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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

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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