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

(1)P1(T13) = P3(T13),
(2)P1(T14) = P4(T14),
(3)P2(T23) = P3(T23),
(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.

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