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2010 Euclid Contest Discussion

  1. Apr 8, 2010 #1
    As you might know, the 2010 Euclid Contest was officially taken yesterday. So lets discuss!

    I thought it wasnt too bad. #10 was hard though (the triangle one). Here was the question:

    For each positive integer n, let T(n) be the number of triangles with integer side lengths, positive area, and perimeter n. For example, T(6) = 1 since only such triangle with a perimeter of 6 has side lengths 2, 2 and 2.
    (b) If m is a positive integer with m >=(greater than or equal to) 3, prove that T(2m) = T(2m-3).
    (c) Determine the smallest positive integer n such that T(n) > 2010.

    Also, #9 was hard:

    (b) In triangle ABC, BC = a, AC = b, AB = c, and a < .5(b+c).
    Prove that angle BAC < .5 (angle ABC + angle ACB).

    Can you please help me with those problems?

    Also, if anyone wants to share how they solved #7 and 8, that would be appreciated :)
     
    Last edited: Apr 8, 2010
  2. jcsd
  3. Apr 8, 2010 #2
    2010 Euclid Contest , very easy. just 10(b)(c) i cannot do it.

    9(b):
    use this formula:A/SINA=B/SINB=C/SINC

    7A:too easy .cannot remember the answer
    7B:i just remenbered there are 2 points.
    8A:cannot remember
    8b:110
    9A(i):too easy
    9A(ii):120
     
  4. Apr 8, 2010 #3
    oh ****. 9A(ii)should be 4. i misunderstand the queston.
    but anyway, i can get at least 80:)
     
  5. Apr 10, 2010 #4
    Does the fact that [tex]{\sin}BAC<\frac{1}{2}({\sin}ABC+{\sin}BCA)[/tex] imply the desired result in problem 9b?
     
  6. Apr 10, 2010 #5
    have you not got the answer 9(B)? very very easy!!!!

    here is my answer:
    A/SINA=B/SINB=C/SINC
    so 2SINA<SINB+SINC

    because this is permanent.
    we know SINB+SINC is bigger or equal 2(root SINB*SINC) only when SINB=SINC , it can be equal.
    so 2SINA<2(root SINB*SINC)
    SINCE SINB=SINC
    2SINA<2SINB=2SINC
    SINA<SINB=SINC
    SINCE A+B+C=180
    SO WHEN SINA=SINB=SINC,A=B=C=60, THEN A MUST LESS THAN 60
    SO 0<A<60
    120<B+C<180

    A<0.5(B+C)
     
    Last edited: Apr 10, 2010
  7. Apr 10, 2010 #6
    Doesn't that proof lose generality when you assume that [tex]{\sin}B={\sin}C[/tex]?
     
  8. Apr 10, 2010 #7
    oh yes
    i should rethink it
     
    Last edited: Apr 11, 2010
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