Epicycloids and Hypocycloids

If the large circle has radius $Nr$ and the small circle has radius $r$ , then the epicycloid has parametric equation $\mathbf{r} \; = \; (N+1)r\binom{\cos t}{\sin t} - r \cos Nt \binom{\cos t}{\sin t} + r\sin Nt \binom{\sin t}{-\cos t}$ so that $x \; = \; (N +1)r\cos t - r\cos(N+1)t \hspace{2cm} y \; = \; (N+1)r\sin t - r\sin(N+1)t \hspace{3cm} 0 \le t \le 2\pi$ It is easy to calculate that the arc-length $s$ satisfies $\dot{s} \; = \; \sqrt{\dot{x}^2 + \dot{y}^2} \; = \; 2(N+1)r|\sin\tfrac12Nt| \hspace{2cm} 0 \le t \le 2\pi$ and hence $L_A \; = \; \int_0^{2\pi} \dot{s}\, do \; = \; N\int_0^{\frac{2\pi}{N}} 2(N+1)r\sin \tfrac12Nt\,dt \; = \; 8(N+1)r$ Similarly, the hypocycloid has equation $\mathbf{r} \; = \; (N-1)r\binom{\cos t}{\sin t} + r \cos Nt \binom{\cos t}{\sin t} + r\sin Nt \binom{\sin t}{-\cos t}$ so that $x \; = \; (N -1)r\cos t + r\cos(N-1)t \hspace{2cm} y \; = \; (N-1)r\sin t - r\sin(N-1)t \hspace{3cm} 0 \le t \le 2\pi$ It is easy to calculate that the arc-length $s$ satisfies $\dot{s} \; = \; \sqrt{\dot{x}^2 + \dot{y}^2} \; = \; 2(N-1)r|\sin\tfrac12Nt| \hspace{2cm} 0 \le t \le 2\pi$ and hence $L_B \; = \; \int_0^{2\pi} \dot{s} \, dt\; = \; N\int_0^{\frac{2\pi}{N}} 2(N-1)r\sin \tfrac12Nt\,dt \; = \; 8(N-1)r$ So that $L_A - L_B =16r$ , irrespective of the value of $N$ .

In this case $(N=12$ , $r=1$ ), we deduce that the answer is $\boxed{16}$ .

If the larger circle has a radius of $a$ and the smaller circle has a radius of $b$ , then the length of the epicycloid is $L_A = 8mb$ , where $a = (m - 1)b$ . Combining these equations, we find that $L_A = 8a + 8b$ .

If the larger circle has a radius of $a$ and the smaller circle has a radius of $b$ , then the length of the hypocycloid is $L_B = 8nb$ , where $a = (n + 1)b$ . Combining these equations, we find that $L_B = 8a - 8b$ .

This means that $L_A - L_B = (8a + 8b) - (8a - 8b)$ , or $L_A - L_B = 16b$ .

In this problem, the smaller circle has a radius of $1$ , so $b = 1$ , and $L_A - L_B = \boxed{16}$ .

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For the first bonus question, the radius of the larger circle has no effect on $L_A - L_B$ , so it's still $16$ .

For the second bonus question, $b = r$ , so $L_A - L_B = 16r$ .