1990 AMC 12 Problem 21

Question

Consider a pyramid

P\text{-}ABCD

whose base

ABCD

is square and whose vertex

P

is equidistant from

A,

B,

C,

and

D.

If

AB=1

and

\angle APB=2\theta,

then the volume of the pyramid is

\frac{\sin\theta}{6}

\frac{\cot\theta}{6}

\frac1{6\sin\theta}

\frac{1-\sin2\theta}{6}

\frac{\sqrt{\cos2\theta}}{6\sin\theta}

In isosceles triangle

APB,

express

PA

using the chord

AB

Relate

PA

to the pyramid height and the center-to-vertex distance of the square

Let

R=PA=PB.

In isosceles triangle

APB,

1=AB=2R\sin\theta,

so

R=\frac{1}{2\sin\theta}.

If

h

is the pyramid height, the horizontal distance from the square’s center to

A

is

\frac{1}{\sqrt2},

hence

\begin{aligned} h^2 &=R^2-\frac12\\ &=\frac{\cos2\theta}{4\sin^2\theta}. \end{aligned}

Thus

h=\frac{\sqrt{\cos2\theta}}{2\sin\theta},

and the volume is

\frac13(1)h=\frac{\sqrt{\cos2\theta}}{6\sin\theta}.

Thus the correct answer is E.

Problem 21 in other years