Messenger Riddle 
A column of troops one km long is moving along a straight road at a
  uniform pace. A messenger is sent from the head of the column,
  delivers a message at the rear of the column and returns. He also
  moves at a uniform pace and arrives back at the head of the column
  when it has just covered its own length. How far did the messenger
  travel?

I can't get any ideas on how to start. 
Thanks for any help.
 A: Let us assume that the speed of the column is $1$ km per unit of time. For convenience, call that unit an hour. The column took $1$ hour to cover its own length.
Let $v$ be the speed of the messenger. When she is travelling to the back, the combined speed of approach of  the messenger and the column rear is $v+1$, so the time it takes is $\frac{1}{v+1}$.  Going the other way, the speed at which the messenger gains on the head is $v-1$, so the time it takes to gain the whole $1$ km is $\frac{1}{v-1}$.  The whole task took $1$ hour, and therefore
$$\frac{1}{v+1}+\frac{1}{v-1}=1.$$
This gives $v=1+\sqrt{2}$. The time taken is an hour, so the distance travelled is $1+\sqrt{2}$.
A: Let the speed of troop is $u$ kmph and that of messenger is $v$ kmph.
So the relative speed when messenger is going backward is $v+u$ kmph.
And relative speed while going forward is $v-u$ kmph $(v>u)$.
So the total time taken is $$t=\frac{1}{v+u}+\frac{1}{v-u}$$ 
but in this time troop had moved $1$km so the $t=\frac{1}{u}$
$$\frac{1}{u}=\frac{1}{v+u}+\frac{1}{v-u}$$
$$v^2-2uv-u^2=0$$
which give us the following relation
$$v=u(1+\sqrt2)$$
so the distance traveled is 
$$\begin{align*} d= &\underbrace{\frac{v}{v+u}}_{backward}+\underbrace{\frac{v}{v-u}}_{forward}\\
&=\frac{2v^2}{v^2-u^2}\\
&=\frac{2v^2}{2uv}\\ &=\frac{v}{u}=1+\sqrt{2}\text{ km}\end{align*}$$
A: Let $x$ be the troops' speed and $y$ be the messenger's speed.
Total messenger travelling time is $1/x$ (as the troops moved 1km forward).
In the troops frame of reference, the messenger first moved backwards with the speed of $y+x$, and then moved forward with the speed of $y-x$ so that in the $1/x$ total time he returned to the initial position.
Let $z$ be the time messenger spent to reach the rear of the column. Obviously, $z = 1/(y+x)$ (as messenger moves with the speed of $y+x$ relative to the column and has to travel $1$ km relative to the column to reach its rear).
From the other side, we have $z \times (y+x) + (1/x - z) \times (y-x) = 0$, from which follows $2zx + y/x - 1 = 2/(y/x+1) + y/x - 1 = 0$.
Note that the total messenger travelling distance is $d = y/x$. From the previous equation we get $2/(d+1) + d - 1 = 0$, $d^2-3 = 0$, and thus $d = \sqrt{3}$.
A: Since the time the column travels is equal to the time the messenger travels, then D= rt makes the distance equal to the rate if we let t = 1.
Suppose the distance the column moves until the messenger overtakes it is x, then the distance the messenger travels is 1+x, making the comparison 1+x/ x. On the messenger's return, the distance is x while the column moves 1-x.
So: 1+x/x = x/1-x since both column and messenger move at constant rates.
x then is equal to 1/2 the square root of 2, so the messenger travels 1+ half the square root of 2 out and another half the square root of two back. Total distance of 1 plus the square root of two.
