Difference between revisions of "2015 AIME I Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | Let <math>f(x)</math> = ax^3+bx^2+cx+d. | + | Let <math>f(x)</math> = <math>ax^3+bx^2+cx+d</math>. |
| − | 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. | + | Since <math>f(x)</math> 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. | 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. | ||
| − | + | <cmath> a + b + c + d = 12</cmath> | |
| − | + | <cmath>8a + 4b + 2c + d = -12</cmath> | |
| − | + | <cmath> 27a + 9b + 3c + d = -12</cmath> | |
| − | 125a + 25b + 5c + d = 12 | + | <cmath>125a + 25b + 5c + d = 12</cmath> |
| − | 216a + 36b + 6c + d = 12 | + | <cmath>216a + 36b + 6c + d = 12</cmath> |
| − | 343a + 49b + 7c + d = -12 | + | <cmath>343a + 49b + 7c + d = -12</cmath> |
| − | Using any four of these functions as a system of equations yields f(0) = 072 | + | Using any four of these functions as a system of equations yields <math>f(0) = 072</math> |
==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 21:11, 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 = 
.
Since 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.
![\[a + b + c + d = 12\]](http://latex.artofproblemsolving.com/3/1/9/3196472c2defbbe3508f342d660983b2f7d75aaf.png)
![\[8a + 4b + 2c + d = -12\]](http://latex.artofproblemsolving.com/5/6/4/56405fc36b2a09d2590e0fff854acc1e3309754f.png)
![\[27a + 9b + 3c + d = -12\]](http://latex.artofproblemsolving.com/9/3/c/93c7a606f24ba035e2c6db5a2eab61e38d9527f1.png)
![\[125a + 25b + 5c + d = 12\]](http://latex.artofproblemsolving.com/4/b/0/4b031ff14c2f11bd36c3bfb04bdd71838ad02d43.png)
![\[216a + 36b + 6c + d = 12\]](http://latex.artofproblemsolving.com/c/a/1/ca17f15d180459c9cd2a0d9aa1cf4226bf02b8bc.png)
![\[343a + 49b + 7c + d = -12\]](http://latex.artofproblemsolving.com/e/e/4/ee43b79d15ef3b9c805f837e16b3ff27c13588a9.png)
Using any four of these functions as a system of equations yields 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
