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

(→‎Solution 5 (fast))
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Using any four of these functions as a system of equations yields <math>|f(0)| = \boxed{072}</math>
 
Using any four of these functions as a system of equations yields <math>|f(0)| = \boxed{072}</math>
  
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==Solution 2==
+
 
−
Express <math>f(x)</math> in terms of powers of <math>(x-4)</math>:
+
==Solution 2 (fast)==
−
<cmath>f(x) = a(x-4)^3 + b(x-4)^2 + c(x-4) + d</cmath>
+
By drawing the function, and similar to Solution 1, WLOG let <math>f(1)=f(5)=f(6)=12</math>. Then, <math>f(2)=f(3)=f(7)</math>.
−
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):
+
Set <math>g(x)+12=f(x)</math>. Then the roots of <math>g(x)</math> are <math>1,5,6</math>. So, <math>g(x)=a(x-1)(x-5)(x-6)</math>. Plug in <math>x=2</math> to find a. We know <math>-24=-12-12=f(2)-12=g(2)=a(1)(-3)(-4)=12a</math>. So, <math>a=-2</math>. Thus, <math>f(x)=g(x)+12=-2(x-1)(x-5)(x-6)+12</math>, and then <math>|f(0)|=60+12=\boxed{072}</math>.
−
<cmath>a(1)^3 + c(1) = -12</cmath>
+
 
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<cmath>a(2)^3 + c(2) = -12</cmath>
 
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<math>a=2</math> and <math>c=-14</math>
 
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<cmath>|f(0)| = |2(-4)^3 - 14(-4)| = \boxed{072}</cmath>
 
  
 
==Solution 3==
 
==Solution 3==
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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 (fast)==
+
 
−
By drawing the function, and similar to Solution 1, WLOG let <math>f(1)=f(5)=f(6)=12</math>. Then, <math>f(2)=f(3)=f(7)</math>.
+
==Solution 5==
−
Set <math>g(x)+12=f(x)</math>. Then the roots of <math>g(x)</math> are <math>1,5,6</math>. So, <math>g(x)=a(x-1)(x-5)(x-6)</math>. Plug in <math>x=2</math> to find a. We know <math>-24=-12-12=f(2)-12=g(2)=a(1)(-3)(-4)=12a</math>. So, <math>a=-2</math>. Thus, <math>f(x)=g(x)+12=-2(x-1)(x-5)(x-6)+12</math>, and then <math>|f(0)|=60+12=\boxed{072}</math>.
+
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)| = \boxed{072}</cmath>
 +
 
  
 
==Solution 6 (Finite differences)==
 
==Solution 6 (Finite differences)==

Revision as of 11:19, 14 December 2020

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 (fast)

By drawing the function, and similar to Solution 1, WLOG let $f(1)=f(5)=f(6)=12$. Then, $f(2)=f(3)=f(7)$. Set $g(x)+12=f(x)$. Then the roots of $g(x)$ are $1,5,6$. So, $g(x)=a(x-1)(x-5)(x-6)$. Plug in $x=2$ to find a. We know $-24=-12-12=f(2)-12=g(2)=a(1)(-3)(-4)=12a$. So, $a=-2$. Thus, $f(x)=g(x)+12=-2(x-1)(x-5)(x-6)+12$, and then $|f(0)|=60+12=\boxed{072}$.


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

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 6 (Finite differences)

Because a cubic must come in a "wave form" with two "humps" (Called points of inflections) , we can see that $f(1)=f(5)=f(6)$, and $f(2)=f(3)=f(7)$. By symmetry, $f(4)=0$. Now, WLOG let $f(1)=12$, and $f(2)=f(3)=-12$. Then, we can use finite differences to get that the third (constant) difference is $12$, and therefore $f(0)=12+(24+(24+12))=\boxed{072}$.

Solution 7 (Like solution 1 without annoying systems)

We can rewrite our function as two different cubics, $f(x)=k(x-a)(x-b)(x-c)\pm12=k(x-d)(x-e)(x-f)\mp12$. Note that $k$ is the same in both because they must be equal, so their leading terms must be. We must then, following Vieta's, choose our roots such that $a+b+c=d+e+f$ and verify that the other Vieta's formulas hold. Additionally, a cubic must only cross the x-axis thrice, restricting our choices for roots further. Choosing $a=1$, $b=5$, $c=6$, $d=2$, $e=3$, $f=7$ yields: \[kx^3-12kx^2+41kx-30k\pm12=kx^3-12kx^2+41kx-42k\mp12\] For the constant terms to have a difference of 24 ($|\pm12-\mp12|$), we must have $k=\pm2$, so the constant term of our polynomial is $\pm72$, the absolute value of which is $\boxed{072}$. -- Solution by eiis1000

Solution 8(godspeed)

Trivially, assume the function can be written as $f(x) = (x-4)(g(x))$, where $g(x)$ is quadratic. Then we set up the following equations using the coefficients of $g(x)$.

$a + b + c = 4$

$4a + 2b + c = -6$

$9a + 3b + c = -12$

Subtract the first equation from the second and the second from the third to get a system of two variables. Solve and plug it in to get $g(x) = 2x^2-16x+18$.$|f(0)| = |18*-4| = | -72| = \boxed{072}$

See Also

2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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