Draw the circumscribing parallelogram, G L H E M,
whose sides are respectively parallel to P S and P R. Join L M. By the
conditions of this problem, the path must somewhere cut the circle E D F;
and since L M cuts L H, which is a tangent to it, it is clear it must cut
every path--such as a a, parallel to L H, or to P R--that cuts the
circle. Similarly, the same line, L M, must cut every path parallel to P
S, such as b b. Now if L M cuts every path that is parallel to either of
the extreme directions, P R or P S, it is obvious that it must also cut
every path that is parallel to an intermediate direction, such as c c,
but
PL = PH/cos HPL = PD/cos 1/2 RPS;
The consequence of which is that P L exceeds P D by one-sixth, one-half
as much again, or twice as much again, according as R P S = 60 degrees,
90 degrees degrees, or 140 degrees.
The traveller who can only answer the questions A and B, but not C, must
be prepared to travel from P to L, and back again through P to M, a
distance equal to 3 P L. If, however, he can answer the question C, he
knows at once whether to travel towards L or towards M, and he has no
return journey to fear. At the worst, he has simply to travel the
distance P L.
The probable distance, as distinguished from the utmost possible distance
that a man may have to travel in the three cases, can be calculated
mathematically.
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