AMC 12B 2021 Spring (Problem 16)Let g(x)g(x)g(x) be a polynomial with leading coefficient 111, whose three roots are the reciprocals of the three roots of f(x)=x3+ax2+bx+cf(x)=x^3+ax^2+bx+cf(x)=x3+ax2+bx+c, where 1<a<b<c1<a<b<c1<a<b<c. What is g(1)g(1)g(1) in terms of aaa, bbb, and ccc?(A) 1+a+b+cc\text{(A)}\;\frac{1+a+b+c}{c}(A)c1+a+b+c(B) 1+a+b+c\text{(B)}\;\text{1+a+b+c}(B)1+a+b+c(C) 1+a+b+cc2\text{(C)}\;\frac{1+a+b+c}{c^2}(C)c21+a+b+c(D) a+b+cc2\text{(D)}\;\frac{a+b+c}{c^2}(D)c2a+b+c(E) 1+a+b+ca+b+c\text{(E)}\;\frac{1+a+b+c}{a+b+c}(E)a+b+c1+a+b+cRelated TopicsCoreToolkit 30 — Vieta's FormulaHints (4)Hint 1If r1,r2,r3r_1,r_2,r_3r1,r2,r3 are the roots of f(x)f(x)f(x), then by Toolkit 30 — Vieta's Formula,r1+r2+r3=−a,r_1+r_2+r_3=-a,r1+r2+r3=−a,r1r2+r1r3+r2r3=b,r_1r_2+r_1r_3+r_2r_3=b,r1r2+r1r3+r2r3=b,r1r2r3=−c.r_1r_2r_3=-c.r1r2r3=−c.Hint 2The polynomial g(x)g(x)g(x) has reciprocal roots, sog(x)=(x−1r1)(x−1r2)(x−1r3).g(x)=\left(x-\frac1{r_1}\right)\left(x-\frac1{r_2}\right)\left(x-\frac1{r_3}\right).g(x)=(x−r11)(x−r21)(x−r31).Hint 3g(1)=(1−1r1)(1−1r2)(1−1r3)g(1)=\left(1-\frac1{r_1}\right)\left(1-\frac1{r_2}\right)\left(1-\frac1{r_3}\right)g(1)=(1−r11)(1−r21)(1−r31)=(r1−1)(r2−1)(r3−1)r1r2r3=\frac{(r_1-1)(r_2-1)(r_3-1)}{r_1r_2r_3}=r1r2r3(r1−1)(r2−1)(r3−1)=r1r2r3−(r1r2+r1r3+r2r3)+(r1+r2+r3)−1r1r2r3.=\frac{r_1r_2r_3-(r_1r_2+r_1r_3+r_2r_3)+(r_1+r_2+r_3)-1}{r_1r_2r_3}.=r1r2r3r1r2r3−(r1r2+r1r3+r2r3)+(r1+r2+r3)−1.Hint 4By Hints 1 and 3,g(1)=−c−b−a−1−cg(1)=\frac{-c-b-a-1}{-c}g(1)=−c−c−b−a−1=1+a+b+cc.=\frac{1+a+b+c}{c}.=c1+a+b+c.Final Answer(A) 1+a+b+cc\frac{1+a+b+c}{c}c1+a+b+cRelated Problems (8)AMC 10/12A 2022 (Problem 16)AMC 10/12A Spring 2021 (Problem 14)AMC 12B 2023 (Problem 14)AMC 12B 2022 (Problem 4)AMC 10A/12A 2025 (Problem 18/12)AMC 12A 2025 (Problem 19)AMC 12A 2024 (Problem 15)AMC 12B 2024 (Problem 17)