2001 AMC 12 Problems/Problem 17
A point is selected at random from the interior of the pentagon with vertices , , , , and . What is the probability that is obtuse?
The angle is obtuse if and only if lies inside the circle with diameter . (This follows for example from the fact that the inscribed angle is half of the central angle for the same arc.)
The area of is , and the area of is .
From the Pythagorean theorem the length of is , thus the radius of the circle is , and the area of the half-circle that is inside is .
Therefore the probability that is obtuse is .
Remember that an obtuse angle is any angle greater than a right angle (90 degrees). This means that if we can figure out all of the points at which creates a right angle with and , any points within these bounds will yield an obtuse angle. Notice that by the Pythagorean theorem, the length of is . Also notice that when there is a right angle at , becomes the hypotenuse. Therefore, the sum of the squares of lengths and (or the "legs") must equal , or . This can be written as an equation.
If Point is situated at , then the distance from to is as follows: . Accordingly, the distance from to is . Since absolute value and squaring (because we will be squaring the lengths) both eliminate negatives, we can ignore the absolute value and solve. Setting the sum of the squares of these lengths equal to 20 yields . Completing the square for both and , we get , or the equation of a circle with center and radius . Graphing both the pentagon and the circle, we see that the area enclosed is a semicircle. This is or . Calculating the area of the pentagon yields . The probability of point P being within the bounds of the semicircle is then or .
Since if and only if lies on the semicircle with center and radius , the angle is obtuse if and only if the point lies inside this semicircle. The semicircle lies entirely inside the pentagon, since the distance, 3, from to is greater than the radius of the circle. Thus the probability that the angle is obtuse is the ratio of the area of the semicircle to the area of the pentagon.
Let , , , , , and . Then the area of the pentagon isand the area of the semicircle isThe probability is
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