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Difference between revisions of "2015 AIME I Problems/Problem 10"
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<cmath>216a + 36b + 6c +  d =  12</cmath>
 
<cmath>216a + 36b + 6c +  d =  12</cmath>
 
<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>
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Using any four of these functions as a system of equations yields <math>|f(0)| = \boxed{072}</math>
  
  
Line 23: Line 23:
 
<cmath>a(2)^3 + c(2) = -12</cmath>
 
<cmath>a(2)^3 + c(2) = -12</cmath>
 
<math>a=2</math> and <math>c=-14</math>
 
<math>a=2</math> and <math>c=-14</math>
−
<cmath>|f(0)| = |2(-4)^3 - 14(-4)| = 072</cmath>
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<cmath>|f(0)| = |2(-4)^3 - 14(-4)| = \boxed{072}</cmath>
  
 
==Solution 3==
 
==Solution 3==

Revision as of 11:15, 8 June 2017

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 $|f(0)| = \boxed{072}$


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 $c=-14$ \[|f(0)| = |2(-4)^3 - 14(-4)| = \boxed{072}\]

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.


Solution 4

The following solution is similar to solution 3, but assumes nothing. Let $g(x)=(f(x))^2-144$. Since $f$ has degree 3, $g$ has degree 6 and has roots 1,2,3,5,6, and 7. Therefore, $g(x)=k(x-1)(x-2)(x-3)(x-5)(x-6)(x-7)$ for some $k$. Hence $|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}$. Note that $g(x)=(f(x)+12)(f(x)-12)$. Since $f$ has degree 3, so do $f(x)+12$ and $f(x)-12$; and both have the same leading coefficient. Hence $f(x)+12=a(x-q)(x-r)(x-s)$ and $f(x)-12=a(x-t)(x-u)(x-v)$ for some $a\neq 0$ (else $f$ is not cubic) where $\{q,r,s,t,u,v\}$ is the same as the set $\{1,2,3,5,6,7\}$. Subtracting the second equation from the first, expanding, and collecting like terms, we have that \[24=a((t+u+v-(q+r+s))x^2-a(tu+uv+tv-(qr+qs+rs))x+a(tuv-qrs)\] which must hold for all $x$. Since $a\neq 0$ we have that (1) $t+u+v=q+r+s$, (2) $tu+uv+tv=qr+qs+rs$ and (3) $a(tuv-qrs)=24$. Since $q+r+s+t+u+v$ is the sum of 1,2,3,5,6, and 7, we have $q+r+s+t+u+v=24$ so that by (1) we have $q+r+s=12$ and $t+u+v=12$. We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be $\geq 12$ with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be $\{2,3,7\}$ and $\{1,5,6\}$. Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that $|a|=\frac{24}{|tuv-qrs|}=\frac{24}{12}=2$. Since $a$ is the leading coefficient of $f(x)$, the leading coefficient of $(f(x))^2$ is $a^2=|a|^2=2^2=4$. Thus the leading coefficient of $g(x)$ is 4, i.e. $k=4$. Then from earlier, $|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}=\sqrt{1260\cdot4+144}=\sqrt{5184}=72$ so that the answer is $\boxed{072}$.

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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