the $8$ by $8$ square passing through the points $(-4,-4), (-4, 4), (4,4)$ and $(4, -4)$. For the points to be at most $2$ units from the origin, the points must lie in a circle of radius $2$ centered at the origin. Thus, the probability is the area of the circle over the area of the square, or \[\frac{2^2 \pi}{8^2} = \frac{4 \pi}{64} = \frac{\pi}{16}\] $\boxed{D}$

~JustinLee2017

Invalid username
Login to AoPS