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

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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>
−
simply asks for the absolute value of the sum of the product of the roots of the polynomial taken two at a time or
+
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
 
<cmath>(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}.</cmath> Call this sum <math>S</math>.  
 
<cmath>(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}.</cmath> Call this sum <math>S</math>.  
  

Revision as of 02:13, 15 March 2019

The 2019 AIME I takes place on March 13, 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

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 $(-1)^1 * \dfrac{20}{1}=-20=z_1+z_1+z_1+z_2+z_2+z_2+ \dots + z_{673}+z_{673}+z_{673}=3(z_1+z_2+z_3+ \dots +z_{673})$. 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 $(-1)^2 * \dfrac{19}{1} = 19.$ 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, $m+n = 343+9 = \fbox{352}$.

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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