Difference between revisions of "2019 AIME I Problems/Problem 10"
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By Vieta's formulas, we have that the sum of the roots of <math>P</math> is <math>(-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})</math>. Thus, <math>z_1+z_2+z_3+ \dots +z_{673}=- \dfrac{20}{3}.</math> | By Vieta's formulas, we have that the sum of the roots of <math>P</math> is <math>(-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})</math>. Thus, <math>z_1+z_2+z_3+ \dots +z_{673}=- \dfrac{20}{3}.</math> | ||
− | Similarly, we also have that the the sum of the roots of <math>P</math> taken two at a time | + | Similarly, we also have that the the sum of the roots of <math>P</math> taken two at a time is <math>(-1)^2 * \dfrac{19}{1} = 19.</math> This is equal to <math>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.</math> |
Now we need to find and expression for <math>z_1^2+z_2^2+ \dots + z_{673}^2</math> in terms of <math>S</math>. We note that <math>(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.</math> | Now we need to find and expression for <math>z_1^2+z_2^2+ \dots + z_{673}^2</math> in terms of <math>S</math>. We note that <math>(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.</math> |
Revision as of 01:30, 15 March 2019
The 2019 AIME I takes place on March 13, 2019.
Problem 10
For distinct complex numbers , the polynomial
can be expressed as
, where
is a polynomial with complex coefficients and with degree at most
. The value of
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 roots of the polynomial taken two at a time or
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 prime,
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.