1984 AIME Problems/Problem 15

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

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

Postscript 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. So instead of expanding it fully, we will find what the coefficients of the $t^4$ and $t^3$ terms are, so we can eventually apply Vieta's. We can write the long equation as \[(x^2 + y^2 + z^2 + w^2)t^3 + \dots = t^4 - (1^2 + 3^2 + 5^2 + 7^2)t^3 + \dots\] Rearranging gives us \[t^4 - (1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2)t^3 \dots = 0.\] By Vieta's, we know that the sum of the roots of this equation is \[1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2.\] (recall that the roots of the original and this manipulated form of it had roots $2^2, 4^2, 6^2,$ and $8^2$). Thus, \[x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2 - 1^2 - 3^2 - 5^2 - 7^2 = \boxed{36}.\]

Solution 3 (Highly Unrecommended)

Before starting this solution, I highly recommend never following this unless you have no idea what to do with an hour of your time. Even so, learning the above solutions will be more beneficial.

\begin{align*} \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\\ \end{align*} can be rewritten as \begin{align*} \frac{x^2}{3}-\frac{y^2}{5}-\frac{z^2}{21}-\frac{w^2}{45}=1\\ \frac{x^2}{15}+\frac{y^2}{7}-\frac{z^2}{9}-\frac{w^2}{33}=1\\ \frac{x^2}{35}+\frac{y^2}{27}+\frac{z^2}{11}-\frac{w^2}{13}=1\\ \frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\ \end{align*} You might be able to see where this is going. First off, find $\text{lcm}(3,5,21,45),\text{lcm}(15,7,9,33), \text{lcm}(35,27,11,13),$ and $\text{lcm}(63,55,39,15)$. Then, multiply by the respective lcm to clear all of the denominators. Once you do that, maniuplate the equations to solve for $w^2+x^2+y^2+z^2$.

Now, most of this is just a brainless bash, and reemphasizing, please try to learn the above solutions. This is only a last resort and only to be used if you have too much time left. The exact amount of time this bash takes depends on the person and how quickly they can manipulate the equations.

See also

1984 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Question
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
Invalid username
Login to AoPS