If AB and CD are perpendicular diameters of circle Q, P in AQ, and ∠ QPC=60°, then the length of PQ divided by the length of AQ is

1980 AMC 12 Problem 5

Attempt Problem 5 of the 1980 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1980 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

AB

and

CD

are perpendicular diameters of circle

Q,

P

AQ,

and

\angle QPC=60^\circ,

then the length of

PQ

divided by the length of

AQ

\frac{\sqrt3}{2}

\frac{\sqrt3}{3}

\frac{\sqrt2}{2}

\frac12

\frac23

Use the perpendicular diameters to identify a right angle in

\triangle PQC

Relate the short and long legs of the resulting

30^\circ

60^\circ

90^\circ

triangle

Triangle

PQC

is right at

Q

and has

\angle P=60^\circ,

so it is a

30^\circ

60^\circ

90^\circ

triangle. Since

QC=AQ,

\frac{PQ}{AQ}=\frac{PQ}{QC}=\frac1{\sqrt3} =\frac{\sqrt3}{3}.

Therefore, the correct answer is B.