Difference between revisions of "2007 AIME I Problems/Problem 8"
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The first two equations show that <math>m - n = 29 - \frac{43}{2} = \frac{15}{2}</math>. The last two equations show that <math>\frac{m}{n} = -2</math>. Solving these show that <math>m = 5</math> and that <math>n = -\frac{5}{2}</math>. Substituting back into the equations, we eventually find that <math>k = 30</math>. | The first two equations show that <math>m - n = 29 - \frac{43}{2} = \frac{15}{2}</math>. The last two equations show that <math>\frac{m}{n} = -2</math>. Solving these show that <math>m = 5</math> and that <math>n = -\frac{5}{2}</math>. Substituting back into the equations, we eventually find that <math>k = 30</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Since <math>Q_1(x)</math> and <math>Q_2(x)</math> are both factors of <math>P(x)</math>, which is cubic, we know the other factors associated with each of <math>Q_1(x)</math> and <math>Q_2(x)</math> must be linear. Let <math>Q_1(x)R(x) = Q_2(x)S(x) = P(x)</math>, where <math>R(x) = ax + b</math> and <math>S(x) = cx + d</math>. Then we have that <math>((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)</math>. Equating coefficients, we get the following system of equations: | ||
+ | |||
+ | <cmath>\begin{align} | ||
+ | a = 2c \\ | ||
+ | b = -d \\ | ||
+ | 2c(k - 29) - d = c(2k - 43) + 2d \\ | ||
+ | -d(k - 29) - 2ck = d(2k - 43) + ck | ||
+ | \end{align}</cmath> | ||
+ | |||
+ | Using equations <math>(1)</math> and <math>(2)</math> to make substitutions into equation <math>(3)</math>, we see that the <math>k</math>'s drop out and we're left with <math>d = -5c</math>. Substituting this expression for <math>d</math> into equation <math>(4)</math> and solving, we see that <math>k</math> must be <math>\boxed {30}</math>. | ||
+ | |||
+ | ~ anellipticcurveoverq | ||
== See also == | == See also == |
Revision as of 13:09, 4 July 2020
Problem
The polynomial is cubic. What is the largest value of for which the polynomials and are both factors of ?
Solution
Solution 1
We can see that and must have a root in common for them to both be factors of the same cubic.
Let this root be .
We then know that is a root of , so .
We then know that is a root of so we get: or , so is the highest.
We can trivially check into the original equations to find that produces a root in common, so the answer is .
Solution 2
Again, let the common root be ; let the other two roots be and . We can write that and that .
Therefore, we can write four equations (and we have four variables), , , , and .
The first two equations show that . The last two equations show that . Solving these show that and that . Substituting back into the equations, we eventually find that .
Solution 3
Since and are both factors of , which is cubic, we know the other factors associated with each of and must be linear. Let , where and . Then we have that . Equating coefficients, we get the following system of equations:
Using equations and to make substitutions into equation , we see that the 's drop out and we're left with . Substituting this expression for into equation and solving, we see that must be .
~ anellipticcurveoverq
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.