Difference between revisions of "1996 AIME Problems/Problem 10"

(The first link was not needed.)
Line 3: Line 3:
  
 
== 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>.
 +
 
 +
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.
 +
 
 +
Thus, <math>19x \equiv 141 \pmod{180}</math>.
 +
 
 +
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>.
 +
 
 +
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 19:58, 14 June 2008

Problem

Find the smallest positive integer solution to $\tan{19x^{\circ}}=\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}}$.

Solution

$\dfrac{\cos{96^{\circ}}+\sin{96^{\circ}}}{\cos{96^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{186^{\circ}}+\sin{96^{\circ}}}{\sin{186^{\circ}}-\sin{96^{\circ}}} =$ $\dfrac{\sin{(141^{\circ}+45^{\circ})}+\sin{(141^{\circ}-45^{\circ})}}{\sin{(141^{\circ}+45^{\circ})}-\sin{(141^{\circ}-45^{\circ})}} =$ $\dfrac{2\sin{141^{\circ}}\cos{45^{\circ}}}{2\cos{141^{\circ}}\sin{45^{\circ}}} = \tan{141^{\circ}}$.

The period of the tangent function is $180^\circ$, and the tangent function is one-to-one over each period of its domain.

Thus, $19x \equiv 141 \pmod{180}$.

Since $19^2 \equiv 361 \equiv 1 \pmod{180}$, multiplying both sides by $19$ yields $x \equiv 141 \cdot 19 \equiv (140+1)(18+1) \equiv 0+140+18+1 \equiv 159 \pmod{180}$.

Therefore, the smallest positive solution is $x = \boxed{159}$.

See also

1996 AIME (ProblemsAnswer KeyResources)
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