Three points, $A, B$ and $C$ travel along a straight line, each having the same origin.

Points $A$ and $B$ travel in the same direction at a speed of $5$ km/hr and $10$ km/hr, respectively. Point $C$ traverses between points $A$ and $B$ at a speed of 15 km/hr. Whenever $C$ reaches to $A$, he starts to move back to $B$, then again move back to $A$. After $1$ hour, how far from the origin is $C$ located?

I have no solution to the problem, this is a spin-off from a much easier problem.

Note: I have changed the original of the question to make it solvable and interesting. (Oray)

  • 2
    $\begingroup$ Did you get this problem from somebody or some place, or did you make it up after being inspired by another problem?  If it’s the first case, you must identify your source.  If it’s the second case, tracing the history of the problem is still a good thing to do. $\endgroup$ Aug 31 '17 at 5:04
  • $\begingroup$ When you say "Point C traverses between points A and B", does this mean that point C pendles back and forth between these two points (starts in A, moves to B, back to A, to B, to A, etc)? $\endgroup$
    – Gamow
    Sep 5 '17 at 5:44
  • 1
    $\begingroup$ @Oray it remains unsolvable, Wen1now's answer explains why $\endgroup$
    – Miff
    Sep 5 '17 at 14:05

In theory, point C could be

anywhere from 5 to 10 km away

Think about it this way:

Imagine playing the points moving backwards. Point C will always remain between A and B, up to t=0. Now reverse, and we have a valid scenario of C reaching any distance between 5 and 10km.

I think the problem is

C moves in an undefined way at t=0 so bad stuff happens.

Or maybe I'm completely wrong.

  • $\begingroup$ I had troubles about how to approach the origin, any close approximations are welcome. (For instance, if B's origin was just a hair ahead of A and C.) $\endgroup$
    – Casey
    Aug 31 '17 at 2:09
  • $\begingroup$ What do you mean? @Casey $\endgroup$
    – Wen1now
    Aug 31 '17 at 3:13
  • $\begingroup$ I mean to assume that points A and C originate at 0, and B originates at some finite yet "small" distance from 0, so as to eliminate any possibility of the problem being unsolvable. $\endgroup$
    – Casey
    Aug 31 '17 at 3:36
  • $\begingroup$ After I read the question (before I looked at this answer), I felt myself being drawn down the rabbit hole of some form of Zeno’s Paradox.  I’m reassured to see that somebody else had similar misgivings. $\endgroup$ Aug 31 '17 at 4:36
  • $\begingroup$ ... and I’m beginning to believe that, even if we stipulate that B starts at point ε, which is an arbitrarily small (positive) number, it’s still impossible to solve.  If we stipulate that B starts at point B₀, which is small but known, then it becomes solvable — and collapses into a tedious math problem rather than an interesting math-based puzzle.  I look forward to being proved wrong. $\endgroup$ Aug 31 '17 at 5:00

Here is a diagram that demonstrates why point C could be anywhere between A and B.   Start at...

... t = 1 hour, place C anywhere between A and B, then run time backward.   (In this diagram, backward in time goes downward; forward motion goes leftward in the left side example and rightward on the right side.)

Regardless of the position or direction of C at 1 hour, a reverse path to t = 0 exists such that C’s time-line always has a slope of ±15 km/hr.

Therefore any position and direction of C at 1 hour is consistent with the puzzle statement.


Assuming that B and C both start a small distance from the origin, and getting a computer to do all the work, I get a distance of 6.24km from the origin for point C at the end. However, this varies hugely depending on the initial conditions, to the point where there's clearly no one single answer

Python code below for anyone who'd like to try it out - I'd suggest experimenting with the initial distances for b and c (s_b and s_c) to see how it affects the results. Try varying the time step (dt) too.

v_a = 5  
v_b = 10  
v_c = 15  

s_a = 0  
s_b = 0.001  
s_c = 0.0001  

dt = 0.0001  
for t in xrange(0,int(1/dt),1):  
    s_a = v_a * t*dt  
    s_b = v_b * t*dt  
    s_c = s_c + v_c * dt  

    if s_c >= s_b:  
        v_c = -15  

    if s_c <= s_a:  
        v_c = 15  

print 'A dist = ' + `s_a`  
print 'B dist = ' + `s_b`  
print 'C dist = ' + `s_c`  

I think the question is somewhat incomplete, opening up for loophole answers.

Shouldn't there be an information when A, B and C are travelling? The term origin (of the lines) can be just geographical, and positions A=B=C at t=0 is just an assumption.

A and B could be days apart and C would always travel in between the spots (because A is at 0 and B at his way-off-destination).


The answer is:

There is no solution to this problem.

To see why consider:

The rules of the problem can be ran in reverse. A starts 5 km from the origin, B starts 10 km from the origin, C starts somewhere between A and B. A and B travel towards the origin at their previous speeds, and C bounces between A and B with the same speed it had. No matter where you place C initially between A and B, you will always end up with all 3 located at the origin after 1 hour. Therefore, back to the original problem, C could be anywhere between 5 and 10 km. The problem lies with the infinite number of bounces which occur near 0.

  • $\begingroup$ This answer is almost identical to Wen1now's. Can you please explain how you came to a different conclusion from the same reasoning? $\endgroup$
    – boboquack
    Sep 5 '17 at 6:02
  • $\begingroup$ You could add that in the reversed problem not only a starting position, but also the initial direction and speed of $C$ does not matter (as long as it exceeds the speed of $B$). $\endgroup$
    – CiaPan
    Sep 5 '17 at 10:37

My take on the answer

C is 5km from the origin, i.e. where A is located at the moment


Plain & Simple Time and Distance Approach:

A travels 5km (1 unit of distance) in an hour(1 unit of time), B travels 10 km (2 unit) in an hour, C travels 15 km (3 unit) in the same hour.

Also since C shuttles between A & B, after passage of one unit of time C will invariably be tied with A. (2 units forward + 1 unit backward)

  • 1
    $\begingroup$ See this. $\endgroup$ Sep 1 '17 at 4:42
  • $\begingroup$ Hmm.. assumed all three points to travel along the same straight line.. My bad.. $\endgroup$
    – Kalaivanan
    Sep 1 '17 at 5:24
  • $\begingroup$ They do; that's a chart of location vs. time. $\endgroup$ Sep 1 '17 at 5:34

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