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

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Each side of this equation is a [[polynomial]] in <math>t</math> of degree at most 3, and they are equal for 4 values of <math>t</math> (when <math>t=4,16,36,64</math>). Therefore, the polynomials must be equal.*
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Each side of this equation is a [[polynomial]] in <math>t</math> of degree at most 3, and they are equal for 4 values of <math>t</math> (when <math>t=4,16,36,64</math>). Therefore, the polynomials must be equal for all <math>t</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
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/*Lengthy proof that any two cubic polynomials in <math>t</math> which are equal at 4 values of <math>t</math> are themselves equivalent:
 
/*Lengthy proof that any two cubic polynomials in <math>t</math> which are equal at 4 values of <math>t</math> are themselves equivalent:
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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.
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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.
 
'''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.

Revision as of 22:00, 29 December 2013

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 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 $x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)$ $+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)$. When $t=4,16,36,64$, a $(t-4)(t-16)(t-36)(t-64)$ term can be subtracted from the right-hand side because it equals 0. Thus we have the following equation which holds for $t=4,16,36,64$:

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

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

Each side of this equation is a polynomial in $t$ of degree at most 3, and they are equal for 4 values of $t$ (when $t=4,16,36,64$). Therefore, the polynomials must be equal for all $t$.*

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

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

/*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 (Problems • Answer Key • Resources)
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