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

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Given that <math>\displaystyle (1+\sin t)(1+\cos t)=5/4</math> and
 
Given that <math>\displaystyle (1+\sin t)(1+\cos t)=5/4</math> and
 
:<math>(1-\sin t)(1-\cos t)=\frac mn-\sqrt{k},</math>
 
:<math>(1-\sin t)(1-\cos t)=\frac mn-\sqrt{k},</math>
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where <math>\displaystyle k, m,</math> and <math>\displaystyle n_{}</math> are positive integers with <math>\displaystyle m_{}</math> and <math>\displaystyle n_{}</math> relatively prime, find <math>\displaystyle k+m+n.</math>
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where <math>\displaystyle k, m,</math> and <math>\displaystyle n_{}</math> are [[positive integer]]s with <math>\displaystyle m_{}</math> and <math>\displaystyle n_{}</math> [[relatively prime]], find <math>\displaystyle k+m+n.</math>
  
 
== Solution ==
 
== Solution ==
 +
From the givens,
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<math>2\sin t \cos t + 2 \sin t + 2 \cos t = \frac{1}{2}</math>, and adding <math>\sin^2 t + \cos^2t = 1</math> to both sides gives <math>(\sin t + \cos t)^2 + 2(\sin t + \cos t) = \frac{3}{2}</math>.  Completing the square on the left in the variable <math>(\sin t + \cos t)</math> gives <math>\sin t + \cos t = -1 \pm \sqrt{\frac{5}{2}}</math>.  Since <math>|\sin t + \cos t| \leq \sqrt 2 < 1 + \sqrt{\frac{5}{2}}</math>, we have <math>\sin t + \cos t = \sqrt{5}{2} - 1</math>.  Subtracting twice this from our original equation gives <math>(\sin t - 1)(\cos t - 1) = \sin t \cos t - \sin t - \cos t + 1 = \frac{13}{4} - \sqrt{10}</math>, so the answer is <math>13 + 4 + 10 = 027</math>.
  
 
== See also ==
 
== See also ==
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* [[1995 AIME Problems]]
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{{AIME box|year=1995|num-b=6|num-a=8}}
  
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{{AIME box|year=1995|num-b=6|num-a=8}}
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[[Category:Intermediate Trigonometry Problems]]

Revision as of 22:18, 8 February 2007

Problem

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

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

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

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 $\sin t + \cos t = \sqrt{5}{2} - 1$. 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 $13 + 4 + 10 = 027$.

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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