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2021 AMC 12B Problem 16

Problem 16 of 25IntermediateAlgebra

Let g(x)g(x) be a polynomial with leading coefficient 1,1, whose three roots are the reciprocals of the three roots of f(x)=x3+ax2+bx+c,f(x)=x^3+ax^2+bx+c, where 1<a<b<c.1\lt a\lt b\lt c. What is g(1)g(1) in terms of a,a, b,b, and c?c?

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Solution

Let ff have roots r,s,t.r,s,t. Since gg is monic with roots 1r,1s,1t,\tfrac1r,\tfrac1s,\tfrac1t, g(1)=(1−1r)(1−1s)⋅(1−1t)=(r−1)(s−1)(t−1)rst. \begin{aligned} g(1) &= \left(1-\tfrac1r\right)\left(1-\tfrac1s\right) \\ &\quad {}\cdot \left(1-\tfrac1t\right) \\ &= \dfrac{(r-1)(s-1)(t-1)}{rst}. \end{aligned} Now f(1)=(1−r)(1−s)(1−t)f(1)=(1-r)(1-s)(1-t) =1+a+b+c,=1+a+b+c, so (r−1)(s−1)(t−1)(r-1)(s-1)(t-1) =−(1+a+b+c).=-(1+a+b+c). Also rst=−c.rst=-c. Therefore g(1)g(1) =−(1+a+b+c)−c=\dfrac{-(1+a+b+c)}{-c} =1+a+b+cc.=\dfrac{1+a+b+c}{c}. Thus, the correct answer is A.

More practice

Concepts: Vieta’s Formulas · polynomial

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.