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Difference between revisions of "1995 AIME Problems/Problem 7"

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From this we have <math>\sin x \cos x = \frac{y^2-1}{2}</math> and <math>\sin x + \cos x = y</math>
 
From this we have <math>\sin x \cos x = \frac{y^2-1}{2}</math> and <math>\sin x + \cos x = y</math>
  
−
Substituting this into the first equation we have <math>2y^2+4x-3=0</math>, <math>y = \frac {-2 \pm \sqrt{10}}{2}</math>
+
Substituting this into the first equation we have <math>2y^2+4y-3=0</math>, <math>y = \frac {-2 \pm \sqrt{10}}{2}</math>
  
 
<math>\frac{5}{4} - 2(\frac{-2+\sqrt{10}}{2}) = \frac{13}{4}-\sqrt{10} \rightarrow 13+10+4=\boxed{027}</math>
 
<math>\frac{5}{4} - 2(\frac{-2+\sqrt{10}}{2}) = \frac{13}{4}-\sqrt{10} \rightarrow 13+10+4=\boxed{027}</math>

Revision as of 20:38, 3 January 2019

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, $0 \leq x < 4$. Therefore, $x = \frac{13}{4} - \sqrt{10}$, and the answer is $13 + 4 + 10 = \boxed{027}$.

== Solution 3 == (-synergy)

We have $1+\sin x\cos x+\sin x+\cos x = \frac{5}{4}$

We want to find $1+\sin x \cos x-\sin x-\cos x$

If we find $\sin x+\cos x$, we will be done with the problem.

Let $y = \sin x+\cos x$

Squaring, we have $y^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2\sin x \cos x$

From this we have $\sin x \cos x = \frac{y^2-1}{2}$ and $\sin x + \cos x = y$

Substituting this into the first equation we have $2y^2+4y-3=0$, $y = \frac {-2 \pm \sqrt{10}}{2}$

$\frac{5}{4} - 2(\frac{-2+\sqrt{10}}{2}) = \frac{13}{4}-\sqrt{10} \rightarrow 13+10+4=\boxed{027}$

See also

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

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