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

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By Vieta's, <cmath>3z_1+3z_2=-20</cmath> and <cmath>3z_1^2+3z_2^2+9z_1z_2 = 19.</cmath>
 
By Vieta's, <cmath>3z_1+3z_2=-20</cmath> and <cmath>3z_1^2+3z_2^2+9z_1z_2 = 19.</cmath>
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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 =\dfrac{19}{3}</math>.
+
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>
 
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>

Revision as of 20:45, 4 October 2019

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 value of \[\left| \sum_{1 \le j <k \le 673} z_jz_k \right|\]can be expressed in the form $\tfrac{m}{n}$, 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 $\dfrac{400}{3} + 3z_1z_2 = \dfrac{57}{3}$ $3z_1z_2 = \dfrac{-343}{3}$, $z_1z_2 = \dfrac{-343}{9}$, $|\dfrac{-343}{9}|=\dfrac{343}{9}$, \[m+n = 343+9 = \boxed{352}.\] ~Ish_Sahh

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 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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