How many triangle can be drawn with those points?

Question

There are 7 points on the circumference of a circle.How many acute triangle can be drawn with those points. please help me to solve this problem.

Answer 1

Assuming that the points are distributed evenly across the circle:

Note that three points on a circle form a non-acute triangle exactly when they lie within the same semicircle. It follows that the set of acute triangles that we can make from the seven points is the same as all of the triangles except for those that can be made from $4$ adjacent vertices.

Label the seven points on the circle as $1,2,\dots,7$.

There are $\binom 73 = \frac{7\times 6 \times 5}{3 \times 2 \times 1} = 35$ triangles total.

Of those, the $7$ can be made by rotating $\{1,2,3\}$ about the center of the circle (that is, $\{2,3,4\}, \{3,4,5\}$ and so forth) are not acute. Those that can be made by rotating $\{1,2,4\}$ are also non-acute, as are those can be made by rotating $\{1,3,4\}$.

Thus, there are $7 \times 3 = 21$ non-acute triangles. It follows that there are exactly $14$ acute triangles.