2020 AIME II Problem 2

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2.

Let PP be a point chosen uniformly at random in the interior of the unit square with vertices at (0,0),(0, 0), (1,0),(1, 0), (1,1),(1, 1), and (0,1).(0, 1). The probability that the slope of the line determined by PP and the point (58,38)\left(\frac{5}{8}, \frac{3}{8}\right) is greater than or equal to 12\frac{1}{2} can be written as mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Answer: 171
Concepts:geometric probabilityslopetrapezoid
Difficulty rating: 2110
Solution:

Let Q=(58,38)Q = \left(\frac{5}{8}, \frac{3}{8}\right) and P=(x,y).P = (x, y). The slope condition y3/8x5/812\frac{y - 3/8}{x - 5/8} \ge \frac{1}{2} becomes yx2+116y \ge \frac{x}{2} + \frac{1}{16} when x>58,x \gt \frac{5}{8}, and yx2+116y \le \frac{x}{2} + \frac{1}{16} when x<58x \lt \frac{5}{8} (multiplying by the negative quantity x58x - \frac{5}{8} reverses the inequality).

For x>58,x \gt \frac{5}{8}, the region above the line y=x2+116y = \frac{x}{2} + \frac{1}{16} inside the square is a trapezoid with parallel vertical sides of lengths 58\frac{5}{8} (at x=58x = \frac{5}{8}) and 716\frac{7}{16} (at x=1x = 1) and width 38,\frac{3}{8}, with area 385/8+7/162=51256.\frac{3}{8} \cdot \frac{5/8 + 7/16}{2} = \frac{51}{256}. For x<58,x \lt \frac{5}{8}, the region below the line is a trapezoid with parallel sides 116\frac{1}{16} (at x=0x = 0) and 38\frac{3}{8} (at x=58x = \frac{5}{8}) and width 58,\frac{5}{8}, with area 581/16+3/82=35256.\frac{5}{8} \cdot \frac{1/16 + 3/8}{2} = \frac{35}{256}.

The probability is 51256+35256=86256=43128,\frac{51}{256} + \frac{35}{256} = \frac{86}{256} = \frac{43}{128}, so m+n=43+128=171.m + n = 43 + 128 = 171.

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