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. We can calculate this using the combination formula:

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

where n is the total number of points, and r is the number of points to choose (which is 3 in this case).

Thus, we have:

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

Next, we must consider cases where the chosen points might be collinear. We analyze each side individually:

• On side AB: Choosing any 3 points will not result in a triangle, so it contributes 0.
• On side AC: Similarly, this contributes 0.
• On side BC: This also contributes 0.

Thus, there are no collinear combinations that need to be subtracted from the total.

Since all combinations of 3 points from the total of 12 points yield a valid triangle, the total number of triangles formed is 220.