[SOLVED by @adxpoi ] show that there are $\dfrac{n(n+1)(2n+1)}{6}$ triangles made from the vertices of a regular $2n+1$-gon such that the center of inscribed circle lies interior to the triangle

Question

[SOLVED by @adxpoi ] show that there are $\dfrac{n(n+1)(2n+1)}{6}$  triangles made from the vertices of a regular $2n+1$-gon such that the center of inscribed circle lies interior to the triangle

parthkohli · Answer

So does that mean we want acute triangles?

ganeshie8 · Answer

Precicely! https://i.gyazo.com/201ec61d12a823629c7fee212b70b2ee.png

ganeshie8 · Answer

https://oeis.org/search?q=5%2C14%2C30&language=english&go=Search

empty · Answer

Each n adds a new square that much I know. :P

ganeshie8 · Answer

Haha, suppose we don't know the formula..

empty · Answer

So making them odd sided removes any conflict with a point lying on the boundary of our triangle that's helpful at least.

ganeshie8 · Answer

I think this way of counting also gives a new way to derive formula for sum of first n squares

parthkohli · Answer

So first of all, each interior angle in a regular 2n+1-gon is$$\frac{180(2n-1 )}{2n+1} $$which is greater than $90 $ when $n\ge \frac{3}{2}$ so basically it's always obtuse except for triangles. 

So one thing we're sure about is that no interior angle can be a part of the triangles we make. That means that the vertices cannot be adjacent to each other.

parthkohli · Answer

Now I think it boils down to counting and stuff.

ganeshie8 · Answer

right, another trivial observation  : 

If you fix the first vertex of triangle, then the other two vertices must chosen  diametrically opposite such that the triangle captures the center

ganeshie8 · Answer

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