The diagram shows triangle ABC with points chosen on each of the sides - HSC - SSCE Mathematics Extension 1 - Question 7 - 2022 - Paper 1

To form a triangle, we need to choose 3 points from the total of 12 points. The number of ways to choose 3 points from 12 is given by the combination formula:

$C(n, r) = \frac{n!}{r!(n-r)!}$

So, we compute:

$C(12, 3) = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$

However, we need to subtract the cases where the selected points are collinear.

• For side AB (3 points): The number of triangles that can be formed using all points from AB is:

$C(3, 3) = 1$

• For side AC (4 points): The cases are:

$C(4, 3) = 4$

• For side BC (5 points): The cases are:

$C(5, 3) = 10$

Total collinear combinations = 1 + 4 + 10 = 15.

Therefore, the total number of triangles that can be formed using the chosen points as vertices is:

$Total\ Triangles = C(12, 3) - Total\ Collinear\ Points = 220 - 15 = 205$