1984 AIME Problems/Problem 15

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Problem

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

Solution 1

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

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

$\begin{align*}y^2&=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}\\z^2&=...$

Now adding them up,

$\begin{align*}z^2+w^2&=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}\\x^2+y^2&=\frac{3^2\cd...$

with a sum of

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

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 $1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}$ .

1984 AIME (Problems • Resources)
Preceded by Problem 14	Followed by Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1984 AIME Problems/Problem 15

From AoPSWiki

Contents

Problem

Solution 1

Solution 2

See also