2008 AIME I Problems/Problem 13
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
$\begin{align*} p(0,0) &= a_0 = 0\\ p(1,0) &= a_0 + a_1 + a_3 + a_6 = a_1 + a_3 + a_6 = 0\\ p(-1,0) &= -a_1 + a_3 - a_6 = 0\end{align*}$ (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
![$x(1-x)(1+x) = y(1-y)(1+y)$](http://latex.artofproblemsolving.com/8/4/4/8441c3c87aea15dfd987d273cd22864e0ce424ea.png)
and
![$x(1-x)(1+x) = 1.5 xy (1-x).$](http://latex.artofproblemsolving.com/e/9/c/e9ca0f9bd6511fccd9398392a365b7f5c3cb7aef.png)
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 |
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