1996 AMC 8 Problem 25

Attempt Problem 25 of the 1996 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1996 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

25.

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?
14\dfrac{1}{4}
13\dfrac{1}{3}
12\dfrac{1}{2}
23\dfrac{2}{3}
34\dfrac{3}{4}
Answer: A
Concepts:geometric probabilityarea ratio
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 12\tfrac12 to the whole circle of radius 11
Solution:
Take the radius to be 1.1. A point at distance rr from the center is closer to the center than to the boundary when r<1−r,r \lt 1 - r, i.e. r<12.r \lt \tfrac12.
The favorable region is a circle of radius 12,\tfrac12, with area π(12)2=π4,\pi(\tfrac12)^2 = \tfrac{\pi}{4}, out of the total area π.\pi. The probability is π4π=14.\dfrac{\frac{\pi}{4}}{\pi} = \dfrac14.
Thus, the correct answer is A .
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