The starships Enterprise and Intrepid, after refits, are undergoing flight tests.

The starships Enterprise and Intrepid, after refits, are undergoing flight tests.

One such test involves them flying, in opposite directions, around a large circular course.

Enterprise and Intrepid are at opposite ends of a diameter, when, after James T KIrk has said ready steady go,  Enterprise sets off in a clockwise direction and Intrepid in an anticlockwise direction. Flying at constant speeds, they first pass each other when the Enterprise has travelled 20 light years. They pass each other again when the Intrepid has travelled on a further 10 light years.

What is the circumference of the circular course ?

I. First pass each other:

$\small{ \begin{array}{llcl} & \text{Way Enterprise } = w_1 &=& 20\ \text{Ly} \\ & \text{Way Intrepid} = w_2 \\ & \text{Speed Enterprise } = v_1 \\ & \text{Speed Intrepid} = v_2 \\ & \text{Circumference } = 2\pi r \\ \end{array}\\ \begin{array}{llcl} \text{velocity time rule }& \\ & w_1 &=& v_1\cdot t_1 \\ & w_2 &=& v_2\cdot t_1 \\ & t_1 &=& \frac{w_1}{v_1} = \frac{w_2}{v_2} \\ \hline (1)& w_1+w_2 &=& \frac{ 2\pi r }{2} = \pi r \\ & w_2 &=& \pi r - w_1 \\ (2)& w_1 &=& \frac{v_1}{v_2} \cdot w_2 \\ & w_1 &=& \frac{v_1}{v_2} \cdot (\pi r - w_1) \\ & \dfrac{v_1}{v_2} &=& \dfrac{w_1}{\pi r - w_1} \\ \end{array} }$

II. Second pass each other:

$\small{ \begin{array}{llcl} & \text{Way Enterprise } = W_1 \\ & \text{Way Intrepid} = W_2 &=& 10\ \text{Ly} \\ \end{array}\\ \begin{array}{llcl} \text{velocity time rule }& \\ & W_1 &=& v_1\cdot t_2 \\ & W_2 &=& v_2\cdot t_2 \\ & t_2 &=& \frac{W_1}{v_1} = \frac{W_2}{v_2} \\ \hline (3)& W_1+W_2 &=& 2\pi r \\ & W_1 &=& 2\pi r - W_2 \\ (4)& W_2 &=& \frac{v_2}{v_1} \cdot W_1 \\ & W_2 &=& \frac{v_2}{v_1} \cdot (2\pi r - W_2) \\ & \dfrac{v_1}{v_2} &=& \dfrac{2\pi r - W_2}{W_2} \\ \end{array} }$

III. conclusion

$\small{ \begin{array}{rcl} \dfrac{v_1}{v_2} = \dfrac{w_1}{\pi r - w_1} &=& \dfrac{2\pi r - W_2}{W_1} \\\\ \dfrac{w_1}{\pi r - w_1} &=& \dfrac{2\pi r - W_2}{W_2} \\\\ w_1\cdot W_2 &=& (2\pi r - W_2)(\pi r - w_1) \\ w_1\cdot W_2 &=& 2\pi^2 r^2 - 2\pi r\cdot w_1+ W_2\pi r+ w_1\cdot W_2\\ w_1\cdot W_2 &=& 2\pi^2 r^2 -r(2\pi w_1+W_2\pi)+ w_1\cdot W_2\\ 2\pi^2 r^2 -r(2\pi w_1+W_2\pi) &=& 0\\ 2\pi^2 r^2 &=& r(2\pi w_1+W_2\pi) \qquad | \qquad : r\\ 2\pi^2 r &=& 2\pi w_1+W_2\pi \qquad | \qquad : \pi\\ 2\pi r &=& 2 \cdot w_1+W_2 \qquad | \qquad w_1 = 20\ \text{Ly} \qquad W_2 = 10\ \text{Ly}\\ 2\pi r &=& 2 \cdot (20\ \text{Ly} )+10\ \text{Ly} \\ \mathbf{ 2\pi r } & \mathbf{=} & \mathbf{ 50\ \text{Ly} } \\ \end{array} }$

The circumference of the circular course is 50 light years.

The question and answer both look really interesting.  Thanks Heureka :D

Nice solution heureka, however I much prefer Sam Lloyds method.

This is a futuristic version of the Sam Lloyd ferries problem that was posted a week or two ago, and Lloyd's method works equally well with this problem.

When the ships first pass each other, their combined distances travelled will be equal to a half of the circumference, call this H, say.

If Intrepid's distance travelled is X, then X + 20 = H.

When they pass for a second time their combined distances travelled will be 3H, and, multiplying the earlier equation by 3,

3H = 3X + 60.

The 3X will be made up of the original X plus another 10.

So, 3X = X + 10,

2X = 10,

X = 5.

Therefore, H = 5 + 20 = 25, meaning that the circumference of the circle is 50 lt yrs.