Polygons and Polynomials

Algebra Level 3

Let $f(x)=x^3-2n x^2+\left(\frac 12 n+3m\right)x+\sqrt n$ , where $m$ and $n$ are reals. If the real roots of $f(x)$ are $a$ , $b$ , and $c$ , then the centroid of a triangle with coordinates $(a,b)$ , $(b,c)$ , and $(c,a)$ is $(4m, m+n-1)$ . Determine the area of that triangle.

The answer is 5.

1 solution

Yashas Ravi
Jul 27, 2019

The centroid of the triangle would be $(0.333(a+b+c), 0.333(a+b+c))$ . Thus, $4m=m+n-1$ so $n=3m+1$ . Also, $a+b+c=2n$ by Vieta's formulas. As a result, $12m=2n$ so $n=6m$ . Then, $6m=3m+1$ so $m=0.3333$ and $n=2$ .

The area of the triangle, using the coordinate formula, is $0.5|(ab+bc+ac)-(a^2+b^2+c^2)|$ , which is derived by substitution using the given coordinates. Then, $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)$ and $(ab+bc+ac)=0.5n+3m=2$ , so the area is $0.5|2-(4^2-2(2))|=5$ which is the final answer.

Polygons and Polynomials

The answer is 5.

1 solution

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