Difference between revisions of "2008 AIME I Problems/Problem 13"
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== Solution == | == Solution == | ||
− | < | + | <cmath>\begin{align*} |
p(0,0) &= a_0 = 0\\ | p(0,0) &= a_0 = 0\\ | ||
p(1,0) &= a_0 + a_1 + a_3 + a_6 = a_1 + a_3 + a_6 = 0\\ | p(1,0) &= a_0 + a_1 + a_3 + a_6 = a_1 + a_3 + a_6 = 0\\ | ||
− | p(-1,0) &= -a_1 + a_3 - a_6 = 0\end{align*}</ | + | p(-1,0) &= -a_1 + a_3 - a_6 = 0\end{align*}</cmath> |
Adding the above two equations gives <math>a_3 = 0</math>, and so we can deduce that <math>a_6 = -a_1</math>. | Adding the above two equations gives <math>a_3 = 0</math>, and so we can deduce that <math>a_6 = -a_1</math>. | ||
Similarly, plugging in <math>(0,1)</math> and <math>(0,-1)</math> gives <math>a_5 = 0</math> and <math>a_9 = -a_2</math>. Now, | Similarly, plugging in <math>(0,1)</math> and <math>(0,-1)</math> gives <math>a_5 = 0</math> and <math>a_9 = -a_2</math>. Now, | ||
− | < | + | <cmath>\begin{align*}p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9\\ |
&= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 = a_4 + a_7 + a_8 = 0\\ | &= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 = a_4 + a_7 + a_8 = 0\\ | ||
− | p(1,-1) &= a_0 + a_1 - a_2 + 0 - a_4 + 0 - a_1 - a_7 + a_8 + a_2\\ &= -a_4 - a_7 + a_8 = 0\end{align*}</ | + | p(1,-1) &= a_0 + a_1 - a_2 + 0 - a_4 + 0 - a_1 - a_7 + a_8 + a_2\\ &= -a_4 - a_7 + a_8 = 0\end{align*}</cmath> |
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, | ||
<math>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</math> | <math>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</math> |
Revision as of 17:31, 13 March 2015
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.