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 T₁₂ = 0. We wish to demonstrate the existence of T₃₄ (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 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.

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny