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 2)
(→‎Solution 2)
Line 13: Line 13:
 
SOLUTION 3 (-synergy)
 
SOLUTION 3 (-synergy)
  
−
We have <math>1+\sinx\cosx+\sinx+\cosx = \frac{5}{4}</math>
+
We have <math>1+\sin x\cos x+\sin x+\cos x = \frac{5}{4}</math>
  
−
We want to find <math>1+\sinx\cosx-\sinx-\cosx</math>
+
We want to find <math>1+\sin x \cos x-\sin x-\cos x</math>
  
−
If we find \sinx+\cosx, we will be done with the problem.
+
If we find \sin x+\cos x, we will be done with the problem.
  
−
Let <math>y = \sinx+\cosx</math>
+
Let <math>y = \sin x+\cos x</math>
  
−
Squaring, we have <math>y^2 = \sin^2 x + \cos^2 x + 2\sinx\cosx = 1 + 2\sinx\cosx</math>
+
Squaring, we have <math>y^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2\sin x \cos x</math>
  
−
From this we have <math>\sinx\cosx = y</math> and <math>\sinx + \cos x = \frac{y^2-1}{2}</math>
+
From this we have <math>\sin x \cos x = y</math> and <math>\sin x + \cos x = \frac{y^2-1}{2}</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+4x-3=0</math>, <math>y = \frac {-2 \pm \sqrt{10}}{2}</math>

Revision as of 11:51, 16 June 2018

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

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

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

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

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=95193"