Difference between revisions of "1990 AIME Problems/Problem 15"
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== Solution 4 == | == Solution 4 == | ||
− | We first let the answer to this problem be k. Multiplying the first equation by <math>x</math> gives <math>ax^2 + bxy=3x</math> Subtracting this equation from the second equation gives <math>by^2-bxy=7-3x</math>. Similarily, doing the same for the other equations, we obtain: | + | We first let the answer to this problem be k. Multiplying the first equation by <math>x</math> gives <math>ax^2 + bxy=3x</math>. |
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+ | Subtracting this equation from the second equation gives <math>by^2-bxy=7-3x</math>. Similarily, doing the same for the other equations, we obtain: | ||
<math>by^2-bxy=7-3x</math>, <math>by^3-bxy^2=16-7x</math>, <math>by^4-bxy^3=42-16x</math>, and <math>by^5-bxy^4=k-42x</math> | <math>by^2-bxy=7-3x</math>, <math>by^3-bxy^2=16-7x</math>, <math>by^4-bxy^3=42-16x</math>, and <math>by^5-bxy^4=k-42x</math> | ||
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Now lets take the first equation. Multiplying this by y and subtracting this from the second gives us <math>by^3-bxy^2=(7-3x)y</math>. We can also obtain <math>by^4-bxy^3=(16-7x)y</math>. | Now lets take the first equation. Multiplying this by y and subtracting this from the second gives us <math>by^3-bxy^2=(7-3x)y</math>. We can also obtain <math>by^4-bxy^3=(16-7x)y</math>. | ||
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Now we can solve for x and y! <math>(7-3x)y = 16-7x</math> and <math>(16-7x)y=42-16x</math>. Solving for x and y gives us <math>(-7+\sqrt{87},-7-\sqrt{87})</math>.(It can be switched, but since the given equations are symmetric, it doesn't matter). <math>k-42x= (42-16x)y</math>, and solving for k gives us <math>k= \boxed{020}</math>. | Now we can solve for x and y! <math>(7-3x)y = 16-7x</math> and <math>(16-7x)y=42-16x</math>. Solving for x and y gives us <math>(-7+\sqrt{87},-7-\sqrt{87})</math>.(It can be switched, but since the given equations are symmetric, it doesn't matter). <math>k-42x= (42-16x)y</math>, and solving for k gives us <math>k= \boxed{020}</math>. | ||
Revision as of 18:02, 25 May 2020
Problem
Find if the real numbers , , , and satisfy the equations
Solution 1
Set and . Then the relationship
can be exploited:
Therefore:
Consequently, and . Finally:
Solution 2
A recurrence of the form will have the closed form , where are the values of the starting term that make the sequence geometric, and are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.
Suppose we have such a recurrence with and . Then , and .
Solving these simultaneous equations for and , we see that and . So, .
Solution 3
Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting:
.
Similarly take the first two terms, yielding:
.
Lastly take an alternating three-term sum,
.
Now to get the solution, let the answer be , so
.
Multiplying out this expression gets the answer term and then a lot of other expressions, which can be eliminated as done in the first solution.
Solution 4
We first let the answer to this problem be k. Multiplying the first equation by gives .
Subtracting this equation from the second equation gives . Similarily, doing the same for the other equations, we obtain:
, , , and
Now lets take the first equation. Multiplying this by y and subtracting this from the second gives us . We can also obtain .
Now we can solve for x and y! and . Solving for x and y gives us .(It can be switched, but since the given equations are symmetric, it doesn't matter). , and solving for k gives us .
pi_is_3.141
See also
1990 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.