## Friday, January 19, 2007

### Triangularly Speaking. Topic: Algebra/Geometry. Level: AIME.

Problem: Suppose the roots of the cubic polynomial $P(x) = x^3-2rx^2+sx+t$ are the sides of a triangle (assume they are all positive reals and satisfy the triangle inequality). What is the area of the triangle in terms of the coefficients?

Solution: Well, let's think about the formulas for the area of a triangle given its side lengths. The only one that comes to mind that uses only side lengths is Heron's formula. Why don't we try that?

$A = \sqrt{s(s-a)(s-b)(s-c)}$,

where $a, b, c$ are the side lengths and $s = \frac{a+b+c}{2}$. From $P(x)$, by Vieta's, we know that $r_1+r_2+r_3 = 2r$, where $r_i$ are the roots of the polynomial. So $s$ in this case is $r$. How do we get the rest, though? We could expand and use Newton sums, but that sounds like a pain. Note that

$(s-a)(s-b)(s-c)$ looks a lot like $(x-a)(x-b)(x-c)$.

But wait, that's just a polynomial with roots $a, b, c$! So if we write $P(x) = (x-r_1)(x-r_2)(x-r_3)$, we have $(s-a)(s-b)(s-c) = (s-r_1)(s-r_2)(s-r_3) = P(s) = P(r)$. Thus we can say that the area of the triangle is

$\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{rP(r)}$,

which can clearly be evaluated in terms of the coefficients. QED.

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Comment: This is probably one of the neatest tricks that Vieta's can help you with, as well as a cool link between algebra and geometry. Before I thought of this problem, I had never realized the now-clear relationship between Heron's formula and a cubic polynomial with sides as the roots. A nice discovery.

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Practice Problem: (2007 Mock AMC 12 - #24) The lengths of the sides of a triangle are the roots of the equation $7x^3+35x = 28x^2+13$. If the square of the area of the triangle is $\displaystyle \frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers, what is $p+q$?