[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

- ganeshie8

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- ParthKohli

So does that mean we want acute triangles?

- ganeshie8

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

- ganeshie8

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

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## More answers

- Empty

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

- ganeshie8

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

- Empty

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

- ganeshie8

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

- ParthKohli

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

Now I think it boils down to counting and stuff.

- ganeshie8

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

|dw:1439631577726:dw|

- ganeshie8

The other two vertices of triangle must go on opposite sides of that diameter

- ParthKohli

Why do you think squares are being summed up? Maybe we're dividing this polygon into smaller triangles or something?

- ParthKohli

Wait, never mind.

- ParthKohli

That was such a dumb question.

- ParthKohli

Could you draw me a regular pentagon here?

- ganeshie8

|dw:1439632213799:dw|

- ganeshie8

regular 7-gon
|dw:1439632244889:dw|

- ParthKohli

|dw:1439632232991:dw|

- ParthKohli

We include all triangles that include that vertex except one. If we fix one vertex, there are \(2n \) others. The number of triangles is \(2n\cdot (2n-1) - 1 = 4n^2 - 2n - 1 \). This is only for one vertex. We can multiply this by \(2n+1\). But we'll also have to divide by the number of times we count a triangle.

- ParthKohli

We'll be counting a triangle thrice, right?

- ParthKohli

\[\frac{1}{3}(2n+1)(2n(2n-1) - 1)\]\[= \frac{1}{3}(2n+1) (4n^2 - 2n -1 )\]Nope... :(

- ganeshie8

two triangles don't include that fixed vertex right ?
|dw:1439632648793:dw|

- ParthKohli

|dw:1439632744506:dw|

- ParthKohli

|dw:1439632796315:dw|

- ParthKohli

Also the star isn't supposed to be complete because then that'd be the triangle which is obtuse :P

- ParthKohli

OK, let's see again.
If I fix a vertex \(A\), then I have \(2n\) remaining vertices. I want to choose two out of them, so I have \(2n(2n-1)\) sets of vertices \(\{(A', A'')\}\). Then I have to divide by two because I am counting \((A',A'') ; (A'', A')\) separately. Then I have \(n(2n-1)\) such triangles. Now I have to subtract one because you know why. That's \(2n^2 - n - 1 = 2n^2 - 2n + n - 1 = (2n+1)(n-1)\) triangles for a vertex. And I'm still getting the wrong answer.

- ParthKohli

I've never studied counting my entire life. Where's the problem?

- ParthKohli

So you mean we have to actually subtract \(n+1\) triangles?

- ganeshie8

not so sure, im still thinking
btw idk solution for this problem..

- imqwerty

nice prb :)

- anonymous

i was thinking.. and somehow it should come to 1/4... but im not sure how...

- ganeshie8

The probabiltiy for randomly picking an acute triangle approaches \(\dfrac{1}{4}\) as the number of sides increase so that polygon becomes a circle :
http://www.wolframalpha.com/input/?i=lim%28n%5Cto%5Cinfty%29++%5Cdfrac%7Bn%28n%2B1%29%282n%2B1%29%2F6%7D%7B%282n%2B1%29+choose+3%7D

- anonymous

See the figure attached.
Let's first fix a vertex , say, L. Now draw the largest
chord that can be drawn with one end L and the other end a vertex of the polygon.
Say, it's LQ.
Draw the diameter through Q. This bisects the opposite side to Q
perpendicularly (possible since it's a regular polygon with odd number of sides).
Let's name this side LJ and the perpendicular bisector as QT.
Now, we try to find all triangles of the required nature with one vertex L.
Note that the other two vertices apart from L must belong to opposite sides of QT.
Let LME be one such triangle.
Let LM makes an angle \(k*2\pi /2n+1\) and let LE makes the
angle \(l*2\pi/2n+1\) at the center.
We assume this form because, each side makes an angle of \(2\pi/n\) at the
center, and the diagonals LM and LE are assumed to encompass k and l
sides respectively.
Then for the triangle LME to contain the center, it must be
that angles LUM and LUe must be strictly greater than \(\pi\)
that is, \(k*2\pi /(2n+1) + l*2\pi/(2n+1) > \pi\), that is,
\(k + l > n + 1/2\) , or,\(k + l > n+1\)
Also note that there are n vertices on each side of QT,
so , \(k \leq n\) and \(l \leq n\)
So, when k =1, l must be equal to n
when k =2, l can be n, n-1
....
....when k =a, l can be n, n-1,.., n-(a-1) -> a values
....
....when k =n, l can be n, ... , 1 - > n values
So , basically the total number of triangles containing
the center and with L as one vertex is
\(1+2+...+n = n(n+1)/2 \)
Now there are 2n+1 choices for L.
But as we navigate through all these choices for L, each
triangle is counted thrice, once for each vertex, so we
take a third of (2n+1)n(n+1)/2, that is,
\(n(n+1)(2n+1)/6\)
[Not very confident or happy about the solution :P ]

##### 1 Attachment

- ganeshie8

Very nice!
so basically, each vertex involves in exactly \(\dfrac{n(n+1)}{2}\) acute triangles. Since there are \(2n+1\) vertices, the total count is \(\dfrac{n(n+1)}{2}(2n+1)\). However the same triangle appears in \(3\) vertices count. accounting for these repititions, the magic formula follows! Looks pretty neat to me! thank you so much :)

- ganeshie8

that is like my summary in my head of your proof..

- anonymous

Yup that's basically what it means :)

- anonymous

you've both actually forgotten something... you must subtract (n-1) triangles from n(n+1)/2 as those are congruent to some triangles you have already made with one point. I have deduced this through a lot of trials, and the probability is still 1/4.

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