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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.

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marked as duplicate by Macavity, user127.0.0.1, Daryl, Stefan Hansen, mau Feb 3 '14 at 8:57

This question was marked as an exact duplicate of an existing question.

Are the points distributed evenly across the circle? Certainly if all points lie within the same semicircle, then no such triangle can be drawn. – Omnomnomnom Feb 3 '14 at 2:26
As the different distribution of the points may result some very different answers. You need to add the condition here. – LorenMt Feb 3 '14 at 2:35
Same as… – Macavity Feb 3 '14 at 3:15

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.

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