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  1. #1
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    Triangle divided...

    A vertical line divides the triangle whose vertices are at (0,0), (1,1), and (9,1) into two parts; a triangle and a quadrilateral. If the areas of the triangle and the quadrilateral are equal, what is the equation of the vertical line?
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    for vertical line x=a make the areas of the shapes the same.

    You can try to solve \displaystyle \int_a^9 f(x)~dx = \int_0^a g(x)~dx

    All you have to do is determine f(x), g(x) which are linear combinations of the lines joining the triangle.

    Can you find the equations of the 3 lines that form the triangle?

    Spoiler:


    \displaystyle y=x,y=1,y=\frac{x}{9}

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  3. #3
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    graphical approach

    from the diagram,

    \frac{d}{1}=\frac{(9-a)}{9}
    d=\frac{(9-a)}{9}

    considering the area of the small triangle,find the area of it

    If A is the area of the larger triangle,

    A=(A+A') - A'

    A1 = \frac{A}{2}

    substitute A, A1, (A+A') and A' with d and a

    this is when vertical line meets AB
    do the same thing when the vertical line meets CA
    Attached Thumbnails Attached Thumbnails Triangle divided...-math.bmp  
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  4. #4
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    Quote Originally Posted by MATNTRNG View Post
    A vertical line divides the triangle whose vertices are at (0,0), (1,1), and (9,1) into two parts; a triangle and a quadrilateral. If areas of the triangle and quadrilateral are equal, what is the equation of the vertical line?
    Ok, now that you've been shown how, try this similar one:

    A vertical line divides the triangle whose vertices are at (0,0), (7,24), and (32,24) into two parts;
    a triangle and a quadrilateral. If the areas of the triangle and the quadrilateral are equal,
    what is the LENGTH of the vertical line?

    You'll find that all side lengths (plus the areas) are integers;
    smallest case, I believe, for an all-integer.

    SOLUTION (since no response!):
    A(0,0), B(7,24), C(32,24)
    u=7,v=32,w=24 : so A(0,0), B(u,w), C(v,w)
    Vertical line length = wSQRT[(v - u) / (2v)] = 15

    Smallest all-integer case is smaller than I thought: A(0,0), B(4,15), C(36,15) ; vertical line = 10
    Last edited by Wilmer; November 10th 2010 at 05:36 AM. Reason: Added solution
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  5. #5
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    Might be easier to just find the area of the full triangle

    \displaystyle A = \frac{1}{2}\times 8\times \sqrt{2}\times \sin {135}\approx 0.4998

    Take half of that and solve for the smaller triangle section.

    \displaystyle \int_a^9 1-\frac{x}{9}~dx = \frac{0.4998}{2}
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  6. #6
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    Quote Originally Posted by pickslides View Post
    Might be easier to just find the area of the full triangle
    \displaystyle A = \frac{1}{2}\times 8\times \sqrt{2}\times \sin {135}\approx 0.4998
    Pickslides, that should be 4 exactly, not .4998.

    Easier if we let point (1,1) = (u,w) and point (9,1) = (v,w);
    then A = w(v - u) / 2 = 1(9 - 1) / 2 = 4 ; agree?
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