2008 AIME I Problems/Problem 13
Problem
Let
![\[p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.\]](http://latex.artofproblemsolving.com/3/c/2/3c2f869a894828abc1f1f9da90ed74de9ee70964.png)
Suppose that
![\[p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.\]](http://latex.artofproblemsolving.com/9/3/7/93754a4657c1bd6861887da1a6ffbf68a98fb7ff.png)
There is a point 
for which 
for all such polynomials, where 
, 
, and 
are positive integers, and are relatively prime, and 
. Find 
.
Solution
$\begin{align*} p(0,0) &= a_0 = 0\\ p(1,0) &= a_0 + a_1 + a_3 + a_6 = a_1 + a_3 + a_6 = 0\\ p(-1,0) &= -a_1 + a_3 - a_6 = 0\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Adding the above two equations gives 
, and so we can deduce that 
.
Similarly, plugging in 
and 
gives 
and 
. Now,
$\begin{align*}p(1,1) &= a_0 + a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9\\
&= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 = a_4 + a_7 + a_8 = 0\\
p(1,-1) &= a_0 + a_1 - a_2 + 0 - a_4 + 0 - a_1 - a_7 + a_8 + a_2\\ &= -a_4 - a_7 + a_8 = 0\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
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. 
