Tuesday, January 12, 2010

GP Question 7


Newzad submitted this question for our consideration. It is more related to numbers theory than geometry, and looks very complicated. Does anybody have any idea? Please do comment.

6 comments:

  1. Thank you very much for publishing this question 8foxes
    After final exams (nearly two weeks), i will be back to follow your great questions.

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  2. answer is yes. For any N, there exists a way to build a right triangle with horizontal leg divided in N segments such that all triangles have integer sides.

    Here is a (non-optimal) way to build such a triangle: for any m,n such that m>n>1 (2mn, m^2-n^2, m^2+n^2) form a pythagorean triad. Take the following (m,n) pairs:
    (2^(2N-1),1), (2^(2N-2),2), ... , (2^N, 2^(N-1))
    they generate N distinct triads with common leg 2^2N. Corresponding triangles can be drawn leading to requested figure.

    Optimal height leg for N=3 is 12. The 3 triads are (12,5,13), (12,16,20), (12,35,37). Horizontal segments are 5, 11, 19

    Interestingly, 24 (as depicted in example) is the optimal leg for N=5. Therefore the example given can be extended with the following segments: 7, 3, 22, 38, 73. Corresponding triads: (24,7,25), (24,10,26), (24,32,40), (24,70,74), (24,143,145)

    bleaug

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  3. oops, I forgot that (m,n) generators do not build all pythagorean triads.

    Actually 12 is optimal for N=3 and N=4. Triad (12,9,15) was missing. Segments would be 5, 4, 7, 19.

    24 is optimal for N=5, N=6 and N=7! The following triads were missing: (24,18,30) and (24, 45, 51). Segments would be: 7, 3, 8, 14, 13, 25, 73.

    bleaug

    ReplyDelete
  4. Good work for triads.
    Are there any quadriads?
    Pentads?
    Now it really gets ugly.

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  5. Solution was finded bras.

    http://geometri-problemleri.blogspot.com/2010/08/gp-question-7-solution.html

    ReplyDelete
  6. Also found the the following solution from Aykut.
    Link: http://www.metu.edu.tr/~baykut/Newzad_Triangles.pdf

    Text:
    For a finite fixed a, can we draw infinitely many Newzad Triangles by extending the hypotenuse?
    First notice that if we can draw infinitely many triangles for a fixed a, then there are infinitely
    many b, c, d integers that satisfy the following equations. (Since we can rename b + c as b and d as c and e as d; or b + c + d as b, e as c ...)
    a^2 + b^2 = c^2 (1)
    a^2 + (b + c)^2 = d^2 (2)

    Then
    a^2 = (c − b)(c + b) (3)
    a^2 = (d − c − b)(d + c + b) (4)
    Let a1 and a2 be two integers such that
    a1 × a2= a^2 and a1 = c − b and a2 = c + b.
    Then for a unique fixed (a1, a2) pair b and c are unique. Then from equation (4) d is also unique.
    Which means that for a unique (a1, a2) pair we have unique integers b, c and d. Then since a finite
    fixed a has a finite number of integer multipliar pairs (a1, a2), we have finite number of integers b,
    c and d. Then for a finite fixed a, one can not draw infinitely many Newzad Triangles.

    ReplyDelete