2021 AIME II Problems/Problem 4

Revision as of 01:40, 23 March 2021 by Arnigam2007 (talk | contribs) (Solution 3 (Somewhat Bashy))

Problem

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

Solution 1

Conjugate root theorem

Solution in progress

~JimY

Solution 2

Solution 3 (Somewhat Bashy)

$(-20)^{3} + (-20)a + b = 0$, hence $-20a + b = 8000$

Also, $(-21)^{3} + c(-21)^{2} + d = 0$, hence $441c + d = 9261$

$m + i \sqrt{n}$ satisfies both $\Rightarrow$ we can put it in both equations and equate to 0.

In the first equation, we get $(m + i \sqrt{n})^{3} + a(m + i \sqrt{n}) + b = 0$ Simplifying this further, we get $(m^{3} - 3mn + am + b) + i(3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n}) = 0$

Hence, $m^{3} - 3mn + am + b = 0$ and $3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n} = 0 \Rightarrow 3m^{2} - n + a = 0 \rightarrow (1)$

In the second equation, we get $(m + i \sqrt{n})^{3} + c(m + i \sqrt{n})^{2} + d = 0$ Simplifying this further, we get $(m^{3} + m^{2}c - nc - 3mn + d) + i(3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n}) = 0$

Hence, $m^{3} + m^{2}c - nc - 3mn + d = 0$ and $3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n} = 0 \Rightarrow 3m^{2} - n + 2mc = 0 \rightarrow (2)$

-Arnav Nigam

See also

2021 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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