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Difference between revisions of "2015 AIME I Problems/Problem 10"

m (→‎Solution)
(→‎Solution)
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<cmath>343a + 49b + 7c +  d = -12</cmath>
 
<cmath>343a + 49b + 7c +  d = -12</cmath>
 
Using any four of these functions as a system of equations yields <math>f(0) = 072</math>
 
Using any four of these functions as a system of equations yields <math>f(0) = 072</math>
 +
 +
 +
==Solution 2==
 +
Express <math>f(x)</math> in terms of powers of <math>(x-4)</math>:
 +
<cmath>f(x) = a(x-4)^3 + b(x-4)^2 + c(x-4) + d</cmath>
 +
By the same argument as in the first Solution, we see that <math>f(x)</math> is an odd function about the line <math>x=4</math>, so its coefficients <math>b</math> and <math>d</math> are 0. From there it is relatively simple to solve <math>f(2)=f(3)=-12</math> (as in the above solution, but with a smaller system of equations):
 +
<cmath>a(1)^3 + c(1) = -12</cmath>
 +
<cmath>a(2)^3 + c(2) = -12</cmath>
 +
<math>a=2</math> and <math>c=-14</math>
 +
<cmath>|f(0)| = |2(-4)^3 - 14(-4)| = 072</cmath>
  
 
==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 21:23, 20 March 2015

Problem

Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$.

Solution

Let $f(x)$ = $ax^3+bx^2+cx+d$. 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 $f(0) = 072$


Solution 2

Express $f(x)$ in terms of powers of $(x-4)$: \[f(x) = a(x-4)^3 + b(x-4)^2 + c(x-4) + d\] By the same argument as in the first Solution, we see that $f(x)$ is an odd function about the line $x=4$, so its coefficients $b$ and $d$ are 0. From there it is relatively simple to solve $f(2)=f(3)=-12$ (as in the above solution, but with a smaller system of equations): \[a(1)^3 + c(1) = -12\] \[a(2)^3 + c(2) = -12\] $a=2$ and $c=-14$ \[|f(0)| = |2(-4)^3 - 14(-4)| = 072\]

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

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