1995 AIME Problems/Problem 7

Problem

Given that $(1+\sin t)(1+\cos t)=5/4$ and

$(1-\sin t)(1-\cos t)=\frac mn-\sqrt{k},$

where $k, m,$ and $n_{}$ are positive integers with $m_{}$ and $n_{}$ relatively prime, find $k+m+n.$

Solution

From the givens, $2\sin t \cos t + 2 \sin t + 2 \cos t = \frac{1}{2}$ , and adding $\sin^2 t + \cos^2t = 1$ to both sides gives $(\sin t + \cos t)^2 + 2(\sin t + \cos t) = \frac{3}{2}$ . Completing the square on the left in the variable $(\sin t + \cos t)$ gives $\sin t + \cos t = -1 \pm \sqrt{\frac{5}{2}}$ . Since $|\sin t + \cos t| \leq \sqrt 2 < 1 + \sqrt{\frac{5}{2}}$ , we have $\sin t + \cos t = \sqrt{\frac{5}{2}} - 1$ . Subtracting twice this from our original equation gives $(\sin t - 1)(\cos t - 1) = \sin t \cos t - \sin t - \cos t + 1 = \frac{13}{4} - \sqrt{10}$ , so the answer is $13 + 4 + 10 = \boxed{027}$ .

Solution 2

Let $(1 - \sin t)(1 - \cos t) = x$ . Multiplying $x$ with the given equation, $\frac{5x}{4} = (1 - \sin^2 t)(1 - \cos^2 t) = \sin^2 t \cos ^2 t$ , and $\frac{\sqrt{5x}}{2} = \sin t \cos t$ . Simplifying and rearranging the given equation, $\sin t + \cos t = \frac{5}{4} - (\sin^2 t + \cos^2 t) - \sin t \cos t = \frac{1}{4} - \frac{\sqrt{5x}}{2}$ . Notice that $(1 + \sin t)(1 + \cos t) - 2(\sin t + \cos t) = x$ , and substituting, $x = \frac{5}{4} - 2( \frac{1}{4} - \frac{\sqrt{5x}}{2}) = \frac{3}{4} + \sqrt{5x}$ . Rearranging and squaring, $5x = x^2 - \frac{3}{2} x + \frac{9}{16}$ , so $x^2 - \frac{13}{2} x + \frac{9}{16} = 0$ , and $x = \frac{13}{4} \pm \sqrt{10}$ , but as both parts of the product $x$ doesn't exceed $2$ , $x < 4$ . Therefore, $x = \frac{13}{4} - \sqrt{10}$ , and the answer is $13 + 4 + 10 = \boxed{027}$ .

1995 AIME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

1995 AIME Problems/Problem 7

Contents

Problem

Solution

Solution 2

See also