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

Revision as of 19:26, 26 March 2007

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

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

@@ Line 8: / Line 8: @@
 == Solution ==
-{{solution}}
+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>
+<math>=(t-1)(t-9)(t-25)(t-49)-(t-4)(t-16)(t-36)(t-64)</math>
+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
+<math>x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)</math>
+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>
+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>
+<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>
+<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>
+Now add them up...
+<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 ==
 * [[1984 AIME Problems/Problem 14 | Previous problem]]
 * [[1984 AIME Problems]]

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

Revision as of 19:26, 26 March 2007

Problem

Solution

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