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First of all we need to assign letters to the important variables in the problem.
Let $v$ be the speed of the column.
Let $u$ be the speed of the messenger.
Let $L$ be the length of the column.
We can assume that the messenger travelled at a constant speed because he was ordered to deliver the message and return to his former position "without delay". Assume that the speed $u$ is the messenger's maximum speed which is constant (he does not get tired, for instance). We need to find ${u\over v}$.
When delivering the message, the messenger travels at a speed $u-v$ relative to the column. He travels a distance $L$ relative to the column. Therefore, the time to reach the head of column is $${L\over u-v}\;.$$
When returning to his position, the messenger travels at a speed $u+v$ relative to the column. Again, he travels a distance $L$ relative to the column. Therefore, the time to return to the end of column is $${L\over u+v}\;.$$
While the messenger is carrying out his mission, the end of the column travels a distance $L$.
The time for the column to travel this distance is $${L \over v}\;.$$
The time for the messenger to complete the mission is equal to the time for the column to travel the distance $L$.
$$\begin{eqnarray} {L \over u-v} + {L \over u+v} &=& {L\over v}\\ \Rightarrow {1\over u-v} + {1\over u+v} &=& {1\over v}\qquad\, \mathrm{[divide\ by\ L]}\\ \Rightarrow {v\over u-v} + {v\over u+v} &=& 1\qquad\;\; \mathrm{[multiply\ by\ v]}\\ \end{eqnarray}$$
As stated above, we need to find ${u\over v}$. If we divide top and bottom of the fractions on the left hand side by $v$ and then substitute $x$ for ${u\over v}$, we get the following equation.
$$\begin{align} {1\over x-1} + {1\over x+1} &= 1\\ \Rightarrow (x+1) + (x-1) &= (x+1)(x-1)\qquad \mathrm{[multiply\ through\ by\ (x+1)(x-1)]}\\ \Rightarrow 2x &= x^2 - 1\qquad\qquad\quad\; \mathrm{[expand\ the\ brackets\ and\ gather\ terms]}\\ \Rightarrow 0 &= x^2 - 2x - 1 \end{align}$$
We now have a quadratic equation, for which we can use the standard formula to solve.
$$\begin{align} x &= \frac{2 \pm \sqrt{4 + 4}}{2}\\ x &= \frac{2 \pm \sqrt{8}}{2}\\ x &= 1 \pm \sqrt{2} \end{align}$$
as $\sqrt{8}$ is equal to $2\sqrt{2}$.
We now have to consider the two possible solutions. The ratio cannot be negative, because the velocity of the messenger is in the same direction as the velocity of the column. This means that the ratio cannot be $1 - \sqrt{2}$. Thus, the ratio of the messenger's speed to that of the column is $x=1 + \sqrt{2}.$
Take a Message Soldier
Age 14 to 18
Challenge Level 
First of all we need to assign letters to the important variables in the problem.
Let $v$ be the speed of the column.
Let $u$ be the speed of the messenger.
Let $L$ be the length of the column.
We can assume that the messenger travelled at a constant speed because he was ordered to deliver the message and return to his former position "without delay". Assume that the speed $u$ is the messenger's maximum speed which is constant (he does not get tired, for instance). We need to find ${u\over v}$.
When delivering the message, the messenger travels at a speed $u-v$ relative to the column. He travels a distance $L$ relative to the column. Therefore, the time to reach the head of column is $${L\over u-v}\;.$$
When returning to his position, the messenger travels at a speed $u+v$ relative to the column. Again, he travels a distance $L$ relative to the column. Therefore, the time to return to the end of column is $${L\over u+v}\;.$$
While the messenger is carrying out his mission, the end of the column travels a distance $L$.
The time for the column to travel this distance is $${L \over v}\;.$$
The time for the messenger to complete the mission is equal to the time for the column to travel the distance $L$.
$$\begin{eqnarray} {L \over u-v} + {L \over u+v} &=& {L\over v}\\ \Rightarrow {1\over u-v} + {1\over u+v} &=& {1\over v}\qquad\, \mathrm{[divide\ by\ L]}\\ \Rightarrow {v\over u-v} + {v\over u+v} &=& 1\qquad\;\; \mathrm{[multiply\ by\ v]}\\ \end{eqnarray}$$
As stated above, we need to find ${u\over v}$. If we divide top and bottom of the fractions on the left hand side by $v$ and then substitute $x$ for ${u\over v}$, we get the following equation.
$$\begin{align} {1\over x-1} + {1\over x+1} &= 1\\ \Rightarrow (x+1) + (x-1) &= (x+1)(x-1)\qquad \mathrm{[multiply\ through\ by\ (x+1)(x-1)]}\\ \Rightarrow 2x &= x^2 - 1\qquad\qquad\quad\; \mathrm{[expand\ the\ brackets\ and\ gather\ terms]}\\ \Rightarrow 0 &= x^2 - 2x - 1 \end{align}$$
We now have a quadratic equation, for which we can use the standard formula to solve.
$$\begin{align} x &= \frac{2 \pm \sqrt{4 + 4}}{2}\\ x &= \frac{2 \pm \sqrt{8}}{2}\\ x &= 1 \pm \sqrt{2} \end{align}$$
as $\sqrt{8}$ is equal to $2\sqrt{2}$.
We now have to consider the two possible solutions. The ratio cannot be negative, because the velocity of the messenger is in the same direction as the velocity of the column. This means that the ratio cannot be $1 - \sqrt{2}$. Thus, the ratio of the messenger's speed to that of the column is $x=1 + \sqrt{2}.$
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