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

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== Problem ==
 
== Problem ==
Determine <math>\displaystyle w^2+x^2+y^2+z^2</math> if
+
Determine <math>x^2+y^2+z^2+w^2</math> if
  
 
<div style="text-align:center;"><math> \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 </math><br /><math> \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 </math><br /><math> \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 </math><br /><math> \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 </math></div>
 
<div style="text-align:center;"><math> \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 </math><br /><math> \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 </math><br /><math> \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 </math><br /><math> \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 </math></div>
  
== Solution ==
+
== Solution 1 ==
For each of the values <math>t=4,16,36,64</math>, we have the equation
+
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.
 +
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>
 +
                         
 +
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
  
<math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)</math>
+
<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>
 
+
</div>
<math>=(t-1)(t-9)(t-25)(t-49)-(t-4)(t-16)(t-36)(t-64)</math>
 
 
 
However, each side of the equation is a polynomial in <math>t</math> of degree at most 3, and they have 4 common roots. Therefore, the polynomials must be equal.
 
  
 +
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>.
  
 
Now we can plug in <math>t=1</math> into the polynomial equation. Most terms drop, and we end up with
 
Now we can plug in <math>t=1</math> into the polynomial equation. Most terms drop, and we end up with
  
<math>x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)</math>
+
<cmath>x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)</cmath>
  
 
so that
 
so that
  
<math>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}}</math>
+
<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
 
Similarly, we can plug in <math>t=9,25,49</math> and get
  
<math>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}}</math>
+
<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>
  
<math>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}}</math>
+
Now adding them up,
  
<math>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}}</math>
+
<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
  
Now add them up...
+
<cmath>\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.</cmath>
  
<math>z^2+w^2=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}</math>
+
/*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>.
  
<math>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}</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.
  
with a sum of
+
== 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>.
  
<math>\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=36</math>
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=1984|num-b=14|after=Last Question}}
 
{{AIME box|year=1984|num-b=14|after=Last Question}}
* [[AIME Problems and Solutions]]
+
 
* [[American Invitational Mathematics Examination]]
+
[[Category:Intermediate Algebra Problems]]
* [[Mathematics competition resources]]
 

Revision as of 20:15, 6 August 2020

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 $t = 4,16,36,64$, as then the equation is equivalent to the given equations. After clearing fractions, for each of the values $t=4,16,36,64$, we have the equation $x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)$ $+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)$. We can move the expression $(t-1)(t-9)(t-25)(t-49)$ to the left hand side to obtain the difference of the polynomials: $x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)$ $+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)$ and $(t-1)(t-9)(t-25)(t-49)$

Since the polynomials are equal at $t=4,16,36,64$, we can express the difference of the two polynomials with a quartic polynomial that has roots at $t=4,16,36,64$, so

$x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)$ $+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = -(t-4)(t-16)(t-36)(t-64)$

Note the leading coefficient of the RHS is $-1$ because it must match the leading coefficient of the LHS, which is $-1$.

Now we can plug in $t=1$ into the polynomial equation. Most terms drop, and we end up with

\[x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)\]

so that

\[x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}\]

Similarly, we can plug in $t=9,25,49$ and get

\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*}

Now adding them up,

\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*}

with a sum of

\[\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.\]

/*Lengthy proof that any two cubic polynomials in $t$ which are equal at 4 values of $t$ are themselves equivalent: Let the two polynomials be $A(t)$ and $B(t)$ and let them be equal at $t=a,b,c,d$. Thus we have $A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0$. Also the polynomial $A(t) - B(t)$ is cubic, but it equals 0 at 4 values of $t$. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into $(t-a)(t-b)(t-c)(t-d)($some nonzero polynomial$)$ which would have a degree greater than or equal to 4, contradicting the statement that $A(t) - B(t)$ is cubic. Because $A(t) - B(t) = 0, A(t)$ and $B(t)$ are equivalent and must be equal for all $t$.

Post script 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

$(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)$ $-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)$

$=(t-4)(t-16)(t-36)(t-64)$

Now the coefficient of $t^3$ on both sides must be equal. Therefore we have $1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}$.

See also

1984 AIME (ProblemsAnswer KeyResources)
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