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 triangles, we need to select 3 points from the total of 12. We use the combination formula:

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

Where n is the total points and r is the points selected:

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

Now, we must exclude the combinations of points that are collinear.

• For side AB (3 points), only 1 triangle can be formed: $C(3, 3) = 1$
• For side AC (4 points), we can form: $C(4, 3) = 4$
• For side BC (5 points), we can form: $C(5, 3) = 10$

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

Finally, we subtract the collinear combinations from the total combinations of points:

Total triangles = Total combinations - Collinear combinations

Total triangles = 220 - 15 = 205.

Therefore, the number of triangles that can be formed using the chosen points as vertices is 205.