Difference between revisions of "2021 AMC 10B Problems/Problem 15"
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Revision as of 08:52, 2 March 2021
Contents
Problem
The real number satisfies the equation . What is the value of
Solution 1
We square to get . We subtract 2 on both sides for and square again, and see that so . We can divide our original expression of by to get that it is equal to . Therefore because is 7, it is equal to .
Solution 2
Multiplying both sides by and using the quadratic formula, we get . We can assume that it is , and notice that this is also a solution the equation , i.e. we have . Repeatedly using this on the given (you can also just note Fibonacci numbers),
~Lcz
Solution 3
We can immediately note that the exponents of are an arithmetic sequence, so they are symmetric around the middle term. So, . We can see that since , and therefore . Continuing from here, we get , so . We don't even need to find what is! This is since is evidently , which is our answer.
~sosiaops
Solution 4
We begin by multiplying by , resulting in . Now we see this equation: . The terms all have in common, so we can factor that out, and what we're looking for becomes . Looking back to our original equation, we have , which is equal to . Using this, we can evaluate to be , and we see that there is another , so we put substitute it in again, resulting in . Using the same way, we find that is . We put this into , resulting in , so the answer is .
~purplepenguin2
Video Solution by OmegaLearn (Algebraic Manipulations and Symmetric Polynomials)
~ pi_is_3.14
Video Solution by Interstigation (Simple Silly Bashing)
~ Interstigation
Video Solution by TheBeautyofMath
Not the most efficient method, but gets the job done.
https://youtu.be/L1iW94Ue3eI?t=1468
~IceMatrix
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.