Difference between revisions of "1984 AIME Problems/Problem 15"
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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}}\\ | 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}}\\ | 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}}</cmath> | + | 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*}</cmath> |
Now adding them up, | Now adding them up, |
Revision as of 15:39, 13 March 2015
Contents
Problem
Determine if
Solution 1
Rewrite the system of equations as This equation is satisfied when , as then the equation is equivalent to the given equations. After clearing fractions, for each of the values , we have the equation . When , a term can be subtracted from the right-hand side because it equals 0. Thus we have the following equation which holds for :
Each side of this equation is a polynomial in of degree at most 3, and they are equal for 4 values of (when ). Therefore, the polynomials must be equal for all .*
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
Now adding them up,
with a sum of
/*Lengthy proof that any two cubic polynomials in which are equal at 4 values of are themselves equivalent: Let the two polynomials be and and let them be equal at . Thus we have . Also the polynomial is cubic, but it equals 0 at 4 values of . Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into some nonzero polynomial which would have a degree greater than or equal to 4, contradicting the statement that is cubic. Because and are equivalent and must be equal for all .
Post script for the puzzled: This solution which is seemingly unnecessarily redundant in that it computes and separately before adding them to obtain the final answer is appealing because it gives the individual values of and which can be plugged into the given equations to check.
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 |