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Difference between revisions of "1952 AHSME Problems/Problem 43"

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== Solution ==
 
== Solution ==
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<math>\boxed{}</math>
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Note that the circumference of a circle is <math>\pi*d</math>.
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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 
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<math>\boxed{A}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 17:45, 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

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