Difference between revisions of "2007 AMC 10A Problems/Problem 20"
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<math>\text{(A)}\ 164 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 192 \qquad \text{(D)}\ 194 \qquad \text{(E)}\ 212</math> | <math>\text{(A)}\ 164 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 192 \qquad \text{(D)}\ 194 \qquad \text{(E)}\ 212</math> | ||
− | + | == Solution 1 (Generalized) == | |
− | + | Notice that for all real numbers <math>k,</math> we have <math>a^{2k} + a^{-2k} + 2 = \left(a^{k} + a^{-k}\right)^2,</math> from which <cmath>a^{2k} + a^{-2k} = \left(a^{k} + a^{-k}\right)^2-2.</cmath> We apply this result twice to get the answer: | |
− | + | <cmath>\begin{align*} | |
− | Notice that <math>(a^{k} + a^{-k})^2 | + | a^4 + a^{-4} &= \left(a^2 + a^{-2}\right)^2 - 2 \ |
+ | &= \left[\left(a + a^{-1}\right)^2 - 2\right]^2 - 2 \ | ||
+ | &= \boxed{\text{(D)}\ 194}.</cmath> | ||
+ | ~Azjps (Fundamental Logic) | ||
− | + | ~MRENTHUSIASM (Reconstruction) | |
+ | |||
+ | == Solution 2 == | ||
Notice that <math>(a^{4} + a^{-4}) = (a^{2} + a^{-2})^{2} - 2</math>. Since D is the only option 2 less than a perfect square, that is correct. | Notice that <math>(a^{4} + a^{-4}) = (a^{2} + a^{-2})^{2} - 2</math>. Since D is the only option 2 less than a perfect square, that is correct. | ||
PS: Because this is a multiple choice test, this works. | PS: Because this is a multiple choice test, this works. | ||
− | + | == Solution 3 == | |
<math>4a = a^2 + 1</math>. We apply the [[quadratic formula]] to get <math>a = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}</math>. | <math>4a = a^2 + 1</math>. We apply the [[quadratic formula]] to get <math>a = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}</math>. | ||
Thus <math>a^4 + a^{-4} = (2+\sqrt{3})^4 + \frac{1}{(2+\sqrt{3})^4} = (2+\sqrt{3})^4 + (2-\sqrt{3})^4</math> (so it doesn't matter which root of <math>a</math> we use). Using the [[binomial theorem]] we can expand this out and collect terms to get <math>194</math>. | Thus <math>a^4 + a^{-4} = (2+\sqrt{3})^4 + \frac{1}{(2+\sqrt{3})^4} = (2+\sqrt{3})^4 + (2-\sqrt{3})^4</math> (so it doesn't matter which root of <math>a</math> we use). Using the [[binomial theorem]] we can expand this out and collect terms to get <math>194</math>. | ||
− | + | == Solution 4 == | |
(similar to Solution 1) | (similar to Solution 1) | ||
We know that <math>a+\frac{1}{a}=4</math>. We can square both sides to get <math>a^2+\frac{1}{a^2}+2=16</math>, so <math>a^2+\frac{1}{a^2}=14</math>. Squaring both sides again gives <math>a^4+\frac{1}{a^4}+2=14^2=196</math>, so <math>a^4+\frac{1}{a^4}=\boxed{194}</math>. | We know that <math>a+\frac{1}{a}=4</math>. We can square both sides to get <math>a^2+\frac{1}{a^2}+2=16</math>, so <math>a^2+\frac{1}{a^2}=14</math>. Squaring both sides again gives <math>a^4+\frac{1}{a^4}+2=14^2=196</math>, so <math>a^4+\frac{1}{a^4}=\boxed{194}</math>. | ||
− | + | == Solution 5 == | |
We let <math>a</math> and <math>1/a</math> be roots of a certain quadratic. Specifically <math>x^2-4x+1=0</math>. We use [[Newton's Sums]] given the coefficients to find <math>S_4</math>. | We let <math>a</math> and <math>1/a</math> be roots of a certain quadratic. Specifically <math>x^2-4x+1=0</math>. We use [[Newton's Sums]] given the coefficients to find <math>S_4</math>. | ||
<math>S_4=\boxed{194}</math> | <math>S_4=\boxed{194}</math> | ||
− | + | == 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. | ||
Revision as of 22:32, 24 June 2021
Contents
[hide]Problem
Suppose that the number satisfies the equation . What is the value of ?
Solution 1 (Generalized)
Notice that for all real numbers we have from which We apply this result twice to get the answer:
\begin{align*} a^4 + a^{-4} &= \left(a^2 + a^{-2}\right)^2 - 2 \\ &= \left[\left(a + a^{-1}\right)^2 - 2\right]^2 - 2 \\ &= \boxed{\text{(D)}\ 194}. (Error compiling LaTeX. Unknown error_msg)
~Azjps (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2
Notice that . Since D is the only option 2 less than a perfect square, that is correct.
PS: Because this is a multiple choice test, this works.
Solution 3
. 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 4
(similar to Solution 1) We know that . We can square both sides to get , so . Squaring both sides again gives , so .
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.
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.