2016 AMC 12A Problems/Problem 21
Contents
Problem
A quadrilateral is inscribed in a circle of radius Three of the sides of this quadrilateral have length What is the length of its fourth side?
Solution 1
Let . Let be the center of the circle. Then is twice the altitude of . Since is isosceles we can compute its area to be , hence .
Now by Ptolemy's Theorem we have .
Solution 2
Using trig. Since all three sides equal , they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is . Similarly, the cosine is . Since there are three sides, and since ,we seek to find . First, and by Pythagorean.
See Also
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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All AMC 12 Problems and Solutions |
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