Difference between revisions of "1984 AIME Problems/Problem 15"
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For each of the values <math>t=4,16,36,64</math>, we have the [[equation]] | For each of the values <math>t=4,16,36,64</math>, we have the [[equation]] | ||
− | <div style="text-align:center;"><math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)</math> | + | <div style="text-align:center;"><math>x^2(t-1)(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+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> | <math>=(t-1)(t-9)(t-25)(t-49)-(t-4)(t-16)(t-36)(t-64)</math> | ||
</div> | </div> |
Revision as of 15:34, 2 October 2011
Contents
Problem
Determine if
Solution 1
For each of the values , we have the equation
However, each side of the equation is a polynomial in of degree at most 3, and they have 4 common roots. Therefore, the polynomials must be equal.
Now we can plug in into the polynomial equation. Most terms drop, and we end up with
so that
Similarly, we can plug in 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. ! LaTeX Error: \begin{align*} on input line 20 ended by \end{document}.)
Now adding them up,
with a sum of
Solution 2
As in Solution 1, we have
Now the coefficient of on both sides must be equal. Therefore we have .
See also
1984 AIME (Problems • Answer Key • Resources) | ||
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 |