Difference between revisions of "1997 AIME Problems/Problem 9"
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Looking at the properties of the number, it is immediately guess-able that <math>a = \phi = \frac{1+\sqrt{5}}2</math> (the [[phi|golden ratio]]) is the answer. The following is the way to derive that: | Looking at the properties of the number, it is immediately guess-able that <math>a = \phi = \frac{1+\sqrt{5}}2</math> (the [[phi|golden ratio]]) is the answer. The following is the way to derive that: | ||
− | Since <math>\sqrt{2} < a < \sqrt{3}</math>, <math>0 < \frac{1}{\sqrt{3}} < a^{-1} < \frac{1}{\sqrt{2}} < 1</math>. Thus <math>\langle a^2 \rangle = a^{-1}</math>, and it follows that <math>a^2 - 2 = a^{-1} \Longrightarrow a^3 - 2a - 1 = 0</math>. Noting that <math>-1</math> is a root, this factors to <math>(a+1)(a^2 - a - 1) = 0</math>, so <math>a = \frac{1 | + | Since <math>\sqrt{2} < a < \sqrt{3}</math>, <math>0 < \frac{1}{\sqrt{3}} < a^{-1} < \frac{1}{\sqrt{2}} < 1</math>. Thus <math>\langle a^2 \rangle = a^{-1}</math>, and it follows that <math>a^2 - 2 = a^{-1} \Longrightarrow a^3 - 2a - 1 = 0</math>. Noting that <math>-1</math> is a root, this factors to <math>(a+1)(a^2 - a - 1) = 0</math>, so <math>a = \frac{1 + \sqrt{5}}{2}</math> (we discard the negative root). |
Our answer is <math>(a^2)^{6}-144a^{-1} = \left(\frac{3+\sqrt{5}}2\right)^6 - 144\left(\frac{2}{1 + \sqrt{5}}\right)</math>. [[Complex conjugate]]s reduce the second term to <math>-72(\sqrt{5}-1)</math>. The first term we can expand by the [[binomial theorem]] to get <math>\frac 1{2^6}\left(3^6 + 6\cdot 3^5\sqrt{5} + 15\cdot 3^4 \cdot 5 + 20\cdot 3^3 \cdot 5\sqrt{5} + 15 \cdot 3^2 \cdot 25 + 6 \cdot 3 \cdot 25\sqrt{5} + 5^3\right)</math> <math>= \frac{1}{64}\left(10304 + 4608\sqrt{5}\right) = 161 + 72\sqrt{5}</math>. The answer is <math>161 + 72\sqrt{5} - 72\sqrt{5} + 72 = \boxed{233}</math>. | Our answer is <math>(a^2)^{6}-144a^{-1} = \left(\frac{3+\sqrt{5}}2\right)^6 - 144\left(\frac{2}{1 + \sqrt{5}}\right)</math>. [[Complex conjugate]]s reduce the second term to <math>-72(\sqrt{5}-1)</math>. The first term we can expand by the [[binomial theorem]] to get <math>\frac 1{2^6}\left(3^6 + 6\cdot 3^5\sqrt{5} + 15\cdot 3^4 \cdot 5 + 20\cdot 3^3 \cdot 5\sqrt{5} + 15 \cdot 3^2 \cdot 25 + 6 \cdot 3 \cdot 25\sqrt{5} + 5^3\right)</math> <math>= \frac{1}{64}\left(10304 + 4608\sqrt{5}\right) = 161 + 72\sqrt{5}</math>. The answer is <math>161 + 72\sqrt{5} - 72\sqrt{5} + 72 = \boxed{233}</math>. | ||
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== Solution 2 == | == Solution 2 == | ||
+ | |||
+ | Find <math>a</math> as shown above. Note that, since <math>a</math> is a root of the equation <math>a^3 - 2a - 1 = 0</math>, <math>a^3 = 2a + 1</math>, and <math>a^{12} = (2a + 1)^4</math>. Also note that, since <math>a</math> is a root of <math>a^2 - a - 1 = 0</math>, <math>\frac{1}{a} = a - 1</math>. The expression we wish to calculate then becomes <math>(2a + 1)^4 - 144(a - 1)</math>. Plugging in <math>a = \frac{1 + \sqrt{5}}{2}</math>, we plug in to get an answer of <math>(161 + 72\sqrt{5}) + 72 - 72\sqrt{5} = 161 + 72 = \boxed{233}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
Find <math>a</math> as shown above. Note that <math>a</math> satisfies the equation <math>a^2 = a+1</math> (this is the equation we solved to get it). Then, we can simplify <math>a^{12}</math> as follows using the fibonacci numbers: | Find <math>a</math> as shown above. Note that <math>a</math> satisfies the equation <math>a^2 = a+1</math> (this is the equation we solved to get it). Then, we can simplify <math>a^{12}</math> as follows using the fibonacci numbers: | ||
− | <math>a^{12} = a^{11}+a^{10}= 2a^{10} + a^{9} = 3a^ | + | <math>a^{12} = a^{11}+a^{10}= 2a^{10} + a^{9} = 3a^9+ 2a^8 = ... = 144a^1+89a^0 = 144a+89</math> |
So we want <math>144(a-\frac1a)+89 = 144(1)+89 = \boxed{233}</math> since <math>a-\frac1a = 1</math> is equivalent to <math>a^2 = a+1</math>. | So we want <math>144(a-\frac1a)+89 = 144(1)+89 = \boxed{233}</math> since <math>a-\frac1a = 1</math> is equivalent to <math>a^2 = a+1</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | As Solution 1 stated, <math>a^3 - 2a - 1 = 0</math>. <math>a^3 - 2a - 1 = a^3 - a^2 -a + a^2 -a -1 = (a+1)(a^2 - a - 1)</math>. So, <math>a^2 - a - 1 = 0</math>, <math>1 = a^2 - a</math>, <math>\frac1a = a-1</math>, <math>a^3 = 2a+1</math>, <math>a^2 = a+1</math>. | ||
+ | |||
+ | <math>a^6 = (a^3)^2 = (2a+1)^2= 4a^2 + 4a +1= 4(a+1) + 4a + 1= 8a+5</math> | ||
+ | |||
+ | <math>a^{12} = (a^6)^2 = (8a+5)^2 = 64a^2 + 80a + 25 = 64(a+1) + 80a + 25 = 144a + 89</math> | ||
+ | |||
+ | Therefore, <math>a^{12} - 144 a^{-1} = 144a + 89 - 144(a-1) = 89 + 144 = \boxed{233}</math> | ||
+ | |||
+ | |||
+ | Another way to factor <math>a^3 - 2a - 1</math>: | ||
+ | |||
+ | <math>a^3 - 2a - 1 = a^3 + 1 -2a -2 = (a+1)(a^2 - a + 1) - 2(a+1) = (a + 1)(a^2 - a - 1)</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:53, 30 December 2023
Problem
Given a nonnegative real number , let
denote the fractional part of
; that is,
, where
denotes the greatest integer less than or equal to
. Suppose that
is positive,
, and
. Find the value of
.
Solution 1
Looking at the properties of the number, it is immediately guess-able that (the golden ratio) is the answer. The following is the way to derive that:
Since ,
. Thus
, and it follows that
. Noting that
is a root, this factors to
, so
(we discard the negative root).
Our answer is . Complex conjugates reduce the second term to
. The first term we can expand by the binomial theorem to get
. The answer is
.
Note that to determine our answer, we could have also used other properties of like
.
Solution 2
Find as shown above. Note that, since
is a root of the equation
,
, and
. Also note that, since
is a root of
,
. The expression we wish to calculate then becomes
. Plugging in
, we plug in to get an answer of
.
Solution 3
Find as shown above. Note that
satisfies the equation
(this is the equation we solved to get it). Then, we can simplify
as follows using the fibonacci numbers:
So we want since
is equivalent to
.
Solution 4
As Solution 1 stated, .
. So,
,
,
,
,
.
Therefore,
Another way to factor :
See also
1997 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.