1. A plank of wood has one end, $A$, against a vertical wall. Its other end, $B$, is on horizontal ground. When end $A$ slips down $8 \mathrm{~cm}$, end $B$ moves $4 \mathrm{~cm}$ further away from the wall. When end $A$ slips down a further $9 \mathrm{~cm}$, end $B$ moves a further $3 \mathrm{~cm}$ away from the wall. Find the length of the plank.
SOlUTION
Let the original distances from $A$ and $B$ to the corner where the ground meets the wall be $a \mathrm{~cm}$ and $b \mathrm{~cm}$ respectively.

Then, using Pythagoras' theorem, we have
$$
a^{2}+b^{2}=(a-8)^{2}+(b+4)^{2}=(a-17)^{2}+(b+7)^{2}
$$
. After simplification, this produces the equations below.
$$
\begin{array}{c}
2 a-b=10 \\
17 a-7 b=169
\end{array}
$$

Solving these, we obtain $a=33$ and $b=56$.
The length of the plank is then $65 \mathrm{~cm}$.
This problem relies on three Pythagorean triples $(3,4,5),(5,12,13)$ and $(33,56,65)$ where the first two are scaled up to give the same hypotenuse.