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

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The 2019 AIME I takes place on March 13, 2019.
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==Problem==
 +
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 sum <cmath> \left| \sum_{1 \le j <k \le 673} z_jz_k \right| </cmath> 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>.
 +
==Video Solution & More by MegaMath==
 +
Video #2 (Vieta's Formulas):
 +
https://www.youtube.com/watch?v=6-kcdBsmCmc
  
−
==Problem 10==
+
==Solution 1==
−
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> \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==
 
 
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>  
+
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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Plugging this into our other Vieta equation, we have <math> 3 \left( \dfrac{400}{9} -2S \right) +9S = 19</math>. This gives <math>S = - \dfrac{343}{9} \Rightarrow \left| S \right| = \dfrac{343}{9}</math>. Since 343 is relatively prime to 9, <math>m+n = 343+9 = \fbox{352}</math>.
 
Plugging this into our other Vieta equation, we have <math> 3 \left( \dfrac{400}{9} -2S \right) +9S = 19</math>. This gives <math>S = - \dfrac{343}{9} \Rightarrow \left| S \right| = \dfrac{343}{9}</math>. Since 343 is relatively prime to 9, <math>m+n = 343+9 = \fbox{352}</math>.
  
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==Solution 2==
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==Solution 2 (Fake Solve)==
−
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> .
+
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>.
 +
 
 +
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>
 +
 
 +
==Solution 3==
 +
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,
 +
<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\\
 +
\implies x&=-\frac{343}9,
 +
\end{align*}
 +
</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>
 +
<cmath> 19 = 3a^2+3b </cmath>
 +
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>
  
−
By Vieta's, <cmath>3q_1+3q_2=-20</cmath> and <cmath>3q_1^2+3q_2^2+9q_1q_2 = 19</cmath>.
+
==Solution 5 (Newton's Sums)==
−
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>.
+
In the problem statement, we're given that the polynomial is <math>(x-z_1)^3(x-z_2)^3(x-z_3)^2\cdots (x-z_{673})^3</math> which can be expressed as <math>x^{2019}+20x^{2018}+19x^{2017}+g(x)</math>. So it has 673 roots of multiplicity 3 each, for a total of 2019 roots.
  
−
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>
+
We start by calling <math> \left| \sum_{1 \le j <k \le 673} z_jz_k \right|  = S</math>. Next, let the sum of the roots of the polynomial be <math>P_1</math>, which equals <math>3(x_1+x_2+\cdots+x_{673})</math>, and the sum of the square of the roots be <math>P_2</math>, which equals <math>3(z_1^2+z_2^2+z_3^2+\dots+z_{673}^2)</math>.
−
<cmath>3q_1q_2 = \dfrac{-343}{3}</cmath> <cmath>q_1q_2 = \dfrac{-343}{9}</cmath> <cmath>|\dfrac{-343}{9}|=\dfrac{343}{9}</cmath> <cmath>m+n = 343+9 = \fbox{352}.</cmath>
+
 
 +
By Vieta's,
 +
<cmath> -20 = 3(z_1+z_2+z_3+z_4 \dots+z_{673}) </cmath>
 +
<cmath>19 = 3(z_1^2+z_2^2+z_3^2+\dots+z_{673}^2) + 9S </cmath>
 +
 
 +
The latter can be easily verified by using a combinatorics approach. <math>19</math> is the sum of all the possible pairs of two roots of the polynomial. Which has <math>\binom{2019}{2}</math> without simplification. Now looking at the latter above, there are <math>3\cdot673</math> terms in the first part and <math>9\cdot\binom{673}{2}</math>.
 +
 
 +
With some computation, we see
 +
<math>\binom{2019}{2}</math> <math>= 3\cdot673+</math> <math>9\cdot\binom{673}{2}</math>.
 +
This step was simply done to check that we missed no steps.
 +
 
 +
Now using [[Newton's Sums | Newton Sums]], where <math>P_2 = z_1^2+z_2^2+z_3^2+\dots+z_{673}^2</math>. We first have that <math>P_1\cdot a_n + 1\cdot a_{n-1}=0</math> and plugging in values, we get that <math>P_1\cdot 1 + 1\cdot 20 = 0 \implies P_1=-20.</math> Then,
 +
<math>P_2\cdot a_n + P_1 \cdot a_{n-1} + 2\cdot a_{n-2}</math> and plugging in the values we know, we get that
 +
 
 +
<cmath>P_2 + -20\cdot20 + 19\cdot2 = 0 \implies P_2 = 362</cmath>
 +
 
 +
<cmath>19 = 362 + 9S \implies S = \frac{343}{9}</cmath>
 +
Thus, <math>\frac{343}{9}</math> leads to the answer <math>\boxed{352}</math>.
 +
 
 +
~YBSuburbanTea
 +
 
 +
~minor edits by BakedPotato66
 +
 
 +
==Solution 6 (Official MAA 1)==
 +
Because each root of the polynomial appears with multiplicity <math>3,</math> Viète's Formulas show that <cmath>z_1+z_2+\cdots+z_{673}=-\frac{20}3</cmath> and <cmath>z_1^2+z_2^2+\cdots+z_{673}^2 =\frac13\left((-20)^2-2\cdot19\right)=\frac{362}3.</cmath> Then the identity <cmath>\left(\sum_{i=1}^{673}z_i\right)^2=\sum_{i=1}^{673}z_i^2+2\left(\sum_{1\le j<k\le 673}z_jz_k\right)</cmath> shows that <cmath>\sum_{1\le j<k\le 673}z_jz_k=\frac{\left(-\frac{20}3\right)^2-\frac{362}3}{2}=-\frac{343}9.</cmath> The requested sum is <math>343+9=352.</math>
 +
 
 +
Note that such a polynomial does exist. For example, let <math>z_{673}=-\tfrac{20}3,</math> and for <math>i=1,2,3,\dots,336,</math> let <cmath>z_i=\sqrt{\frac{343i}{9\sum_{j=1}^{336}j}}\qquad \text{and}\qquad z_{i+336}=-z_i.</cmath> Then <cmath>\sum_{i=1}^{673}z_i=-\frac{20}3\qquad\text{and}\qquad\sum_{i=1}^{673}z_i^2=2\sum_{i=1}^{336}\frac{343i}{9\sum_{i=1}^{336}j}+\left(\frac{20}3\right)^2=\frac{362}3,</cmath> as required.
 +
 
 +
==Solution 7 (Official MAA 2)==
 +
There are constants <math>a</math> and <math>b</math> such that <cmath>(x-z_1)(x-z_2)(x-z_3)\cdots(x-z_{673})=x^{673}+ax^{672}+bx^{671}+\cdots.</cmath> Then <cmath>(x^{673}+ax^{672}+bx^{671}+\cdots)^3=x^{2019}+20x^{2018}+19x^{2017}+\cdots.</cmath> Comparing the <math>x^{2018}</math> and <math>x^{2017}</math> coefficients shows that <math>3a=20</math> and <math>3a^2+3b=19.</math> Solving this system yields <math>a=\tfrac{20}3</math> and <math>b=-\tfrac{343}9.</math> Viète's Formulas then give <math>\left|\sum_{1\le j<k\le 673}z_jz_k\right|=|b|=\tfrac{343}9,</math> as above.
 +
 
 +
==Solution 8==
 +
 
 +
Note that we are trying to find the sum of the products of the pairwise distinct <math>z_k</math>. First, disregard the repeated roots and call the roots of the polynomial <math>x_1, x_2\ldots x_{2019}</math> for simplicity's sake.
 +
 
 +
Then, <math>(z_1+z_2\ldots z_{2019})^2=2\sum_{1\leq j<k\leq2019}{z_jz_k}+\sum_{1\leq k\leq 2019}{z_k^2}</math>. Notice that <math>\sum_{1\leq j<k\leq2019}
 +
{z_jz_k}=19</math> and <math>(x_1+ x_2\ldots +x_{2019})^2=20^2=400</math> by Vietas so <math>\sum_{1\leq k\leq 2019}{z_k^2}=362</math>. Also, since every root is repeated 3 times, <math>\sum_{1\leq k\leq 2019}{z_k^2}=3\sum_{1\leq k\leq673}{z_k^2} \Longrightarrow \sum_{1\leq k\leq673}{z_k^2}=\frac{362}{3}</math> where the <math>z_k</math> in the second sum are all distinct.
 +
 
 +
Call <math>|\sum_{1 \le j <k \le 673} z_jz_k|=|s|</math>. Taking into account repeated roots, if we pair any two roots together, there will be <math>\binom{3}{1}=3</math> copies for each <math>z_k^2, 1\leq z\leq 673</math> and <math>3\cdot 3=9</math> copies for each <math>z_jz_k</math> where <math>z_j</math> and <math>z_k</math> are distinct.
 +
 
 +
Since the sum of any two pairs of roots is 19, <math>9s+3\sum_{1\leq k\leq673}{z_k^2}=19</math> or <math>9s+362=19 \Longrightarrow s=\frac{-343}{9}, |s|=\frac{343}{9}</math>. <math>m+n=343+9=352</math>.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/Dp-pw6NNKRo?t=776
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution by The Power of Logic==
 +
https://www.youtube.com/watch?v=7SFKuEdgwMA
 +
 
 +
~The Power of Logic
  
 
==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}}

Latest revision as of 00:13, 25 December 2024

Problem

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 $m+n$.

Video Solution & More by MegaMath

Video #2 (Vieta's Formulas): https://www.youtube.com/watch?v=6-kcdBsmCmc

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 (Fake Solve)

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 $|b|=\frac{343}{9}=\frac{m}{n}$. $m+n=343+9=\boxed{352}$

Solution 5 (Newton's Sums)

In the problem statement, we're given that the polynomial is $(x-z_1)^3(x-z_2)^3(x-z_3)^2\cdots (x-z_{673})^3$ which can be expressed as $x^{2019}+20x^{2018}+19x^{2017}+g(x)$. So it has 673 roots of multiplicity 3 each, for a total of 2019 roots.

We start by calling $\left| \sum_{1 \le j <k \le 673} z_jz_k \right| = S$. Next, let the sum of the roots of the polynomial be $P_1$, which equals $3(x_1+x_2+\cdots+x_{673})$, and the sum of the square of the roots be $P_2$, which equals $3(z_1^2+z_2^2+z_3^2+\dots+z_{673}^2)$.

By Vieta's, \[-20 = 3(z_1+z_2+z_3+z_4 \dots+z_{673})\] \[19 = 3(z_1^2+z_2^2+z_3^2+\dots+z_{673}^2) + 9S\]

The latter can be easily verified by using a combinatorics approach. $19$ is the sum of all the possible pairs of two roots of the polynomial. Which has $\binom{2019}{2}$ without simplification. Now looking at the latter above, there are $3\cdot673$ terms in the first part and $9\cdot\binom{673}{2}$.

With some computation, we see $\binom{2019}{2}$ $= 3\cdot673+$ $9\cdot\binom{673}{2}$. This step was simply done to check that we missed no steps.

Now using Newton Sums, where $P_2 = z_1^2+z_2^2+z_3^2+\dots+z_{673}^2$. We first have that $P_1\cdot a_n + 1\cdot a_{n-1}=0$ and plugging in values, we get that $P_1\cdot 1 + 1\cdot 20 = 0 \implies P_1=-20.$ Then, $P_2\cdot a_n + P_1 \cdot a_{n-1} + 2\cdot a_{n-2}$ and plugging in the values we know, we get that

\[P_2 + -20\cdot20 + 19\cdot2 = 0 \implies P_2 = 362\]

\[19 = 362 + 9S \implies S = \frac{343}{9}\] Thus, $\frac{343}{9}$ leads to the answer $\boxed{352}$.

~YBSuburbanTea

~minor edits by BakedPotato66

Solution 6 (Official MAA 1)

Because each root of the polynomial appears with multiplicity $3,$ Viète's Formulas show that \[z_1+z_2+\cdots+z_{673}=-\frac{20}3\] and \[z_1^2+z_2^2+\cdots+z_{673}^2 =\frac13\left((-20)^2-2\cdot19\right)=\frac{362}3.\] Then the identity \[\left(\sum_{i=1}^{673}z_i\right)^2=\sum_{i=1}^{673}z_i^2+2\left(\sum_{1\le j<k\le 673}z_jz_k\right)\] shows that \[\sum_{1\le j<k\le 673}z_jz_k=\frac{\left(-\frac{20}3\right)^2-\frac{362}3}{2}=-\frac{343}9.\] The requested sum is $343+9=352.$

Note that such a polynomial does exist. For example, let $z_{673}=-\tfrac{20}3,$ and for $i=1,2,3,\dots,336,$ let \[z_i=\sqrt{\frac{343i}{9\sum_{j=1}^{336}j}}\qquad \text{and}\qquad z_{i+336}=-z_i.\] Then \[\sum_{i=1}^{673}z_i=-\frac{20}3\qquad\text{and}\qquad\sum_{i=1}^{673}z_i^2=2\sum_{i=1}^{336}\frac{343i}{9\sum_{i=1}^{336}j}+\left(\frac{20}3\right)^2=\frac{362}3,\] as required.

Solution 7 (Official MAA 2)

There are constants $a$ and $b$ such that \[(x-z_1)(x-z_2)(x-z_3)\cdots(x-z_{673})=x^{673}+ax^{672}+bx^{671}+\cdots.\] Then \[(x^{673}+ax^{672}+bx^{671}+\cdots)^3=x^{2019}+20x^{2018}+19x^{2017}+\cdots.\] Comparing the $x^{2018}$ and $x^{2017}$ coefficients shows that $3a=20$ and $3a^2+3b=19.$ Solving this system yields $a=\tfrac{20}3$ and $b=-\tfrac{343}9.$ Viète's Formulas then give $\left|\sum_{1\le j<k\le 673}z_jz_k\right|=|b|=\tfrac{343}9,$ as above.

Solution 8

Note that we are trying to find the sum of the products of the pairwise distinct $z_k$. First, disregard the repeated roots and call the roots of the polynomial $x_1, x_2\ldots x_{2019}$ for simplicity's sake.

Then, $(z_1+z_2\ldots z_{2019})^2=2\sum_{1\leq j<k\leq2019}{z_jz_k}+\sum_{1\leq k\leq 2019}{z_k^2}$. Notice that $\sum_{1\leq j<k\leq2019} {z_jz_k}=19$ and $(x_1+ x_2\ldots +x_{2019})^2=20^2=400$ by Vietas so $\sum_{1\leq k\leq 2019}{z_k^2}=362$. Also, since every root is repeated 3 times, $\sum_{1\leq k\leq 2019}{z_k^2}=3\sum_{1\leq k\leq673}{z_k^2} \Longrightarrow \sum_{1\leq k\leq673}{z_k^2}=\frac{362}{3}$ where the $z_k$ in the second sum are all distinct.

Call $|\sum_{1 \le j <k \le 673} z_jz_k|=|s|$. Taking into account repeated roots, if we pair any two roots together, there will be $\binom{3}{1}=3$ copies for each $z_k^2, 1\leq z\leq 673$ and $3\cdot 3=9$ copies for each $z_jz_k$ where $z_j$ and $z_k$ are distinct.

Since the sum of any two pairs of roots is 19, $9s+3\sum_{1\leq k\leq673}{z_k^2}=19$ or $9s+362=19 \Longrightarrow s=\frac{-343}{9}, |s|=\frac{343}{9}$. $m+n=343+9=352$.

~Magnetoninja

Video Solution by OmegaLearn

https://youtu.be/Dp-pw6NNKRo?t=776

~ pi_is_3.14

Video Solution by The Power of Logic

https://www.youtube.com/watch?v=7SFKuEdgwMA

~The Power of Logic

See Also

2019 AIME I (Problems • Answer Key • Resources)
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