2019 AIME I Problems/Problem 10
The 2019 AIME I takes place on March 13, 2019.
Contents
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
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, 
and 
.
Rearranging gives 
and 
and 
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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