Difference between revisions of "1952 AHSME Problems/Problem 43"
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The diameter of a circle is divided into <math>n</math> equal parts. On each part a semicircle is constructed. As <math>n</math> becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length: | The diameter of a circle is divided into <math>n</math> equal parts. On each part a semicircle is constructed. As <math>n</math> becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length: | ||
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<math>\textbf{(A) } \qquad</math> equal to the semi-circumference of the original circle | <math>\textbf{(A) } \qquad</math> equal to the semi-circumference of the original circle | ||
<math>\textbf{(B) } \qquad</math> equal to the diameter of the original circle | <math>\textbf{(B) } \qquad</math> equal to the diameter of the original circle | ||
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== Solution == | == Solution == | ||
| − | Note that the circumference of a circle is <math>\pi*d</math>. | + | Note that the half the circumference of a circle with diameter <math>d</math> is <math>\frac{\pi*d}{2}</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} | + | 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>, and thus each semicircle measures <math>\frac{D*pi}{n*2}</math>. The total sum of those is <math>n*\frac{D*pi}{n*2}=\frac{D*pi}{2}</math>, and since that is the exact expression for the semi-circumference of the original circle, the answer is <math>\boxed{A}</math>. |
| − | <math>\boxed{A}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 12:28, 29 December 2024
Problem
The diameter of a circle is divided into 
equal parts. On each part a semicircle is constructed. As becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length:
equal to the semi-circumference of the original circle
equal to the diameter of the original circle
greater than the diameter, but less than the semi-circumference of the original circle
that is infinite
greater than the semi-circumference
Solution
Note that the half the circumference of a circle with diameter 
is 
.
Let's call the diameter of the circle D. Dividing the circle's diameter into n parts means that each semicircle has diameter 
, and thus each semicircle measures 
. The total sum of those is 
, and since that is the exact expression for the semi-circumference of the original circle, the answer is 
.
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
