2019 AIME I Problems/Problem 10
Problem 10
For distinct complex numbers 
, the polynomial
![\[(x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3\]](http://latex.artofproblemsolving.com/3/5/b/35b2995ac0fe946045c887ed51b6c5a4920ade87.png)
can be expressed as 
, where 
is a polynomial with complex coefficients and with degree at most 
. The value of
![\[\left| \sum_{1 \le j <k \le 673} z_jz_k \right|\]](http://latex.artofproblemsolving.com/8/f/4/8f41f0be45a01d496350a366e307c73e63020126.png)
can be expressed in the form 
, where 
and 
are relatively prime positive integers. Find 
.
Solution 1
In order to begin this problem, we must first understand what it is asking for. The notation
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
![\[(z_1z_2+z_1z_3+ \dots + z_1z_{672}+z_1z_{673})+(z_2z_3+z_2z_4+ \dots +z_2z_{673}) + (z_3z_4+z_3z_5+ \dots +z_3z_{673}) + \dots +z_{672}z_{673}.\]](http://latex.artofproblemsolving.com/0/2/6/0263dcdfd19a3abff591649aafb8e681f852cc81.png)
Call this sum 
.
Now we can begin the problem. Rewrite the polynomial as 
. Then we have that the roots of 
are 
.
By Vieta's formulas, we have that the sum of the roots of is 
. Thus, 
Similarly, we also have that the the sum of the roots of taken two at a time is 
This is equal to 
Now we need to find and expression for 
in terms of . We note that 
Thus, 
.
Plugging this into our other Vieta equation, we have 
. This gives 
. Since 343 is relatively prime to 9, 
.
Solution 2
This is a quick fake solve using 
where 
and only 
.
By Vieta's, ![\[3q_1+3q_2=-20\]](http://latex.artofproblemsolving.com/5/1/e/51e9e947d75b016cbd4130c5f3e194fc4e96150a.png)
and ![\[3q_1^2+3q_2^2+9q_1q_2 = 19.\]](http://latex.artofproblemsolving.com/a/2/c/a2c9782a7e7076aa0b1bb99633f06eaefa660b45.png)
Rearranging gives 
and 
giving 
.
Substituting gives 
which simplifies to 
, 
, 
, 
~Ish_Sahh
Solution 3
Let 
. By Vieta's, ![\[3\sum_{i=1}^{673}z_i=-20\implies \sum_{i=1}^{673}z_i=-\frac{20}3.\]](http://latex.artofproblemsolving.com/f/b/6/fb6bacb13f883bd60251f72465a72c57535440fd.png)
Then, consider the 
term. To produce the product of two roots, the two roots can either be either 
for some 
, or 
for some 
. In the former case, this can happen in 
ways, and in the latter case, this can happen in 
ways. Hence,
and the requested sum is 
.
(Solution by TheUltimate123)
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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