Solution to 4 Travelers problem
Let Tij denote the time of meeting of persons i and j. Without
loss of generality, assume
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
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
Thus the lines along which persons 3 and 4 travel would be parallel, contradicting the problem statement.
Copyright © 1996-2017 Alexander Bogomolny