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Difference between revisions of "2008 AIME I Problems/Problem 13"

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== Solution ==
 
== Solution ==
−
<cmath>\begin{align*}
+
<center><math>\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*}</cmath>
+
p(-1,0) &= -a_1 + a_3 - a_6 = 0\end{align*}</math></center>
  
 
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,
−
 
+
<center><math>\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\\
−
<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*}</cmath>
+
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*}</math></center>
−
 
+
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
+
<center><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></center>
−
<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>3a_1 + 3a_2 + 2a_4 = 0</math>.
 
So <math>3a_1 + 3a_2 + 2a_4 = 0</math>.
  
Line 31: Line 29:
  
 
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>
+
<center><math>x(1-x)(1+x) = y(1-y)(1+y)</math></center>
 
and
 
and
−
<cmath>x(1-x)(1+x) = 1.5 xy (1-x).</cmath> Canceling terms on the second equation gives us
+
<center><math>x(1-x)(1+x) = 1.5 xy (1-x).</math></center>
−
<math>1+x = 1.5 y \Longrightarrow x = 1.5 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>.
+
Canceling terms on the second equation gives us <math>1+x = 1.5 y \Longrightarrow x = 1.5 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 14:08, 19 April 2008

Problem

Let

\[p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.\]

Suppose that

\[p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.\]

There is a point $\left(\frac {a}{c},\frac {b}{c}\right)$ for which $p\left(\frac {a}{c},\frac {b}{c}\right) = 0$ for all such polynomials, where $a$, $b$, and $c$ are positive integers, $a$ and $c$ are relatively prime, and $c > 1$. Find $a + b + c$.

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 $a_3 = 0$, and so we can deduce that $a_6 = -a_1$.

Similarly, plugging in $(0,1)$ and $(0,-1)$ gives $a_5 = 0$ and $a_9 = -a_2$. 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 $a_8 = 0$ and $a_7 = -a_4$. Finally,

$p(2,2) = 0 + 2a_1 + 2a_2 + 0 + 4a_4 + 0 - 8a_1 - 8a_4 +0 - 8a_2 = -6 a_1 - 6 a_2 - 4 a_4 = 0$

So $3a_1 + 3a_2 + 2a_4 = 0$.

Now $p(x,y) = 0 + x a_1 + y a_2 + 0 + xy a_4 + 0 - x^3 a_1 - x^2 y a_4 + 0 - y^3 a_2$ $= x(1-x)(1+x) a_1 + y(1-y)(1+y) a_2 + xy (1-x) a_4$.

In order for the above to be zero, we must have

$x(1-x)(1+x) = y(1-y)(1+y)$

and

$x(1-x)(1+x) = 1.5 xy (1-x).$

Canceling terms on the second equation gives us $1+x = 1.5 y \Longrightarrow x = 1.5 y - 1$. Plugging that into the first equation and solving yields $x = 5/19, y = 16/19$, and $5+16+19 = \boxed{040}$.

See also

2008 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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All AIME Problems and Solutions
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