Page 2 of 2 FirstFirst 12
Results 16 to 24 of 24

Math Help - Geometry

  1. #16
    Super Member
    Joined
    Apr 2009
    Posts
    677
    Quote Originally Posted by Drexel28 View Post
    Question:

    Spoiler:
    What do you mean "lying in \mathbb{Q}^2? Do you mean that the verticies of the triangle are rational points or that literally every point of the triangle is rational? Because I have a problem witht the second one...
    Guess, you can't have that. Points in a line sengment are not countable - so every point can't be rational. Correct?

    Also, what do you mean by tangent of the angle?
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Member
    Joined
    Nov 2009
    Posts
    106
    Quote Originally Posted by aman_cc View Post
    Guess, you can't have that. Points in a line sengment are not countable - so every point can't be rational. Correct?

    Also, what do you mean by tangent of the angle?
    Sure you're right, I meant the vertices are in , my mistake for no paying enough attention to wording.
    If a is an interior angle of the triangle, tan(a) is what I meant by the tangent of the angle, and the proposition claims that it is rational.
    In the original case of equilateral triangle, it turns out no equilateral triangle
    exist with vertices in , as tan(60 deg) = sqrt(3)
    Last edited by Unbeatable0; December 2nd 2009 at 09:21 AM. Reason: Typo
    Follow Math Help Forum on Facebook and Google+

  3. #18
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by aman_cc View Post
    Guess, you can't have that. Points in a line sengment are not countable - so every point can't be rational. Correct?

    Also, what do you mean by tangent of the angle?
    That is precisely why! When speaking of triangles, we must first talk about the edges of the triangle. These edges are connected and may be interpreted as being homeomorphic to some interval in \mathbb{R} and since any interval in \mathbb{R} is uncountable we can't speak of lines in a countable spac such as \mathbb{Q}^2

    P.S. I didn't check that my language above was precise, but you get the idea.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Member
    Joined
    Nov 2009
    Posts
    106
    I almost forgot about the first problem you stated.
    The solution:

    Spoiler:

    Let O be the center of the regular polygon (the center of the circumcenter), and let R be the distance from the O to the vertices.
    The angle between two lines beginning in O and ending in two adjacent vertices is thus \frac{2\pi}{n}, and from the sinus law we get that the area of a triangle formed by such two lines is given by \frac{1}{2}R^2\sin{\frac{2\pi}{n}} = R^2\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}
    But there are n such triangles in the regular polygon. Using our last result this gives us n R^2\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}
    Furthermore, if we drop an altitude to the base in one of those isosceles triangles, we come up with R = \frac{\frac{1}{2} a}{\sin{\frac{\pi}{n}}} , and after subtitution in the expression for the area we get  n (\frac{\frac{1}{2} a}{\sin{\frac{\pi}{n}}})^2\sin{\frac{\pi}{n}}\cos{  \frac{\pi}{n}} = \frac{1}{4} a^2 n \cot{\frac{\pi}{n}} .
    Q.E.D
    Last edited by Unbeatable0; December 3rd 2009 at 11:10 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #20
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Unbeatable0 View Post
    I almost forgot about the first problem you stated.
    The solution:

    Spoiler:

    Let O be the center of the regular polygon (the center of the circumcenter), and let R be the distance from the O to the vertices.
    The angle between two lines beginning in O and ending in two adjacent vertices is thus \frac{2\pi}{n}, and from the sinus law we get that the area of a triangle formed by such two lines is given by \frac{1}{2}R^2\sin{\frac{2\pi}{n}} = R^2\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}
    But there are n such triangles in the regular polygon. Using our last result this gives us n R^2\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}
    Furthermore, if we drop an altitude to the base in one of those isosceles triangles, we come up with R = \frac{\frac{1}{2} a}{\sin{\frac{\pi}{n}}} , and after subtitution in the expression for the area we get  n (\frac{\frac{1}{2} a}{\sin{\frac{\pi}{n}}})^2\sin{\frac{\pi}{n}}\cos{  \frac{\pi}{n}} = \frac{1}{4} a^2 n \cot{\frac{\pi}{n}} .
    Q.E.D
    Note now that with this and Pick's theorem we can say any n such that \cot\left(\frac{\pi}{n}\right)\notin\mathbb{Q} cannot have all vertices on lattice points. I just came to that conclusion! Sweet! Suck that pentagon!
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Member
    Joined
    Nov 2009
    Posts
    106
    Quote Originally Posted by Drexel28 View Post
    Note now that with this and Pick's theorem we can say any n such that \cot\left(\frac{\pi}{n}\right)\notin\mathbb{Q} cannot have all vertices on lattice points. I just came to that conclusion! Sweet! Suck that pentagon!
    Actually, from the proposition I posted before, it follows that for any regular n-sided polygon, if \cot{\frac{\pi}{n}} is irrational, then it can't have three vertices in , two of which are adjacent.
    Nobody is up to prove the claim in post #14?
    Follow Math Help Forum on Facebook and Google+

  7. #22
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Unbeatable0 View Post
    Actually, from the proposition I posted before, it follows that for any regular n-sided polygon, if \cot{\frac{\pi}{n}} is irrational, then it can't have three vertices in , two of which are adjacent.
    Nobody is up to prove the claim in post #14?
    I would like to give it a try, but I am swamped right now. Give me, or someone else, a day or so and then post up the solution otherwise!
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Member
    Joined
    Nov 2009
    Posts
    106
    Quote Originally Posted by Unbeatable0 View Post
    Prove that in any triangle in \mathbb{R}^2 whose vertices are in \mathbb{Q}^2, the tangents
    of the angles are rational.

    I allow myself to post a solution (not mine), after seeing that the problem has not caught too much interest. The proof is in the attached image.
    Attached Thumbnails Attached Thumbnails Geometry-rationaltangents.png  
    Follow Math Help Forum on Facebook and Google+

  9. #24
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Unbeatable0 View Post
    I allow myself to post a solution (not mine), after seeing that the problem has not caught too much interest. The proof is in the attached image.
    That was close to my solution! Which I actually logged onto to post! I didn't have a picture though..so this one is decidedly better. Sorry about not doing it. I had Putnam, finals, etc. Thanks for the solution!
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. [SOLVED] Finite Geometry: Young's Geometry
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: September 15th 2010, 07:20 AM
  2. geometry
    Posted in the Geometry Forum
    Replies: 2
    Last Post: December 21st 2009, 03:18 AM
  3. Modern Geometry: Taxicab Geometry Problems
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 30th 2009, 07:32 PM
  4. geometry
    Posted in the Geometry Forum
    Replies: 3
    Last Post: November 17th 2008, 05:29 PM
  5. geometry
    Posted in the Geometry Forum
    Replies: 13
    Last Post: June 8th 2008, 07:31 AM

Search Tags


/mathhelpforum @mathhelpforum