2016 AMC 12A Problem 21

Question

A quadrilateral is inscribed in a circle of radius

200\sqrt{2}.

Three of the sides of this quadrilateral have length

200.

What is the length of its fourth side?

200

200\sqrt{2}

200\sqrt{3}

300\sqrt{2}

500

Each side of length

200

subtends a central angle

\theta

with

\cos\theta=\frac34

The fourth chord has the same length as one subtending

3\theta;

use

\cos 3\theta=4\cos^3\theta-3\cos\theta

Let

\theta

be the central angle subtending a side of length

200,

with radius

R=200\sqrt2.

By the law of cosines on the isosceles triangle from the center,

\begin{gathered} 200^2=2R^2(1-\cos\theta)\\ =160000(1-\cos\theta), \end{gathered}

so

\cos\theta=\dfrac34.

The three equal sides use three consecutive arcs of angle

\theta,

so the fourth arc has angle

2\pi-3\theta.

Its chord has the same length as a chord with central angle

3\theta,

and

\begin{gathered} \cos 3\theta=4\cos^3\theta-3\cos\theta\\ =4\cdot\dfrac{27}{64}-\dfrac94\\ =-\dfrac{9}{16}. \end{gathered}

Its length squared is

\begin{gathered} 2R^2(1-\cos 3\theta)\\ =160000\left(1+\dfrac{9}{16}\right)\\ =160000\cdot\dfrac{25}{16}\\ =250000, \end{gathered}

so the fourth side is

500.

Thus, the correct answer is E.

Problem 21 in other years