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 "1952 AHSME Problems/Problem 43"

(→‎Solution)
(→‎Solution)
Line 9: Line 9:
  
 
== Solution ==
 
== Solution ==
−
<math>\boxed{}</math>
+
Note that the circumference of a circle is <math>\pi*d</math>.
 +
 
 +
Let's call the diameter of the circle D. Dividing the circle's diameter into n parts means that each semicircle has diameter <math>\frac{D}{n}.</math> Since there are n circles, each with diameter <math>\frac{D}{n}</math>, the sum of the circumferences of the small circles is D. Since we are only drawing semicircles and not full circles, the requested sum is 
 +
<math>\boxed{A}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 18:45, 20 April 2020

Problem

The diameter of a circle is divided into $n$ equal parts. On each part a semicircle is constructed. As $n$ becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length: $\textbf{(A) } \qquad$ equal to the semi-circumference of the original circle $\textbf{(B) } \qquad$ equal to the diameter of the original circle $\textbf{(C) } \qquad$ greater than the diameter, but less than the semi-circumference of the original circle $\textbf{(D) } \qquad$ that is infinite $\textbf{(E) }$ greater than the semi-circumference

Solution

Note that the circumference of a circle is $\pi*d$.

Let's call the diameter of the circle D. Dividing the circle's diameter into n parts means that each semicircle has diameter $\frac{D}{n}.$ Since there are n circles, each with diameter $\frac{D}{n}$, the sum of the circumferences of the small circles is D. Since we are only drawing semicircles and not full circles, the requested sum is $\boxed{A}$.

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 42 		Followed by
Problem 44
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=1952_AHSME_Problems/Problem_43&oldid=121286"