Difference between revisions of "2008 AIME I Problems/Problem 13"
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Therefore <math>a_8 = 0</math> and <math>a_7 = -a_4</math>. Finally | Therefore <math>a_8 = 0</math> and <math>a_7 = -a_4</math>. Finally | ||
− | <cmath>p(2,2) = 0 + | + | <cmath>p(2,2) = 0 + 2a_1 + 2a_2 + 0 + 4a_4 + 0 - 8a_1 - 8a_4 +0 - 8a_2 = -6 a_1 - 6 a_2 - 4 a_4 = 0</cmath> |
− | So <math> | + | So <math>3a_1 + 3a_2 + 2a_4 = 0</math>. |
− | Now <math>p(x,y) = 0 + x | + | Now <math>p(x,y) = 0 + x a_1 + y a_2 + 0 + xy a_4 + 0 - x^3 a_1 - x^2 y a_4 + 0 - y^3 a_2</math> <math>= x(1-x)(1+x) a_1 + y(1-y)(1+y) a_2 + xy (1-x) a_4</math>. |
In order for the above to be zero, we must have | In order for the above to be zero, we must have | ||
<cmath>x(1-x)(1+x) = y(1-y)(1+y)</cmath> | <cmath>x(1-x)(1+x) = y(1-y)(1+y)</cmath> | ||
and | and | ||
− | <cmath>x(1-x)(1+x) = 3/2 xy (1-x)</cmath> | + | <cmath>x(1-x)(1+x) = 3/2 xy (1-x).</cmath> Canceling terms on the second equation gives us |
− | <math>1+x = 3/2 y \Longrightarrow x = 3/2 y - 1</math>. Plugging that into the first equation and solving | + | <math>1+x = 3/2 y \Longrightarrow x = 3/2 y - 1</math>. Plugging that into the first equation and solving yields <math>x = 5/19, y = 16/19</math>, and <math>5+16+19 = \boxed{040}</math>. |
== See also == | == See also == |
Revision as of 13:49, 23 March 2008
Problem
Let
Suppose that
There is a point for which for all such polynomials, where , , and are positive integers, and are relatively prime, and . Find .
Solution
Adding the above two equations gives , and so we can deduce that .
Similarly, plugging in and gives and . Now,
Therefore and . Finally So .
Now .
In order for the above to be zero, we must have and Canceling terms on the second equation gives us . Plugging that into the first equation and solving yields , and .
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |