QuoteOriginally posted by: outrunQuoteOriginally posted by: Traden4AlphaMy first, pre-morning-coffee guess is 8. Each pair of cities defines one edge of the triangle and we have a choice of two lines -- the shortest great-circle and it's complement.That's a pre-coffee answer to a different question. It's missing the orderingGood point! If ABC != BCA !=CAB !=ACB !=CBA !=BAC, then we have a factor of 6 more triangles based on ordering of the vertices.So, now we are up to 48 ways = (3 first city choices) * (2 second city choices) * (2 direction choices to city 2) * (2 direction choices to city 3) * (2 direction choices to city 1) ? P.S. the "around the globe" phrase in the problem statement seems to make this a problem in spherical geometry with each triangle edge having two "around the globe" alternatives.
Last edited by Traden4Alpha
on March 13th, 2012, 11:00 pm, edited 1 time in total.