Art of Problem Solving
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

1986 AHSME Problems/Problem 21: Difference between revisions

Timneh (talk | contribs)
2,460 edits
Created page with "==Problem== In the configuration below, <math>\theta</math> is measured in radians, <math>C</math> is the center of the circle, <math>BCD</math> and <math>ACE</math> are line ..."
 
m Fixed LaTeX
 
(One intermediate revision by one other user not shown)
Line 31: Line 31:
==Solution==
==Solution==


Well, the shaded sector's area is basically <math>\text{(ratio of } \theta \text{ to total angle of circle)} \times \text{(total area)} = \frac{\theta}{2\pi} \cdot (\pi r^2) = \frac{\theta}{2} \cdot (AC)^2</math>.
In addition, if you let <math>\angle{ACB} = \theta</math>, then <cmath>\tan \theta = \frac{AB}{AC}</cmath><cmath>AB = AC\tan \theta = r\tan \theta</cmath><cmath>[ABC] = \frac{AB \cdot AC}{2} = \frac{r^2\tan \theta}{2}</cmath>Then the area of that shaded thing on the left becomes <cmath>\frac{r^2\tan \theta}{2} - \frac{\theta \cdot r^2}{2}</cmath>We want this to be equal to the sector area so <cmath>\frac{r^2\tan \theta}{2} - \frac{\theta \cdot r^2}{2} = \frac{\theta \cdot r^2}{2}</cmath><cmath>\frac{r^2\tan \theta}{2} = \theta \cdot r^2</cmath><cmath>\tan \theta = 2\theta</cmath>


== See also ==
== See also ==

Latest revision as of 17:50, 1 April 2018

Problem

In the configuration below, $\theta$ is measured in radians, $C$ is the center of the circle, $BCD$ and $ACE$ are line segments and $AB$ is tangent to the circle at $A$.

[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair A=(0,-1), E=(0,1), C=(0,0), D=dir(10), F=dir(190), B=(-1/sin(10*pi/180))*dir(10); fill(Arc((0,0),1,10,90)--C--D--cycle,mediumgray); fill(Arc((0,0),1,190,270)--B--F--cycle,mediumgray); draw(unitcircle); draw(A--B--D^^A--E); label("$A$",A,S); label("$B$",B,W); label("$C$",C,SE); label("$\theta$",C,SW); label("$D$",D,NE); label("$E$",E,N); [/asy]

A necessary and sufficient condition for the equality of the two shaded areas, given $0 < \theta < \frac{\pi}{2}$, is

$\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad \textbf{(C)}\ \tan\theta = 4\theta\qquad \textbf{(D)}\ \tan 2\theta =\theta\qquad\\ \textbf{(E)}\ \tan\frac{\theta}{2}=\theta$


Solution

Well, the shaded sector's area is basically $\text{(ratio of } \theta \text{ to total angle of circle)} \times \text{(total area)} = \frac{\theta}{2\pi} \cdot (\pi r^2) = \frac{\theta}{2} \cdot (AC)^2$.

In addition, if you let $\angle{ACB} = \theta$, then \[\tan \theta = \frac{AB}{AC}\]\[AB = AC\tan \theta = r\tan \theta\]\[[ABC] = \frac{AB \cdot AC}{2} = \frac{r^2\tan \theta}{2}\]Then the area of that shaded thing on the left becomes \[\frac{r^2\tan \theta}{2} - \frac{\theta \cdot r^2}{2}\]We want this to be equal to the sector area so \[\frac{r^2\tan \theta}{2} - \frac{\theta \cdot r^2}{2} = \frac{\theta \cdot r^2}{2}\]\[\frac{r^2\tan \theta}{2} = \theta \cdot r^2\]\[\tan \theta = 2\theta\]

See also

1986 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AHSME_Problems/Problem_21&oldid=93702"