Maclaurin 2017 Q2

Method 1

There are 17 dots in the array, and we must choose 3 of them to obtain a triangle. This can be done in \(\binom{17}{3} = 680\) ways. However, some of these triangles have zero area.

The triangle will have zero area if we choose all three dots on the same line. Hence the number of triangles of zero area is \(2 \times \binom{9}{3} = 2 \times 84 = 168\).

So there are 512 triangles of non-zero area.

Method 2

We may choose two points on the line across the page in \(\binom{9}{2} = 36\) ways, and one point on the line up the page (but not at the centre) in 8 ways. These choices give rise to \(8 \times 36 = 288\) triangles of non-zero area.

Similarly we obtain another 288 triangles from two points on the line up the page and one on the line across the page.

But we have counted twice triangles with a vertex at the point where the lines meet, and there are \(8 \times 8 = 64\) of these. So altogether we have \(2 \times 288 - 64 = 512\) triangles.