Difference between revisions of "2011 AIME II Problems/Problem 8"
(→Solution 2) |
m (→Solution 2) |
||
Line 13: | Line 13: | ||
<cmath>(z^6-2^{18})(z^6+2^{18})=0</cmath> | <cmath>(z^6-2^{18})(z^6+2^{18})=0</cmath> | ||
<cmath>(z^2-2^6)(z^2+2^6)({(z^2+2^6)}^2-z^2\cdot2^6)({(z^2-2^6)}^2+z^2\cdot2^6)=0</cmath> | <cmath>(z^2-2^6)(z^2+2^6)({(z^2+2^6)}^2-z^2\cdot2^6)({(z^2-2^6)}^2+z^2\cdot2^6)=0</cmath> | ||
− | <cmath>(z^2-2^6)(z^2+2^6)(z^2+2^6-z\cdot2^3)(z^2+2^6+z\cdot2^3) (z^2-2^6- | + | <cmath>(z^2-2^6)(z^2+2^6)(z^2+2^6-z\cdot2^3)(z^2+2^6+z\cdot2^3) (z^2-2^6-iz\cdot2^3)(z^2-2^6+i z\cdot2^3)=0</cmath> |
Since this is a 12th degree equation, there are 12 roots. Also, since each term in the equation is even, the positive or negative value of each root is another root. That would mean there are 6 roots that can be multiplied by <math>-1</math> and since we have 6 factors, that’s 1 root per factor. We just need to solve for <math>z</math> in each factor and pick whether or not to multiply by <math>i</math> and <math>-1</math> for each one depending on the one that yields the highest real value. After that process, we get <math>8+8+2((4\sqrt{3}+4)+(4\sqrt{3}-4))</math> Adding the values up yields <math>16+16\sqrt{3}</math>, or <math>16+\sqrt{768}</math>, and <math>16+768=\boxed{784}</math>. | Since this is a 12th degree equation, there are 12 roots. Also, since each term in the equation is even, the positive or negative value of each root is another root. That would mean there are 6 roots that can be multiplied by <math>-1</math> and since we have 6 factors, that’s 1 root per factor. We just need to solve for <math>z</math> in each factor and pick whether or not to multiply by <math>i</math> and <math>-1</math> for each one depending on the one that yields the highest real value. After that process, we get <math>8+8+2((4\sqrt{3}+4)+(4\sqrt{3}-4))</math> Adding the values up yields <math>16+16\sqrt{3}</math>, or <math>16+\sqrt{768}</math>, and <math>16+768=\boxed{784}</math>. |
Revision as of 15:22, 12 December 2019
Contents
Problem
Let be the 12 zeroes of the polynomial . For each , let be one of or . Then the maximum possible value of the real part of can be written as where and are positive integers. Find .
Solution
The twelve dots above represent the roots of the equation . If we write , then the real part of is and the real part of is . The blue dots represent those roots for which the real part of is greater than the real part of , and the red dots represent those roots for which the real part of is greater than the real part of . Now, the sum of the real parts of the blue dots is easily seen to be and the negative of the sum of the imaginary parts of the red dots is easily seen to also be . Hence our desired sum is , giving the answer .
Solution 2
The equation can be factored as follows:
Since this is a 12th degree equation, there are 12 roots. Also, since each term in the equation is even, the positive or negative value of each root is another root. That would mean there are 6 roots that can be multiplied by and since we have 6 factors, that’s 1 root per factor. We just need to solve for in each factor and pick whether or not to multiply by and for each one depending on the one that yields the highest real value. After that process, we get Adding the values up yields , or , and .
-Solution by Someonenumber011.
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.