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

Revision as of 09:41, 5 March 2015

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

$ $\begin{aligned} p (0, 0) & = a_{0} = 0 \\ p (1, 0) & = a_{0} + a_{1} + a_{3} + a_{6} = a_{1} + a_{3} + a_{6} = 0 \\ p (- 1, 0) & = - a_{1} + a_{3} - a_{6} = 0 \end{aligned}$ $ (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, $ $\begin{aligned} p (1, 1) & = a_{0} + a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} + a_{8} + a_{9} \\ = 0 + a_{1} + a_{2} + 0 + a_{4} + 0 - a_{1} + a_{7} + a_{8} - a_{2} = a_{4} + a_{7} + a_{8} = 0 \\ p (1, - 1) & = a_{0} + a_{1} - a_{2} + 0 - a_{4} + 0 - a_{1} - a_{7} + a_{8} + a_{2} \\ = - a_{4} - a_{7} + a_{8} = 0 \end{aligned}$ $ (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 $5+16+19 = \boxed{040}$ .

@@ Line 11: / Line 11: @@
 == Solution ==
-<center><math>\begin{align*}
+<math>\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*}</math></center>
+p(-1,0) &= -a_1 + a_3 - a_6 = 0\end{align*}</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,
-<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\
+<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\
 &= 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*}</math></center>
+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>
 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>
+<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>
 So <math>3a_1 + 3a_2 + 2a_4 = 0</math>.

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

Revision as of 09:41, 5 March 2015

Problem

Solution

See also