# 1952 AHSME Problems/Problem 43

## 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 half the circumference of a circle with diameter $d$ is $\frac{\pi*d}{2}$.

Let's call the diameter of the circle D. Dividing the circle's diameter into n parts means that each semicircle has diameter $\frac{D}{n}$, and thus each semicircle measures $\frac{D*pi}{n*2}$. The total sum of those is $n*\frac{D*pi}{n*2}=\frac{D*pi}{2}$, and since that is the exact expression for the semi-circumference of the original circle, the answer is $\boxed{A}$.

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