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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Finding smallest positive Integer | AIME I, 1996 Problem 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.

Finding smallest positive Integer – AIME I, 1996


Find the smallest positive integer solution to \(tan19x=\frac{cos96+sin96}{cos96-sin96}\).

  • is 107
  • is 159
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Trigonometry

Integers

Check the Answer


Answer: is 159.

AIME I, 1996, Question 10

Plane Trigonometry by Loney

Try with Hints


\(\frac{cos96+sin96}{cos96-sin96}\)

=\(\frac{sin(90+96)+sin96}{sin(90+96)-sin96}\)

=\(\frac{sin186+sin96}{sin186-sin96}\)

=\(\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}\)

=\(\frac{2sin141cos45}{2cos141sin45}\)

=tan141

here \(tan(180+\theta)\)=\(tan\theta\)

\(\Rightarrow 19x=141+180n\) for some integer n is first equation

multiplying equation with 19 gives

\(x \equiv 141\times 19 \equiv 2679 \equiv 159(mod180)\) [since 2679 divided by 180 gives remainder 159]

\(\Rightarrow x=159\).

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