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AMC 8 USA Math Olympiad

Perimeter of a circle : AMC 8 2013 Problem 25

This is a beautiful problem form AMC 8 which involves the concept of calculating the perimeter of a semicircle. We provide sequential hints.

What is the area and perimeter of a circle?


A circle is a curve which maintains same distance from a fixed point called center.

The perimeter of a circle is the length of the curve and area of a circle is portion of a plane bounded by the curve.

Try the problem


A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are $R_1 = 100$ inches, $R_2 = 60$ inches, and $R_3 = 80$ inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?

$\textbf{(A)}\ 238\pi \qquad \textbf{(B)}\ 240\pi \qquad \textbf{(C)}\ 260\pi \qquad \textbf{(D)}\ 280\pi \qquad \textbf{(E)}\ 500\pi$

AMC 8 2013 Problem 25

Geometry : Perimeter of a circle

7 out of 10

Mathematical Circles.

Knowledge Graph


Perimeter of a circle- knowledge graph

Use some hints


First I want to give you the formula required.

You can clearly notice that we have to find the perimeters of all of the semicircles

The perimeter of a circle of radius $r$ unit can be obtained by the formula $2\pi r$. Then can you find perimeter of the semicircles ?!!!

So using the formula, the perimeters of

Semicircle 1 =$\frac{2\pi\times 100}{2}$ inches.

Semicircle 2 =$\frac{2\pi\times 60}{2}$ inches.

Semicircle 3 =$\frac{2\pi\times 80}{2}$ inches.

So the total path covered by the ball is

$\pi(100+60+80)=240\pi$ inches.

Is it the final answer??? Or have we ignored something ?

OK !!! please notice that they have asked for the distance covered by the center of the ball.

And the ball is of radius \(2\) inches.

So for the \(1^{st}\) and \(3^{rd}\) semicircle : The center will roll along a semicircular path of radius \(R_1-2\) and \(R_3-2\).

See this image :

And for the \(2^{nd}\) semicircle : The center will roll along a semicircular path of radius \(R_2+2\).

See the image below :

So the length of the path covered by the center of the ball is

\([\pi(100-2)+\pi(60+2)+\pi(80-2)] \quad \text{inches} \\=\pi(98+62+78) \quad \text{inches}\\=238\pi \quad \text{inches}\).

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