Categories
AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Trigonometry and positive integers | AIME I, 1995 | Question 7

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Trigonometry and positive integers.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Trigonometry and positive integers.

Trigonometry and positive integers – AIME I, 1995


Given that (1+sint)(1+cost)=\(\frac{5}{4}\) and (1-sint)(1-cost)=\(\frac{m}{n}-k^\frac{1}{2}\) where k,m,n are positive integers with m and n relatively prime, find k+m+n.

  • is 107
  • is 27
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Algebra

Trigonometry

Check the Answer


Answer: is 27.

AIME I, 1995, Question 7

Plane Trigonometry by Loney

Try with Hints


Let (1-sint)(1-cost)=x

\(\Rightarrow \frac{5x}{4}=(1-sin^{2}t)(1-cos^{2}t)=sin^{2}tcos^{2}t\)

\(\Rightarrow\frac{(5x)^\frac{1}{2}}{2}=sintcost\)

from given equation sint+cost=\(\frac{5}{4}-(sin^{2}t+cos^{2}t)-sintcost\)=\(\frac{1}{4}-\frac{(5x)^\frac{1}{2}}{2}\)

again (1+sint)(1+cost)-2sintcost=x

\(\Rightarrow x=\frac{5}{4}-2(\frac{1}{4}-\frac{(5x)^\frac{1}{2}}{2})=\frac{3}{4}+(5x)^\frac{1}{2}\)

\(\Rightarrow (x-\frac{3}{4})^{2}=5x\)

\(\Rightarrow x=\frac{13}{4}-(10)^\frac{1}{2}\) for \(x \geq 0\)

\(\Rightarrow 13+4+10=27\).

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *