Difference between revisions of "1996 AIME Problems/Problem 10"
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Multiplying by <math>19</math> gives <math>x \equiv 141 \cdot 19 \equiv 2679 \equiv 159 \pmod{180}</math>. | Multiplying by <math>19</math> gives <math>x \equiv 141 \cdot 19 \equiv 2679 \equiv 159 \pmod{180}</math>. | ||
The smallest positive solution of this is <math>x = \boxed{159}</math> | The smallest positive solution of this is <math>x = \boxed{159}</math> | ||
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+ | == Solution 3 (Only sine and cosine sum formulas) == | ||
+ | It seems reasonable to assume that <math>\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} = \tan{\theta}</math> for some angle <math>\theta</math>. This means <cmath>\dfrac{\alpha (\cos{96^{\circ}}+\sin{96^{\circ}})}{\alpha (\cos{96^{\circ}}-\sin{96^{\circ}})} = \frac{\sin{\theta}}{\cos{\theta}}</cmath> for some constant <math>\alpha</math>. We can set <math>\alpha (\cos{96^{\circ}}+\sin{96^{\circ}}) = \sin{\theta}</math>.Note that if we have <math>\alpha</math> equal to both the sine and cosine of an angle, we can use the sine sum formula (and the cosine sum formula on the denominator). So, since <math>\sin{45^{\circ}} = \cos{45^{\circ}} = \tfrac{\sqrt{2}}{2}</math>, if <math>\alpha = \tfrac{\sqrt{2}}{2}</math> we have | ||
+ | <cmath>\alpha (\cos{96^{\circ}} + \sin{96^{\circ}}) = \cos{96^{\circ}} \frac{\sqrt{2}}{2} + \sin{96^{\circ}} \frac{\sqrt{2}}{2} = \cos{96^{\circ}} \sin{45^{\circ}} + \sin{96^{\circ}} \cos{45^{\circ}} = \sin({45^{\circ} + 96^{\circ}}) = \sin{141^{\circ}}</cmath> | ||
+ | from the sine sum formula. For the denominator, from the cosine sum formula, we have | ||
+ | <cmath>\alpha (\cos{96^{\circ}} - \sin{96^{\circ}}) = \cos{96^{\circ}} \frac{\sqrt{2}}{2} + \sin{96^{\circ}} \frac{\sqrt{2}}{2} = \cos{96^{\circ}} \cos{45^{\circ}} + \sin{96^{\circ}} \sin{45^{\circ}} = \cos({96^{\circ} + 45^{\circ}}) = \cos{141^{\circ}}.</cmath> | ||
+ | This means <math>\theta = 141^{\circ},</math> so <math>19x = 141 + 180k</math> for some positive integer <math>k</math> (since the period of tangent is <math>180^{\circ}</math>), or <math>19 x \equiv 141 \pmod{180}</math>. Note that the inverse of <math>19</math> modulo <math>180</math> is itself as <math>19^2 \equiv 361 \equiv 1 \pmod {180}</math>, so multiplying this congruence by <math>19</math> on both sides gives <math>x \equiv 2679 \equiv 159 \pmod{180}.</math> For the smallest possible <math>x</math>, we take <math>x = \boxed{159}.</math> | ||
== See Also == | == See Also == | ||
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[[Category:Intermediate Trigonometry Problems]] | [[Category:Intermediate Trigonometry Problems]] | ||
− | {{MAA | + | {{MAA Notice}} |
Latest revision as of 15:46, 3 August 2022
Contents
Problem
Find the smallest positive integer solution to .
Solution
.
The period of the tangent function is , and the tangent function is one-to-one over each period of its domain.
Thus, .
Since , multiplying both sides by yields .
Therefore, the smallest positive solution is .
Solution 2
which is the same as .
So , for some integer . Multiplying by gives . The smallest positive solution of this is
Solution 3 (Only sine and cosine sum formulas)
It seems reasonable to assume that for some angle . This means for some constant . We can set .Note that if we have equal to both the sine and cosine of an angle, we can use the sine sum formula (and the cosine sum formula on the denominator). So, since , if we have from the sine sum formula. For the denominator, from the cosine sum formula, we have This means so for some positive integer (since the period of tangent is ), or . Note that the inverse of modulo is itself as , so multiplying this congruence by on both sides gives For the smallest possible , we take
See Also
1996 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.