During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

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

(Solution 4 (Synthetic Division))
(Added another solution)
 
(12 intermediate revisions by 3 users not shown)
Line 2: Line 2:
 
There are real numbers <math>a, b, c,</math> and <math>d</math> such that <math>-20</math> is a root of <math>x^3 + ax + b</math> and <math>-21</math> is a root of <math>x^3 + cx^2 + d.</math> These two polynomials share a complex root <math>m + \sqrt{n} \cdot i,</math> where <math>m</math> and <math>n</math> are positive integers and <math>i = \sqrt{-1}.</math> Find <math>m+n.</math>
 
There are real numbers <math>a, b, c,</math> and <math>d</math> such that <math>-20</math> is a root of <math>x^3 + ax + b</math> and <math>-21</math> is a root of <math>x^3 + cx^2 + d.</math> These two polynomials share a complex root <math>m + \sqrt{n} \cdot i,</math> where <math>m</math> and <math>n</math> are positive integers and <math>i = \sqrt{-1}.</math> Find <math>m+n.</math>
  
==Solution 1 (Complex Conjugate Root Theorem)==
+
==Solution 1 (Complex Conjugate Root Theorem and Vieta's Formulas)==
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 Complex Conjugate Root Theorem, the imaginary roots for each of <math>x^3+ax+b</math> and <math>x^3+cx^2+d</math> are 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>
  
 
We know that
 
We know that
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
z+\overline{z}&=2m, \hspace{10mm} & (1) \\
+
z+\overline{z}&=2m, & (1) \\
 
z\overline{z}&=m^2+n. & (2)
 
z\overline{z}&=m^2+n. & (2)
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
Applying Vieta's Formulas to <math>x^3+ax+b,</math> we have <math>-20+z+\overline{z}=0.</math> Substituting <math>(1)</math> into this equation, we get <math>m=10.</math>
  
Applying Vieta's Formulas to <math>x^3+ax+b,</math> we have <math>-20+z+\overline{z}=0,</math> from which
+
Applying Vieta's Formulas to <math>x^3+cx^2+d,</math> we have <math>-21z-21\overline{z}+z\overline{z}=0,</math> or <math>-21(z+\overline{z})+z\overline{z}=0.</math> Substituting <math>(1)</math> and <math>(2)</math> into this equation, we get <math>n=320.</math>
<cmath>\begin{align*}
 
z+\overline{z}&=20 \hspace{12.5mm} & (3) \\
 
\underbrace{2m}_{\text{by }(1)}&=20 & \\
 
m&=10. & (4)
 
\end{align*}</cmath>
 
  
Applying Vieta's Formulas to <math>x^3+cx^2+d,</math> we have <math>-21z-21\overline{z}+z\overline{z}=0,</math> from which
+
Finally, the answer is <math>m+n=\boxed{330}.</math>
<cmath>\begin{align*}
 
-21\left(z+\overline{z}\right)+z\overline{z}&=0 \\
 
-21\underbrace{\left(20\right)}_{\text{by }(3)}+\underbrace{\left(m^2+n\right)}_{\text{by }(2)}&=0 \\
 
m^2+n&=420 \\
 
{\underbrace{10}_{\text{by }(4)}}^2+n&=420 \\
 
n&=320. \hspace{5.75mm} (5)
 
\end{align*}</cmath>
 
Finally, we get <math>m+n=\boxed{330}</math> by <math>(4)</math> and <math>(5).</math>
 
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
Line 86: Line 74:
  
 
==Solution 4 (Synthetic Division)==
 
==Solution 4 (Synthetic Division)==
We note that <math>x^3 + ax + b = (x+20)P(x)</math> and <math>x^3 + cx^2 + d = (x+21)Q(x)</math> for some polynomials <math>P(x)</math> and <math>Q(x)</math>. Through synthetic division (ignoring the remainder as we can set <math>b</math> and <math>d</math> to constant values such that the remainder is zero), <math>P(x) = x^2 - 20x + (400+a)</math>, and <math>Q(x) = x^2 + (c-21)x + (441 - 21c)</math> By the complex conjugate root theorem, we know that <math>P(x)</math> and <math>Q(x)</math> share the same roots, and they share the same leading coefficient, so <math>P(x) = Q(x)</math>. Therefore, <math>c-21 = -20</math> and <math>441-21c = 400 + a</math>. Solving the system equations, we get <math>a = 20</math> and <math>c = 1</math>, so <math>P(x) = Q(x) = x^2 - 20x + 420</math>.
+
We note that <math>x^3 + ax + b = (x+20)P(x)</math> and <math>x^3 + cx^2 + d = (x+21)Q(x)</math> for some polynomials <math>P(x)</math> and <math>Q(x)</math>.
 +
 
 +
Through synthetic division (ignoring the remainder as we can set <math>b</math> and <math>d</math> to constant values such that the remainder is zero), <math>P(x) = x^2 - 20x + (400+a)</math>, and <math>Q(x) = x^2 + (c-21)x + (441 - 21c)</math>.
 +
 
 +
By the complex conjugate root theorem, we know that <math>P(x)</math> and <math>Q(x)</math> share the same roots, and they share the same leading coefficient, so <math>P(x) = Q(x)</math>.  
 +
 
 +
Therefore, <math>c-21 = -20</math> and <math>441-21c = 400 + a</math>. Solving the system of equations, we get <math>a = 20</math> and <math>c = 1</math>, so <math>P(x) = Q(x) = x^2 - 20x + 420</math>.
 +
 
 
Finally, by the quadratic formula, we have roots of <math>\frac{20 \pm \sqrt{400 - 1680}}{2} = 10 \pm \sqrt{320}i</math>, so our final answer is <math>10 + 320 = \boxed{330}</math>
 
Finally, by the quadratic formula, we have roots of <math>\frac{20 \pm \sqrt{400 - 1680}}{2} = 10 \pm \sqrt{320}i</math>, so our final answer is <math>10 + 320 = \boxed{330}</math>
  
 
-faefeyfa
 
-faefeyfa
+
 
 
==Solution 5 (Fast and Easy)==
 
==Solution 5 (Fast and Easy)==
 +
We plug -20 into the equation obtaining <math>(-20)^3-20a+b</math>, likewise, plugging -21 into the second equation gets <math>(-21)^3+441c+d</math>.
 +
 +
Both equations must have 3 solutions exactly, so the other two solutions must be <math>m + \sqrt{n} \cdot i</math> and <math>m - \sqrt{n} \cdot i</math>.
 +
 +
By Vieta's, the sum of the roots in the first equation is <math>0</math>, so <math>m</math> must be <math>10</math>.
 +
 +
Next, using Vieta's theorem on the second equation, you get <math>x1x2+x2x3+x1x3 = 0</math>. However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with -21, so <math>-21*(20)+(m^2+n)=0</math>.
 +
 +
Given that <math>m</math> is <math>10</math>, then <math>n</math> is equal to <math>320</math>.
 +
 +
Therefore, the answer to the equation is <math>\boxed{330}</math>
 +
 +
==Solution 6 (solution by '''integralarefun''')==
 +
Since <math>m+i\sqrt{n}</math> is a common root and all the coefficients are real, <math>m-i\sqrt{n}</math> must be a common root, too.
 +
 +
Now that we know all three roots of both polynomials, we can match coefficients(or more specifically, the zero coefficients).
 +
 +
First, however, the product of the two common roots is:
 +
<cmath>\begin{align*}
 +
&&&(x-m-i\sqrt{n})(x-m+i\sqrt{n})\\
 +
&=&&x^2-x(m+i\sqrt{n}+m-i\sqrt{n})+(m+i\sqrt{n})(m-i\sqrt{n})\\
 +
&=&&x^2-2xm+(m^2-i^2n)\\
 +
&=&&x^2-2xm+m^2+n
 +
\end{align*}</cmath>
 +
 +
Now, let's equate the two forms of both the polynomials:
 +
<cmath>x^3+ax+b=(x^2-2xm+m^2+n)(x+20)</cmath>
 +
<cmath>x^3+cx^2+d=(x^2-2xm+m^2+n)(x+21)</cmath>
 +
Now we can match the zero coefficients.
 +
<cmath>-2m+20=0\to m=10\text{ and}</cmath>
 +
<cmath>-42m+m^2+n=0\to-420+100+n=0\to n=320\text{.}</cmath>
 +
Thus, <math>m+n=10+320=\boxed{330}</math>.
  
 
==Video Solution==
 
==Video Solution==
Line 99: Line 126:
 
https://www.youtube.com/watch?v=HJ0EldshLuE
 
https://www.youtube.com/watch?v=HJ0EldshLuE
  
==See also==
+
==See Also==
 
{{AIME box|year=2021|n=II|num-b=3|num-a=5}}
 
{{AIME box|year=2021|n=II|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:24, 27 November 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 and Vieta's Formulas)

By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are 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 $-21,z,\overline{z}.$

We know that \begin{align*} z+\overline{z}&=2m, & (1) \\ z\overline{z}&=m^2+n. & (2) \end{align*} Applying Vieta's Formulas to $x^3+ax+b,$ we have $-20+z+\overline{z}=0.$ Substituting $(1)$ into this equation, we get $m=10.$

Applying Vieta's Formulas to $x^3+cx^2+d,$ we have $-21z-21\overline{z}+z\overline{z}=0,$ or $-21(z+\overline{z})+z\overline{z}=0.$ Substituting $(1)$ and $(2)$ into this equation, we get $n=320.$

Finally, the answer is $m+n=\boxed{330}.$

~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 of 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

Solution 5 (Fast and Easy)

We plug -20 into the equation obtaining $(-20)^3-20a+b$, likewise, plugging -21 into the second equation gets $(-21)^3+441c+d$.

Both equations must have 3 solutions exactly, so the other two solutions must be $m + \sqrt{n} \cdot i$ and $m - \sqrt{n} \cdot i$.

By Vieta's, the sum of the roots in the first equation is $0$, so $m$ must be $10$.

Next, using Vieta's theorem on the second equation, you get $x1x2+x2x3+x1x3 = 0$. However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with -21, so $-21*(20)+(m^2+n)=0$.

Given that $m$ is $10$, then $n$ is equal to $320$.

Therefore, the answer to the equation is $\boxed{330}$

Solution 6 (solution by integralarefun)

Since $m+i\sqrt{n}$ is a common root and all the coefficients are real, $m-i\sqrt{n}$ must be a common root, too.

Now that we know all three roots of both polynomials, we can match coefficients(or more specifically, the zero coefficients).

First, however, the product of the two common roots is: \begin{align*} &&&(x-m-i\sqrt{n})(x-m+i\sqrt{n})\\ &=&&x^2-x(m+i\sqrt{n}+m-i\sqrt{n})+(m+i\sqrt{n})(m-i\sqrt{n})\\ &=&&x^2-2xm+(m^2-i^2n)\\ &=&&x^2-2xm+m^2+n \end{align*}

Now, let's equate the two forms of both the polynomials: \[x^3+ax+b=(x^2-2xm+m^2+n)(x+20)\] \[x^3+cx^2+d=(x^2-2xm+m^2+n)(x+21)\] Now we can match the zero coefficients. \[-2m+20=0\to m=10\text{ and}\] \[-42m+m^2+n=0\to-420+100+n=0\to n=320\text{.}\] Thus, $m+n=10+320=\boxed{330}$.

Video Solution

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

Video Solution by TheCALT

https://www.youtube.com/watch?v=HJ0EldshLuE

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

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

Invalid username
Login to AoPS