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

(→‎Solution 1 (Complex Conjugate Root Theorem))
m (→‎Solution 1 (Complex Conjugate Root Theorem))
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By the <b>Complex Conjugate Root Theorem</b>, the imaginary roots for each of <math>x^3+ax+b</math> and <math>x^3+cx^2+d</math> are a pair of complex conjugates. Let <math>z=m+\sqrt{n}\cdot i</math> and <math>\overline{z}=m-\sqrt{n}\cdot i.</math> It follows that the roots of <math>x^3+ax+b</math> are <math>-20,z,\overline{z},</math> and the roots of <math>x^3+cx^2+d</math> are <math>-21,z,\overline{z}.</math>
 
By the <b>Complex Conjugate Root Theorem</b>, the imaginary roots for each of <math>x^3+ax+b</math> and <math>x^3+cx^2+d</math> are a pair of complex conjugates. Let <math>z=m+\sqrt{n}\cdot i</math> and <math>\overline{z}=m-\sqrt{n}\cdot i.</math> It follows that the roots of <math>x^3+ax+b</math> are <math>-20,z,\overline{z},</math> and the roots of <math>x^3+cx^2+d</math> are <math>-21,z,\overline{z}.</math>
  
−
By Vieta's Formulas on <math>x^3+ax+b,</math> we have <math>-20+z+\overline{z}=0,</math> from which
+
Applying Vieta's Formulas on <math>x^3+ax+b,</math> we have <math>-20+z+\overline{z}=0,</math> from which
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
z+\overline{z}&=20 \ \ \ \ \ \ \ \ \ (1) \\
 
z+\overline{z}&=20 \ \ \ \ \ \ \ \ \ (1) \\
Line 13: Line 13:
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
−
By Vieta's Formulas on <math>x^3+cx^2+d,</math> we have <math>-21z-21\overline{z}+z\overline{z}=0,</math> from which
+
Applying Vieta's Formulas on <math>x^3+cx^2+d,</math> we have <math>-21z-21\overline{z}+z\overline{z}=0,</math> from which
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
-21\left(z+\overline{z}\right)+z\overline{z}&=0 \\
 
-21\left(z+\overline{z}\right)+z\overline{z}&=0 \\

Revision as of 15:35, 30 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 (Complex Conjugate Root Theorem)

By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are a pair of complex conjugates. Let $z=m+\sqrt{n}\cdot i$ and $\overline{z}=m-\sqrt{n}\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\overline{z}.$

Applying Vieta's Formulas on $x^3+ax+b,$ we have $-20+z+\overline{z}=0,$ from which \begin{align*} z+\overline{z}&=20 \ \ \ \ \ \ \ \ \ (1) \\ \left(m+\sqrt{n}\cdot i\right)+\left(m-\sqrt{n}\cdot i\right)&=20 \\ 2m&=20 \\ m&=10. \hspace{9.5mm} (2) \end{align*}

Applying Vieta's Formulas on $x^3+cx^2+d,$ we have $-21z-21\overline{z}+z\overline{z}=0,$ from which \begin{align*} -21\left(z+\overline{z}\right)+z\overline{z}&=0 \\ -21\underbrace{\left(20\right)}_{\text{by }(1)}+\underbrace{|z|^2}_{\substack{z\overline{z}=|z|^2 \\ \text{ Identity}}}&=0 \\ |z|^2&=420 \\ m^2+n&=420 \\ {\underbrace{10}_{\text{by }(2)}}^2+n&=420 \\ n&=320. \hspace{24.5mm} (3) \end{align*} Finally, we get $m+n=\boxed{330}$ by $(2)$ and $(3).$

~MRENTHUSIASM

Solution 2 (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 $3m^{2} \sqrt{n} - n\sqrt{n} + 2mc\sqrt{n} = 0 \Rightarrow 3m^{2} - n + 2mc = 0 \rightarrow (2)$

Comparing (1) and (2),

$a = 2mc$ and $am + b = m^{2}c - nc + d \rightarrow (3)$

$b = 8000 + 20a \Rightarrow b = 40mc + 8000$; $d = 9261 - 441c$

Substituting these in $(3)$ gives, $2m^{2}c + 8000 + 40mc = m^{2}c - nc + 9261 - 441c$

This simplifies to $m^{2}c + nc + 40mc + 441c = 1261 \Rightarrow c(m^{2} + n + 40m + 441) = 1261$

Hence, $c|1261 \Rightarrow c \in {1,13,97,1261}$


Consider case of $c = 1$:

$c = 1 \Rightarrow d = 8820$ Also, $a = 2m, b = 8000 + 40m$

$am + b = m^{2} - n + 8820$ (because c = 1) Also, $m^{2} + n + 40m = 820 \rightarrow (4)$ Also, Equation (2) gives $3m^{2} - n + 2m = 0 \rightarrow (5)$

Solving (4) and (5) simultaneously gives $m = 10, n = 320$

[AIME can not have more than one answer, so we can stop here also 😁... Not suitable for Subjective exam]

Hence, $m + n = 10 + 320 = \boxed{330}$

-Arnav Nigam

Solution 3 (Heavy Calculation Solution)

start off by applying vieta's and you will find that $a=m^2+n-40m$ $b=20m^2+20n$ $c=21-2m$ and $d=21m^2+21n$. After that, we have to use the fact that $-20$ and $-21$ are roots of $x^3+ax+b$ and $x^3+cx^2+d$, respectively. Since we know that if you substitute the root of a function back into the function, the output is zero, therefore $(-20)^3-20(a)+b=0$ and $(-21)^3+c*(-21)^2+d=0$ and you can set these two equations equal to each other while also substituting the values of $a$, $b$, $c$, and $d$ above to give you $21m^2+21n-1682m+8000=0$, then you can rearrange the equation into $21n = -21m^2+1682m-8000$. With this property, we know that $-21m^2+1682m-8000$ is divisible by $21$ therefore that means $1682m-8000=0(mod 21)$ which results in $2m-20=0(mod 21)$ which finally gives us m=10 mod 21. We can test the first obvious value of $m$ which is $10$ and we see that this works as we get $m=10$ and $n=320$. That means your answer will be $m + n = 10 + 320 = \boxed{330}$

-Jske25

Solution 4 (Synthetic Division)

We note that $x^3 + ax + b = (x+20)P(x)$ and $x^3 + cx^2 + d = (x+21)Q(x)$ for some polynomials $P(x)$ and $Q(x)$. Through synthetic division (ignoring the remainder as we can set $b$ and $d$ to constant values such that the remainder is zero), $P(x) = x^2 - 20x + (400+a)$, and $Q(x) = x^2 + (c-21)x + (441 - 21c)$ By the complex conjugate root theorem, we know that $P(x)$ and $Q(x)$ share the same roots, and they share the same leading coefficient, so $P(x) = Q(x)$. Therefore, $c-21 = -20$ and $441-21c = 400 + a$. Solving the system equations, we get $a = 20$ and $c = 1$, so $P(x) = Q(x) = x^2 - 20x + 420$. Finally, by the quadratic formula, we have roots of $\frac{20 \pm \sqrt{400 - 1680}}{2} = 10 \pm \sqrt{320}i$, so our final answer is $10 + 320 = \boxed{330}$

-faefeyfa

Video Solution

https://www.youtube.com/watch?v=sYRWWQayNyQ

See also

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

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