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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.

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

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 $x \geq 0$

$\Rightarrow 13+4+10=27$.