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Difference between revisions of "1989 AIME Problems/Problem 8"

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== Solution ==
 
== Solution ==
 
=== Solution 1===
 
=== Solution 1===
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Let us try to derive a way to find the last expression in terms of the three given expressions. The coefficients of <math>x_i</math> in the first equation can be denoted as <math>y^2</math>, making its coefficients in the second equation as <math>(y+1)^{2}</math> and the third as <math>(y+2)^2</math>. We need to find a way to sum them up to make <math>(y+3)^2</math>.  
+
Notice that because we are given a system of <math>3</math> equations with <math>7</math> unknowns, the values <math>(x_1, x_2, \ldots, x_7)</math> are not fixed; indeed one can take any four of the variables and assign them arbitrary values, which will in turn fix the last three.  
  
−
Thus, we can write that <math>ay^2 + b(y+1)^2 + c(y+2)^2 = (y + 3)^2</math>. [[FOIL]]ing out all of the terms, we get <math>ay^2 + by^2 + cy^2 + 2by + 4cy + b + 4c = y^2 + 6y + 9</math>. We can set up the three equation system:
 
  
−
<center>
+
Given this, we suspect there is a way to derive the last expression as a linear combination of the three given expressions. Let the coefficent of <math>x_i</math> in the first equation be <math>y_i^2</math>; then its coefficients in the second equation is <math>(y_i+1)^{2}</math> and the third as <math>(y_i+2)^2</math>. We need to find a way to sum these to make <math>(y_i+3)^2</math> [this is in fact a specific approach generalized by the next solution below].
−
<math>a + b + c = 1</math>
+
 
 +
Thus, we hope to find constants <math>a,b,c</math> satisfying <math>ay_i^2 + b(y_i+1)^2 + c(y_i+2)^2 = (y_i + 3)^2</math>. [[FOIL]]ing out all of the terms, we get
 +
 
 +
<center><math>[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.</math></center>  
  
−
<math>2b + 4c = 6</math>
+
Comparing coefficents gives us the three equation system:
  
−
<math>b + 4c = 9</math>
+
<center>
 +
<cmath>\begin{align*}a + b + c &= 1 \\ 2b + 4c &= 6 \\ b + 4c &= 9 \end{align*}</cmath>
 
</center>  
 
</center>  
  
−
Subtracting the second and third equations yields that <math>e = -3</math>, so <math>f = 3</math> and <math>d = 1</math>. Thus, we have to add <math>d \cdot 1 + e \cdot 12 + f \cdot 123 = 1 - 36 + 369 = \boxed{334}</math>.
+
Subtracting the second and third equations yields that <math>b = -3</math>, so <math>c = 3</math> and <math>a = 1</math>. It follows that the desired expression is <math>a \cdot (1) + b \cdot (12) + c \cdot (123)  = 1 - 36 + 369 = \boxed{334}</math>.
  
 
=== Solution 2===
 
=== Solution 2===
−
Notice that we may rewrite the equations in the more compact form:
+
Notice that we may rewrite the equations in the more compact form as:
 +
 
 +
<math>\sum_{i=1}^{7}i^2x_i=c_1,\ \ \sum_{i=1}^{7}(i+1)^2x_i=c_2,\ \ \sum_{i=1}^{7}(i+2)^2x_i=c_3,</math> and <math>\sum_{i=1}^{7}(i+3)^2x_i=c_4,</math>
  
−
<math>\sum_{i=1}^{7}(2k_1+1)^2x_i=c_1, \sum_{i=1}^{7}(2k_2+1)^2x_i=c_2, \sum_{i=1}^{7}(2k_3+1)^2x_i=c_3,</math> and <math>\sum_{i=1}^{7}(2k_4+1)^2x_i=c_4</math>
+
where <math>c_1=1, c_2=12, c_3=123,</math> and <math>c_4</math> is what we're trying to find.
  
−
, where <math>k_1=i-1, \ k_2=i, \ k_3=i+1, \ k_4=i+2</math> and <math>c_1=1, c_2=12, c_3=123,</math> and <math>c_4</math> is what we're trying to find.
+
Now consider the polynomial given by <math> f(z) := \sum_{i=1}^7 (z+i)^2x_i </math> (we are only treating the <math>x_i</math> as coefficients).
 +
Notice that <math>f</math> is in fact a quadratic. We are given <math>f(0), \ f(1), \ f(2)</math> as <math>c_1, \ c_2, \ c_3</math> and are asked to find <math>c_4</math>. Using the concept of [[finite differences]] (a prototype of [[differentiation]]) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find <math>c_4=334</math>. Indeed <math>c_4 = f(2)+[((f(2) - f(1))-(f(1)-f(0)))+(f(2)-f(1))]={3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334</math>.
  
−
Now undergo a paradigm shift: consider the polynomial in <math>k: \ f(k):= \sum_{i=1}^7 (2(k+i)-1)^2x_i </math> (we are only treating the <math>x_i</math> as coefficients).
 
−
Notice that the degree of <math>f</math> must be <math>2</math>; it is a quadratic. We are given <math>f(1), \ f(2), \ f(3)</math> as <math>c_1, \ c_2, \ c_3</math> and are asked to find <math>c_4</math>. Using the concept of [[finite differences]] (a prototype of [[differentiation]]) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find <math>c_4=334</math>.
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=1989|num-b=7|num-a=9}}
 
{{AIME box|year=1989|num-b=7|num-a=9}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 19:20, 5 August 2009

Problem

Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that

$x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7=1$
$4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7=12$
$9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7=123$

Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$.

Solution

Solution 1

Notice that because we are given a system of $3$ equations with $7$ unknowns, the values $(x_1, x_2, \ldots, x_7)$ are not fixed; indeed one can take any four of the variables and assign them arbitrary values, which will in turn fix the last three.


Given this, we suspect there is a way to derive the last expression as a linear combination of the three given expressions. Let the coefficent of $x_i$ in the first equation be $y_i^2$; then its coefficients in the second equation is $(y_i+1)^{2}$ and the third as $(y_i+2)^2$. We need to find a way to sum these to make $(y_i+3)^2$ [this is in fact a specific approach generalized by the next solution below].

Thus, we hope to find constants $a,b,c$ satisfying $ay_i^2 + b(y_i+1)^2 + c(y_i+2)^2 = (y_i + 3)^2$. FOILing out all of the terms, we get

$[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.$

Comparing coefficents gives us the three equation system:

\begin{align*}a + b + c &= 1 \\ 2b + 4c &= 6 \\ b + 4c &= 9 \end{align*}

Subtracting the second and third equations yields that $b = -3$, so $c = 3$ and $a = 1$. It follows that the desired expression is $a \cdot (1) + b \cdot (12) + c \cdot (123) = 1 - 36 + 369 = \boxed{334}$.

Solution 2

Notice that we may rewrite the equations in the more compact form as:

$\sum_{i=1}^{7}i^2x_i=c_1,\ \ \sum_{i=1}^{7}(i+1)^2x_i=c_2,\ \ \sum_{i=1}^{7}(i+2)^2x_i=c_3,$ and $\sum_{i=1}^{7}(i+3)^2x_i=c_4,$

where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we're trying to find.

Now consider the polynomial given by $f(z) := \sum_{i=1}^7 (z+i)^2x_i$ (we are only treating the $x_i$ as coefficients). Notice that $f$ is in fact a quadratic. We are given $f(0), \ f(1), \ f(2)$ as $c_1, \ c_2, \ c_3$ and are asked to find $c_4$. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find $c_4=334$. Indeed $c_4 = f(2)+[((f(2) - f(1))-(f(1)-f(0)))+(f(2)-f(1))]={3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334$.

See also

1989 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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