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Math Help - The area of a triangle within a triangle

  1. #1
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    The area of a triangle within a triangle

    I've been confronted by what seems to be a very difficult geometry problem.

    The questions asks how much smaller triangle DEF is from triangle ABC. Lines are constructed on each side at the 1/3 point (of the side) and go to a corresponding vertex. The lines essentially split each side of the triangle into 2 corresponding line segments, 1/3 and 2/3 the length of the entire side. The lines meet up to form the triangle DEF. ABC is not an equilateral triangle.

    I attempted to use Heron's formula to find the area of ABC, but xyz for the sides didn't get me too far. I included an illustration of this problem in the attachment.

    Thanks in advance for any help.
    Attached Thumbnails Attached Thumbnails The area of a triangle within a triangle-problem.jpg  
    Last edited by Dirtyeagle; April 10th 2008 at 10:33 AM.
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  2. #2
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    . . . . . . . . . .
    (I have re-labelled the coordinates of the inner triangle, which becomes PQR instead of DEF. Also, L, M and N are the points of trisection on BC, CA and AB.)

    The answer is that the area of PQR is one-seventh of the area of ABC.

    The only way I can see to prove this is by using vectors. Let a, b and c be position vectors of the points A, B and C. The point L has position vector (1/3)(2b+c), and similarly (1/3)(2c+a) for M, and (1/3)(2a+b) for N.

    The point P lies on both BM and CN, so its position vector p must be of the form \textstyle \lambda\mathbf{b} + (1-\lambda)\frac13(2\mathbf{c}+\mathbf{a}) and also of the form \textstyle \mu\mathbf{c} + (1-\mu)\frac13(2\mathbf{a}+\mathbf{b}). Solve these equations to see that \textstyle \mathbf{p} = \frac17(2\mathbf{a}+\mathbf{b}+4\mathbf{c}). Similarly, the position vectors of Q and R are \textstyle \mathbf{q} = \frac17(4\mathbf{a}+2\mathbf{b}+\mathbf{c}) and \textstyle \mathbf{r} = \frac17(\mathbf{a}+4\mathbf{b}+2\mathbf{c}).

    Now think of all these vectors as being in 3-dimensional space (with z-coordinate 0). The standard formula for the area of a triangle is given by a cross product in the form area(ABC) = \textstyle \frac12|\mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a} + \mathbf{a}\times\mathbf{b}|. If you apply the same formula to the triangle PQR then you get its area as \textstyle \frac12|\mathbf{q}\times\mathbf{r} + \mathbf{r}\times\mathbf{p} + \mathbf{p}\times\mathbf{q}| = \frac1{14}|\mathbf{b}\times\mathbf{c} + \mathbf{c}\times\mathbf{a} + \mathbf{a}\times\mathbf{b}| (using the above formulas for p, q and r and simplifying the result. This is 1/7th of the area of ABC.
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  3. #3
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    Thanks for your help.

    Your answer corresponds with Routh's Theorem
    <http://en.wikipedia.org/wiki/Routh%27s_theorem>
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  4. #4
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    Quote Originally Posted by Opalg View Post
    . . . . . . . . . .
    The point P lies on both BM and CN, so its position vector p must be of the form \textstyle \lambda\mathbf{b} + (1-\lambda)\frac13(2\mathbf{c}+\mathbf{a}) and also of the form \textstyle \mu\mathbf{c} + (1-\mu)\frac13(2\mathbf{a}+\mathbf{b}). Solve these equations to see that \textstyle \mathbf{p} = \frac17(2\mathbf{a}+\mathbf{b}+4\mathbf{c}).
    I do not know how to solve vector equations like these. How did you solve them?
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  5. #5
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    Quote Originally Posted by Opalg View Post
    .The point P lies on both BM and CN, so its position vector p must be of the form \textstyle \lambda\mathbf{b} + (1-\lambda)\frac13(2\mathbf{c}+\mathbf{a}) and also of the form \textstyle \mu\mathbf{c} + (1-\mu)\frac13(2\mathbf{a}+\mathbf{b}). Solve these equations to see that \textstyle \mathbf{p} = \frac17(2\mathbf{a}+\mathbf{b}+4\mathbf{c}).
    Quote Originally Posted by sael View Post
    I do not know how to solve vector equations like these. How did you solve them?
    In the equation \textstyle \lambda\mathbf{b} + (1-\lambda)\frac13(2\mathbf{c}+\mathbf{a}) = \mu\mathbf{c} + (1-\mu)\frac13(2\mathbf{a}+\mathbf{b}), compare the coefficients of a, b and c. This gives
    \textstyle \frac13(1-\lambda) = \frac23(1-\mu) and hence -\lambda+2\mu = 1 (for the coefficient of a), and similarly
    3\lambda+\mu = 1 (coefficient of b), and
    2\lambda+3\mu = 2 (coefficient of c).
    Now solve those equations for λ and μ, to get λ=1/7 and μ=4/7.
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