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ganeshie8
 one year ago
[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
 one year ago
[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

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1So does that mean we want acute triangles?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Precicely! https://i.gyazo.com/201ec61d12a823629c7fee212b70b2ee.png

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4https://oeis.org/search?q=5%2C14%2C30&language=english&go=Search

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Each n adds a new square that much I know. :P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Haha, suppose we don't know the formula..

Empty
 one year ago
Best ResponseYou've already chosen the best response.0So making them odd sided removes any conflict with a point lying on the boundary of our triangle that's helpful at least.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I think this way of counting also gives a new way to derive formula for sum of first n squares

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1So first of all, each interior angle in a regular 2n+1gon is\[\frac{180(2n1 )}{2n+1} \]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
 one year ago
Best ResponseYou've already chosen the best response.1Now I think it boils down to counting and stuff.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4right, 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
 one year ago
Best ResponseYou've already chosen the best response.4dw:1439631577726:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4The other two vertices of triangle must go on opposite sides of that diameter

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Why do you think squares are being summed up? Maybe we're dividing this polygon into smaller triangles or something?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1That was such a dumb question.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Could you draw me a regular pentagon here?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4dw:1439632213799:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4regular 7gon dw:1439632244889:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439632232991:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1We include all triangles that include that vertex except one. If we fix one vertex, there are \(2n \) others. The number of triangles is \(2n\cdot (2n1)  1 = 4n^2  2n  1 \). 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
 one year ago
Best ResponseYou've already chosen the best response.1We'll be counting a triangle thrice, right?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{1}{3}(2n+1)(2n(2n1)  1)\]\[= \frac{1}{3}(2n+1) (4n^2  2n 1 )\]Nope... :(

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4two triangles don't include that fixed vertex right ? dw:1439632648793:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439632744506:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439632796315:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Also the star isn't supposed to be complete because then that'd be the triangle which is obtuse :P

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1OK, 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 \(2n(2n1)\) sets of vertices \(\{(A', A'')\}\). Then I have to divide by two because I am counting \((A',A'') ; (A'', A')\) separately. Then I have \(n(2n1)\) such triangles. Now I have to subtract one because you know why. That's \(2n^2  n  1 = 2n^2  2n + n  1 = (2n+1)(n1)\) triangles for a vertex. And I'm still getting the wrong answer.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1I've never studied counting my entire life. Where's the problem?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1So you mean we have to actually subtract \(n+1\) triangles?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4not so sure, im still thinking btw idk solution for this problem..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i was thinking.. and somehow it should come to 1/4... but im not sure how...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4The 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
 one year ago
Best ResponseYou've already chosen the best response.0See 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, n1 .... ....when k =a, l can be n, n1,.., n(a1) > 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 ]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Very 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
 one year ago
Best ResponseYou've already chosen the best response.4that is like my summary in my head of your proof..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yup that's basically what it means :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you've both actually forgotten something... you must subtract (n1) 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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