# $7$ points inside a circle at equal distances

BdMO 2014

There are $7$ points on a circle.Any 2 consecutive points are at equal distance from one another.How many acute angled triangles can you form taking any 3 of these points?

I believe there are 5 possible triangles that we can form using any three points.Then we can find out which of these are acute-angled.Then we can compute the number one by one and sum them up.Am I right?

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I just learned that the circumcentre must always be inside the circle if it is acute.The converse is true as well.Surely we can apply that here. –  rah4927 Feb 2 '14 at 8:18

Let the points be at $\theta_n = \frac{2\pi}7 n$ for $n = 0, 1, 2, ...6$ on the unit circle. Note that the triangle will be acute as long as all three vertices are not on the same semicircle.

So if we choose the first vertex to be $n_1=0$, then we can choose the second vertex to be any other of $n_2=1, 2, 3$ points. Now if $n_2=1$, then $n_3=4$, if $n_2=2$ then $n_3 = 4, 5$ and if $n_2=3$, we can have $n_3 = 4, 5, 6$. As this covers all cases, there are $6$ triangles possible with $n_1=0$.

As there are $7$ choices for $n_1$, but we would end up counting each solution thrice, we have a total of $6\times7 \div 3 = 14$ possible distinct acute triangles.

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Thanks.Sorry I wasn't able to reply earlier but I didn't receive any notification for some reason. –  rah4927 Feb 3 '14 at 4:42

We have a regular polygon. The triangle will be obtuse iff we connect one vertex with another vertex that is 4 or more vertices away, because then the opposite angle from that side would intercept an arc of greater than half the circle.

So this can be similarly stated as how many ordered partitions of $7$ intro three are there such that none of the partitions exceed $3$ and each is at least $1$. This is the same as how many solutions are there of $x_1 + x_2 + x_3 = 4$ with $x_i < 3$. There are ${6 \choose 4} = 15$ ways to do so; of these, $9$ of them do not fit the second restriction ($(3,1,0) \times 6; (4, 0, 0) \times 3$), giving us $6$.

Then we rotate each of these $6$ valid configurations $6$ times, giving $7 \times 6 = 42$ acute triangles. We triple count each, so there are $14$.

Alternate method:

There are $7 \choose 3$ triangles in total. If the three points lie on the same semi-circle, they will be obtuse. There are $7$ sets of $4$ consecutive points which lie on the same semi-circle, and for each set there are $3$ obtuse triangles we can make (from vertices $1-2-3, 1-2-4, 1-3-4$ -- note that $2-3-4$ is counted in the next set). Thus our answer is ${7 \choose 3} - 7(3) = 14$

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