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 "1991 AIME Problems/Problem 9"

(→‎Solution)
(fmtt)
Line 2: Line 2:
 
Suppose that <math>\sec x+\tan x=\frac{22}7</math> and that <math>\csc x+\cot x=\frac mn,</math> where <math>\frac mn</math> is in lowest terms.  Find <math>m+n^{}_{}.</math>
 
Suppose that <math>\sec x+\tan x=\frac{22}7</math> and that <math>\csc x+\cot x=\frac mn,</math> where <math>\frac mn</math> is in lowest terms.  Find <math>m+n^{}_{}.</math>
  
 +
__TOC__
 
== Solution ==
 
== Solution ==
−
== Solution 1 ==
+
=== Solution 1 ===
 
Use the two [[Trigonometric identities#Pythagorean Identities|trigonometric Pythagorean identities]] <math>1 + \tan^2 x = \sec^2 x</math> and <math>1 + \cot^2 x = \csc^2 x</math>.  
 
Use the two [[Trigonometric identities#Pythagorean Identities|trigonometric Pythagorean identities]] <math>1 + \tan^2 x = \sec^2 x</math> and <math>1 + \cot^2 x = \csc^2 x</math>.  
  
 
If we square <math>\sec x = \frac{22}{7} - \tan x</math>, we find that <math>\sec^2 x = \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x</math>, so <math>1 = \left(\frac{22}7\right)^2 - \frac{44}7 \tan x</math>. Solving shows that <math>\tan x = \frac{435}{308}</math>.  
 
If we square <math>\sec x = \frac{22}{7} - \tan x</math>, we find that <math>\sec^2 x = \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x</math>, so <math>1 = \left(\frac{22}7\right)^2 - \frac{44}7 \tan x</math>. Solving shows that <math>\tan x = \frac{435}{308}</math>.  
  
−
Call <math>y = \frac mn</math>. Rewrite the second equation in a similar fashion: <math>1 = y^2 - 2y\cot x</math>. Substitute in <math>\cot x = \frac{1}{\tan x} = \frac{308}{435}</math> to get a [[quadratic equation|quadratic]]: <math>0 = y^2 - \frac{616}{435} - 1</math>. The quadratic is [[factor]]able (though somewhat ugly); <math>(15y - 29)(29y + 15) = 0</math>. It turns out that only the [[positive]] root will work, so the value of <math>y = \frac{29}{15}</math> and <math>m + n = 044</math>.
+
Call <math>y = \frac mn</math>. Rewrite the second equation in a similar fashion: <math>1 = y^2 - 2y\cot x</math>. Substitute in <math>\cot x = \frac{1}{\tan x} = \frac{308}{435}</math> to get a [[quadratic equation|quadratic]]: <math>0 = y^2 - \frac{616}{435}y - 1</math>. This factors as <math>(15y - 29)(29y + 15) = 0</math>. It turns out that only the [[positive]] root will work, so the value of <math>y = \frac{29}{15}</math> and <math>m + n = \boxed{044}</math>.
−
 
 
−
== Solution 2==
 
−
 
 
−
This solution is fast for calculus students. Make the substitution <math>u = \tan \frac x2</math>. <math>\tan \frac x2 = \frac{\sin x}{1+\cos x}</math>, so <math>\csc x + \cot x = \frac{1+\cos x}{\sin x} = \frac1u = \frac mn</math>. <math>\sec x + \tan x = \frac{1 + \sin x}{\cos x}.</math> Now note the following:
 
−
 
 
−
<math>\sin x = \frac{2u}{1+u^2}</math>;
 
−
<math>\cos x = \frac{1-u^2}{1+u^2}</math>
 
  
 +
=== Solution 2===
 +
Make the substitution <math>u = \tan \frac x2</math> (a substitution commonly used in calculus). <math>\tan \frac x2 = \frac{\sin x}{1+\cos x}</math>, so <math>\csc x + \cot x = \frac{1+\cos x}{\sin x} = \frac1u = \frac mn</math>. <math>\sec x + \tan x = \frac{1 + \sin x}{\cos x}.</math> Now note the following:
 +
<cmath>\begin{align*}\sin x &= \frac{2u}{1+u^2}\\
 +
\cos x &= \frac{1-u^2}{1+u^2}\end{align*}</cmath>
 
Plugging these into our equality gives:
 
Plugging these into our equality gives:
−
 
+
<cmath>\frac{1+\frac{2u}{1+u^2}}{\frac{1-u^2}{1+u^2}} = \frac{22}7</cmath>
−
<math>\frac{1+\frac{2u}{1+u^2}}{\frac{1-u^2}{1+u^2}} = \frac{22}7</math>
 
  
 
This simplifies to <math>\frac{1+u}{1-u} = \frac{22}7</math>, and solving for <math>u</math> gives <math>u = \frac{15}{29}</math>, and <math>\frac mn = \frac{29}{15}</math>. Finally, <math>m+n = 044</math>.
 
This simplifies to <math>\frac{1+u}{1-u} = \frac{22}7</math>, and solving for <math>u</math> gives <math>u = \frac{15}{29}</math>, and <math>\frac mn = \frac{29}{15}</math>. Finally, <math>m+n = 044</math>.

Revision as of 21:21, 11 April 2008

Problem

Suppose that $\sec x+\tan x=\frac{22}7$ and that $\csc x+\cot x=\frac mn,$ where $\frac mn$ is in lowest terms. Find $m+n^{}_{}.$

Solution

Solution 1

Use the two trigonometric Pythagorean identities $1 + \tan^2 x = \sec^2 x$ and $1 + \cot^2 x = \csc^2 x$.

If we square $\sec x = \frac{22}{7} - \tan x$, we find that $\sec^2 x = \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x$, so $1 = \left(\frac{22}7\right)^2 - \frac{44}7 \tan x$. Solving shows that $\tan x = \frac{435}{308}$.

Call $y = \frac mn$. Rewrite the second equation in a similar fashion: $1 = y^2 - 2y\cot x$. Substitute in $\cot x = \frac{1}{\tan x} = \frac{308}{435}$ to get a quadratic: $0 = y^2 - \frac{616}{435}y - 1$. This factors as $(15y - 29)(29y + 15) = 0$. It turns out that only the positive root will work, so the value of $y = \frac{29}{15}$ and $m + n = \boxed{044}$.

Solution 2

Make the substitution $u = \tan \frac x2$ (a substitution commonly used in calculus). $\tan \frac x2 = \frac{\sin x}{1+\cos x}$, so $\csc x + \cot x = \frac{1+\cos x}{\sin x} = \frac1u = \frac mn$. $\sec x + \tan x = \frac{1 + \sin x}{\cos x}.$ Now note the following: \begin{align*}\sin x &= \frac{2u}{1+u^2}\\ \cos x &= \frac{1-u^2}{1+u^2}\end{align*} Plugging these into our equality gives: \[\frac{1+\frac{2u}{1+u^2}}{\frac{1-u^2}{1+u^2}} = \frac{22}7\]

This simplifies to $\frac{1+u}{1-u} = \frac{22}7$, and solving for $u$ gives $u = \frac{15}{29}$, and $\frac mn = \frac{29}{15}$. Finally, $m+n = 044$.

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_9&oldid=24675"