2018 AMC 10B Problem 22

Below is the professionally curated solution for Problem 22 of the 2018 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 10B solutions, or check the answer key.

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Concepts:geometric probabilitylaw of cosines

Difficulty rating: 2100

22.

Real numbers xx and yy are chosen independently and uniformly at random from the interval [0,1].[0, 1]. Which of the following numbers is closest to the probability that x,x, y,y, and 11 are the side lengths of an obtuse triangle?

0.210.21

0.250.25

0.290.29

0.500.50

0.790.79

Solution:

The three lengths x,y,1x, y, 1 make a triangle iff x+y>1.x + y > 1. Since 11 is the longest side, that triangle is obtuse iff x2+y2<1.x^2 + y^2 < 1. So in the unit square we want the region inside the quarter circle x2+y2=1x^2 + y^2 = 1 but above the line x+y=1.x + y = 1. That's the quarter disk with the right triangle under the chord removed: π4120.285.\tfrac{\pi}{4} - \tfrac12 \approx 0.285. The closest choice is 0.29.0.29. Therefore, the answer is C.

Problem 22 in Other Years