Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
m
Line 26: Line 26:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Solution 3 (Somewhat Bashy)==
+
==Solution 2 (Somewhat Bashy)==
 
<math>(-20)^{3} + (-20)a + b = 0</math>, hence <math>-20a + b = 8000</math>
 
<math>(-20)^{3} + (-20)a + b = 0</math>, hence <math>-20a + b = 8000</math>
  

Revision as of 03:03, 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

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 $-21,z,\overline{z}.$

By 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. \ \ \ \ \ \ \ \ (2) \end{align*}

By 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

See also

2021 AIME II (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_4&oldid=149960"