



 
David, "What is the maximum distance that the car can be from town A?" should be "What is the maximum distance that the car can be from town C?", right?
Where the heck do you drink, mate, the Ramanujan Arms? This sure beats darts & sports trivia! In general the distance to the third town is at most the sum of the other two distances. I cannot find this result in Bottema et al, which is a classic on geometric inequalities. I don't see any slick solution, though I can't shake the feeling that there might be one. If you hang with a crowd like that, I don't need to draw you a picture, so here's an outline. Let the car be at D, let $\theta = \angle DAB,$ and let $AB = BC = CA = x$
Good Hunting! RD
In fact, there is an easier approach to the problem if you know Ptolemy's theorem (or what Wikipedia calls Ptolemy's inequality): Using your notation, the theorem says that if you have 4 points $A, P, B, C$ in the plane, then the product of the "diagonals" $AB \times PC$ is always less than or equal to the sum of the products of opposite sides. In symbols, \[B \times PC \le AC \times PB + AP \times BC.\] Furthermore, you get equality exactly when all four points lie on a circle in the order APBC. In your problem, the towns are an equal distance apart, call it $x:$ If the car is at the point P, then we are given $AP = 3$ and $PB = 4.$ Let us denote the unknown distance of the car from town $C$ by $d: d = PC.$ Ptolemy tells us that dx can be at most $3x + 4x;$ that is, d is at most $3+4,$ and it equals $7$ when the car is at a point between$A$ and $B$ on the circumcircle of triangle $ABC.$ Chris  


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