Difference between revisions of "2019 AIME I Problems/Problem 10"
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Now we can begin the problem. Rewrite the polynomial as <math>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})</math>. Then we have that the roots of <math>P</math> are <math>z_1,z_1,z_1,z_2,z_2,z_2, \dots , z_{673},z_{673},z_{673}</math>. | Now we can begin the problem. Rewrite the polynomial as <math>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})</math>. Then we have that the roots of <math>P</math> are <math>z_1,z_1,z_1,z_2,z_2,z_2, \dots , z_{673},z_{673},z_{673}</math>. | ||
− | By Vieta's formulas, we have that the sum of the roots of <math>P</math> is <math>(-1)^1 | + | By Vieta's formulas, we have that the sum of the roots of <math>P</math> is <math>(-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})</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 is <math>(-1)^2 | + | Similarly, we also have that the the sum of the roots of <math>P</math> taken two at a time is <math>(-1)^2 \cdot \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 16:44, 18 March 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 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 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 giving .
Substituting gives which simplifies to , , ,
~Ish_Sahh
Solution 3
Let . By Vieta's, 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)
Solution 4
Let Therefore, . This is also equivalent to for some real coefficients and and some polynomial with degree . We can see that the big summation expression is simply summing the product of the roots of taken two at a time. By Vieta's, this is just the coefficient . The first three terms of can be bashed in terms of and to get Thus, and . That is .
~jakeg314
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.