Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1980 AHSME Problems/Problem 5"
(→‎Solution)
Line 19: Line 19:
 
== Solution ==
 
== 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>.
+
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 ==
 
== See also ==
 
{{AHSME box|year=1980|num-b=4|num-a=6}}
 
{{AHSME box|year=1980|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:29, 14 July 2025

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

Notice that $\angle PCQ=30^\circ$, and therefore $\triangle PCQ$ is a $30^\circ-60^\circ-90^\circ$ right triangle. Let $x = PQ$. Then $CQ=AQ=x\sqrt{3}$ and $\frac{PQ}{AQ}=\frac{x}{x\sqrt{3}}=\boxed{(\textbf{B})\ \frac{\sqrt{3}}{3}}$.

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. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_5&oldid=254201"