Solution to puzzle 139: Three towns
The towns of Alpha, Beta, and Gamma are equidistant from each other. If a car is three miles from Alpha and four miles from Beta, what is the maximum possible distance of the car from Gamma? Assume the land is flat.
Let the point P denote the car, A denote Alpha, B denote Beta, and C denote Gamma.
Clearly P must be on the opposite side of AB to C, for otherwise we could reflect P in AB, thereby increasing CP, while keeping AP and BP the same.
Also, P must be on the same side of AC as B, for otherwise we could reflect P in AC, and then extend AB so that BP = 4, thereby increasing CP.
Similarly, P must be on the same side of AB as C.
Hence quadrilateral APBC is convex, and with diagonals AB and CP, so that we may apply Ptolemy's Inequality, which states that:
AB·CP 
AP·BC + BP·AC, with equality if, and only if, APBC is cyclic.
Since AB = BC = AC, we get CP AP + BP = 7, with equality if P lies on the arc AB of the (unique) circumcircle of 
ABC.
It is clear that equality can occur, as, for any side length, AP/BP increases continuously from 0 without limit as P moves anticlockwise along the arc AB (omitting the end point B.) Hence at some point AP/BP will reach the value 3/4.
Therefore, the maximum possible distance of the car from Gamma is 7 miles.
Remarks
What is the minimum possible distance of the car from Gamma?
Source: To be announced