Difference between revisions of "1984 AIME Problems/Problem 15"
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Plug in <math>t=1^2, 3^2, 5^2, 7^2</math> in succession, into <math>(\dag)</math>. In each case, most terms drop, and we end up with | Plug in <math>t=1^2, 3^2, 5^2, 7^2</math> in succession, into <math>(\dag)</math>. In each case, most terms drop, and we end up with | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | x^2 | + | x^2=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}, \quad y^2=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}},\quad z^2=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}},\quad w^2=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}} |
− | y^2 | ||
− | z^2 | ||
− | w^2 | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | Adding them up we get the sum as | + | Adding them up we get the sum as <math>3^2\cdot 4=\boxed{036}</math>. |
− | < | ||
'''Postscript for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes <math>x^2,y^2,z^2,</math> and <math>w^2</math> separately before adding them to obtain the final answer is appealing because it gives the individual values of <math>x^2,y^2,z^2,</math> and <math>w^2</math> which can be plugged into the given equations to check. | '''Postscript for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes <math>x^2,y^2,z^2,</math> and <math>w^2</math> separately before adding them to obtain the final answer is appealing because it gives the individual values of <math>x^2,y^2,z^2,</math> and <math>w^2</math> which can be plugged into the given equations to check. |
Revision as of 15:33, 9 July 2023
Problem
Determine if
Solution 1
Rewrite the system of equations as This equation is satisfied when . After clearing fractions, for each of the values , we have the equation where and , for .
Since the polynomials on each side are equal at , we can express the difference of the two polynomials by a quartic polynomial that has roots at , so The leading coefficient of the RHS is because the leading coefficient of the LHS is .
Plug in in succession, into . In each case, most terms drop, and we end up with Adding them up we get the sum as .
Postscript 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 where and , for .
Now the coefficient of on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the and terms are, so we can eventually apply Vieta's. We can write the long equation as Rearranging gives us By Vieta's, we know that the sum of the roots of this equation is (recall that the roots of the original and this manipulated form of it had roots and ). Thus,
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.
can be rewritten as You might be able to see where this is going. First off, find and . Then, multiply by the respective lcm to clear all of the denominators. Once you do that, maniuplate the equations to solve for .
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 (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 |