When the Boat Comes In

Using Pythagoras' Theorem
This diagram shows the 'before' and 'after' positions of the boat at B and C respectively. The boat has travelled the distance BC.




In the version below, the sides of the two right-angled triangles have been labelled.




Applying Pythagoras' Theorem to the 'before' triangle, $l^2+h^2=(a+1)^2=a^2+2a+1$.
Applying Pythagoras' Theorem to the 'after' triangle, $k^2+h^2=a^2$.
Subtracting the 'after' equation from the 'before' equation, $l^2-k^2=2a+1$, so $l^2=k^2+2a+1$.

If $l=k+1$, then $l^2$ would equal $(k+1)^2=k^2+2k+1$.
In fact, $l^2=k^2+2a+1$, and since $a$ is the hypotenuse, it is longer than $k$, so $2a>2k.$ So $l^2>k^2+2k+1$, so $l>k+1$. So the boat has moved more than 1 metre.


Using a diagram (and the triangle inequality) 
This diagram shows the 'before' and 'after' positions of the boat at B and C respectively. The boat has travelled the distance BC.


The length AC has been rotated onto the line AC, so that the two blue line segments are the same length. The length marked on is 1 metre because the blue length (equal to AC) is 1 metre shorter than AB, because you have pulled the rope along 1 metre.

Imagine starting at B and travelling to A. It will clearly be shorter to go along the line AB than along the line BC and then along the blue line. So 1 metre must be shorter than BC. So the boat moves more than 1 metre.


Using the triangle inequality and algebra
This diagram shows the 'before' and 'after' positions of the boat at B and C respectively. The boat has travelled the distance BC.


Because the rope has been pulled 1 metre along, AB = AC + 1.
ABC forms a triangle, so AB must be shorter than AC + BC.
So AC + 1 is shorter than AC + BC, which means that 1 must be shorter than BC.
So the boat moves by more than 1 metre.