AIME 2019 I · 第 10 题

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. 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 $j. In the former case, this can happen in$\tbinom 32=3$ways, and in the latter case, this can happen in$3^2=9$ ways. Hence,

$$\begin{aligned} 19=3\sum_{i=1}^{673} z_i^2+9\sum_{1\le j<k\le 673} z_jz_k=3\left(\left(-\frac{20}3\right)^2-2x\right)+9x&=\frac{400}3+3x\\ \implies x&=-\frac{343}9, \end{aligned}$$ 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)

The polynomial has 673 unique roots of multiplicity 3 each, for a total of 2019 roots.

To start, let $S$ be `\left| \sum_{1 \le j . Next, let the sum of the roots of the polynomial be`P_1$, which is equivalent to the expression$3(z_1+z_2+\cdots+z_{673})$, and the sum of squares of these roots be$P_2$, equivalent to$3(z_1^2+z_2^2+z_3^2+\dots+z_{673}^2)$.

By Vieta's,

$$3( z_1 + z_2 + z_3 + z_4 \dots + z_{673}) = -20$$ $$3( z_1^2 + z_2^2 + z_3^2 + \dots + z_{673}^2) + 9S = 19$$ Therefore, $P_1 = -20$. Using the definition of $P_2$ provided above, Newton Sums tells us that $P_2 + -20 \cdot P_1 \cdot + 19 \cdot 2$

This means that $P_2 = 362$. Plugging back into the second Vieta's equation yields

$$362 + 9S = 19 \implies S = -\frac{343}{9}$$ Taking the absolute value, our answer is therefore $343 + 9 = \boxed{352}$

Note: The tripled roots might seem confusing, but it's easily verified that our Vieta's system of equations is correct using combinatorics to find out how many terms we should have. The right hand side of the first equation has $3\cdot673 = 2019$ terms (exactly what we expect). Unsimplified, $P_2$ also has $2019$ terms and $S$ has $\binom{673}{2}$ terms. We want a total of $\binom{2019}{2}$ terms, which happens to be equal to $2019 + 9\cdot\binom{673}{2}$ so this is correct.

~YBSuburbanTea

~edits by BakedPotato66, faliure167

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 as above.

Video Solution by OmegaLearn

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

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Video Solution by The Power of Logic

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

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Video Solution & More by MegaMath

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