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.