Difference between revisions of "1980 AHSME Problems/Problem 5"
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<math> \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 </math> | <math> \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 </math> | ||
| + | |||
| + | == Solution == | ||
| + | |||
| + | We find that <math> m\angle PCQ=30^\circ </math>. Because it is a <math> 30^\circ-60^\circ-90^\circ </math> right triangle, we can let <math> PQ=x </math>, so <math> CQ=AQ=x\sqrt{3} </math>. Thus, <math> \frac{PQ}{AQ}=\frac{x}{x\sqrt{3}}=\frac{\sqrt{3}}{3}\Rightarrow\boxed{(B)} </math>. | ||
| + | |||
| + | |||
| + | == See also == | ||
| + | {{AHSME box|year=1980|num-b=4|num-a=6}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:47, 5 July 2013
Problem
If 
and 
are perpendicular diameters of circle 
, 
in 
, and 
, then the length of 
divided by the length of 
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]](http://latex.artofproblemsolving.com/6/3/e/63ec9372d36bf2cbfb2d46c5081c15178c3f7092.png)
Solution
We find that 
. Because it is a 
right triangle, we can let 
, so 
. Thus, 
.
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
