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

Latest revision as of 11:10, 15 December 2024

Problem

Assume that $x_1,x_2,\ldots,x_7$ are real numbers such that $\begin{align*} x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5 + 36x_6 + 49x_7 &= 1, \\ 4x_1 + 9x_2 + 16x_3 + 25x_4 + 36x_5 + 49x_6 + 64x_7 &= 12, \\ 9x_1 + 16x_2 + 25x_3 + 36x_4 + 49x_5 + 64x_6 + 81x_7 &= 123. \end{align*}$ Find the value of $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7$ .

Solution 1 (Quadratic Function)

Note that each given equation is of the form $f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7$ for some $k\in\{1,2,3\}.$

When we expand $f(k)$ and combine like terms, we obtain a quadratic function of $k:$ $f(k)=ak^2+bk+c,$ where $a,b,$ and $c$ are linear combinations of $x_1,x_2,x_3,x_4,x_5,x_6,$ and $x_7.$

We are given that $\begin{alignat*}{10} f(1)&=\phantom{42}a+b+c&&=1, \\ f(2)&=4a+2b+c&&=12, \\ f(3)&=9a+3b+c&&=123, \end{alignat*}$ and we wish to find $f(4).$

We eliminate $c$ by subtracting the first equation from the second, then subtracting the second equation from the third: $\begin{align*} 3a+b&=11, \\ 5a+b&=111. \end{align*}$ By either substitution or elimination, we get $a=50$ and $b=-139.$ Substituting these back produces $c=90.$

Finally, the answer is $f(4)=16a+4b+c=\boxed{334}.$

~Azjps ~MRENTHUSIASM

Solution 2 (Linear Combination)

For simplicity purposes, we number the given equations $(1),(2),$ and $(3),$ in that order. Let $16x_1+25x_2+36x_3+49x_4+64x_5+81x_6+100x_7=S. \hspace{29.5mm}(4)$ Subtracting $(1)$ from $(2),$ subtracting $(2)$ from $(3),$ and subtracting $(3)$ from $(4),$ we obtain the following equations, respectively: $\begin{align*} 3x_1 + 5x_2 + 7x_3 + 9x_4 + 11x_5 + 13x_6 + 15x_7 &=11, \hspace{20mm}&(5) \\ 5x_1 + 7x_2 + 9x_3 + 11x_4 + 13x_5 + 15x_6 + 17x_7 &=111, &(6) \\ 7x_1 + 9x_2 + 11x_3 + 13x_4 + 15x_5 + 17x_6 + 19x_7 &=S-123. &(7) \\ \end{align*}$ Subtracting $(5)$ from $(6)$ and subtracting $(6)$ from $(7),$ we obtain the following equations, respectively: $\begin{align*} 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=100, &(8) \\ 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=S-234. \hspace{20mm}&(9) \end{align*}$ Finally, applying the Transitive Property to $(8)$ and $(9)$ gives $S-234=100,$ from which $S=\boxed{334}.$

~Duohead ~MRENTHUSIASM

Solution 3 (Finite Differences by Arithmetic)

Note that the second differences of all quadratic sequences must be constant (but nonzero). One example is the following sequence of perfect squares:

$[asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=1; i<=10; ++i) { label("\boldmath{$"+string(i^2)+"$}",(i-1,0)); } for (real i=1; i<=9; ++i) { label("$"+string(1+2*i)+"$",(i-0.5,-0.75)); } for (real i=1; i<=8; ++i) { label("$2$",(i,-1.5)); } for (real i=1; i<=9; ++i) { draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); } for (real i=1; i<=8; ++i) { draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); } for (real i=1; i<=9; ++i) { draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); } for (real i=1; i<=8; ++i) { draw((0.1+i,-1.35)--(0.4+i,-0.9),red); } label("\textbf{First Differences}",(-0.75,-0.75),align=W); label("\textbf{Second Differences}",(-0.75,-1.5),align=W); [/asy]$

Label equations $(1),(2),(3),$ and $(4)$ as Solution 2 does. Since the coefficients of $x_1,x_2,x_3,x_4,x_5,x_6,x_7,$ or $(1,4,9,16),(4,9,16,25),(9,16,25,36),(16,25,36,49),(25,36,49,64),(36,49,64,81),(49,64,81,100),$ respectively, all form quadratic sequences with second differences $2,$ we conclude that the second differences of equations $(1),(2),(3),(4)$ must be constant.

It follows that the second differences of $(1,12,123,S)$ must be constant, as shown below:

$[asy] /* Made by MRENTHUSIASM */ size(10cm); label("\boldmath{$1$}",(0,0)); label("\boldmath{$12$}",(1,0)); label("\boldmath{$123$}",(2,0)); label("\boldmath{$S$}",(3,0)); label("$11$",(0.5,-0.75)); label("$111$",(1.5,-0.75)); label("$d_1$",(2.5,-0.75)); label("$100$",(1,-1.5)); label("$d_2$",(2,-1.5)); for (real i=1; i<=3; ++i) { draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); } for (real i=1; i<=2; ++i) { draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); } for (real i=1; i<=3; ++i) { draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); } for (real i=1; i<=2; ++i) { draw((0.1+i,-1.35)--(0.4+i,-0.9),red); } label("\textbf{First Differences}",(-0.75,-0.75),align=W); label("\textbf{Second Differences}",(-0.75,-1.5),align=W); [/asy]$

Finally, we have $d_2=100,$ from which $\begin{align*} S&=123+d_1 \\ &=123+(111+d_2) \\ &=\boxed{334}. \end{align*}$ ~MRENTHUSIASM

Solution 4 (Finite Differences by Algebra)

Notice that we may rewrite the equations in the more compact form as: $\begin{align*} \sum_{i=1}^{7}i^2x_i&=c_1, \\ \sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\ \sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\ \sum_{i=1}^{7}(i+3)^2x_i&=c_4, \end{align*}$ where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we are trying to find.

Now consider the polynomial given by $f(z) = \sum_{i=1}^7 (i+z)^2x_i$ (we are only treating the $x_i$ as coefficients).

Notice that $f$ is in fact a quadratic. We are given $f(0)=c_1, f(1)=c_2, f(2)=c_3$ and are asked to find $f(3)=c_4$ . 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 $c_4=\boxed{334}$ .

Alternatively, applying finite differences, one obtains $c_4 = {3 \choose 2}f(2) - {3 \choose 1}f(1) + {3 \choose 0}f(0) =334.$

Solution 5 (Assumption)

The idea is to multiply the first, second and third equations by $a,b,$ and $c,$ respectively.

We can only consider the coefficients of $x_1,x_2,$ and $x_3:$ $\begin{align} a+4b+9c&=16, \\ 4a+9b+16c&=25, \\ 9a+16b+25c&=36. \end{align}$ Subtracting $(1)$ from $(2),$ we get $3a+5b+7c=9. \hspace{15mm}(4)$ Subtracting $3\cdot(4)$ from $(3),$ we get $b+4c=9. \hspace{25.5mm}(5)$ Subtracting $(1)$ from $4\cdot(5),$ we get $7c-a=20. \hspace{23mm}(6)$ From $(5)$ and $(6),$ we have $(a,b,c)=(7c-20,9-4c,c).$ Substituting this into $(2)$ gives $(a,b,c)=(1,-3,3).$

Therefore, the answer is $1\cdot1+12\cdot(-3) + 123\cdot3 = \boxed{334}.$

Solution 6 (Assumption)

We let $(x_4,x_5,x_6,x_7)=(0,0,0,0)$ . Thus, we have $\begin{align*} x_1+4x_2+9x_3&=1,\\ 4x_1+9x_2+16x_3&=12,\\ 9x_1+16x_2+25x_3&=123.\\ \end{align*}$ Grinding this out, we have $(x_1,x_2,x_3)=\left(\frac{797}{4},-229,\frac{319}{4}\right)$ which gives $\boxed{334}$ as our final answer.

~Pleaseletmewin

Solution 7 (Similar to Solutions 3 and 4)

Let $s_n = n^2$ be the sequence of perfect squares. By either expanding or via finite differences, one can prove the miraculous recursion $s_n = 3s_{n-1} - 3s_{n-2} + s_{n-3}.$ Hence, the answer is simply $3 \cdot 123 - 3 \cdot 12 + 1 = \boxed{334}.$

I first saw this in a Mathologer video: https://www.youtube.com/watch?v=4AuV93LOPcE

~Ritwin

Note: You can prove this recursion through considering the coefficients for each $x_i$ as $a^2, (a+1)^2, (a+2)^2$ for the equations respectively.

All we are left to do is find $(a+3)^2$ in terms of the above $3$ quadratics. Because we have an $a^2$ as one, we don't have to worry about the $a^2$ term, we can just manipulate it by adding or subtracting any number of the first quadratic.

We are left to find a combination of the second and third quadratics that have the linear term and constant of $(a+3)^2$ . With little experimentation we find that $3(a+2)^2 - 3(a+1)^2 = 6a+9$ . All we have to add back is one $a^2$ .

Thus, the recursion holds. That for any term $x_i$ the coefficients would be recursible by $a^2$ = $3(a-1)^2 - 3(a-2)^2 + (a-3)$ (Where we have set $a$ as $a-3$ )

~Avi_2009

Video Solution

https://www.youtube.com/watch?v=4mOROTEkvWI ~ MathEx

@@ Line 190: / Line 190: @@
 ~Ritwin
+Note: You can prove this recursion through considering the coefficients for each <math>x_i</math> as <math>a^2, (a+1)^2, (a+2)^2</math> for the equations respectively.
+All we are left to do is find <math>(a+3)^2</math> in terms of the above <math>3</math> quadratics. Because we have an <math>a^2</math> as one, we don't have to worry about the <math>a^2</math> term, we can just manipulate it by adding or subtracting any number of the first quadratic.
+We are left to find a combination of the second and third quadratics that have the linear term and constant of <math>(a+3)^2</math>. With little experimentation we find that <math>3(a+2)^2 - 3(a+1)^2 = 6a+9</math>. All we have to add back is one <math>a^2</math>.
+Thus, the recursion holds. That for any term <math>x_i</math> the coefficients would be recursible by <math>a^2</math> = <math>3(a-1)^2 - 3(a-2)^2 + (a-3)</math>(Where we have set <math>a</math> as <math>a-3</math>)
+~Avi_2009
 ==Video Solution==

1989 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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