Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1996 AIME Problems/Problem 10"
(→‎Solution)
Line 11: Line 11:
 
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>.  
 
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>.  
+
Therefore, the smallest positive solution is <math>x = \boxed{159}</math>.
 +
== Solution 2 ==
 +
<math>\dfrac{\cos{96^\circ}}+\sin{96^{\circ}}}{\cos96^{\circ}}-\sin{96^{\circ}}}=<cmath>\dfrac{1+\tan{96^{\circ}}}{1-\tan{96^{\circ}}} = </cmath>\drac{\tan{45^{\circ}}+\tan{96^{\circ}}}{1-\tan{45^{\circ}}\tan{96^{\circ}}} = </math><math> \tan{141^{\circ}}</math>
 +
 
 +
So <math>19x = 141 +180n</math>, for some integer <math>n</math>.
 +
Then <math>0 \equiv 8 + 9n \pmod{19}</math>, from which <math>n \equiv 16 \pmod{19}</math> after some manipulation. The smallest suitable value of <math>n</math> is therefore 16 from which <math>x = \boxed{159}</math>
  
 
== See also ==
 
== See also ==

Revision as of 19:00, 17 August 2015

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 $x = \boxed{159}$.

Solution 2

$\dfrac{\cos{96^\circ}}+\sin{96^{\circ}}}{\cos96^{\circ}}-\sin{96^{\circ}}}=<cmath>\dfrac{1+\tan{96^{\circ}}}{1-\tan{96^{\circ}}} = </cmath>\drac{\tan{45^{\circ}}+\tan{96^{\circ}}}{1-\tan{45^{\circ}}\tan{96^{\circ}}} =$ (Error compiling LaTeX. Unknown error_msg)$\tan{141^{\circ}}$

So $19x = 141 +180n$, for some integer $n$. Then $0 \equiv 8 + 9n \pmod{19}$, from which $n \equiv 16 \pmod{19}$ after some manipulation. The smallest suitable value of $n$ is therefore 16 from which $x = \boxed{159}$

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1996_AIME_Problems/Problem_10&oldid=71676"