# 1990 AHSME Problems/Problem 21

## Problem

Consider a pyramid $P-ABCD$ 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 $\text{(A) } \frac{\sin(\theta)}{6}\quad \text{(B) } \frac{\cot(\theta)}{6}\quad \text{(C) } \frac{1}{6\sin(\theta)}\quad \text{(D) } \frac{1-\sin(2\theta)}{6}\quad \text{(E) } \frac{\sqrt{\cos(2\theta)}}{6\sin(\theta)}$

## Solution

As the base has area $1$, the volume will be one third of the height. Drop a line from $P$ to $AB$, bisecting it at $Q$. $[asy] import three;unitsize(1cm);size(200);real h = 0.7; //currentprojection=perspective(1/3,-1,1/2); triple P = (.5,.5,h); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--P--(1,1,0)^^(1,0,0)--P--(0,1,0)); draw((.5,.5,0)--P--(.5,0,0)--cycle,dotted); dot((0,0,0));dot((1,0,0)); label("P",P,N);label("Q",(.5,0,0),S);label("B",(1,0,0),S);label("A",(0,0,0),S); [/asy]$ Then $\angle QPB=\theta$, so $\cot\theta=\frac{PQ}{BQ}=2PQ$. Therefore $PQ=\tfrac12\cot\theta$.

Now turning to the dotted triangle, by Pythagoras, the square of the pyramid's height is $$PQ^2-(\tfrac12)^2=\frac{\cos^2\theta}{4\sin^2\theta}-\frac14=\frac{\cos^2\theta-\sin^2\theta}{4\sin^2\theta}=\frac{\cos 2\theta}{4\sin^2\theta}$$ and after taking the square root and dividing by three, the result is $\fbox{E}$

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