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Gretings.

Trying the "hard solution" for the puzzle below (which has been discussed, with a different angle, elsewhere on this forum) I got to a point where I have three seemingly valid solutions, and two do not match. I have checked the procedure several times, both by hand and with Maxima, and it appears correct - but I am obviously missing something.I cannot sort this out, maybe someone here can. (Sorry for the ASCII math, I am not MathJax-enabled)

The story and the puzzle:

A colleague approached one day John Von Neumann with a puzzle that had two paths to a solution, a laborious, complicated calculation and an elegant, Aha!-type solution. This colleague had a theory: in such a case, mathematicians work out the laborious solution while the (lazier, but smarter) physicists pause and find the quick-and-easy solution. Which solution would von Neumann find? You know the sort of puzzle: Two trains, 100 miles apart, are approaching each other on the same track, one going 30 miles per hour, the other going 20 miles per hour. A bird flying 120 miles per hour starts at train A (when they are 100 miles apart), flies to train B, turns around and flies back to the approaching train A, and so forth, until the two trains collide. How far has the bird flown when the collision occurs? “Two hundred and forty miles,” von Neumann answered almost instantly. “Darn,” replied his colleague, “I predicted you’d do it the hard way, summing the infinite series.” “Ay!” von Neumann cried in embarrassment, smiting his forehead. “There’s an easy way!”

Solution 1 - Easy:

The trains will collide in $T_t=\frac{D_0}{V_1+V_2}$. Hence $D_b=V_b\cdot\frac{D_0}{V_1+V_2}=120\cdot\frac{100}{50}=240$

Hard Solution preliminary:

On the first leg, the bird travels the distance $D_0$ in time $T_0=\frac{D_0}{V_b+V_2}$;

On the second leg, distance and duration amount to: $D_1 = D_0-T_0 \cdot (V_1+V_2) = D_0 \cdot \frac{V_b-V_1}{V_b+V_2}; T_1=D_0 \cdot \frac{V_1-V_b}{V_1+V_b}*(V_2+V_b))=T_0 \cdot \frac{V_b-V_1}{V_b+V_1}$

On the third leg, a recurrence emerges:

$D_2 = D_0 \cdot \frac{(V_1-V_b)\cdot(V_2-V_b)}{(V_1+V_b)\cdot(V_2+V_b)} ; T_2 = T_0 \cdot \frac{(V_1-V_b)\cdot(V_2-V_b)}{(V_1+V_b)\cdot(V_2+V_b)}$

The same recurrence is found to hold between $D_3$ and $D_1$ and $T_3$ and $T_1$.

If we set:

$$R=\frac{(V_1-V_b)\cdot(V_2-V_b)}{(V_1+V_b)\cdot(V_2+V_b)}$$

We see that both the total duration and the total distance traveled by the bird can be computed summing a geometrical series with coefficient $R$ and respective first terms $(T_0+T_1)$ and $(D_0+D_1)$. $R< 1$ holds, so convergence is assured.

Therefore we have.

Solution 2:

$$T_t= \sum (T_0+T_1)R^n = \frac{D_0}{V_1+V_2}; D_b= V_b\cdot\frac{D_0}{V_1+V_2}$$

This matches solution (1) as it should.

Solution 3:

$$D_b= \sum(D_0+D_1)R^n ) = D_0 \cdot \frac{(V_1+V_b)\cdot(V_2-V_1+2 V_b)}{2 V_b\cdot(V_2+V_1)}$$

I am stymied.

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2  
See this solution: math.stackexchange.com/questions/346384/… – Ron Gordon May 7 '14 at 12:03
    
Numerically what are the answers you are getting? – MNKY May 7 '14 at 12:34
    
Right, so the mistake is that my solution (3) is not in fact a solution to anything. I am adding the distances that separate the trains at each leg, not the distances flown by the bird on that leg... – Alien Life Form May 7 '14 at 12:36
    
Here's another related problem that was discussed recently: math.stackexchange.com/questions/779107/… – Paul Safier May 7 '14 at 14:35

And of course there is no paradox. (As mentioned by Ron Gordon above, The same problem/same angle is discussed on: math.stackexchange.com/questions/346384/)

Solution 3 - above - is no solution at all. It just sums the distances of the trains at the start of each leg, but that is not the distance flown by the bird on that leg: that is given by (obviously) the bird's velocity times the leg duration. Whoops. The fixed solution is below.

Solution 1 - Easy:

The trains will collide in $T_t=\frac{D_0}{V_1+V_2}$. Hence $D_b=V_b\cdot\frac{D_0}{V_1+V_2}=120\cdot\frac{100}{50}=240$

**Hard Solution **

On the first leg, the ** trains are separated by** the distance $D_0$, which the birds covers in time $T_0=\frac{D_0}{V_b+V_2}$.

The distance flown by the bird is $S_0=V_B\cdot\frac{D_0}{V_b+V_2}$

On the second leg, distance between trains, duration and distance covered by the bird amount to: $D_1 = D_0-T_0 \cdot (V_1+V_2) = D_0 \cdot \frac{V_b-V_1}{V_b+V_2};$

$T_1=D_0 \cdot \frac{V_1-V_b}{V_1+V_b}*(V_2+V_b))=T_0 \cdot \frac{V_b-V_1}{V_b+V_1};$

$S_1=V_b \cdot T_1$

On the third leg, a recurrence emerges:

$D_2 = D_0 \cdot \frac{(V_1-V_b)\cdot(V_2-V_b)}{(V_1+V_b)\cdot(V_2+V_b)} ;$

$T_2 = T_0 \cdot \frac{(V_1-V_b)\cdot(V_2-V_b)}{(V_1+V_b)\cdot(V_2+V_b)}; $

$S_2=V_b \cdot T_1$

The same recurrence is found to hold between $D_3$ and $D_1$, $T_3$ and $T_1$, $S_3$ and $S_1$.

If we set:

$$R=\frac{(V_1-V_b)\cdot(V_2-V_b)}{(V_1+V_b)\cdot(V_2+V_b)}$$

We see that both the total duration and the total distance traveled by the bird can be computed summing a geometrical series with coefficient $R$ and respective first terms $(T_0+T_1)$ and $(S_0+S_1)$. $R< 1$ holds, so convergence is assured.

Therefore we have:

Solution 2:

$$T_t= \sum (T_0+T_1)R^n = \frac{T_0+T_1}{1-R} = \frac{D_0}{V_1+V_2}; $$

$$S_b= \sum (S_0+S_1)R^n = V_b\cdot T_t = V_b\cdot\frac{D_0}{V_1+V_2}$$

This matches solution (1) as it should.

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