We are given a triangle $\triangle ABC$ with several points on its sides. Specifically, there are:

• 6 points on side $AB$,
• 7 points on side $BC$,
• 8 points on side $AC$.

We are tasked with finding how many triangles can be formed using one point from side $AB$, one point from side $BC$, and one point from side $AC$.

Approach

We can think of this problem as choosing one point from each of the three sides to form a triangle. The condition "one vertex on $AB$, one vertex on $BC$, and one vertex on $AC$" simplifies the problem, as it guarantees that no three points are collinear.

The total number of such triangles is simply the product of the number of available points on each side:

$\text{Total number of triangles} = \text{(number of points on } AB\text{)} \times \text{(number of points on } BC\text{)} \times \text{(number of points on } AC\text{)}$

Substituting the given values:

$\text{Total number of triangles} = 6 \times 7 \times 8$

$\text{Total number of triangles} = 336$

Thus, the total number of triangles that can be formed with one vertex on $AB$, one vertex on $BC$, and one vertex on $AC$ is $\boxed{336}$.

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Would you like further details on any part of the solution?

Here are 5 related questions for deeper exploration:

1. What if there were additional points on one or more sides?
2. How would the problem change if the points were allowed to coincide with the triangle's vertices?
3. What is the impact if we add constraints on the type of triangle (e.g., equilateral)?
4. Can this combinatorial approach be applied to quadrilaterals instead of triangles?
5. How many triangles could be formed if one vertex must be a triangle vertex?

Tip: In combinatorics problems like this, multiplying the number of choices across independent events (in this case, choosing points from different sides) often leads to the solution!