# Least number of scalene triangles formed from $13$ points in a plane

Given $$13$$ points in a plane with no three on a line, prove that there are at least $$130$$ scalene triangles formed from the points.

I thought the highest number of non-scalene triangles with $$13$$ points would happen in a $$13$$ side polygon, so I tried to calculate this value, and I got $$78$$ so I subtracted it from the number of triangles formable by $$13$$ points, $$13 \choose 3$$ and then I got $$208$$. I know it's wrong but I can't think of anything else.

Any help would be greatly appreciated. Thanks!

• Show your work first. Commented Nov 24, 2020 at 18:58
• What have you tried so far? (This may also include resources found on the internet/methods applied...) Also visit math.stackexchange.com/help/how-to-ask Commented Nov 24, 2020 at 18:59
• What is your definition of a scalene triangle ? All sidelengths different ? Commented Nov 24, 2020 at 19:04
• @JeanMarie Why, what other definition did you have in mind? Commented Nov 24, 2020 at 19:05
• I thought the highest number of non-scalene triangles with 13 points would happen in a 13 side polygon, so I tried to calculate this value, and I got 78 so I subtracted it from the number of triangles formable by 13 points (c(13,3)) and then I got 208. I know it's wrong but I can't think of anything else Commented Nov 24, 2020 at 19:05

There are $${13 \choose 2}$$ couple of points.
Each couple can be the endpoints of the base of at most $$2$$ isosceles triangles, otherwise with three or more triangles, the corresponding three or more vertices opposite to the base would be on a line (the perpendicular bisector of the base). If one or more of the isosceles triangles are instead equilateral, we can choose a base with some criterion, e.g. the one formed with the two points nearest to the origin.
Therefore there are at most $${13 \choose 2} \cdot 2$$ isosceles or equilateral triangles and at least $${13 \choose 3} - {13 \choose 2} \cdot 2 = 130$$ scalene triangles. I believe there are actually much more than that though.
Instead of counting only at the base, we could count at all sides. In that case for each couple of points one can build from the corresponding side at most $$6$$ isosceles or equilateral triangles ($$2$$ triangles with the side as base and $$4$$ triangles with the side as leg), but in this way each triangle would be counted $$3$$ times, so we still have at least $${13 \choose 3} - {13 \choose 2} \cdot \frac{6}{3} = 130$$ scalene triangles.