PK 2015 Q25

Suppose the first of the two special points mentioned has \(a\) points to its left, and \(b\) points to its right. Then the number of line segments it lies on is \(a \times b\) (\(a\) choices for the left end, \(b\) choices for the right end). Also the number of points will be \(a + b + 1\). Similarly if the second point has \(c\) points to its left, and \(d\) to the right, then the number of line segments it lies on is \(c \times d\), and the number of points is \(c + d + 1\).
Hence we need to find integers \(a\), \(b\), \(c\), \(d\) such that \(ab = 80\), \(cd = 90\), \(a + b + 1 = c + d + 1\) (or more simply \(a + b = c + d\)).
The factor pairs of 80 (and hence the possible values of \(a\), \(b\)) are \(1 \times 80\), \(2 \times 40\), \(4 \times 20\), \(5 \times 16\), \(8 \times 10\).
The factor pairs of 90 (and hence possible values of \(c\), \(d\)) are \(1 \times 90\), \(2 \times 45\), \(3 \times 30\), \(5 \times 18\), \(6 \times 15\), \(9 \times 10\).
The only pairs for which we have \(a + b = c + d\) are 5, 16 and 6, 15. Since both of these add to 21, the number of points must be \(a + b + 1 = 21 + 1 = 22\).