Difference between revisions of "1989 AIME Problems/Problem 8"
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== Solution == | == Solution == | ||
=== Solution 1=== | === Solution 1=== | ||
− | + | 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. | |
− | |||
− | < | + | 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 = | + | |
+ | 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> | ||
− | + | Comparing coefficents gives us the three equation system: | |
− | < | + | <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> | + | 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> | + | where <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>. | ||
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== 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 20:20, 5 August 2009
Problem
Assume that are real numbers such that
![$x_1+4x_2+9x_3+16x_4+25x_5+36x_6+49x_7=1$](http://latex.artofproblemsolving.com/b/b/d/bbd4ee126c1e2ba03742a2555dd91ba41c435eb0.png)
![$4x_1+9x_2+16x_3+25x_4+36x_5+49x_6+64x_7=12$](http://latex.artofproblemsolving.com/f/e/a/fea54b9584a9454d2777bb82f2393a5f447dc856.png)
![$9x_1+16x_2+25x_3+36x_4+49x_5+64x_6+81x_7=123$](http://latex.artofproblemsolving.com/7/e/1/7e12b1c45673baa64573cfa54c757a2daba05d66.png)
Find the value of .
Solution
Solution 1
Notice that because we are given a system of equations with
unknowns, the values
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 in the first equation be
; then its coefficients in the second equation is
and the third as
. We need to find a way to sum these to make
[this is in fact a specific approach generalized by the next solution below].
Thus, we hope to find constants satisfying
. FOILing out all of the terms, we get
![$[ay^2 + by^2 + cy^2] + [2by + 4cy] + b + 4c = y^2 + 6y + 9.$](http://latex.artofproblemsolving.com/d/e/0/de0a9256131a0d09b604f6d5b019ae307eff232d.png)
Comparing coefficents gives us the three equation system:
Subtracting the second and third equations yields that , so
and
. It follows that the desired expression is
.
Solution 2
Notice that we may rewrite the equations in the more compact form as:
and
where and
is what we're trying to find.
Now consider the polynomial given by (we are only treating the
as coefficients).
Notice that
is in fact a quadratic. We are given
as
and are asked to find
. 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
. Indeed
.
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 |