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

Revision as of 00:03, 22 August 2021

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 $m+n$ .

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

Solution 5 (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 6 (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.

Video Solution by The Power of Logic

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

~The Power of Logic

2019 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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