Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.2 AMC 10/12 Find the answer

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

Pick one

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 $Q$ .

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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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.