Difference between revisions of "2008 AIME I Problems/Problem 13"
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== Solution == | == Solution == | ||
− | < | + | === Solution 1 === |
− | p(0,0) &= a_0 = 0\\ | + | <cmath>\begin{align*} |
− | p(1,0) &= a_0 + a_1 + a_3 + a_6 = a_1 + a_3 + a_6 = 0\\ | + | p(0,0) &= a_0 \\ |
− | p(-1,0) &= -a_1 + a_3 - a_6 = 0\end{align*}</ | + | &= 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> | ||
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, | ||
− | < | + | <cmath>\begin{align*} |
− | &= 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 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 \\ |
− | 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*}</ | + | &= 0 + a_1 + a_2 + 0 + a_4 + 0 - a_1 + a_7 + a_8 - a_2 \\ |
− | Therefore <math>a_8 = 0</math> and <math>a_7 = -a_4</math>. | + | &= a_4 + a_7 + a_8 \\ |
− | < | + | &= 0 \\ |
− | So <math>3a_1 + 3a_2 + 2a_4 = 0</math>. | + | 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, | ||
+ | <cmath>\begin{align*} | ||
+ | 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 | ||
+ | \end{align*}</cmath> | ||
+ | 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>\left(\frac{a}{c}, \frac{b}{c}\right)</math>, | |
+ | <cmath>a_1x + a_2y + a_4xy - a_1x^3 - a_4x^2y - a_2y^3 = 0 \iff</cmath> | ||
+ | <cmath>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</cmath> | ||
+ | <cmath>a_1x(x - 1)\left(x + 1 - \frac{3}{2}y\right) + a_2y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0</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)\left(x + 1 - \frac{3}{2}y\right) = y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0</math>. | ||
− | + | As the answer format implies that the <math>x</math>-coordinate of the root is non-integral, <math>x(x - 1)\left(x + 1 - \frac{3}{2}y\right) = 0 \iff x + 1 - \frac{3}{2}y = 0 \iff y = \frac{2}{3}(x + 1)\ (1)</math>. The format also implies that <math>y</math> is positive, so <math>y\left(y^2 - 1 - \frac{3}{2}x(x - 1)\right) = 0 \iff y^2 - 1 - \frac{3}{2}x(x - 1) = 0\ (2)</math>. Substituting <math>(1)</math> into <math>(2)</math> and reducing to a 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>. | |
− | < | + | |
− | + | The answer is <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 + 19 = \boxed{040}</math> | ||
== See also == | == See also == |
Revision as of 13:30, 3 January 2021
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 and reducing to a 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 |
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