2018 AMC 10B Problems/Problem 22
Contents
Problem
Real numbers and are chosen independently and uniformly at random from the interval . Which of the following numbers is closest to the probability that and are the side lengths of an obtuse triangle?
Video Solution & More by MegaMath
https://www.youtube.com/watch?v=d6oFfN5N_70
Solution 1
The Pythagorean Inequality tells us that in an obtuse triangle, . The triangle inequality tells us that . So, we have two inequalities: The first equation is of a circle with radius , and the second equation is a line from to . So, the area is which is approximately
Solution 2 (Trig)
Note that the obtuse angle in the triangle has to be opposite the side that is always length . This is because the largest angle is always opposite the largest side, and if two sides of the triangle were , the last side would have to be greater than to make an obtuse triangle. Using this observation, we can set up a law of cosines where the angle is opposite :
where and are the sides that go from and is the angle opposite the side of length .
By isolating , we get:
For to be obtuse, must be negative. Therefore, is negative. Since and must be positive, must be negative, so we must make positive. From here, we can set up the inequality Additionally, to satisfy the definition of a triangle, we need: The solution should be the overlap between the two equations in the first quadrant.
By observing that is the equation for a circle, the amount that is in the first quadrant is . The line can also be seen as a chord that goes from to . By cutting off the triangle of area that is not part of the overlap, we get .
Video Solution by OmegaLearn
https://youtu.be/LwtoLiBwO-E?t=316
~ pi_is_3.14
Video Solution
https://youtu.be/tWkE_c3Fa3I -- Geometric Probability and Inequalities!
https://www.youtube.com/watch?v=GHAMU60rI5c
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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