A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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

  • This Question is Closed
  1. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So does that mean we want acute triangles?

  2. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  3. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  4. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  6. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  7. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  8. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  9. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Now I think it boils down to counting and stuff.

  10. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  11. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    |dw:1439631577726:dw|

  12. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  13. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  14. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Wait, never mind.

  15. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    That was such a dumb question.

  16. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Could you draw me a regular pentagon here?

  17. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    |dw:1439632213799:dw|

  18. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    regular 7-gon |dw:1439632244889:dw|

  19. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1439632232991:dw|

  20. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  21. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    We'll be counting a triangle thrice, right?

  22. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  23. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  24. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1439632744506:dw|

  25. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1439632796315:dw|

  26. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  27. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  28. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  29. ParthKohli
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  30. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  31. imqwerty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    nice prb :)

  32. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  33. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  34. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
  35. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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 :)

  36. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  37. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yup that's basically what it means :)

  38. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  39. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.