Solution to 4 Travelers problem
Ken Ross
Let Tij denote the time of meeting of persons i and j. Without
loss of generality, assume T12 = 0. We wish to demonstrate the
existence of T34 (and obtain its value).
Let Pi(t) denote the position of person i at time t, with
Pi(t) = Ai+ tBi. Here Ai and
Bi are vectors. Note that since no two roads are parallel, vectors
Bi are pairwise linearly independent. Without loss of generality,
assume persons 1 and 2 meet at the origin, so that A1=A2=0.
We know that
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(1) | P1(T13) = P3(T13), |
|
(2) | P1(T14) = P4(T14), |
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(3) | P2(T23) = P3(T23), |
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(4) | P2(T24) = P4(T24). |
Using (1) and (3) we can solve for A3 and B3 in
terms of B1 and B2. Similarly for A4 and
B4 using (2) and (4).
Now, P3(T) = P4(T) gives an equation in T. Since
B1 and B2 are linearly independent, one can equate to
zero coefficients of each of B1 and B2 in this equation:
Fortunately, in both cases, the coefficients equal 0 when T = (T13·T23·T24
+ T13·T14·T24 - T13·T14·T23
- T14·T23·T24)/(T13·T24
- T14·T23).
Finally, we need to verify that (T13·T24 - T14·T23)
is nonzero. If this number were zero, then T13/T14 =
T23/T24, which would imply that the triangles (0/P3(T13)/P3(T23))
and (0/P4(T14)/P4(T24)) are
similar.
Indeed, let S1 and S2 denote the speeds of persons 1 and 2, respectively.
Then the distance d(O, P1(T13)) between the origin (by convention, it's
the point of intersection of roads 1 and 2) and the point P1(T13) equals
S1T13. Similarly d(O, P1(T14)) = S1T14 so that
d(O, P1(T13))/d(O, P1(T14)) = T13/T14. The
ratio T23/T24 is evaluated similarly.
Thus the lines along which persons 3 and 4 travel would be parallel,
contradicting the problem statement.

Copyright © 1996-2009 Alexander Bogomolny
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