2016 AMC 12A Problems/Problem 21
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 triange 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.
\[\sin 3\theta=\sin(2\theta+\theta)=\sin 2\theta\cos\theta+\sin \theta\cos 2\theta=\frac{\sqrt{7}}{4}\left(\frac{\sqrt{14}}{4}\right)+\frac\sqrt{2}}{4}\left(\frac{3}{4}\right)=\frac{7\sqrt{2}+3\sqrt{2}}{16}=\frac{5\sqrt{2}}{8}\] (Error compiling LaTeX. Unknown error_msg)
.
\[2r\sin 3\theta=2\left(200\sqrt{2}\right)\left(\frac5\sqrt{2}{}{8}\right)=400\sqrt{2}\left(\frac{5\sqrt{2}}{8}\right)=\frac{800\cdot 5}{8}=\boxed{\textbf{(C)}\text{ 500}}\] (Error compiling LaTeX. Unknown error_msg)