Difference between revisions of "2007 AMC 10A Problems/Problem 20"
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Notice that <math>(a^{k} + a^{-k})^2 = a^{2k} + a^{-2k} + 2</math>. Thus <math>a^4 + a^{-4} = (a^2 + a^{-2})^2 - 2 = [(a + a^{-1})^2 - 2]^2 - 2 = 194\ \mathrm{(D)}</math>. | Notice that <math>(a^{k} + a^{-k})^2 = a^{2k} + a^{-2k} + 2</math>. Thus <math>a^4 + a^{-4} = (a^2 + a^{-2})^2 - 2 = [(a + a^{-1})^2 - 2]^2 - 2 = 194\ \mathrm{(D)}</math>. | ||
− | === Solution 2 === | + | === Solution 2(LIFEHACK) === |
− | <math>4a = a^2 + 1< | + | Notice that <math>(a^{4} + a^{-4}) = (a^{2} + a^{-2} + 2</math>. Since \ \mathrm{(D)}<math> 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 === | ||
+ | </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< | + | 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 3 === | === Solution 3 === | ||
(similar to Solution 1) | (similar to Solution 1) | ||
− | We know that <math>a+\frac{1}{a}=4< | + | 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 4 === | === Solution 4 === | ||
− | We let <math>a< | + | 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>S_4=\boxed{194}<math> |
=== Solution 5 === | === Solution 5 === | ||
− | Let <math>a< | + | 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)$ and solving gets 194. |
== See also == | == See also == |
Revision as of 22:12, 4 January 2020
Problem
Suppose that the number satisfies the equation . What is the value of ?
Solutions
Solution 1
Notice that . Thus .
Solution 2(LIFEHACK)
Notice that . Since \ \mathrm{(D)}4a = a^2 + 1a = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.
Thus$ (Error compiling LaTeX. Unknown error_msg)a^4 + a^{-4} = (2+\sqrt{3})^4 + \frac{1}{(2+\sqrt{3})^4} = (2+\sqrt{3})^4 + (2-\sqrt{3})^4a194$.
=== Solution 3 === (similar to Solution 1) We know that$ (Error compiling LaTeX. Unknown error_msg)a+\frac{1}{a}=4a^2+\frac{1}{a^2}+2=16a^2+\frac{1}{a^2}=14a^4+\frac{1}{a^4}+2=14^2=196a^4+\frac{1}{a^4}=\boxed{194}$.
=== Solution 4 === We let$ (Error compiling LaTeX. Unknown error_msg)a1/ax^2-4x+1=0S_4S_4=\boxed{194}a\cos(x)i\sin(x)a + a^{-1} = 2\cos(x)\cos(x) = 2a^4 + a^{-4}2\cos(4x)$ 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 |
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