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Solution to puzzle 139: Three towns

Skip restatement of puzzle.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.

Equilateral triangle ABC, with point P such that AP = 3 and BP = 4.

Hence quadrilateral APBC is convex, and with diagonals AB and CP, so that we may apply Ptolemy's Inequality, which states that:
AB·CP less than or equal to AP·BC + BP·AC, with equality if, and only if, APBC is cyclic.

Since AB = BC = AC, we get CP less than or equal to AP + BP = 7, with equality if P lies on the arc AB of the (unique) circumcircle of triangleABC.
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

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