1980 AHSME Problems/Problem 5

Problem

If $AB$ and $CD$ are perpendicular diameters of circle $Q$ , $P$ in $\overline{AQ}$ , and $\measuredangle QPC = 60^\circ$ , then the length of $PQ$ divided by the length of $AQ$ is

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(-1,0), B=(1,0), C=(0,1), D=(0,-1), Q=origin, P=(-0.5,0); draw(P--C--D^^A--B^^Circle(Q,1)); label("$A$", A, W); label("$B$", B, E); label("$C$", C, N); label("$D$", D, S); label("$P$", P, S); label("$Q$", Q, SE); label("$60^\circ$", P+0.0.5*dir(30), dir(30));[/asy]$

$\text{(A)} \ \frac{\sqrt{3}}{2} \qquad \text{(B)} \ \frac{\sqrt{3}}{3} \qquad \text{(C)} \ \frac{\sqrt{2}}{2} \qquad \text{(D)} \ \frac12 \qquad \text{(E)} \ \frac23$

Solution

We find that $m\angle PCQ=30^\circ$ . Because it is a $30^\circ-60^\circ-90^\circ$ right triangle, we can let $PQ=x$ , so $CQ=AQ=x\sqrt{3}$ . Thus, $\frac{PQ}{AQ}=\frac{x}{x\sqrt{3}}=\frac{\sqrt{3}}{3}\Rightarrow\boxed{(B)}$ .

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1980 AHSME Problems/Problem 5

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