not to scale A 10 meter long rope with uniform mass-density and bending module is hanging between two poles. Given the fact that both ends of the rope are 5 meters higher than the lowest point of the rope, find the distance $x$ between the poles.

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    $\begingroup$ Ingeneerious - sometimes the simple puzzles are the best ones! $\endgroup$ – Avigrail Oct 15 '17 at 20:03
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    $\begingroup$ This image detracts from the puzzle a lot. $\endgroup$ – insidesin Oct 16 '17 at 1:59

The distance between the poles is:

0 metres

All of the rope is accounted for if the rope is positioned vertically. 5 metres is used from one end to the centre and another 5 metres from the centre to the other end. There's no more rope left for there to be any horizontal distance between the ends.

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Viewing the rope, $5\ \rm m$ of pole and distance $\frac12x$ as a right angle triangle (approx not including droop of rope) we have a hypotenuse of length $5$ ($\frac12$ of rope length) and adjacent side of length $5\ \rm m$.

Using bit of Pythagorus

$$\begin{align} \text{hypotenuse} &= \text{adjacent} + \text{desired length}\\ 5^2 &= 5^2 + \left(\frac x2\right)^2\\ 5^2 - 5^2 &= \left(\frac x2\right)^2\\ 0 &= \left(\frac x2\right)^2\\ \sqrt0\times2 &= 0 \end{align}$$

Therefore distance between pole and droop point of rope is $0$ and twice that distance is still $\approx 0$. Distance $x = 0$.

Looking now at the error included in the above rough calculation caused by ignoring the droop of the rope. The droop of the rope would make the actual linear distance between pole and droop point (length $\frac x2$) slightly shorter than $5$ resulting in the impossibility that the hypotenuse of a right angle triangle is smaller than the sum of the other two sides squared. The arrangement only works when distance from pole to pole is $0$.

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