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2021 AMC 12B Spring Problem 16

Below is the professionally curated solution for Problem 16 of the 2021 AMC 12B Spring, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Spring solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:Vieta’s Formulaspolynomial

Difficulty rating: 1720

16.

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,b,a, b, and c?c?

1+a+b+cc\dfrac{1+a+b+c}{c}

1+a+b+c1+a+b+c

1+a+b+cc2\dfrac{1+a+b+c}{c^2}

a+b+cc2\dfrac{a+b+c}{c^2}

1+a+b+ca+b+c\dfrac{1+a+b+c}{a+b+c}

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.g(1)=\left(1-\tfrac1r\right)\left(1-\tfrac1s\right)\left(1-\tfrac1t\right)=\dfrac{(r-1)(s-1)(t-1)}{rst}.

Now f(1)=(1−r)(1−s)(1−t)=1+a+b+c,f(1)=(1-r)(1-s)(1-t)=1+a+b+c, so (r−1)(s−1)(t−1)=−(1+a+b+c).(r-1)(s-1)(t-1)=-(1+a+b+c). Also rst=−c.rst=-c.

Therefore g(1)=−(1+a+b+c)−c=1+a+b+cc.g(1)=\dfrac{-(1+a+b+c)}{-c}=\dfrac{1+a+b+c}{c}.

Thus, the correct answer is A.

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