1971 AMC 12 Problem 31

Question

Quadrilateral

ABCD

is inscribed in a circle with side

AD,

a diameter of length

4.

If sides

AB

and

BC

each have length

1,

then side

CD

has length:

\frac{7}{2}

\frac{5\sqrt2}{2}

\sqrt{11}

\sqrt{13}

2\sqrt3

Equal chords

AB

and

BC

subtend equal central angles

If half of either central angle is

t

, then

4\sin t=1

and

CD=4\cos(2t)

The circle has radius

2.

Let the central angles subtending the equal chords

AB

and

BC

each be

2t.

Then

1=4\sin t,

so

\sin t=\frac{1}{4}.

The remaining central angle from

C

to

D

along the semicircle is

\pi-4t,

hence

\begin{aligned} CD&=4\sin\left(\frac{\pi-4t}{2}\right)\\ &=4\cos(2t)\\ &=4\left(1-2\cdot\frac1{16}\right)\\ &=\frac72. \end{aligned}

Therefore, the correct answer is A.

Problem 31 in other years