Difference between revisions of "1996 AIME Problems/Problem 10"
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== Solution == | == Solution == | ||
| − | {{solution} | + | <math>\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} =</math> <math>\dfrac{\sin{186^{\circ}}+\sin{96^{\circ}}}{\sin{186^{\circ}}-\sin{96^{\circ}}} =</math> <math>\dfrac{\sin{(141^{\circ}+45^{\circ})}+\sin{(141^{\circ}-45^{\circ})}}{\sin{(141^{\circ}+45^{\circ})}-\sin{(141^{\circ}-45^{\circ})}} =</math> <math>\dfrac{2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}</math>. |
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| + | The period of the tangent function is <math>180^\circ</math>, and the tangent function is one-to-one over each period of its domain. | ||
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| + | Thus, <math>19x \equiv 141 \pmod{180}</math>. | ||
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| + | Since <math>19^2 \equiv 361 \equiv 1 \pmod{180}</math>, multiplying both sides by <math>19</math> yields <math>x \equiv 141 \cdot 19 \equiv (140+1)(18+1) \equiv 0+140+18+1 \equiv 159 \pmod{180}</math>. | ||
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| + | Therefore, the smallest positive solution is <math>x = \boxed{159}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=1996|num-b=9|num-a=11}} | {{AIME box|year=1996|num-b=9|num-a=11}} | ||
Revision as of 18:58, 14 June 2008
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 
.
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 | ||
