2016 AMC 10A Problem 24

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 the fourth side?

200

200\sqrt{2}

200\sqrt{3}

300\sqrt{2}

500

Equal chords give equal central angles

If a

200

-unit chord subtends angle

2\theta

, use the chord formula and

\sin(3\theta)

Label the quadrilateral

ABCD

, with

AB=BC=CD=200

. Let each of these equal chords subtend central angle

2\theta

. The chord formula gives

200=2(200\sqrt2)\sin\theta,

so

\sin\theta=\frac{1}{2\sqrt2}

. In particular,

\theta<30^\circ

, so the smaller central angle subtended by

AD

is

6\theta<180^\circ

.

Therefore another use of the chord formula gives

AD=2(200\sqrt2)\sin(3\theta).

Using

\sin(3\theta)=3\sin\theta-4\sin^3\theta

, we obtain

\begin{aligned} \sin(3\theta) &=\frac{3}{2\sqrt2}-\frac{1}{4\sqrt2} \\ &=\frac{5}{4\sqrt2}. \end{aligned}

Hence

AD=400\sqrt2\cdot\frac{5}{4\sqrt2}=500.

Thus, the correct answer is E.

Problem 24 in other years