Difference between revisions of "2008 AIME I Problems/Problem 8"
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== Problem == | == Problem == | ||
Find the positive integer <math>n</math> such that | Find the positive integer <math>n</math> such that | ||
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<cmath>\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.</cmath> | <cmath>\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.</cmath> | ||
− | + | == Solution 1 == | |
− | |||
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Since we are dealing with acute angles, <math>\tan(\arctan{a}) = a</math>. | Since we are dealing with acute angles, <math>\tan(\arctan{a}) = a</math>. | ||
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Now, <math>\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \arctan{\dfrac{23}{24}}</math>. | Now, <math>\arctan{\dfrac{7}{11}} + \arctan{\dfrac{1}{5}} = \arctan{\dfrac{23}{24}}</math>. | ||
− | We now have <math>\arctan{\dfrac{23}{24}} + \arctan{\dfrac{1}{n}} = \dfrac{\pi}{4} = \arctan{1}</math>. Thus, <math>\dfrac{\dfrac{23}{24} + \dfrac{1}{n}}{1 - \dfrac{23}{24n}} = 1</math>; and simplifying, <math>23n + 24 = 24n - 23 \Longrightarrow n = \boxed{ | + | We now have <math>\arctan{\dfrac{23}{24}} + \arctan{\dfrac{1}{n}} = \dfrac{\pi}{4} = \arctan{1}</math>. Thus, <math>\dfrac{\dfrac{23}{24} + \dfrac{1}{n}}{1 - \dfrac{23}{24n}} = 1</math>; and simplifying, <math>23n + 24 = 24n - 23 \Longrightarrow n = \boxed{47}</math>. |
− | + | ||
+ | == Solution 2 (generalization) == | ||
From the expansion of <math>e^{iA}e^{iB}e^{iC}e^{iD}</math>, we can see that | From the expansion of <math>e^{iA}e^{iB}e^{iC}e^{iD}</math>, we can see that | ||
<cmath> | <cmath> | ||
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\tan(A + B + C + D) = \frac {1 - \sum \tan A \tan B + \tan A \tan B \tan C \tan D}{\sum \tan A - \sum \tan A \tan B \tan C}, | \tan(A + B + C + D) = \frac {1 - \sum \tan A \tan B + \tan A \tan B \tan C \tan D}{\sum \tan A - \sum \tan A \tan B \tan C}, | ||
</cmath> | </cmath> | ||
− | which makes for more direct, less error-prone computations. Substitution gives the desired answer. | + | which makes for more direct, less error-prone computations. Substitution gives the desired answer. |
+ | |||
+ | == Solution 3: Complex Numbers == | ||
+ | Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, <math>\arctan\frac{1}{n}</math>, is the argument of <math>n+i</math>. The sum of these angles is then just the argument of the product | ||
+ | <cmath>(3+i)(4+i)(5+i)(n+i)</cmath> | ||
+ | and expansion give us <math>(48n-46)+(48+46n)i</math>. Since the argument of this complex number is <math>\frac{\pi}{4}</math>, its real and imaginary parts must be equal; then, we can we set them equal to get | ||
+ | <cmath>48n - 46 = 48 + 46n.</cmath> | ||
+ | Therefore, <math>n=\boxed{47}</math>. | ||
+ | |||
+ | ==Solution 4 Sketch == | ||
+ | You could always just bash out <math>\sin(a+b+c)</math> (where a,b,c,n are the angles of the triangles respectively) using the sum identities again and again until you get a pretty ugly radical and use a triangle to get <math>\cos(a+b+c)</math> and from there you use a sum identity again to get <math>\sin(a+b+c+n)</math> and using what we found earlier you can find <math>\tan(n)</math> by division that gets us <math>\frac{23}{24}</math> | ||
+ | |||
+ | ~YBSuburbanTea | ||
== See also == | == See also == | ||
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[[Category:Intermediate Trigonometry Problems]] | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:36, 28 November 2023
Contents
Problem
Find the positive integer such that
Solution 1
Since we are dealing with acute angles, .
Note that , by tangent addition. Thus,
.
Applying this to the first two terms, we get .
Now, .
We now have . Thus,
; and simplifying,
.
Solution 2 (generalization)
From the expansion of , we can see that
and
If we divide both of these by
, then we have
which makes for more direct, less error-prone computations. Substitution gives the desired answer.
Solution 3: Complex Numbers
Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, , is the argument of
. The sum of these angles is then just the argument of the product
and expansion give us
. Since the argument of this complex number is
, its real and imaginary parts must be equal; then, we can we set them equal to get
Therefore,
.
Solution 4 Sketch
You could always just bash out (where a,b,c,n are the angles of the triangles respectively) using the sum identities again and again until you get a pretty ugly radical and use a triangle to get
and from there you use a sum identity again to get
and using what we found earlier you can find
by division that gets us
~YBSuburbanTea
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.