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Just check my solution to this problem

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the number of of all integer sided isosceles obtuse angle triangles with perimeter 2008.

    3. The attempt at a solution
    I assumed the obtuse angle as 90+2a
    Let the length of the equal sides be L an that of the base = 2X

    Draw the perpendicular to the side opposite to the obtuse angle from the opposite vertex.
    It will divide the triangle in to congruent parts. Using sine rule in one of the to triangles formed.
    I get
    sin (45+a)/X=1/L
    sin (45+a)/X=1/(1004-X)

    X=1004/ (cosec (45+a)+1)

    2<cosec (45+a)+1<2.4142

    as 2X is an integer

    cosec (45+a)+1 should be an integer and a factor of 1004 which is not possible. Therefore, no such triangle is possible!
  2. jcsd
  3. Nov 9, 2008 #2


    Staff: Mentor

    You're given that the perimeter is 2008, not 1004, so the equation should be
    2L + 2X = 2008, or 2(L + X) = 2008, from which you get L + X = 1004
    It's much simpler as sin(45 + a) = X/L
    Replacing L by 1004 - X, this becomes
    sin(45 + a) = X/(1004 - X)
    I don't follow your logic here. It looks like it comes from 1 < csc(45 + a) < sqrt(2), but you'll have to explain why that is true. (It might be, but I don't see it.)

    Also, you can't replace 1 + sqrt(2) with 2.4142; they are only approximately equal.
  4. Nov 11, 2008 #3
    The obtuse angle alpha must indeed be the angle between the equal sides L. Let the side opposite to alpha ideed be 2x. Use of the cosine rule gives
    4*x^2=L^2+L^2 -2L^2*cos(alpha).
    So cos(alpha)=1-2*x^2/L^2. Because the angle is obtuse cos(alpha)<0 so to solve is:
    1-2*x^2/L^2<0, multiply by L^2, since L^2 >0 this gives L^2-2*x^2<0, or L^2<2*x^2,
    L<xSqrt(2)=(1004-L)sqrt(2), L(1+sqrt(2))<1004, L<1004/(1+sqrt(2).
    The number of solutions are: int(1004/(1+sqrt(2).
  5. Nov 21, 2008 #4
    Made an error L has to be greater than x!
  6. Nov 21, 2008 #5
    Just so you know, there is another way that I might prefer to get to the point where L^2<2*x^2. If you remember our Pythagorean Theorem for right triangles:
    [tex]a^{2} + b^{2} = c^{2}[/tex]
    You might also know that for obtuse triangles:
    [tex]a^{2} + b^{2} < c^{2}[/tex]
    Because you chose L for the legs, a and b, and 2x for the largest side, the one opposite alpha or c.
    [tex]L^{2} + L^{2} < (2x)^{2}[/tex]
    [tex]2L^{2} < 4x^{2}[/tex]
    [tex]L^{2} < 2x^{2}[/tex]
    I like this way, and showing it two ways means it's probably right.
  7. Nov 21, 2008 #6
    Ok so x<L<sqrt(2)*x, and how many solutions are there?
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