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

Revision as of 14:00, 21 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 $c=-14$ $|f(0)| = |2(-4)^3 - 14(-4)| = 072$

Solution 3

Without loss of generality, let $f(1) = 12$ . (If $f(1) = -12$ , then take $-f(x)$ as the polynomial, which leaves $|f(0)|$ unchanged.) Because $f$ is third-degree, write $f(x) - 12 = a(x - 1)(x - b)(x - c)$ $f(x) + 12 = a(x - d)(x - e)(x - f)$ where $\{b, c, d, e, f \}$ clearly must be a permutation of $\{2, 3, 5, 6, 7\}$ from the given condition. Thus $b + c + d + e + f = 2 + 3 + 5 + 6 + 7 = 23.$ However, subtracting the two equations gives $-24 = a[(x - 1)(x - b)(x - c) - (x - d)(x - e)(x - f)]$ , so comparing $x^2$ coefficients gives $1 + b + c = d + e + f$ and thus both values equal to $\dfrac{24}{2} = 12$ . As a result, $\{b, c \} = \{5, 6 \}$ . As a result, $-24 = a (12)$ and so $a = -2$ . Now, we easily deduce that $f(0) = (-2) \cdot (-1) \cdot (-5) \cdot (-6) + 12 = 72,$ and so removing the without loss of generality gives $|f(0)| = \boxed{072}$ , which is our answer.

@@ Line 29: / Line 29: @@
 <cmath>f(x) - 12 = a(x - 1)(x - b)(x - c)</cmath>
 <cmath>f(x) + 12 = a(x - d)(x - e)(x - f)</cmath>
-where <math>\{b, c, d, e, f \}</math> clearly must be a permutation of <math>\{2, 3, 5, 6, 7\}</math> from the given condition. Thus <math>b + c + d + e + f = 2 + 3 + 5 + 6 + 7 = 23.</math> However, subtracting the two equations gives <math>-24 = a[(x - 1)(x - b)(x - c) - (x - d)(x - e)(x - f)]</math>, so comparing <math>x^2</math> coefficients gives <math>1 + b + c = d + e + f</math> and thus both values equal to <math>\dfrac{24}{2} = 12</math>. As a result, <math>\{b, c \} = \{5, 6 \}</math>. As a result, <math>-24 = a (12)</math> and so <math>a = -2</math>. Now, we easily deduce that <math>f(0) = (-2) \cdot (-1) \cdot (-5) \cdot (-6) + 12 = 72,</math> and so removing the without loss of generality gives <math>|f(0)| = \boxed(072)</math>, which is our answer.
+where <math>\{b, c, d, e, f \}</math> clearly must be a permutation of <math>\{2, 3, 5, 6, 7\}</math> from the given condition. Thus <math>b + c + d + e + f = 2 + 3 + 5 + 6 + 7 = 23.</math> However, subtracting the two equations gives <math>-24 = a[(x - 1)(x - b)(x - c) - (x - d)(x - e)(x - f)]</math>, so comparing <math>x^2</math> coefficients gives <math>1 + b + c = d + e + f</math> and thus both values equal to <math>\dfrac{24}{2} = 12</math>. As a result, <math>\{b, c \} = \{5, 6 \}</math>. As a result, <math>-24 = a (12)</math> and so <math>a = -2</math>. Now, we easily deduce that <math>f(0) = (-2) \cdot (-1) \cdot (-5) \cdot (-6) + 12 = 72,</math> and so removing the without loss of generality gives <math>|f(0)| = \boxed{072}</math>, which is our answer.
 ==See Also==
 {{AIME box|year=2015|n=I|num-b=9|num-a=11}}
 {{MAA Notice}}

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