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

m ([tex-ed] solution by nr1337)
 
m (corrections?)
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Let
 
Let
  
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<math>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</math>.
+
<cmath>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.</cmath>
  
 
Suppose that  
 
Suppose that  
  
−
<math>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</math>.
+
<cmath>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.</cmath>
  
 
There is a point <math>\left(\frac {a}{c},\frac {b}{c}\right)</math> for which <math>p\left(\frac {a}{c},\frac {b}{c}\right) = 0</math> for all such polynomials, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers, <math>a</math> and <math>c</math> are relatively prime, and <math>c > 1</math>.  Find <math>a + b + c</math>.
 
There is a point <math>\left(\frac {a}{c},\frac {b}{c}\right)</math> for which <math>p\left(\frac {a}{c},\frac {b}{c}\right) = 0</math> for all such polynomials, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers, <math>a</math> and <math>c</math> are relatively prime, and <math>c > 1</math>.  Find <math>a + b + c</math>.
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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
+
<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*}</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
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So <math>3a1 + 3a2 + 2a4 = 0</math>.
 
So <math>3a1 + 3a2 + 2a4 = 0</math>.
  
−
Now <math>p(x,y) = 0 + x a1 + y a2 + 0 + xy a4 + 0 - x^3 a1 - x^2 y a4 + 0 - y^3 a2
+
Now <math>p(x,y) = 0 + x a1 + y a2 + 0 + xy a4 + 0 - x^3 a1 - x^2 y a4 + 0 - y^3 a2</math> <math>= x(1-x)(1+x) a1 + y(1-y)(1+y) a2 + xy (1-x) a4</math>.
−
= x(1-x)(1+x) a1 + y(1-y)(1+y) a2 + xy (1-x) a4</math>.
 
  
 
In order for the above to be zero, we must have
 
In order for the above to be zero, we must have
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and
 
and
 
<cmath>x(1-x)(1+x) = 3/2 xy (1-x)</cmath>.  Canceling terms on the second equation gives us
 
<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. Plugging that into the first equation and solving yieldst </math>x = 5/19, y = 16/19<math>, and </math>5+16+19 = \boxed{040}$.
+
<math>1+x = 3/2 y \Longrightarrow x = 3/2 y - 1</math>. Plugging that into the first equation and solving yieldst <math>x = 5/19, y = 16/19</math>, and <math>5+16+19 = \boxed{040}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 13:47, 23 March 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*}

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*}

Therefore $a_8 = 0$ and $a_7 = -a_4$. Finally \[p(2,2) = 0 + 2a1 + 2a2 + 0 + 4a4 + 0 - 8a1 - 8a4 +0 - 8a2 = -6 a1 - 6 a2 - 4 a4 = 0\] So $3a1 + 3a2 + 2a4 = 0$.

Now $p(x,y) = 0 + x a1 + y a2 + 0 + xy a4 + 0 - x^3 a1 - x^2 y a4 + 0 - y^3 a2$ $= x(1-x)(1+x) a1 + y(1-y)(1+y) a2 + xy (1-x) a4$.

In order for the above to be zero, we must have \[x(1-x)(1+x) = y(1-y)(1+y)\] and \[x(1-x)(1+x) = 3/2 xy (1-x)\]. Canceling terms on the second equation gives us $1+x = 3/2 y \Longrightarrow x = 3/2 y - 1$. Plugging that into the first equation and solving yieldst $x = 5/19, y = 16/19$, 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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