Difference between revisions of "2019 AIME I Problems/Problem 10"
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Plugging this into our other Vieta equation, we have <math> 3 \left( \dfrac{400}{9} -2S \right) +9S = 19</math>. This gives <math>S = - \dfrac{343}{9} \Rightarrow \left| S \right| = \dfrac{343}{9}</math>. Since 343 is relatively prime to 9, <math>m+n = 343+9 = \fbox{352}</math>. | Plugging this into our other Vieta equation, we have <math> 3 \left( \dfrac{400}{9} -2S \right) +9S = 19</math>. This gives <math>S = - \dfrac{343}{9} \Rightarrow \left| S \right| = \dfrac{343}{9}</math>. Since 343 is relatively prime to 9, <math>m+n = 343+9 = \fbox{352}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | This is a quick fake solve using <cmath>z_q = 0</cmath> where <cmath>1 \le q \le 673</cmath> and only <cmath>z_1,z_2 \neq 0</cmath>. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=9|num-a=11}} | {{AIME box|year=2019|n=I|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:18, 15 March 2019
The 2019 AIME I takes place on March 13, 2019.
Contents
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 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 .
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.