# Difference between revisions of "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 clearly, $x < 4$. Therefore, $x = \frac{13}{4} - \sqrt{10}$, and the answer is $13 + 4 + 10 = \boxed{027}$.