Difference between revisions of "1984 AIME Problems/Problem 15"
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Now the coefficient of <math>t^3</math> on both sides must be equal. Therefore we have <math>1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}</math>. | Now the coefficient of <math>t^3</math> on both sides must be equal. Therefore we have <math>1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}</math>. | ||
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+ | == Solution 3 == | ||
+ | |||
+ | Let <math>a + b + c + d = \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</math>, where each of <math>a, b, c, d</math> corresponds to the term in the same position in this original first equation. Dividing the <math>n</math>th term in each equation by the <math>n</math>th term in the equation before it, we find that the ratio between the two is dependent only on the first number in the denominator of each term (namely <math>2, 4, 6,</math> and <math>8</math>), and can be written as a product of linear factors. Computing these ratios, we can rewrite our system linearly as the following: | ||
+ | <div style="text-align:center;"><math>a + b + c + d = 1 </math><br /><math> \frac{1}{5}a-\frac{5}{7}b+\frac{7}{3}c+\frac{15}{11}d=1 </math><br /><math> \frac{3}{35}a-\frac{5}{27}b-\frac{21}{11}c+\frac{45}{13}d=1 </math><br /><math> \frac{1}{21}a-\frac{1}{11}b-\frac{7}{13}c-3d=1</math> | ||
+ | </div> | ||
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+ | Solving this painstaking system, you'll find each of <math>a, b, c, d</math> can be written in reduced form with a denominator of <math>1024</math>, as shown above. Substituting these values back in and solving for <math>x^2, y^2, z^2,</math> and <math>w^2</math> and summing them, we get a final value of <math>\boxed{036}</math>. | ||
+ | |||
+ | ~anellipticcurveoverq | ||
== See also == | == See also == |
Revision as of 13:22, 2 April 2020
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 . We can move the expression to the left hand side to obtain the difference of the polynomials: and
Since the polynomials are equal at , we can express the difference of the two polynomials with a quartic polynomial that has roots at , so
Note the leading coefficient of the RHS is because it must match the leading coefficient of the LHS, which is .
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 .
Solution 3
Let , where each of corresponds to the term in the same position in this original first equation. Dividing the th term in each equation by the th term in the equation before it, we find that the ratio between the two is dependent only on the first number in the denominator of each term (namely and ), and can be written as a product of linear factors. Computing these ratios, we can rewrite our system linearly as the following:
Solving this painstaking system, you'll find each of can be written in reduced form with a denominator of , as shown above. Substituting these values back in and solving for and and summing them, we get a final value of .
~anellipticcurveoverq
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 |