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A point Q is chosen at random inside triangle XYZ which is equilateral. Find the probability that Q is closer to the center of the triangle than to X, Y, or Z. 

(In other words, let O be the center of the triangle. Find the probability that OQ is shorter than all of XQ, YZ and ZQ.)

 Feb 14, 2020
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Let the side length of the equilateral triangle be 3.  Then the points that are closer to the center of the triangle than to X, Y, Z is the circle centered at O with radius 1, so the probability is (pi)/(sqrt(3)/4*9) = (4*pi*sqrt(3))/27.

 Feb 14, 2020

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