Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1995 AIME Problems/Problem 7"
(→‎Solution)
Line 7: Line 7:
 
From the givens,  
 
From the givens,  
 
<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{\frac{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 = \boxed{027}</math>.
 
<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{\frac{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 = \boxed{027}</math>.
 +
 +
==Solution 2==
 +
Let <math>(1 - \sin t)(1 - \cos t) = x</math>. Multiplying <math>x</math> with the given equation, <math>\frac{5x}{4} = (1 - \sin^2 t)(1 - \cos^2 t) = \sin^2 t \cos ^2 t</math>, and <math>\frac{\sqrt{5x}}{2} = \sin t \cos t</math>. Simplifying and rearranging the given equation, <math>\sin t + \cos t = \frac{5}{4} - (\sin^2 t + \cos^2 t) - \sin t \cos t = \frac{1}{4} - \frac{\sqrt{5x}}{2}</math>. Notice that <math>(1 + \sin t)(1 + \cos t) - 2(\sin t + \cos t) = x</math>, and substituting, <math>x = \frac{5}{4} - 2( \frac{1}{4} - \frac{\sqrt{5x}}{2}) = \frac{3}{4} + \sqrt{5x}</math>. Rearranging, <math>\sqrt{5x} = x - \frac{3}{4}</math>, and squaring, <math>5x = x^2 - \frac{3}{2} x + \frac{9}{16}</math>, so <math>x^2 - \frac{13}{2} + \frac{9}{16} = 0</math>, and <math>x = \frac{13}{4} \pm \sqrt{10}</math>, but as both parts of the product <math>x</math> doesn't exceed <math>2</math>, <math>x < 4</math>. Therefore, <math>x = \frac{13}{4} - \sqrt{10}</math>, and the answer is <math> 13 + 4 + 10 = \boxed{027}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 22:57, 10 January 2014

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, $\sqrt{5x} = x - \frac{3}{4}$, and squaring, $5x = x^2 - \frac{3}{2} x + \frac{9}{16}$, so $x^2 - \frac{13}{2} + \frac{9}{16} = 0$, 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 $13 + 4 + 10 = \boxed{027}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_7&oldid=58677"