1991 AIME Problems/Problem 9


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 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 the given $\sec x = \frac{22}{7} - \tan x$, we find that

\begin{align*} \sec^2 x &= \left(\frac{22}7\right)^2 - 2\left(\frac{22}7\right)\tan x + \tan^2 x \\ 1 &= \left(\frac{22}7\right)^2 - \frac{44}7 \tan x \end{align*}

This yields $\tan x = \frac{435}{308}$.

Let $y = \frac mn$. Then squaring,

\[\csc^2 x = (y - \cot x)^2 \Longrightarrow 1 = y^2 - 2y\cot x.\]

Substituting $\cot x = \frac{1}{\tan x} = \frac{308}{435}$ yields a quadratic equation: $0 = 435y^2 - 616y - 435 = (15y - 29)(29y + 15)$. It turns out that only the positive root will work, so the value of $y = \frac{29}{15}$ and $m + n = \boxed{044}$.

Note: The problem is much easier computed if we consider what $\sec (x)$ is, then find the relationship between $\sin( x)$ and $cos (x)$ (using $\tan (x) = \frac{435}{308}$, and then computing $\csc x + \cot x$ using $1/\sin x$ and then the reciprocal of $\tan x$.

Solution 2

Recall that $\sec^2 x - \tan^2 x = 1$, from which we find that $\sec x - \tan x = 7/22$. Adding the equations

\begin{eqnarray*} \sec x + \tan x & = & 22/7 \\ \sec x - \tan x & = & 7/22\end{eqnarray*}

together and dividing by 2 gives $\sec x = 533/308$, and subtracting the equations and dividing by 2 gives $\tan x = 435/308$. Hence, $\cos x = 308/533$ and $\sin x = \tan x \cos x = (435/308)(308/533) = 435/533$. Thus, $\csc x = 533/435$ and $\cot x = 308/435$. Finally,

\[\csc x + \cot x = \frac {841}{435} = \frac {29}{15},\]

so $m + n = 044$.

Solution 3 (least computation)

By the given, $\frac {1}{\cos x} + \frac {\sin x}{\cos x} = \frac {22}{7}$ and $\frac {1}{\sin x} + \frac {\cos x}{\sin x} = k$.

Multiplying the two, we have

\[\frac {1}{\sin x \cos x} + \frac {1}{\sin x} + \frac {1}{\cos x} + 1 = \frac {22}{7}k\]

Subtracting both of the two given equations from this, and simpliyfing with the identity $\frac {\sin x}{\cos x} + \frac {\cos x}{\sin x} = \frac{\sin ^2 x + \cos ^2 x}{\sin x \cos x} = \frac {1}{\sin x \cos x}$, we get

\[1 = \frac {22}{7}k - \frac {22}{7} - k.\]

Solving yields $k = \frac {29}{15}$, and $m+n = 044$

Solution 4

Make the substitution $u = \tan \frac x2$ (a substitution commonly used in calculus). By the half-angle identity for tangent, $\tan \frac x2 = \frac{\sin x}{1+\cos x}$, so $\csc x + \cot x = \frac{1+\cos x}{\sin x} = \frac1u = \frac mn$. Also, we have $\sec x + \tan x = \frac{1 + \sin x}{\cos x}.$ 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, $m+n = 044$.

Solution 5

We are given that $\frac{1+\sin x}{\cos x}=\frac{22}7\implies\frac{1+\sin x}{\cos x}\cdot\frac{1-\sin x}{1-\sin x}=\frac{1-\sin^2x}{\cos x(1-\sin x)}=\frac{\cos^2x}{\cos x(1-\sin x)}$ $=\frac{\cos x}{1-\sin x}$, or equivalently, $\cos x=\frac{7+7\sin x}{22}=\frac{22-22\sin x}7\implies\sin x=\frac{22^2-7^2}{22^2+7^2}$ $\implies\cos x=\frac{2\cdot22\cdot7}{22^2+7^2}$. Note that what we want is just $\frac{1+\cos x}{\sin x}=\frac{1+\frac{2\cdot22\cdot7}{22^2+7^2}}{\frac{22^2-7^2}{22^2+7^2}}=\frac{22^2+7^2+2\cdot22\cdot7}{22^2-7^2}=\frac{(22+7)^2}{(22-7)(22+7)}=\frac{22+7}{22-7}$ $=\frac{29}{15}\implies m+n=29+15=\boxed{044}$.

Solution 6

Assign a right triangle with angle $x$, hypotenuse $c$, adjacent side $a$, and opposite side $b$. Then, through the given information above, we have that..

$\frac{c}{a}+\frac{b}{a}=\frac{22}{7}\implies \frac{c+b}{a}=\frac{22}{7}$

$\frac{c}{b}+\frac{a}{b}=\frac{m}{n}\implies \frac{a+c}{b}=\frac{m}{n}$

Hence, because similar right triangles can be scaled up by a factor, we can assume that this particular right triangle is indeed in simplest terms.

Hence, $a=7$, $b+c=22$

Furthermore, by the Pythagorean Theorem, we have that

$a^2+b^2=c^2\implies 49+b^2=c^2$

Solving for $c$ in the first equation and plugging in into the second equation...

$49+b^2=(22-b)^2\implies 49+b^2=484-44b+b^2\implies 44b=435\implies b=\frac{435}{44}$

Hence, $c=22-\frac{435}{44}=\frac{533}{44}$

Now, we want $\frac{a+c}{b}$

Plugging in, we find the answer is $\frac{\frac{7\cdot{44}}{44}+\frac{533}{44}}{\frac{435}{44}}=\frac{841}{435}=\frac{29}{15}$

Hence, the answer is $29+15=\boxed{044}$

Solution 7

We know that $\sec(x) = \frac{h}{a}$ and that $\tan(x) = \frac{o}{a}$ where $h$, $a$, $o$ represent the hypotenuse, adjacent, and opposite (respectively) to angle $x$ in a right triangle. Thus we have that $\sec(x) + \tan(x) = \frac{h+o}{a}$. We also have that $\csc(x) + \cot(x) = \frac{h}{o} + \frac{a}{o} = \frac{h+a}{o}$. Set $\sec(x) + \tan(x) = \alpha$ and csc(x)+cot(x) = $\beta$. Then, notice that $\alpha + \beta = \frac{h+o}{a} + \frac{h+a}{o} = \frac{oh+ah+o^2 + a^2}{oa} = \frac{h(o+a+h)}{oa}$ ( This is because of the Pythagorean Theorem, recall $o^2 +a^2 = h^2$). But then notice that $\alpha \cdot \beta = \frac{(o+h)(a+h)}{oa} = \frac{oa +oh +ha +h^2}{oa} = 1+ \frac{h(o+a+h)}{oa} = 1+ \alpha + \beta$. From the information provided in the question, we can substitute $\alpha$ for $\frac{22}{7}$. Thus, $\frac{22 \beta}{7}= \beta + \frac{29}{7} \Longrightarrow 22 \beta = 7 \beta + 29 \Longrightarrow 15 \beta = 29 \Longrightarrow \beta = \frac{29}{15}$. Since, essentially we are asked to find the sum of the numerator and denominator of $\beta$, we have $29 + 15 = \boxed{044}$.


Solution 8

Firstly, we write $\sec x+\tan x=a/b$ where $a=22$ and $b=7$. This will allow us to spot factorable expressions later. Now, since $\sec^2x-\tan^2x=1$, this gives us \[\sec x-\tan x=\frac{b}{a}\] Adding this to our original expressions gives us \[2\sec x=\frac{a^2+b^2}{ab}\] or \[\cos x=\frac{2ab}{a^2+b^2}\] Now since $\sin^2x+\cos^2x=1$, $\sin x=\sqrt{1-\cos^2x}$ So we can write \[\sin x=\sqrt{1-\frac{4a^2b^2}{(a^2+b^2)^2}}\] Upon simplification, we get \[\sin x=\frac{a^2-b^2}{a^2+b^2}\] We are asked to find $1/\sin x+\cos x/\sin x$ so we can write that as \[\csc x+\cot x=\frac{1}{\sin x}+\frac{\cos x}{\sin x}\] \[\csc x+\cot x=\frac{a^2+b^2}{a^2-b^2}+\frac{2ab}{a^2+b^2}\frac{a^2+b^2}{a^2-b^2}\] \[\csc x+\cot x=\frac{a^2+b^2+2ab}{a^2-b^2}\] \[\csc x+\cot x=\frac{(a+b)^2}{(a-b)(a+b)}\] \[\csc x+\cot x=\frac{a+b}{a-b}\] Now using the fact that $a=22$ and $b=7$ yields, \[\csc x+\cot x=\frac{29}{15}=\frac{p}{q}\] so $p+q=15+29=\boxed{44}$


Solution 9

Rewriting $\sec{x}$ and $\tan{x}$ in terms of $\sin{x}$ and $\cos{x}$, we know that $\frac{1+\sin{x}}{\cos{x}}=\frac{22}{7}.$

Clearing fractions, \[22\cos{x}=7+7\sin{x}.\]

Squaring to get an expression in terms of $\sin^2{x}$ and $\cos^2{x}$, \[484\cos^2{x}=49+49\sin^2{x}+98\sin{x}.\]

Substituting $\cos^2{x}=1-\sin^2{x},$


Expanding then collecting terms yields a quadratic in $\sin{x}:$


To make calculations easier, let $y=\sin{x}.$


Upon inspection, $y=-1$ is a root. Dividing by $y+1$,


Substituting $y=\sin{x},$ we see that $\sin{x}=-1$ doesn't work, as $\cos{x}=0$, leaving $\tan{x}$ undefined.

We conclude that $\sin{x}=\frac{435}{533}.$

Since $\sin^2{x}+\cos^2{x}=1,$

\[\cos{x}=\pm \sqrt{\frac{533^2-435^2}{533^2}}.\] \[=\pm \frac{308}{533}.\]

After checking via the given equation, we know that only the positive solution works.


\[\csc{x}+\cot{x}=\frac{1}{\sin{x}}+\frac{\cos{x}}{\sin{x}}\] \[=\frac{533}{435}+\frac{308}{435}\] \[=\frac{29}{15}=\frac{m}{n}.\]

Adding $m$ and $n$, our answer is $\boxed{\textbf{044}}.$

-Benedict T (countmath1)

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

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