1952 AHSME Problems/Problem 43
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 | |
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