1996 AIME Problem 10

Attempt Problem 10 of the 1996 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1996 AIME solutions, or check the answer key.

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10.

Find the smallest positive integer solution to tan(19x)=cos96+sin96cos96sin96.\tan(19x^\circ)=\frac{\cos96^\circ+\sin96^\circ}{\cos96^\circ-\sin96^\circ}.

Answer: 159
Concepts:trigonometric identitymodular arithmeticlinear equation
Difficulty rating: 1900
Small Hint:

Recognize the right-hand side using the tangent addition formula

Big Hint:

Solve the resulting congruence modulo 180180

Solution:

The tangent addition formula gives cos96+sin96cos96sin96=tan(45+96)=tan141.\begin{gathered}\frac{\cos96^\circ+\sin96^\circ}{\cos96^\circ-\sin96^\circ}\\=\tan(45^\circ+96^\circ)\\=\tan141^\circ.\end{gathered} Hence 19x141(mod180).19x\equiv141\pmod {180}. Since 192=3611(mod180),19^2=361\equiv1\pmod {180}, multiplying by 1919 gives x19141159(mod180).x\equiv19\cdot141\equiv159\pmod {180}. The smallest positive solution is 159.159.

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