Difference between revisions of "2008 AIME I Problems/Problem 13"
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== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
p(0,0) &= a_0 = 0\\ | p(0,0) &= a_0 = 0\\ | ||
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<center><math>x(1-x)(1+x) = 1.5 xy (1-x).</math></center> | <center><math>x(1-x)(1+x) = 1.5 xy (1-x).</math></center> | ||
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>. | 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>. | ||
+ | === Solution 2 === | ||
+ | Consider the cross section of <math>z = p(x, y)</math> on the plane <math>z = 0</math>. We realize that we could construct the lines/curves in the cross section such that their equations multiply to match the form of <math>p(x, y)</math> and they go over the eight given points. A simple way to do this would be to use the equations <math>x = 0</math>, <math>x = 1</math>, and <math>y = \frac{2}{3}x + \frac{2}{3}</math>, giving us | ||
+ | |||
+ | <math>p_1(x, y) = x\left(x - 1\right)\left( \frac{2}{3}x - y + \frac{2}{3}\right) = \frac{2}{3}x + xy + \frac{2}{3}x^3-x^2y</math>. | ||
+ | |||
+ | Another way to do this would to use the line <math>y = x</math> and the ellipse, <math>x^2 + xy + y^2 = 1</math>. This would give | ||
+ | |||
+ | <math>p_2(x, y) = \left(x - y\right)\left(x^2 + xy + y^2 - 1\right) = -x + y + x^3 - y^3</math>. | ||
+ | |||
+ | At this point, we see that <math>p_1</math> and <math>p_2</math> both must have <math>\left(\frac{a}{c}, \frac{b}{c}\right)</math> as a zero. A quick graph of the 4 lines and the ellipse used to create <math>p_1</math> and <math>p_2</math> gives nine intersection points. Eight of them are the given ones, and the ninth is <math>\left(\frac{5}{9}, \frac{16}{9}\right)</math>. The last intersection point can be found by finding the intersection points of <math>y = \frac{2}{3}x + \frac{2}{3}</math> and <math>x^2 + xy + y^2 = 1</math>. | ||
+ | Finally, just add the values of <math>a</math>, <math>b</math>, and <math>c</math> to get <math>5 + 16 + 9 = \boxed{040}</math> | ||
== See also == | == See also == |
Revision as of 19:50, 17 January 2018
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
Solution 1
Adding the above two equations gives , and so we can deduce that .
Similarly, plugging in and gives and . Now, 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 .
Solution 2
Consider the cross section of on the plane . We realize that we could construct the lines/curves in the cross section such that their equations multiply to match the form of and they go over the eight given points. A simple way to do this would be to use the equations , , and , giving us
.
Another way to do this would to use the line and the ellipse, . This would give
.
At this point, we see that and both must have as a zero. A quick graph of the 4 lines and the ellipse used to create and gives nine intersection points. Eight of them are the given ones, and the ninth is . The last intersection point can be found by finding the intersection points of and . Finally, just add the values of , , and to get
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.