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Difference between revisions of "1984 AIME Problems/Problem 15"

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== Solution 1 ==
 
== Solution 1 ==
−
Rewrite the system of equations as <math> \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. </math> This equation is satisfied when <math>t = 4,16,36,64</math>, as then the equation is equivalent to the given equations.
+
Rewrite the system of equations as <cmath>\frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.</cmath>  
−
After clearing fractions, for each of the values <math>t=4,16,36,64</math>, we have the [[equation]] <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)</math>. We can move the expression <math>(t-1)(t-9)(t-25)(t-49)</math> to the left hand side to obtain the difference of the polynomials: <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)</math> and <math>(t-1)(t-9)(t-25)(t-49)</math>
+
This equation is satisfied when <math>t \in \{4, 16, 36, 64\}</math>. After clearing fractions, for each of the values <math>t=4,16,36,64</math>, we have the equation  
−
                         
+
<cmath>x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),</cmath>where <math>F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)</math> and <math>P_k(t)=F(t)/(t-k^2)</math>, for <math>k=1,3,5,7</math>.
−
Since the polynomials are equal at <math>t=4,16,36,64</math>, we can express the difference of the two polynomials with a quartic polynomial that has roots at <math>t=4,16,36,64</math>, so
 
  
−
<div style="text-align:center;"><math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = -(t-4)(t-16)(t-36)(t-64) </math>
+
Since the polynomials on each side are equal at <math>t=4,16,36,64</math>, we can express the difference of the two polynomials by a quartic polynomial that has roots at <math>t=4,16,36,64</math>, so
−
</div>
+
<cmath>\begin{align} \tag{\dag}x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)-F(t) = -(t-4)(t-16)(t-36)(t-64)
 +
\end{align}</cmath>
 +
The leading coefficient of the RHS is <math>-1</math> because the leading coefficient of the LHS is <math>-1</math>.
  
−
Note the leading coefficient of the RHS is <math>-1</math> because it must match the leading coefficient of the LHS, which is <math>-1</math>.  
+
Plug in <math>t=1^2, 3^2, 5^2, 7^2</math> in succession, into <math>(\dag)</math>. In each case, most terms drop, and we end up with
 +
<cmath>\begin{align*}
 +
x^2=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}, \quad y^2=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}},\quad z^2=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}},\quad w^2=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}
 +
\end{align*}</cmath>
 +
Adding them up we get the sum as <math>3^2\cdot 4=\boxed{036}</math>.
  
−
Now we can plug in <math>t=1</math> into the polynomial equation. Most terms drop, and we end up with
+
'''Postscript for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes <math>x^2,y^2,z^2,</math> and <math>w^2</math> separately before adding them to obtain the final answer is appealing because it gives the individual values of <math>x^2,y^2,z^2,</math> and <math>w^2</math> which can be plugged into the given equations to check.
  
−
<cmath>x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)</cmath>
+
== Solution 2 ==
−
 
+
As in Solution 1, we have <cmath>x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),</cmath>where <math>F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)</math> and <math>P_k(t)=F(t)/(t-k^2)</math>, for <math>k=1,3,5,7</math>.
−
so that
 
−
 
 
−
<cmath>x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}</cmath>
 
−
 
 
−
Similarly, we can plug in <math>t=9,25,49</math> and get
 
−
 
 
−
<cmath>\begin{align*}
 
−
y^2&=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\
 
−
z^2&=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}\\
 
−
w^2&=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}\end{align*}</cmath>
 
−
 
 
−
Now adding them up,
 
−
 
 
−
<cmath>\begin{align*}z^2+w^2&=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\
 
−
x^2+y^2&=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}\end{align*}</cmath>
 
−
 
 
−
with a sum of
 
  
−
<cmath>\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.</cmath>
+
Now the coefficient of <math>t^3</math> on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the <math>t^4</math> and <math>t^3</math> terms are, so we can eventually apply Vieta's. We can write the long equation as <cmath>(x^2 + y^2 + z^2 + w^2)t^3 + \dots = t^4 - (1^2 + 3^2 + 5^2 + 7^2)t^3 + \dots</cmath> Rearranging gives us <cmath>t^4 - (1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2)t^3 \dots = 0.</cmath> By Vieta's, we know that the sum of the roots of this equation is <cmath>1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2.</cmath> (recall that the roots of the original and this manipulated form of it had roots <math>2^2, 4^2, 6^2,</math> and <math>8^2</math>). Thus, <cmath>x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2 - 1^2 - 3^2 - 5^2 - 7^2 = \boxed{36}.</cmath>
−
 
 
−
/*Lengthy proof that any two cubic polynomials in <math>t</math> which are equal at 4 values of <math>t</math> are themselves equivalent:
 
−
Let the two polynomials be <math>A(t)</math> and <math>B(t)</math> and let them be equal at <math>t=a,b,c,d</math>. Thus we have <math>A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0</math>. Also the polynomial <math>A(t) - B(t)</math> is cubic, but it equals 0 at 4 values of <math>t</math>. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into <math>(t-a)(t-b)(t-c)(t-d)(</math>some nonzero polynomial<math>)</math> which would have a degree greater than or equal to 4, contradicting the statement that <math>A(t) - B(t)</math> is cubic. Because <math>A(t) - B(t) = 0, A(t)</math> and <math>B(t)</math> are equivalent and must be equal for all <math>t</math>.
 
−
 
 
−
'''Post script for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes <math>x^2,y^2,z^2,</math> and <math>w^2</math> separately before adding them to obtain the final answer is appealing because it gives the individual values of <math>x^2,y^2,z^2,</math> and <math>w^2</math> which can be plugged into the given equations to check.
 
−
 
 
−
== Solution 2 ==
 
−
As in Solution 1, we have
 
−
<div style="text-align:center;"><math>(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)</math> <math>-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)</math>
 
−
<math>=(t-4)(t-16)(t-36)(t-64)</math>
 
−
</div>
 
−
Now the coefficient of <math>t^3</math> on both sides must be equal. Therefore we have <math>1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}</math>.
 
  
 
== Solution 3 (Highly Unrecommended) ==
 
== Solution 3 (Highly Unrecommended) ==
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\frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\
 
\frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
You might be able to see where this is going. First off, find <math>\text{lcm}(3,5,21,45),\text{lcm}(15,7,9,33), \text{lcm}(35,27,11,13),</math> and <math>\text{lcm}(63,55,39,15)</math>. Then, multiply by the respective lcm to clear all of the denominators. Once you do that, maniuplate the equations to solve for <math>w^2+x^2+y^2+z^2</math>.
+
You might be able to see where this is going. First off, find <math>\text{lcm}(3,5,21,45),\text{lcm}(15,7,9,33), \text{lcm}(35,27,11,13),</math> and <math>\text{lcm}(63,55,39,15)</math>. Then, multiply by the respective lcm to clear all of the denominators. Once you do that, manipulate the equations to solve for <math>w^2+x^2+y^2+z^2</math>.
  
 
Now, most of this is just a brainless bash, and reemphasizing, please try to learn the above solutions. This is only a last resort and only to be used if you have too much time left. The exact amount of time this bash takes depends on the person and how quickly they can manipulate the equations.
 
Now, most of this is just a brainless bash, and reemphasizing, please try to learn the above solutions. This is only a last resort and only to be used if you have too much time left. The exact amount of time this bash takes depends on the person and how quickly they can manipulate the equations.
  
−
== Solution 4 (easy sol)==
+
==Solution 4 (Fast, Efficient)==
−
 
+
Notice how on each line, we have equations of the form <math>\frac{x^2}{a-1^2}+\frac{y^2}{a-3^2}+\frac{z^2}{a-5^2}+\frac{w^2}{a-7^2}=1</math>. We can let this be a polynomial, with respect to <math>a</math>. We can say that <math>w^2</math>, <math>x^2</math>, <math>y^2</math>, and <math>z^2</math> are coefficients, and not variables. So, we can now expand the fractions to get
−
Note that each equation is of the form <cmath>\frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1</cmath> for <math>t=2^2, 4^2, 6^2, 8^2</math>. Removing fractions, we get  
+
<math>(a-1)(a-9)(a-25)(a-49)=x^2(a-9)(a-25)(a-49)</math>
−
 
+
<math>+ y^2(a-1)(a-25)(a-49)</math>
−
<cmath>x^2(t-3^2)(t-5^2)(t-7^2)+y^2(t-1^2)(t-5^2)(t-7^2) + z^2(t-1^2)(t-3^2)(t-7^2)+w^2(t-1^2)(t-3^2)(t-5^2) = (t-1^2)(t-3^2)(t-5^2)(t-7^2).</cmath>  
+
<math>+ z^2(a-1)(a-9)(a-49)</math>
−
 
+
<math>+ w^2(a-1)(a-9)(a-25)</math>.
−
Instead of expanding it fully, we will predict what the coefficients of the <math>t^4</math> and <math>t^3</math> term would be, since that's only what's important to us. So, we can write the long equation as <cmath>(x^2 + y^2 + z^2 + w^2)t^3 + \dots = t^4 - (1^2 + 3^2 + 5^2 + 7^2)t^3 + \dots</cmath>. We rearrange this to get <cmath>t^4 - (1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2) = 0.</cmath> By Vieta's Formulas, we know that the sum of the roots of this equation is <cmath>1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2.</cmath> (recall that the roots of the original and this manipulated form of it had roots <math>2^2, 4^2, 6^2,</math> and <math>8^2.</math>) Thus, <cmath>x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2 - 1^2 - 3^2 - 5^2 - 7^2 = \boxed{36}.</cmath>
 
−
 
 
  
 +
Now, we have arrived at this huge expression, but what do we do with it?
  
−
<math>\text{-thinker123}</math>
+
Well, we can look at what we want to find - <math>x^2+y^2+z^2+w^2</math>. So, we want the sum of <math>x^2</math>, <math>y^2</math>, <math>z^2</math>, and <math>w^2</math>. Looking back to our expression, we can note how on the right hand side, the <math>a^3</math> terms add to <math>x^2+y^2+z^2+w^2</math>. Also, on the left hand side, the <math>a^3</math> coefficient is <math>-84</math> (which is achievable by Vieta's formulas rather than expanding if you want to save a few seconds). So, moving all the <math>a^3</math> terms to the left hand side, then we have that by Vieta's formulas, the sum of the roots is <math>-84-x^2-y^2-z^2-w^2=-(2^2+4^2+6^2+8^2)</math>. Then, we can solve to find that <math>x^2+y^2+z^2+w^2=120-84=\boxed{036}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 18:35, 1 August 2024

Problem

Determine $x^2+y^2+z^2+w^2$ if

$\frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1$
$\frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1$
$\frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1$
$\frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1$

Solution 1

Rewrite the system of equations as \[\frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.\] This equation is satisfied when $t \in \{4, 16, 36, 64\}$. After clearing fractions, for each of the values $t=4,16,36,64$, we have the equation \[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\]where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$, for $k=1,3,5,7$.

Since the polynomials on each side are equal at $t=4,16,36,64$, we can express the difference of the two polynomials by a quartic polynomial that has roots at $t=4,16,36,64$, so \begin{align} \tag{\dag}x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)-F(t) = -(t-4)(t-16)(t-36)(t-64) \end{align} The leading coefficient of the RHS is $-1$ because the leading coefficient of the LHS is $-1$.

Plug in $t=1^2, 3^2, 5^2, 7^2$ in succession, into $(\dag)$. In each case, most terms drop, and we end up with \begin{align*} x^2=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}, \quad y^2=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}},\quad z^2=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}},\quad w^2=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}} \end{align*} Adding them up we get the sum as $3^2\cdot 4=\boxed{036}$.

Postscript for the puzzled: This solution which is seemingly unnecessarily redundant in that it computes $x^2,y^2,z^2,$ and $w^2$ separately before adding them to obtain the final answer is appealing because it gives the individual values of $x^2,y^2,z^2,$ and $w^2$ which can be plugged into the given equations to check.

Solution 2

As in Solution 1, we have \[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\]where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$, for $k=1,3,5,7$.

Now the coefficient of $t^3$ on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the $t^4$ and $t^3$ terms are, so we can eventually apply Vieta's. We can write the long equation as \[(x^2 + y^2 + z^2 + w^2)t^3 + \dots = t^4 - (1^2 + 3^2 + 5^2 + 7^2)t^3 + \dots\] Rearranging gives us \[t^4 - (1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2)t^3 \dots = 0.\] By Vieta's, we know that the sum of the roots of this equation is \[1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2.\] (recall that the roots of the original and this manipulated form of it had roots $2^2, 4^2, 6^2,$ and $8^2$). Thus, \[x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2 - 1^2 - 3^2 - 5^2 - 7^2 = \boxed{36}.\]

Solution 3 (Highly Unrecommended)

Before starting this solution, I highly recommend never following this unless you have no idea what to do with an hour of your time. Even so, learning the above solutions will be more beneficial.

\begin{align*} \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1\\ \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1\\ \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1\\ \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1\\ \end{align*} can be rewritten as \begin{align*} \frac{x^2}{3}-\frac{y^2}{5}-\frac{z^2}{21}-\frac{w^2}{45}=1\\ \frac{x^2}{15}+\frac{y^2}{7}-\frac{z^2}{9}-\frac{w^2}{33}=1\\ \frac{x^2}{35}+\frac{y^2}{27}+\frac{z^2}{11}-\frac{w^2}{13}=1\\ \frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\ \end{align*} You might be able to see where this is going. First off, find $\text{lcm}(3,5,21,45),\text{lcm}(15,7,9,33), \text{lcm}(35,27,11,13),$ and $\text{lcm}(63,55,39,15)$. Then, multiply by the respective lcm to clear all of the denominators. Once you do that, manipulate the equations to solve for $w^2+x^2+y^2+z^2$.

Now, most of this is just a brainless bash, and reemphasizing, please try to learn the above solutions. This is only a last resort and only to be used if you have too much time left. The exact amount of time this bash takes depends on the person and how quickly they can manipulate the equations.

Solution 4 (Fast, Efficient)

Notice how on each line, we have equations of the form $\frac{x^2}{a-1^2}+\frac{y^2}{a-3^2}+\frac{z^2}{a-5^2}+\frac{w^2}{a-7^2}=1$. We can let this be a polynomial, with respect to $a$. We can say that $w^2$, $x^2$, $y^2$, and $z^2$ are coefficients, and not variables. So, we can now expand the fractions to get $(a-1)(a-9)(a-25)(a-49)=x^2(a-9)(a-25)(a-49)$ $+ y^2(a-1)(a-25)(a-49)$ $+ z^2(a-1)(a-9)(a-49)$ $+ w^2(a-1)(a-9)(a-25)$.

Now, we have arrived at this huge expression, but what do we do with it?

Well, we can look at what we want to find - $x^2+y^2+z^2+w^2$. So, we want the sum of $x^2$, $y^2$, $z^2$, and $w^2$. Looking back to our expression, we can note how on the right hand side, the $a^3$ terms add to $x^2+y^2+z^2+w^2$. Also, on the left hand side, the $a^3$ coefficient is $-84$ (which is achievable by Vieta's formulas rather than expanding if you want to save a few seconds). So, moving all the $a^3$ terms to the left hand side, then we have that by Vieta's formulas, the sum of the roots is $-84-x^2-y^2-z^2-w^2=-(2^2+4^2+6^2+8^2)$. Then, we can solve to find that $x^2+y^2+z^2+w^2=120-84=\boxed{036}$.

See also

1984 AIME (Problems • Answer Key • Resources)
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