2018 AMC 10B Problems/Problem 22
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?
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 , which is
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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 .
Solution 3 (Bogus, not legitimate solution)
Similarly to Solution 1, note that The Pythagorean Inequality states that in an obtuse triangle, . We can now complementary count to find the probability by reversing the inequality into: Since it is given that one side is equal to , and the closed interval is from , we can say without loss of generality that .
The probability that and sum to is equal to when both and are (Edit: this is not true, as all the points (x,y) which lie on the unit circle centered at the origin satisfy ). We can estimate to be . Now we know the probability that is just when and/or equal any value between and .
The probability that or lie between and is . This gives us .
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