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

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which must hold for all <math>x</math>. Since <math>a\neq 0</math> we have that (1) <math>t+u+v=q+r+s</math>, (2) <math>tu+uv+tv=qr+qs+rs</math> and (3) <math>a(tuv-qrs)=24</math>. Since <math>q+r+s+t+u+v</math> is the sum of 1,2,3,5,6, and 7, we have <math>q+r+s+t+u+v=24</math> so that by (1) we have <math>q+r+s=12</math> and <math>t+u+v=12</math>. 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 <math>\geq 12</math> 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 <math>\{2,3,7\}</math> and <math>\{1,5,6\}</math>. Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that <math>|a|=\frac{24}{|tuv-qrs|}=\frac{24}{12}=2</math>. Since <math>a</math> is the leading coefficient of <math>f(x)</math>, the leading coefficient of <math>(f(x))^2</math> is <math>a^2=|a|^2=2^2=4</math>. Thus the leading coefficient of <math>g(x)</math> is 4, i.e. <math>k=4</math>. Then from earlier, <math>|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}=\sqrt{1260\cdot4+144}=\sqrt{5184}=72</math> so that the answer is <math>\boxed{072}</math>.
 
which must hold for all <math>x</math>. Since <math>a\neq 0</math> we have that (1) <math>t+u+v=q+r+s</math>, (2) <math>tu+uv+tv=qr+qs+rs</math> and (3) <math>a(tuv-qrs)=24</math>. Since <math>q+r+s+t+u+v</math> is the sum of 1,2,3,5,6, and 7, we have <math>q+r+s+t+u+v=24</math> so that by (1) we have <math>q+r+s=12</math> and <math>t+u+v=12</math>. 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 <math>\geq 12</math> 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 <math>\{2,3,7\}</math> and <math>\{1,5,6\}</math>. Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that <math>|a|=\frac{24}{|tuv-qrs|}=\frac{24}{12}=2</math>. Since <math>a</math> is the leading coefficient of <math>f(x)</math>, the leading coefficient of <math>(f(x))^2</math> is <math>a^2=|a|^2=2^2=4</math>. Thus the leading coefficient of <math>g(x)</math> is 4, i.e. <math>k=4</math>. Then from earlier, <math>|f(0)|=\sqrt{g(0)+144}=\sqrt{1260k+144}=\sqrt{1260\cdot4+144}=\sqrt{5184}=72</math> so that the answer is <math>\boxed{072}</math>.
  
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==Solution 5==
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By drawing the function, WLOG let <math>f(1)=f(5)=f(6)=12</math>. Then, <math>f(2)=f(3)=f(7)</math>. Realize that if we shift <math>f(x)</math> down 12, then this function <math>f(x)-12</math> has roots <math>1,5,6</math> with leading coefficient <math>-2</math> because <math>f(2)-12=-24=-2(1)(-3)(-4)</math>. Therefore <math>f(x)=-2(x-1)(x-5)(x-6)+12</math>, and then <math>|f(0)|=60+12=\boxed{72}</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}}

Revision as of 18:14, 7 October 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 $\boxed{072}$.

Solution 5

By drawing the function, WLOG let $f(1)=f(5)=f(6)=12$. Then, $f(2)=f(3)=f(7)$. Realize that if we shift $f(x)$ down 12, then this function $f(x)-12$ has roots $1,5,6$ with leading coefficient $-2$ because $f(2)-12=-24=-2(1)(-3)(-4)$. Therefore $f(x)=-2(x-1)(x-5)(x-6)+12$, and then $|f(0)|=60+12=\boxed{72}$.

See Also

2015 AIME I (ProblemsAnswer KeyResources)
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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