# Sunday Times Teaser 2978 – Norfolk Flats

*by Andrew Skidmore*

#### Published October 20 2019 (link)

Sam has purchased Norfolk Flats; an area of farmland (less than 100 hectares) bordered by six straight fences. He intends to farm an area which is an equilateral triangle with corners that are the midpoints of three of the existing boundaries. This creates three more distinct areas (one for each of his sons); these areas are identical in shape and size and have two sides that are parallel.

Sam measured the area (in square metres) which each son will farm and also his own area. One of the numbers is a square and the other a cube. If I told you which was which, you should be able to work out the area of Norfolk Flats.

What (in sq metres) is that area?

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The arrangement of the father’s and his son’s fields are shown here:

By inspection it can be seen that the area of each of the son’s fields is a simple fraction of the area of the father’s field. When this configuration is analysed, however, the teaser doesn’t have a unique solution unless the overall area is a regular hexagon with equal sides. This is what is assumed in the following analysis.

The area of a regular hexagon of side \(s\) is given by \(3\sqrt{3}s^2/2\). Defining a unit area \(A\) as \(\sqrt{3}s^2/16\), this becomes \(24A\). The area of the farmers field is then \(9A\) and each of the son’s fields has an area of \(5A\).

Considering the first situation in which the farmer’s field’s area is a square \(x^2\) and the area of each of the sons’ field’s is a cube \(y^3\), we have:\[\begin{array}{l}x^2&=&9A\\y^3&=&5A\\\end{array}\] Eliminating \(A\) and rearranging the result now gives:\[(x/15)^2=(y/5)^3=n^6\] with \(n\) integer (1, 2, …). This \(n^6\) term arises because an integer that is simultaneously a square and a cube must be an integer 6’th power. Hence in this scenario the area of the father’s field is \(x^2 = 225n^6\), that of each of the son’s fields is \(y^3=125n^6\) and the overall area is \(x^2 + 3y^3=600n^6\). There are three solutions with an overall area less than 100 hectares (\(n = 1, 2, 3)\).

Considering the second situation in which the farmer’s field’s area is a cube \(x^3\) and the area of each of the sons’ field’s is a square \(y^2\), we have:\[\begin{array}{l}x^3&=&9A\\y^2&=&5A\\\end{array}\] Eliminating \(A\) and rearranging the result now gives:\[(x/45)^2=(y/225)^3=n^6\] Hence in this scenario the area of the father’s field is \(x^3 = 91125n^6\), that of each of the son’s fields is \(y^2=50625n^6\) and the overall area is \(x^3+3y^2=243000n^6\). In this case there is only one solution with \(n = 1\), giving a Norfolk Flats area of \(243000\) square metres (24.3 hectares).

Here is a solution based on the analysis given above: