# Math Help - The area of a triangle within a triangle

1. ## 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.

2. . . . . . . . . . .
(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.

<http://en.wikipedia.org/wiki/Routh%27s_theorem>

4. Originally Posted by Opalg
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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?

5. Originally Posted by Opalg
.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 $\textstyle \mathbf{p} = \frac17(2\mathbf{a}+\mathbf{b}+4\mathbf{c})$.
Originally Posted by sael
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.