Difference between revisions of "2019 AIME I Problems/Problem 10"

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The 2019 AIME I takes place on March 13, 2019.
 
 
 
==Problem 10==
 
==Problem 10==
 
For distinct complex numbers <math>z_1,z_2,\dots,z_{673}</math>, the polynomial
 
For distinct complex numbers <math>z_1,z_2,\dots,z_{673}</math>, the polynomial
<cmath> (x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3 </cmath>can be expressed as <math>x^{2019} + 20x^{2018} + 19x^{2017}+g(x)</math>, where <math>g(x)</math> is a polynomial with complex coefficients and with degree at most <math>2016</math>. The value of
+
<cmath> (x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3 </cmath>can be expressed as <math>x^{2019} + 20x^{2018} + 19x^{2017}+g(x)</math>, where <math>g(x)</math> is a polynomial with complex coefficients and with degree at most <math>2016</math>. The sum <math> \left| \sum_{1 \le j <k \le 673} z_jz_k \right| </math> can be expressed in the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
<cmath> \left| \sum_{1 \le j <k \le 673} z_jz_k \right| </cmath>can be expressed in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
+
 
==Solution==
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==Solution 1==
 
In order to begin this problem, we must first understand what it is asking for. The notation  
 
In order to begin this problem, we must first understand what it is asking for. The notation  
 
<cmath> \left| \sum_{1 \le j <k \le 673} z_jz_k \right| </cmath>
 
<cmath> \left| \sum_{1 \le j <k \le 673} z_jz_k \right| </cmath>
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Now we can begin the problem. Rewrite the polynomial as <math>P=(x-z_1)(x-z_1)(x-z_1)(x-z_2)(x-z_2)(x-z_2) \dots (x-z_{673})(x-z_{673})(x-z_{673})</math>. Then we have that the roots of <math>P</math> are <math>z_1,z_1,z_1,z_2,z_2,z_2, \dots , z_{673},z_{673},z_{673}</math>.  
 
Now we can begin the problem. Rewrite the polynomial as <math>P=(x-z_1)(x-z_1)(x-z_1)(x-z_2)(x-z_2)(x-z_2) \dots (x-z_{673})(x-z_{673})(x-z_{673})</math>. Then we have that the roots of <math>P</math> are <math>z_1,z_1,z_1,z_2,z_2,z_2, \dots , z_{673},z_{673},z_{673}</math>.  
  
By Vieta's formulas, we have that the sum of the roots of <math>P</math> is <math>(-1)^1 * \dfrac{20}{1}=-20=z_1+z_1+z_1+z_2+z_2+z_2+ \dots + z_{673}+z_{673}+z_{673}=3(z_1+z_2+z_3+ \dots +z_{673})</math>. Thus, <math>z_1+z_2+z_3+ \dots +z_{673}=- \dfrac{20}{3}.</math>  
+
By Vieta's formulas, we have that the sum of the roots of <math>P</math> is <math>(-1)^1 \cdot \dfrac{20}{1}=-20=z_1+z_1+z_1+z_2+z_2+z_2+ \dots + z_{673}+z_{673}+z_{673}=3(z_1+z_2+z_3+ \dots +z_{673})</math>. Thus, <math>z_1+z_2+z_3+ \dots +z_{673}=- \dfrac{20}{3}.</math>  
  
Similarly, we also have that the the sum of the roots of <math>P</math> taken two at a time is <math>(-1)^2 * \dfrac{19}{1} = 19.</math> This is equal to <math>z_1^2+z_1^2+z_1^2+z_1z_2+z_1z_2+z_1z_2+ \dots =  \\ 3(z_1^2+z_2^2+ \dots + z_{673}^2) + 9(z_1z_2+z_1z_3+z_1z_4+ \dots + z_{672}z_{673}) =  3(z_1^2+z_2^2+ \dots + z_{673}^2) + 9S.</math>  
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Similarly, we also have that the the sum of the roots of <math>P</math> taken two at a time is <math>(-1)^2 \cdot \dfrac{19}{1} = 19.</math> This is equal to <math>z_1^2+z_1^2+z_1^2+z_1z_2+z_1z_2+z_1z_2+ \dots =  \\ 3(z_1^2+z_2^2+ \dots + z_{673}^2) + 9(z_1z_2+z_1z_3+z_1z_4+ \dots + z_{672}z_{673}) =  3(z_1^2+z_2^2+ \dots + z_{673}^2) + 9S.</math>  
  
 
Now we need to find and expression for <math>z_1^2+z_2^2+ \dots + z_{673}^2</math> in terms of <math>S</math>. We note that <math>(z_1+z_2+z_3+ \dots +z_{673})^2= (-20/3)^2=\dfrac{400}{9} \\ =(z_1^2+z_2^2+ \dots + z_{673}^2)+2(z_1z_2+z_1z_3+z_1z_4+ \dots + z_{672}z_{673})=(z_1^2+z_2^2+ \dots + z_{673}^2)+2S.</math>  
 
Now we need to find and expression for <math>z_1^2+z_2^2+ \dots + z_{673}^2</math> in terms of <math>S</math>. We note that <math>(z_1+z_2+z_3+ \dots +z_{673})^2= (-20/3)^2=\dfrac{400}{9} \\ =(z_1^2+z_2^2+ \dots + z_{673}^2)+2(z_1z_2+z_1z_3+z_1z_4+ \dots + z_{672}z_{673})=(z_1^2+z_2^2+ \dots + z_{673}^2)+2S.</math>  
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==Solution 2==
 
==Solution 2==
This is a quick fake solve using <math>z_q = 0</math> where <math>3 \le q \le 673</math> and only <math>z_1,z_2 \neq 0</math> .
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This is a quick fake solve using <math>z_i = 0</math> where <math>3 \le i \le 673</math> and only <math>z_1,z_2 \neq 0</math> .
 +
 
 +
By Vieta's, <cmath>3z_1+3z_2=-20</cmath> and <cmath>3z_1^2+3z_2^2+9z_1z_2 = 19.</cmath>
 +
Rearranging gives <math>z_1 + z_2 = \dfrac{-20}{3}</math> and <math>3(z_1^2 + 2z_1z_2 + z_2^2) + 3z_1z_2 = 19</math> giving <math> 3(z_1 + z_2)^2 + 3z_1z_2 = 19</math>.
 +
 
 +
Substituting gives <math>3\left(\dfrac{400}{9}\right) + 3z_1z_2 = 19</math> which simplifies to <math>\dfrac{400}{3} + 3z_1z_2 = \dfrac{57}{3}</math>.
  
By Vieta's, <cmath>3q_1+3q_2=-20</cmath> and <cmath>3q_1^2+3q_2^2+9q_1q_2 = 19.</cmath>
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So, <math>3z_1z_2 = \dfrac{-343}{3}</math>, <math>z_1z_2 = \dfrac{-343}{9}</math>, <math>|\dfrac{-343}{9}|=\dfrac{343}{9}</math>, <cmath>m+n = 343+9 = \boxed{352}.</cmath>
Rearranging gives <math>q_1 + q_2 = \dfrac{-20}{3}</math> and <math>3(q_1^2 + 2q_1q_2 + q_2^2) + 3q_1q_2 = 19</math> giving <math> 3(q_1 + q_2)^2 + 3q_1q_2 =\dfrac{19}{3}</math>.
 
  
Substituting gives <math>3(\dfrac{400}{9}) + 3q_1q_2 = 19</math> which simplifies to <math>\dfrac{400}{3} + 3q_1q_2 = \dfrac{57}{3}</math>
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==Solution 3==
<math>3q_1q_2 = \dfrac{-343}{3}</math>, <math>q_1q_2 = \dfrac{-343}{9}</math>, <math>|\dfrac{-343}{9}|=\dfrac{343}{9}</math>, <math>m+n = 343+9 = \fbox{352}.</math>
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Let <math>x=\sum_{1\le j<k\le 673} z_jz_k</math>. By Vieta's, <cmath>3\sum_{i=1}^{673}z_i=-20\implies \sum_{i=1}^{673}z_i=-\frac{20}3.</cmath>Then, consider the <math>19x^{2017}</math> term. To produce the product of two roots, the two roots can either be either <math>(z_i,z_i)</math> for some <math>i</math>, or <math>(z_j,z_k)</math> for some <math>j<k</math>. In the former case, this can happen in <math>\tbinom 32=3</math> ways, and in the latter case, this can happen in <math>3^2=9</math> ways. Hence,
-Ish_Sahh
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<cmath>
 +
\begin{align*}
 +
19=3\sum_{i=1}^{673} z_i^2+9\sum_{1\le j<k\le 673} z_jz_k=3\left(\left(-\frac{20}3\right)^2-2x\right)+9x&=\frac{400}3+3x\\
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\implies x&=-\frac{343}9,
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\end{align*}
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</cmath>
 +
and the requested sum is <math>343+9=\boxed{352}</math>.
 +
 
 +
==Solution 4==
 +
Let <cmath>(f(x))^3=x^{2019}+20x^{2018}+19x^{2017}+g(x).</cmath> Therefore, <math>f(x)=(x-z_{1})(x-z_{1})\cdots (x-z_{673})</math>. This is also equivalent to <cmath>f(x)=x^{673}+ax^{672}+bx^{671}+h(x)</cmath> for some real coefficients <math>a</math> and <math>b</math> and some polynomial <math>h(x)</math> with degree <math>670</math>. We can see that the big summation expression is simply summing the product of the roots of <math>f(x)</math> taken two at a time. By Vieta's, this is just the coefficient <math>b</math>. The first three terms of <math>(f(x))^3</math> can be bashed in terms of <math>a</math> and <math>b</math> to get
 +
<cmath> 20 = 3a </cmath>
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<cmath> 19 = 3a^2+3b </cmath>
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Thus, <math>a=\frac{20}{3}</math> and <math>b=\frac{1}{3}\left(19-3\left(\frac{20}{3} \right)^2 \right)</math>. That is <math>|b|=\frac{343}{9}=\frac{m}{n}</math>. <math>m+n=343+9=\boxed{352}</math>
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=I|num-b=9|num-a=11}}
 
{{AIME box|year=2019|n=I|num-b=9|num-a=11}}
 +
 +
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:39, 16 October 2020

Problem 10

For distinct complex numbers $z_1,z_2,\dots,z_{673}$, the polynomial \[(x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3\]can be expressed as $x^{2019} + 20x^{2018} + 19x^{2017}+g(x)$, where $g(x)$ is a polynomial with complex coefficients and with degree at most $2016$. The sum $\left| \sum_{1 \le j <k \le 673} z_jz_k \right|$ can be expressed in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

In order to begin this problem, we must first understand what it is asking for. The notation \[\left| \sum_{1 \le j <k \le 673} z_jz_k \right|\] simply asks for the absolute value of the sum of the product of the distinct unique roots of the polynomial taken two at a time or \[(z_1z_2+z_1z_3+ \dots + z_1z_{672}+z_1z_{673})+(z_2z_3+z_2z_4+ \dots +z_2z_{673}) + (z_3z_4+z_3z_5+ \dots +z_3z_{673}) + \dots +z_{672}z_{673}.\] Call this sum $S$.

Now we can begin the problem. Rewrite the polynomial as $P=(x-z_1)(x-z_1)(x-z_1)(x-z_2)(x-z_2)(x-z_2) \dots (x-z_{673})(x-z_{673})(x-z_{673})$. Then we have that the roots of $P$ are $z_1,z_1,z_1,z_2,z_2,z_2, \dots , z_{673},z_{673},z_{673}$.

By Vieta's formulas, we have that the sum of the roots of $P$ is $(-1)^1 \cdot \dfrac{20}{1}=-20=z_1+z_1+z_1+z_2+z_2+z_2+ \dots + z_{673}+z_{673}+z_{673}=3(z_1+z_2+z_3+ \dots +z_{673})$. Thus, $z_1+z_2+z_3+ \dots +z_{673}=- \dfrac{20}{3}.$

Similarly, we also have that the the sum of the roots of $P$ taken two at a time is $(-1)^2 \cdot \dfrac{19}{1} = 19.$ This is equal to $z_1^2+z_1^2+z_1^2+z_1z_2+z_1z_2+z_1z_2+ \dots =  \\ 3(z_1^2+z_2^2+ \dots + z_{673}^2) + 9(z_1z_2+z_1z_3+z_1z_4+ \dots + z_{672}z_{673}) =  3(z_1^2+z_2^2+ \dots + z_{673}^2) + 9S.$

Now we need to find and expression for $z_1^2+z_2^2+ \dots + z_{673}^2$ in terms of $S$. We note that $(z_1+z_2+z_3+ \dots +z_{673})^2= (-20/3)^2=\dfrac{400}{9} \\ =(z_1^2+z_2^2+ \dots + z_{673}^2)+2(z_1z_2+z_1z_3+z_1z_4+ \dots + z_{672}z_{673})=(z_1^2+z_2^2+ \dots + z_{673}^2)+2S.$ Thus, $z_1^2+z_2^2+ \dots + z_{673}^2= \dfrac{400}{9} -2S$.

Plugging this into our other Vieta equation, we have $3 \left( \dfrac{400}{9} -2S \right) +9S = 19$. This gives $S = - \dfrac{343}{9} \Rightarrow \left| S \right| = \dfrac{343}{9}$. Since 343 is relatively prime to 9, $m+n = 343+9 = \fbox{352}$.

Solution 2

This is a quick fake solve using $z_i = 0$ where $3 \le i \le 673$ and only $z_1,z_2 \neq 0$ .

By Vieta's, \[3z_1+3z_2=-20\] and \[3z_1^2+3z_2^2+9z_1z_2 = 19.\] Rearranging gives $z_1 + z_2 = \dfrac{-20}{3}$ and $3(z_1^2 + 2z_1z_2 + z_2^2) + 3z_1z_2 = 19$ giving $3(z_1 + z_2)^2 + 3z_1z_2 = 19$.

Substituting gives $3\left(\dfrac{400}{9}\right) + 3z_1z_2 = 19$ which simplifies to $\dfrac{400}{3} + 3z_1z_2 = \dfrac{57}{3}$.

So, $3z_1z_2 = \dfrac{-343}{3}$, $z_1z_2 = \dfrac{-343}{9}$, $|\dfrac{-343}{9}|=\dfrac{343}{9}$, \[m+n = 343+9 = \boxed{352}.\]

Solution 3

Let $x=\sum_{1\le j<k\le 673} z_jz_k$. By Vieta's, \[3\sum_{i=1}^{673}z_i=-20\implies \sum_{i=1}^{673}z_i=-\frac{20}3.\]Then, consider the $19x^{2017}$ term. To produce the product of two roots, the two roots can either be either $(z_i,z_i)$ for some $i$, or $(z_j,z_k)$ for some $j<k$. In the former case, this can happen in $\tbinom 32=3$ ways, and in the latter case, this can happen in $3^2=9$ ways. Hence, \begin{align*} 19=3\sum_{i=1}^{673} z_i^2+9\sum_{1\le j<k\le 673} z_jz_k=3\left(\left(-\frac{20}3\right)^2-2x\right)+9x&=\frac{400}3+3x\\ \implies x&=-\frac{343}9, \end{align*} and the requested sum is $343+9=\boxed{352}$.

Solution 4

Let \[(f(x))^3=x^{2019}+20x^{2018}+19x^{2017}+g(x).\] Therefore, $f(x)=(x-z_{1})(x-z_{1})\cdots (x-z_{673})$. This is also equivalent to \[f(x)=x^{673}+ax^{672}+bx^{671}+h(x)\] for some real coefficients $a$ and $b$ and some polynomial $h(x)$ with degree $670$. We can see that the big summation expression is simply summing the product of the roots of $f(x)$ taken two at a time. By Vieta's, this is just the coefficient $b$. The first three terms of $(f(x))^3$ can be bashed in terms of $a$ and $b$ to get \[20 = 3a\] \[19 = 3a^2+3b\] Thus, $a=\frac{20}{3}$ and $b=\frac{1}{3}\left(19-3\left(\frac{20}{3} \right)^2 \right)$. That is $|b|=\frac{343}{9}=\frac{m}{n}$. $m+n=343+9=\boxed{352}$

See Also

2019 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AIME Problems and Solutions

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