Question

A point is chosen at random from within a circular region. What is the probability that the point is closer to the center of the region than it is to the boundary of the region?

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

Answer: A

Concepts:geometric probability area ratio

Difficulty rating: 1260

Small Hint:

A point is closer to the center than the boundary exactly when its distance from the center is less than half the radius

Big Hint:

Compare the area of the inner circle of radius $\tfrac12$ to the whole circle of radius $1$

Solution:

Take the radius to be $1.$ A point at distance $r$ from the center is closer to the center than to the boundary when $r \lt 1 - r,$ i.e. $r \lt \tfrac12.$

The favorable region is a circle of radius $\tfrac12,$ with area $\pi(\tfrac12)^2 = \tfrac{\pi}{4},$ out of the total area $\pi.$ The probability is $\dfrac{\frac{\pi}{4}}{\pi} = \dfrac14.$

Thus, the correct answer is A .

1996 AMC 8 Problem 25

Answer: A

Difficulty rating: 1260

Small Hint:

Big Hint:

Solution:

Problem 25 in Other Years