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 "1953 AHSME Problems/Problem 27"
m
 
(One intermediate revision by the same user not shown)
Line 1: Line 1:
 +
==Problem==
 
The radius of the first circle is <math>1</math> inch, that of the second <math>\frac{1}{2}</math> inch, that of the third <math>\frac{1}{4}</math> inch and so on indefinitely. The sum of the areas of the circles is:
 
The radius of the first circle is <math>1</math> inch, that of the second <math>\frac{1}{2}</math> inch, that of the third <math>\frac{1}{4}</math> inch and so on indefinitely. The sum of the areas of the circles is:
  
 
<math>\textbf{(A)}\ \frac{3\pi}{4} \qquad \textbf{(B)}\ 1.3\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ \frac{4\pi}{3}\qquad \textbf{(E)}\ \text{none of these}</math>
 
<math>\textbf{(A)}\ \frac{3\pi}{4} \qquad \textbf{(B)}\ 1.3\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ \frac{4\pi}{3}\qquad \textbf{(E)}\ \text{none of these}</math>
  
−
Note the areas of these triangles is <math>1\pi</math>, <math>\frac{\pi}{4}</math>, <math>\frac{\pi}{16}, \dots</math>. The sum of these areas will thus be <math>\pi\left(1+\frac{1}{4}+\frac{1}{16}+\dots\right)</math>. We use the formula for an infinite geometric series to get the sum of the areas will be <math>\pi\left(\frac{1}{1-\frac{1}{4}}\right)</math>, or <math>\frac{4\pi}{3}</math>.  <math>\boxed{D}</math>
+
==Solution==
 +
Note the areas of these circles is <math>1\pi</math>, <math>\frac{\pi}{4}</math>, <math>\frac{\pi}{16}, \dots</math>. The sum of these areas will thus be <math>\pi\left(1+\frac{1}{4}+\frac{1}{16}+\dots\right)</math>. We use the formula for an infinite geometric series to get the sum of the areas will be <math>\pi\left(\frac{1}{1-\frac{1}{4}}\right)</math>, or <math>\frac{4\pi}{3}</math>.  <math>\boxed{D}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 00:38, 4 February 2020

Problem

The radius of the first circle is $1$ inch, that of the second $\frac{1}{2}$ inch, that of the third $\frac{1}{4}$ inch and so on indefinitely. The sum of the areas of the circles is:

$\textbf{(A)}\ \frac{3\pi}{4} \qquad \textbf{(B)}\ 1.3\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ \frac{4\pi}{3}\qquad \textbf{(E)}\ \text{none of these}$

Solution

Note the areas of these circles is $1\pi$, $\frac{\pi}{4}$, $\frac{\pi}{16}, \dots$. The sum of these areas will thus be $\pi\left(1+\frac{1}{4}+\frac{1}{16}+\dots\right)$. We use the formula for an infinite geometric series to get the sum of the areas will be $\pi\left(\frac{1}{1-\frac{1}{4}}\right)$, or $\frac{4\pi}{3}$. $\boxed{D}$

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_27&oldid=116813"