Difference between revisions of "1980 AHSME Problems/Problem 5"
Mrdavid445 (talk | contribs) (Created page with "==Problem== If <math>AB</math> and <math>CD</math> are perpendicular diameters of circle <math>Q</math>, <math>P</math> in <math>\overline{AQ}</math>, and <math>\measuredangle Q...") |
J314andrews (talk | contribs) (Fixed wording to match original problem. Shrank diagram because it was huge.) |
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==Problem== | ==Problem== | ||
− | If <math>AB</math> and <math>CD</math> are perpendicular diameters of circle <math>Q | + | If <math>AB</math> and <math>CD</math> are perpendicular diameters of circle <math>Q</math>, and <math>\measuredangle QPC = 60^\circ</math>, then the length of <math>PQ</math> divided by the length of <math>AQ</math> is |
<asy> | <asy> | ||
+ | size(150); | ||
defaultpen(linewidth(0.7)+fontsize(10)); | 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); | pair A=(-1,0), B=(1,0), C=(0,1), D=(0,-1), Q=origin, P=(-0.5,0); | ||
Line 13: | Line 14: | ||
label("$P$", P, S); | label("$P$", P, S); | ||
label("$Q$", Q, SE); | label("$Q$", Q, SE); | ||
− | label("$60^\circ$", P+0. | + | label("$60^\circ$", P+0.05*dir(30), dir(30));</asy> |
<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 == | ||
+ | |||
+ | Notice that <math>\angle PCQ=30^\circ</math>, and therefore <math>\triangle PCQ</math> is a <math> 30^\circ-60^\circ-90^\circ </math> right triangle. Let <math>x = PQ</math>. Then <math> CQ=AQ=x\sqrt{3}</math> and <math> \frac{PQ}{AQ}=\frac{x}{x\sqrt{3}}=\boxed{(\textbf{B})\ \frac{\sqrt{3}}{3}} </math>. | ||
+ | |||
+ | == See also == | ||
+ | {{AHSME box|year=1980|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Latest revision as of 02:03, 23 July 2025
Problem
If and
are perpendicular diameters of circle
, and
, then the length of
divided by the length of
is
Solution
Notice that , and therefore
is a
right triangle. Let
. Then
and
.
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.