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

(Solution 2)
Line 5: Line 5:
 
<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 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>.
 
<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==
+
 
 +
==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>
Line 32: Line 33:
  
 
~Ish_Sahh
 
~Ish_Sahh
 +
 +
==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 by TheUltimate123)
  
 
==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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:02, 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 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 $(-1)^1 * \dfrac{20}{1}=-20=z_1+z_1+z_1+z_2+z_2+z_2+ \dots + z_{673}+z_{673}+z_{673}=3(z_1+z_2+z_3+ \dots +z_{673})$. 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, $m+n = 343+9 = \fbox{352}$.

Solution 2

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

By Vieta's, \[3q_1+3q_2=-20\] and \[3q_1^2+3q_2^2+9q_1q_2 = 19.\] Rearranging gives $q_1 + q_2 = \dfrac{-20}{3}$ and $3(q_1^2 + 2q_1q_2 + q_2^2) + 3q_1q_2 = 19$ giving $3(q_1 + q_2)^2 + 3q_1q_2 =\dfrac{19}{3}$.

Substituting gives $3(\dfrac{400}{9}) + 3q_1q_2 = 19$ which simplifies to $\dfrac{400}{3} + 3q_1q_2 = \dfrac{57}{3}$ $3q_1q_2 = \dfrac{-343}{3}$, $q_1q_2 = \dfrac{-343}{9}$, $|\dfrac{-343}{9}|=\dfrac{343}{9}$, $m+n = 343+9 = \fbox{352}.$

~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 by TheUltimate123)

See Also

2019 AIME I (ProblemsAnswer KeyResources)
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

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