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1984 AIME Problems/Problem 15: Difference between revisions

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== Problem ==
== Problem ==
Determine <math>\displaystyle w^2+x^2+y^2+z^2</math> if
Determine <math>w^2+x^2+y^2+z^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 ==
For each of the values <math>t=4,16,36,64</math>, we have the equation
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)+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)</math>
<math>=(t-1)(t-9)(t-25)(t-49)-(t-4)(t-16)(t-36)(t-64)</math>
<math>=(t-1)(t-9)(t-25)(t-49)-(t-4)(t-16)(t-36)(t-64)</math>
</div>


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.
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.
 


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}}</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>
 
<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>
 
with a sum of


<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 11:53, 1 January 2008

Problem

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

For each of the values $t=4,16,36,64$, we have the equation

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

$=(t-1)(t-9)(t-25)(t-49)-(t-4)(t-16)(t-36)(t-64)$

However, each side of the equation is a polynomial in $t$ of degree at most 3, and they have 4 common roots. Therefore, the polynomials must be equal.

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}} (Error compiling LaTeX. Unknown error_msg)

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

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

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
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Last Question
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All AIME Problems and Solutions
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