2005 AMC 12B Problems/Problem 18
Problem
Let and be points in the plane. Define as the region in the first quadrant consisting of those points such that is an acute triangle. What is the closest integer to the area of the region ?
Solution
For angle and to be acute, must be between the two lines that are perpendicular to and contain points and . For angle to be acute, first draw a triangle with as the hypotenuse. Note cannot be inside this triangle's circumscribed circle or else . Hence, the area of is the area of the large triangle minus the area of the small triangle minus the area of the circle, which is , which is approximately . The answer is .
See also
2005 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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All AMC 12 Problems and Solutions |
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