Locating a point where the two persons meet.

Question

Please see the following figure. Suppose two persons start moving from A to B, person 1 takes the route direct from A towards B while the person 2 takes the route from A to C and then to B. The speed of person 2 is twice as much as person 1. Suppose person 1 reaches point B before person 2 and waits for him there. Is it possible for the two persons to meet at a point between A and B on line AB such that none of the persons has to wait for the other one? Is there a way I can mathematically prove that in the conditions given the point will always be outside the range AB, on the right side of B and can never be inside AB.

What I have tried: I have tried different sets of coordinates but can never locate a point in between where the two persons meet without one waiting for the other. Thanks a lot!

Answer 1

We're told that, if person $P$ (traveling on edges $p_i$) and person $Q$ (on edges $q_i$, at twice $P$'s speed) were to meet at $B$, then $P$ would have to wait (presumably, a strictly-positive amount of time). We can write $$2(p_1+p_2) < q_1 + q_2 + q_3 \tag{1}$$ If $Q$ were take a detour ($D$-tour?) at $D$ to meet $P$ simultaneously at $E$, then $$2p_1 = q_1+q_2+q_3^\prime \tag{2}$$ Finally, by the Triangle Inequality (even in its lenient form), we have $$q_3 \leq p_2 + q_3^\prime \tag{3}$$ We deduce $$(q_1+q_2+q_3^\prime) + 2 p_2 < q_1 + q_2 + (p_2 + q_3^\prime) \quad\to\quad 2 p_2 < p_2 \tag{4}$$

A contradiction. Therefore, $P$ and $Q$ cannot meet simultaneously anywhere along $\overline{AB}$ (... well ... except for the starting point $A$). $\square$

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Note that there's nothing special about $Q$'s relative speed, except that it exceeds $P$'s. We can replace all multiplied "$2$"s above with any $k > 1$, and the contradiction stands.

Answer 2

Is it possible for the two persons to meet at a point between A and B on line AB such that none of the persons has to wait for the other one?

Yes, if $C$, $A$, and $B$ be collinear points, in that order, with $\lvert C A \rvert \lt \frac{1}{4} \lvert A B \rvert$. Then, the faster person will catch up with the slower person from behind.

Since the person walking via $C$ is twice as fast as the person walking directly to $B$, they meet at a point $D$ on the line between $A$ and $B$, at distance $2 \lvert CA \rvert$ from $A$ to $B$.

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What if there is an additional restriction, that $A$, $B$, and $C$ are such that the slower person reaches $B$ before the faster one does?

The restriction can be written as $$2 \lvert A B \rvert \lt \lvert A C \rvert + \lvert C B \rvert$$ but note that the three points must still be collinear for the persons to meet. Without loss of generality, we can choose $A$ to be at origin, $B$ to be at $(b, 0)$ on the positive $x$ axis (so $b \gt 0$), and $C$ to be at $(c, 0)$ on the positive or negative $x$ axis. The restriction can now be written as $$2 b < \lvert c \rvert + \lvert b - c \rvert$$ There are two solutions to the restriction: $$c \lt -\frac{b}{2} \quad \text{or} \quad c \gt \frac{3 b}{2}$$

When $c \lt -\frac{b}{2}$, the faster walker will chase the slower walker, but never catch up. (When the faster walker is at $C$, the slower walker is already at least a quarter of the way from $A$ to $B$. When the faster walker is back at $A$, the slower walker is already over half the way from $A$ to $B$.)

When $c \gt \frac{3 b}{2}$, the faster walker will walk initially in front of the slower walker, well past $B$. When the faster walker is at $C$, the slower walker is already at least three quarters of the way from $A$ to $B$. The two will then approach $B$ from opposite sides, and therefore will not meet before both arrive and stop at $B$.