## Solution to 4 Travelers problem

**Ken Ross**

Let T_{ij} denote the time of meeting of persons i and j. Without
loss of generality, assume _{12} = 0._{34} (and obtain its value).

Let P_{i}(t) denote the position of person i at time t, with
P_{i}(t) = A_{i}+ tB_{i}. Here A_{i }and
B_{i }are vectors. Note that since no two roads are parallel, vectors
B_{i} are pairwise linearly independent. Without loss of generality,
assume persons 1 and 2 meet at the origin, so that A_{1}=A_{2}=0.

We know that

(1) | P_{1}(T_{13}) = P_{3}(T_{13}), |

(2) | P_{1}(T_{14}) = P_{4}(T_{14}), |

(3) | P_{2}(T_{23}) = P_{3}(T_{23}), |

(4) | P_{2}(T_{24}) = P_{4}(T_{24}). |

Using (1) and (3) we can solve for A_{3 }and B_{3 }in
terms of B_{1 }and B_{2}. Similarly for A_{4 }and
B_{4 }using (2) and (4).

Now, P_{3}(T) = P_{4}(T) gives an equation in T. Since
B_{1} and B_{2} are linearly independent, one can equate to
zero coefficients of each of B_{1} and B_{2} in this equation:
Fortunately, in both cases, the coefficients equal 0 when T = (T_{13}·T_{23}·T_{24
}+ T_{13}·T_{14}·T_{24 }- T_{13}·T_{14}·T_{23
}- T_{14}·T_{23}·T_{24})/(T_{13}·T_{24
}- T_{14}·T_{23}).

Finally, we need to verify that (T_{13}·T_{24 }- T_{14}·T_{23})
is nonzero. If this number were zero, then _{13}/T_{14} =
T_{23}/T_{24},_{3}(T_{13})/P_{3}(T_{23}))
and (0/P_{4}(T_{14})/P_{4}(T_{24})) are
similar.

Indeed, let S_{1} and S_{2} denote the speeds of persons 1 and 2, respectively.
Then the distance d(O, P_{1}(T_{13})) between the origin (by convention, it's
the point of intersection of roads 1 and 2) and the point P_{1}(T_{13}) equals
_{1}T_{13}. Similarly d(O, P_{1}(T_{14})) = S_{1}T_{14}_{1}(T_{13}))/d(O, P_{1}(T_{14})) = T_{13}/T_{14}._{23}/T_{24} is evaluated similarly.

Thus the lines along which persons 3 and 4 travel would be parallel, contradicting the problem statement.

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