# 196 geometry/smuggler.p

## Description

This article is from the Puzzles FAQ,
by Chris Cole chris@questrel.questrel.com and Matthew Daly
mwdaly@pobox.com with numerous contributions by others.

# 196 geometry/smuggler.p

Somewhere on the high sees smuggler S is attempting, without much

luck, to outspeed coast guard G, whose boat can go faster than S's. G

is one mile east of S when a heavy fog descends. It's so heavy that

nobody can see or hear anything further than a few feet. Immediately

after the fog descends, S changes course and attempts to escape at

constant speed under a new, fixed course. Meanwhile, G has lost track

of S. But G happens to know S's speed, that it is constant, and that S

is sticking to some fixed heading, unknown to G.

How does G catch S?

G may change course and speed at will. He knows his own speed and

course at all times. There is no wind, G does not have radio or radar,

there is enough space for maneuvering, etc.

geometry/smuggler.s

One way G can catch S is as follows (it is not the fastest way).

G waits until he knows that S has traveled for one mile. At that time, both

S and G are somewhere on a circle with radius one mile, and with its center

at the original position of S. G then begins to travel with a velocity that

has a radially outward component equal to that of S, and with a tangential

component as large as possible, given G's own limitation of total speed. By

doing so, G and S will always both be on an identical circle having its

center at the original position of S. Because G has a tangential component

whereas S does not, G will always catch S (actually, this is not proven

until you solve the o.d.e. associated with the problem).

If G can go at 40 mph and S goes at 20 mph, you can work out that it will

take G at most 1h 49m 52s to catch S. On average, G will catch S in:

( -2pi + sqrt(3) ( exp(2pi/sqrt(3)) - 1 )) / 40pi hours,

which is, 27 min and 17 sec.

Continue to: