2021 AIME II Problem 4

Below is the professionally curated solution for Problem 4 of the 2021 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AIME II solutions, or check the answer key.

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Concepts:complex numberVieta’s Formulaspolynomial

Difficulty rating: 2100

4.

There are real numbers a,b,c,a, b, c, and dd such that 20-20 is a root of x3+ax+bx^3 + ax + b and 21-21 is a root of x3+cx2+d.x^3 + cx^2 + d. These two polynomials share a complex root m+ni,m + \sqrt{n} \cdot i, where mm and nn are positive integers and i=1.i = \sqrt{-1}. Find m+n.m + n.

Solution:

Both cubics have real coefficients, so their non-real roots come in conjugate pairs: the roots of the first are 20-20 and m±ni,m \pm \sqrt{n}\,i, and the roots of the second are 21-21 and m±ni.m \pm \sqrt{n}\,i.

The first cubic x3+ax+bx^3 + ax + b has no x2x^2 term, so its roots sum to 0:0: 20+2m=0,-20 + 2m = 0, giving m=10.m = 10. The second cubic x3+cx2+dx^3 + cx^2 + d has no xx term, so the sum of pairwise products of its roots is 0:0: (m+ni)(mni)+(21)(2m)=m2+n42m=0,(m + \sqrt{n}\,i)(m - \sqrt{n}\,i) + (-21)(2m) = m^2 + n - 42m = 0, so n=420100=320.n = 420 - 100 = 320. Then m+n=10+320=330.m + n = 10 + 320 = 330.

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