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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.

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


First hint

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

Second Hint

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 \(x \geq 0\)

Final Step

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

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