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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

### Key Concepts

Integers

Algebra

Trigonometry

But try the problem first…

Source

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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