1984 AIME Problems/Problem 15

Revision as of 19:26, 26 March 2007 by Scorpius119 (talk | contribs) (solution added)

Problem

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

$y^2=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}$

$z^2=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}$

$w^2=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}$


Now add them up...

$z^2+w^2=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}$

$x^2+y^2=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}$

with a sum of

$\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=36$

See also

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