2015 AIME I Problems/Problem 10: Difference between revisions
Created page with "==Problem== Let <math>f(x)</math> be a third-degree polynomial with real coefficients satisfying <cmath>|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.</cmath> Find <math>|f(0)|..." |
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Let <math>f(x)</math> be a third-degree polynomial with real coefficients satisfying | Let <math>f(x)</math> be a third-degree polynomial with real coefficients satisfying | ||
<cmath>|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.</cmath> Find <math>|f(0)|</math>. | <cmath>|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.</cmath> Find <math>|f(0)|</math>. | ||
==Solution== | |||
Let <math>f(x)</math> = ax^3+bx^2+cx+d. | |||
Since f(x) is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. | |||
By drawing a coordinate axis, and two lines representing 12 and -12, it is easy to see that f(1)=f(5)=f(6), and f(2)=f(3)=f(7); otherwise more bends would be required in the graph. Since only the absolute value of f(0) is required, there is no loss of generalization by stating that f(1)=12, and f(2)=-12. This provides the following system of equations. | |||
a + b + c + d = 12 | |||
8a + 4b + 2c + d = -12 | |||
27a + 9b + 3c + d = -12 | |||
125a + 25b + 5c + d = 12 | |||
216a + 36b + 6c + d = 12 | |||
343a + 49b + 7c + d = -12 | |||
Using any four of these functions as a system of equations yields f(0) = 072 | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2015|n=I|num-b=9|num-a=11}} | {{AIME box|year=2015|n=I|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:16, 20 March 2015
Problem
Let 
be a third-degree polynomial with real coefficients satisfying
![\[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\]](http://latex.artofproblemsolving.com/e/e/9/ee9f663562febc0fa6843c525e84235d42aaf6c0.png)
Find 
.
Solution
Let = ax^3+bx^2+cx+d.
Since f(x) is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up.
By drawing a coordinate axis, and two lines representing 12 and -12, it is easy to see that f(1)=f(5)=f(6), and f(2)=f(3)=f(7); otherwise more bends would be required in the graph. Since only the absolute value of f(0) is required, there is no loss of generalization by stating that f(1)=12, and f(2)=-12. This provides the following system of equations.
a + b + c + d = 12 8a + 4b + 2c + d = -12 27a + 9b + 3c + d = -12
125a + 25b + 5c + d = 12 216a + 36b + 6c + d = 12 343a + 49b + 7c + d = -12 Using any four of these functions as a system of equations yields f(0) = 072
See Also
| 2015 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America. 
