Difference between revisions of "2008 AIME I Problems/Problem 13"
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&= 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>. | + | Therefore <math>a_8 = 0</math> and <math>a_7 = -a_4</math>. Finally, |
<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> | <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>. | + | So, <math>3a_1 + 3a_2 + 2a_4 = 0</math>, or equivalently <math>a_4 = -\frac{3(a_1 + a_2)}{2}</math>. |
− | + | Substituting these equations into the original polynomial <math>p</math>, we find that at <math>(\frac{a}{c}, \frac{b}{c})</math>, | |
+ | <cmath>\begin{align*} | ||
+ | a_1x + a_2y + a_4xy - a_1x^3 - a_4x^2y - a_2y^3 &= 0 \iff | ||
+ | a_1x + a_2y + -\frac{3(a_1 + a_2)}{2}xy - a_1x^3 + \frac{3(a_1 + a_2)}{2}x^2y - a_2y^3 &= 0 \iff | ||
+ | a_1x(x - 1)(x + 1 - \frac{3}{2}y) + a_2y(y^2 - 1 - \frac{3}{2}x(x - 1)) &= 0 | ||
+ | \end{align*}</cmath>. | ||
+ | The remaining coefficients <math>a_1</math> and <math>a_2</math> are now completely arbitrary because the original equations impose no more restrictions on them. Hence, for the final equation to hold for all possible <math>p</math>, we must have <math>x(x - 1)(x + 1 - \frac{3}{2}y) = y(y^2 - 1 - \frac{3}{2}x(x - 1)) = 0</math>. | ||
+ | |||
+ | As the answer format implies that the <math>x</math>-coordinate of the root is non-integral, <math>x(x - 1)(x + 1 - \frac{3}{2}y) = 0 \iff x + 1 - \frac{3}{2}y = 0 \iff y = \frac{2}{3}(x + 1)</math>. The format also implies that <math>y</math> is positive, so <math>y(y^2 - 1 - \frac{3}{2}x(x - 1)) = 0 \iff y^2 - 1 - \frac{3}{2}x(x - 1) = 0</math>. Substituting <math>y</math> into this equation and factoring the quadratic yields <math>(19x - 5)(x - 2) = 0</math>, in which the only non-integral root is <math>x = \frac{5}{19}</math>, so <math>y = \frac{16}{19}</math>. | ||
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+ | The answer is <math>a + b + c = \boxed{40}</math>. | ||
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=== Solution 2 === | === 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 | 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 |
Revision as of 12:14, 3 January 2021
Contents
[hide]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, , or equivalently .
Substituting these equations into the original polynomial , we find that at , . The remaining coefficients and are now completely arbitrary because the original equations impose no more restrictions on them. Hence, for the final equation to hold for all possible , we must have .
As the answer format implies that the -coordinate of the root is non-integral, . The format also implies that is positive, so . Substituting into this equation and factoring the quadratic yields , in which the only non-integral root is , so .
The answer is .
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.