1991 AIME Problems/Problem 9

Revision as of 12:47, 15 March 2008 by Duelist (talk | contribs) (Solution)

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: $0 = y^2 - \frac{616}{435} - 1$. 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 $m + n = 044$.

Solution 2

This solution is fast for calculus students. Make the substitution $u = \tan \frac x2$. $\tan \frac x2 = \frac{\sin x}{1+\cos x}$, 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:

$\sin x = \frac{2u}{1+u^2}$; $\cos x = \frac{1-u^2}{1+u^2}$

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 (ProblemsAnswer KeyResources)
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