Two Circles Puzzle

The center of the small circle will move along a circle of radius 5, and the small circle will revolve five times as it moves once around the large circle. (One might at first expect that the small circle revolves four times; but watch the dot carefully!) Thus we can parameterize the "blue curve" as $x=5\cos(t)-\cos(5t)$ and $y=5\sin(t)-\sin(5t)$ for $0\leq t\leq2\pi$ . Using a bit of trigonometry, we find the speed to be $v =10|\sin(2t)|$ , and the arc length is $L=\int_{0}^{2\pi} 10|\sin(2t)|dt=\boxed{40}cm$ .

If R= (m - 1)r, where r=1 and m=5 in this example, then the length of the epicycloid = 8mr, therefore,

Answer=40

I like to have an angle of 0 at the positive x-axis, and start at the tip of a petal, so I have the picture turned over 45°. The center of the orange circle makes a circular motion around the origin, which can be parametrized as (x,y) = 5(cos t, sin t). In the same time, the orange circle makes 5 full circles (4 and an additional one for the revolution). The net motion of the blue dot is described by $\vec{x} = (5 \cos t + \cos 5t, 5 \sin t + \sin 5t)$ . The velocity then is given by the time derivative: the vector $\vec{v} = 5(-\sin t - \sin 5t, \cos t + \cos 5t)$ . The speed then is $|\vec{v}|= 5\sqrt{\sin^2 t + 2\sin t \sin {5t} + \sin^2 {5t} + \cos^2 t +  2 \cos t \cos {5t} + \cos^2 {5t}}$ $= 5\sqrt{1 + 2\sin t \sin {5t} + 2\cos t \cos {5t} +1}$ $= 5\sqrt{2}\sqrt{1+ \cos 5t \cos {(-t)} - \sin {5t} \sin {(-t)} )} = 5\sqrt{2}\sqrt{1 + \cos{(5t-t)} } =  5\sqrt{2}\sqrt{1 + \cos {4t} }$ using $\sin(-t) = -\sin(t)$ and $\cos(-t)=\cos(t)$ .

Integrating the speed over time gives the distance traveled:

$L=5\sqrt{2}\int_{0}^{2\pi}{\sqrt{1+\cos{4t}}}dt$ . Set $x=4t$ and use the range where sin is nonnegative:

$5\sqrt{2}\int_{0}^{8\pi}{\sqrt{1+\cos{x}}}\frac{dx}{4}$ $=\frac{8×5}{4}\sqrt{2}\int_{0}^{\pi}{\sqrt{1+\cos{x}}}dx$

And $\int{\sqrt{1+\cos x}} dx = 2 \sqrt{1- \cos x}$ on the interval $[0,\pi]$

So our expression is $L = 10\sqrt{2} (2\sqrt{1--1} - 2\sqrt{1-1} )= 40$

Having been told the answer was an integer. I carefully drew the shape and examined its length I got 39.91 so the answer was 40.

This kind of shape is called an epicycloid .

Let's say we have an epicycloid whose static circle has radius $a$ and whose turning circle has radius $b$ .

To find the general expression for the epicycloid, consider the movement of the moving circle. Its center goes around another circunference with radius $a+b$ , and along the movement the smaller circle completes $\frac{a}{b}$ rotations corresponding to each curve "petal" plus one full rotation around the static circle, in a total of $\frac{a+b}{b}$ laps.

We can then write the parametric formula as:

$P(t) = (x(t), y(t)) = \left((a+b)\cos t - b\cos \left(\frac{a+b}{b}t \right),(a+b)\sin t - b\sin \left(\frac{a+b}{b}t \right) \right), t \in [0, 2\pi]$

Consider a tiny segment of the curve, short enough we can approximate it as a line segment. We've ${\Delta s}^{2} = {\Delta x}^{2} + {\Delta y}^{2}$ .

So we can find the derivative of P with respect to both $x$ and $y$ and integrate using the above formula to find the total perimeter of the curve.

$( \dot{P_{x}},\dot{P_{y}} ) = \left(-(a+b)\sin t + (a+b)\sin \left(\frac{a+b}{b}t \right),(a+b)\cos t - (a+b)\cos \left(\frac{a+b}{b}t \right) \right)$

$\dot{s}(t)^2 = \dot{P_{x}}^2 + \dot{P_{y}}^2 = (a+b)^2 \left( (\sin^2 t + \cos^2 t +\sin^2 \left(\frac{a+b}{b}t \right) + \cos^2 \left(\frac{a+b}{b}t \right)) - 2 \left( \sin t \cos \left(\frac{a+b}{b}t\right) + \sin t \sin \left(\frac{a+b}{b}t \right) \right) \right)$

$\dot{s}(t)^2 = 2(a+b)^2 \left(1 - \cos \frac{a}{b}t \right)$

$\dot{s}(t)^2 = 4(a+b)^2 \sin^2 \left( \frac{at}{2b} \right)$ since $2\sin^2t=1-\cos2t$

$\implies |\dot{s}(t)| = 2(a+b)|\sin\left( \frac{at}{2b} \right)|$

$\int_{0}^{2\pi}{|\dot{s}(t)|}dt = 2(a+b)\int_{0}^{2\pi}{|\sin\left( \frac{at}{2b}\right)| }dt$ $= 2(a+b)(\frac{a}{b})\int_{0}^{\Large\frac{2\pi b}{a}}{\sin\left( \frac{at}{2b}\right)}dt = 2(a+b)(\frac{a}{b})(\frac{4b}{a}) = 8(a+b)$

Since $a+b = 5$ , the answer is $\boxed{40}$