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Difference between revisions of "1991 AIME Problems/Problem 9"

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(solution, though I'm sure there is a more elegant manner of going about this)
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== Solution ==
 
== Solution ==
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{{solution}}
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Use the two [[Trigonometric identities#Pythagorean Identities|trigonometric Pythagorean identities]] <math>\displaystyle 1 + \tan^2 x = \sec^2 x</math> and <math>\displaystyle 1 + \cot^2 x = \csc^2 x</math>.
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If we square <math>\sec x = \frac{22}{7} - \tan x</math>, we find that <math>\sec^2 x = (\frac{22}7)^2 - 2(\frac{22}7)\tan x + \tan^2 x</math>, so <math>1 = (\frac{22}7)^2 - \frac{44}7 \tan x</math>. Solving shows that <math>\tan x = \frac{435}{308}</math>.
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Call <math>y = \frac mn</math>. Rewrite the second equation in a similar fashion: <math>\displaystyle 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>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=8|num-a=10}}
 
{{AIME box|year=1991|num-b=8|num-a=10}}
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[[Category:Intermediate Trigonometry Problems]]

Revision as of 19:10, 11 March 2007

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

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

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

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

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
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