Difference between revisions of "2007 AMC 10A Problems/Problem 20"
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== Solution 4 == | == Solution 4 == | ||
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== Solution 6 == | == Solution 6 == | ||
Let <math>a</math> = <math>\cos(x)</math> + <math>i\sin(x)</math>. Then <math>a + a^{-1} = 2\cos(x)</math> so <math>\cos(x) = 2</math>. Then by [[De Moivre's Theorem]], <math>a^4 + a^{-4}</math> = <math>2\cos(4x)</math> and solving gets 194. | Let <math>a</math> = <math>\cos(x)</math> + <math>i\sin(x)</math>. Then <math>a + a^{-1} = 2\cos(x)</math> so <math>\cos(x) = 2</math>. Then by [[De Moivre's Theorem]], <math>a^4 + a^{-4}</math> = <math>2\cos(4x)</math> and solving gets 194. | ||
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+ | == Solution 7 (Answer Choices) == | ||
+ | Notice that <math>a^{4} + a^{-4} = \left(a^{2} + a^{-2}\right)^{2} - 2,</math> from which the answer must be <math>2</math> less than a perfect square. The only possibility is <math>\boxed{\text{(D)}\ 194}.</math> | ||
== See also == | == See also == |
Revision as of 23:51, 24 June 2021
Contents
[hide]Problem
Suppose that the number satisfies the equation . What is the value of ?
Solution 1 (Decreases the Powers)
Notice that for all real numbers we have from which We apply this result twice to get the answer: ~Azjps (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2 (Increases the Powers)
Squaring both sides of gives from which
Squaring both sides of gives from which
~Rbhale12 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 3 (Binomial Theorem)
Squaring both sides of gives from which
We raise both sides of to the fourth power, then apply the Binomial Theorem: ~MRENTHUSIASM
Solution 4
. We apply the quadratic formula to get .
Thus (so it doesn't matter which root of we use). Using the binomial theorem we can expand this out and collect terms to get .
Solution 5
We let and be roots of a certain quadratic. Specifically . We use Newton's Sums given the coefficients to find .
Solution 6
Let = + . Then so . Then by De Moivre's Theorem, = and solving gets 194.
Solution 7 (Answer Choices)
Notice that from which the answer must be less than a perfect square. The only possibility is
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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.