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
![[asy] pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D("A", dir(50), dir(50)); pair B = D("B", dir(90), dir(90)); pair C = D("C", dir(130), dir(130)); pair D = D("D", dir(170), dir(170)); pair O = D("O", (0,0), dir(-90)); draw(A--C, red); draw(B--D, blue+dashed); draw(A--B--C--D--cycle); draw(A--O--C); draw(O--B); [/asy]](http://latex.artofproblemsolving.com/b/d/9/bd9ef315c46cee0740c4c3c98bc72bcb682662e7.png)
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 
=\frac{\sqrt{2}}{4}
\frac{100\sqrt{7}}{200\sqrt{2}}=\frac{\sqrt{14}}{4}
\sin\theta=\sin\left(180-\theta\right)
2r\sin 3\theta
\sin 2\theta=2\sin\theta\cos\theta=2\cdot\left(\frac{\sqrt{2}}{4}\right)\left(\frac{\sqrt{14}}{4}\right)=\frac{2\sqrt{2}\sqrt{14}}{16}=\frac{\sqrt{7}}{4}
\cos 2\theta=\frac{3}{4}$ 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)
