Hamilton 2019 Q3

Let the lengths of the sides of the regions be \(k\) cm, \(\ell\) cm, \(a\) cm and \(b\) cm, as shown. We know that each of these is less than 4 cm.

Then the regions have areas \(ka\) cm², \(kb\) cm², \(\ell a\) cm² and \(\ell b\) cm² respectively, and thus each of \(ka\), \(kb\), \(\ell a\) and \(\ell b\) is an integer.

Hence \(ka + kb = k(a + b) = 4k\) is an integer, so that \(k\) is an integer multiple of \(\tfrac{1}{4}\). Similarly, each of \(\ell\), \(a\) and \(b\) is an integer multiple of \(\tfrac{1}{4}\).

However, none of \(k\), \(\ell\), \(a\) and \(b\) is an integer multiple of 4, because each is between 0 and 4. Hence each of them has to be an integer multiple of \(\tfrac{1}{2}\) in order that each of \(ka\), \(kb\), \(\ell a\) and \(\ell b\) is an integer.

If both \(\boldsymbol{a}\) and \(\boldsymbol{k}\) are integers

If one of \(\boldsymbol{a}\) or \(\boldsymbol{k}\) is not an integer (that is, an odd multiple of \(\boldsymbol{\tfrac{1}{2}}\))

Note that once \(a\) and \(k\) have been chosen, then these determine \(b\) and \(\ell\) uniquely.

Thus altogether there are \(9 + 4 + 4 = 17\) ways.

If the grid can be rotated or reflected, so that, for example, \(a = 1\) is counted as being the same as \(k = 1\) or \(a = 3\), then there are 5 ways, given by the following pairs of values of \((a, k)\) or \((k, a)\):

as shown below.