Difference between revisions of "2021 AIME II Problems/Problem 4"

Revision as of 02:31, 23 March 2021

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

@@ Line 24: / Line 24: @@
 Hence, <math>m^{3} - 3mn + am + b = 0</math> and
-  <math>3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n} = 0</math>
+  <math>3m^{2} \sqrt{n} - n\sqrt{n} + a\sqrt{n} = 0 \Rightarrow 3m^{2} - n + a = 0</math>
 ==See also==
 {{AIME box|year=2021|n=II|num-b=3|num-a=5}}
 {{MAA Notice}}

2021 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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