2009 AMC 12A Problem 21

Below is the professionally curated solution for Problem 21 of the 2009 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12A 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:polynomialcomplex numberroots of unity

Difficulty rating: 2170

21.

Let p(x)=x3+ax2+bx+c,p(x) = x^3 + ax^2 + bx + c, where a,a, b,b, and cc are complex numbers. Suppose that p(2009+9002πi)=p(2009)=p(9002)=0.p(2009 + 9002\pi i) = p(2009) = p(9002) = 0. What is the number of nonreal zeros of x12+ax8+bx4+c?x^{12} + ax^8 + bx^4 + c?

44

66

88

1010

1212

Solution:

Since x12+ax8+bx4+c=p(x4),x^{12} + ax^8 + bx^4 + c = p(x^4), a value is a zero exactly when x4x^4 equals one of the roots of p,p, namely 2009+9002πi,2009 + 9002\pi i, 2009,2009, or 9002.9002.

The equation x4=2009+9002πix^4 = 2009 + 9002\pi i has four distinct nonreal roots. Each of x4=2009x^4 = 2009 and x4=9002x^4 = 9002 has two real roots and two nonreal roots.

So the nonreal zeros number 4+2+2=8.4 + 2 + 2 = 8.

Thus, the correct answer is C.

Problem 21 in Other Years