Difference between revisions of "1991 AIME Problems/Problem 9"
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== Problem == | == Problem == | ||
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> | ||
− | + | == 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>. | ||
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Substituting <math>\cot x = \frac{1}{\tan x} = \frac{308}{435}</math> yields a [[quadratic equation]]: <math>0 = 435y^2 - 616y - 435 = (15y - 29)(29y + 15)</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>. | Substituting <math>\cot x = \frac{1}{\tan x} = \frac{308}{435}</math> yields a [[quadratic equation]]: <math>0 = 435y^2 - 616y - 435 = (15y - 29)(29y + 15)</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 | + | Note: The problem is much easier computed if we consider what <math>\sec (x)</math> is, then find the relationship between <math>\sin( x)</math> and <math>cos (x)</math> (using <math>\tan (x) = \frac{435}{308}</math>, and then computing <math>\csc x + \cot x</math> using <math>1/\sin x</math> and then the reciprocal of <math>\tan x</math>. |
+ | |||
+ | == Solution 2 == | ||
Recall that <math>\sec^2 x - \tan^2 x = 1</math>, from which we find that <math>\sec x - \tan x = 7/22</math>. Adding the equations | Recall that <math>\sec^2 x - \tan^2 x = 1</math>, from which we find that <math>\sec x - \tan x = 7/22</math>. Adding the equations | ||
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so <math>m + n = 044</math>. | so <math>m + n = 044</math>. | ||
− | + | == Solution 3 (least computation)== | |
− | (least computation) By the given, | + | By the given, |
<math>\frac {1}{\cos x} + \frac {\sin x}{\cos x} = \frac {22}{7}</math> and | <math>\frac {1}{\cos x} + \frac {\sin x}{\cos x} = \frac {22}{7}</math> and | ||
<math>\frac {1}{\sin x} + \frac {\cos x}{\sin x} = k</math>. | <math>\frac {1}{\sin x} + \frac {\cos x}{\sin x} = k</math>. | ||
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<cmath>1 = \frac {22}{7}k - \frac {22}{7} - k.</cmath> | <cmath>1 = \frac {22}{7}k - \frac {22}{7} - k.</cmath> | ||
− | Solving yields <math>k = \frac {29}{15}</math>, and <math>m+n = 044</math> | + | Solving yields <math>k = \frac {29}{15}</math>, and <math>m+n = 044</math> |
− | + | == Solution 4 == | |
− | 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: | + | Make the substitution <math>u = \tan \frac x2</math> (a substitution commonly used in calculus). By the half-angle identity for tangent, <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>. Also, we have <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}\ | <cmath>\begin{align*}\sin x &= \frac{2u}{1+u^2}\ | ||
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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>. | ||
− | + | == Solution 5 == | |
We are given that <math>\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)}</math> | We are given that <math>\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)}</math> | ||
<math>=\frac{\cos x}{1-\sin x}</math>, or equivalently, <math>\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}</math> | <math>=\frac{\cos x}{1-\sin x}</math>, or equivalently, <math>\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}</math> | ||
<math>\implies\cos x=\frac{2\cdot22\cdot7}{22^2+7^2}</math>. Note that what we want is just <math>\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}</math> | <math>\implies\cos x=\frac{2\cdot22\cdot7}{22^2+7^2}</math>. Note that what we want is just <math>\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}</math> | ||
<math>=\frac{29}{15}\implies m+n=29+15=\boxed{044}</math>. | <math>=\frac{29}{15}\implies m+n=29+15=\boxed{044}</math>. | ||
+ | |||
+ | == Solution 6 == | ||
+ | Assign a right triangle with angle <math>x</math>, hypotenuse <math>c</math>, adjacent side <math>a</math>, and opposite side <math>b</math>. | ||
+ | Then, through the given information above, we have that.. | ||
+ | |||
+ | <math>\frac{c}{a}+\frac{b}{a}=\frac{22}{7}\implies \frac{c+b}{a}=\frac{22}{7}</math> | ||
+ | |||
+ | <math>\frac{c}{b}+\frac{a}{b}=\frac{m}{n}\implies \frac{a+c}{b}=\frac{m}{n}</math> | ||
+ | |||
+ | 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, <math>a=7</math>, <math>b+c=22</math> | ||
+ | |||
+ | Furthermore, by the Pythagorean Theorem, we have that | ||
+ | |||
+ | <math>a^2+b^2=c^2\implies 49+b^2=c^2</math> | ||
+ | |||
+ | Solving for <math>c</math> in the first equation and plugging in into the second equation... | ||
+ | |||
+ | <math>49+b^2=(22-b)^2\implies 49+b^2=484-44b+b^2\implies 44b=435\implies b=\frac{435}{44}</math> | ||
+ | |||
+ | Hence, <math>c=22-\frac{435}{44}=\frac{533}{44}</math> | ||
+ | |||
+ | Now, we want <math>\frac{a+c}{b}</math> | ||
+ | |||
+ | Plugging in, we find the answer is <math>\frac{\frac{7\cdot{44}}{44}+\frac{533}{44}}{\frac{435}{44}}=\frac{841}{435}=\frac{29}{15}</math> | ||
+ | |||
+ | Hence, the answer is <math>29+15=\boxed{044}</math> | ||
+ | |||
+ | == Solution 7 == | ||
+ | |||
+ | We know that <math>\sec(x) = \frac{h}{a} </math> and that <math>\tan(x) = \frac{o}{a} </math> where <math>h</math>, <math>a</math>, <math>o</math> represent the hypotenuse, adjacent, and opposite (respectively) to angle <math>x</math> in a right triangle. Thus we have that <math>\sec(x) + \tan(x) = \frac{h+o}{a}</math>. We also have that <math>\csc(x) + \cot(x) = \frac{h}{o} + \frac{a}{o} = \frac{h+a}{o} </math>. Set <math>\sec(x) + \tan(x) = \alpha</math> and csc(x)+cot(x) = <math>\beta</math>. Then, notice that <math>\alpha + \beta = \frac{h+o}{a} + \frac{h+a}{o} = \frac{oh+ah+o^2 + a^2}{oa} = \frac{h(o+a+h)}{oa}</math> ( This is because of the Pythagorean Theorem, recall <math>o^2 +a^2 = h^2</math>). But then notice that <math>\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</math>. From the information provided in the question, we can substitute <math>\alpha</math> for <math>\frac{22}{7}</math>. Thus, <math>\frac{22 \beta}{7}= \beta + \frac{29}{7} \Longrightarrow 22 \beta = 7 \beta + 29 \Longrightarrow 15 \beta = 29 \Longrightarrow \beta = \frac{29}{15}</math>. Since, essentially we are asked to find the sum of the numerator and denominator of <math>\beta</math>, we have <math>29 + 15 = \boxed{044}</math>. | ||
+ | |||
+ | ~qwertysri987 | ||
+ | |||
+ | |||
+ | == Solution 8 == | ||
+ | |||
+ | Firstly, we write <math>\sec x+\tan x=a/b</math> where <math>a=22</math> and <math>b=7</math>. This will allow us to spot factorable expressions later. Now, since <math>\sec^2x-\tan^2x=1</math>, this gives us <cmath>\sec x-\tan x=\frac{b}{a}</cmath> Adding this to our original expressions gives us <cmath>2\sec x=\frac{a^2+b^2}{ab}</cmath> or <cmath>\cos x=\frac{2ab}{a^2+b^2}</cmath> Now since <math>\sin^2x+\cos^2x=1</math>, <math>\sin x=\sqrt{1-\cos^2x}</math> So we can write <cmath>\sin x=\sqrt{1-\frac{4a^2b^2}{(a^2+b^2)^2}}</cmath> Upon simplification, we get <cmath>\sin x=\frac{a^2-b^2}{a^2+b^2}</cmath> We are asked to find <math>1/\sin x+\cos x/\sin x</math> so we can write that as <cmath>\csc x+\cot x=\frac{1}{\sin x}+\frac{\cos x}{\sin x}</cmath> <cmath>\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}</cmath> <cmath>\csc x+\cot x=\frac{a^2+b^2+2ab}{a^2-b^2}</cmath> <cmath>\csc x+\cot x=\frac{(a+b)^2}{(a-b)(a+b)}</cmath> <cmath>\csc x+\cot x=\frac{a+b}{a-b}</cmath> Now using the fact that <math>a=22</math> and <math>b=7</math> yields, <cmath>\csc x+\cot x=\frac{29}{15}=\frac{p}{q}</cmath> so <math>p+q=15+29=\boxed{44}</math> | ||
+ | |||
+ | ~Chessmaster20000 | ||
+ | |||
+ | == Solution 9 == | ||
+ | Rewriting <math>\sec{x}</math> and <math>\tan{x}</math> in terms of <math>\sin{x}</math> and <math>\cos{x}</math>, we know that <math>\frac{1+\sin{x}}{\cos{x}}=\frac{22}{7}.</math> | ||
+ | |||
+ | Clearing fractions, | ||
+ | <cmath>22\cos{x}=7+7\sin{x}.</cmath> | ||
+ | |||
+ | Squaring to get an expression in terms of <math>\sin^2{x}</math> and <math>\cos^2{x}</math>, | ||
+ | <cmath>484\cos^2{x}=49+49\sin^2{x}+98\sin{x}.</cmath> | ||
+ | |||
+ | Substituting <math>\cos^2{x}=1-\sin^2{x},</math> | ||
+ | |||
+ | <cmath>484(1-\sin^2{x})=49+49\sin^2{x}+98\sin{x}.</cmath> | ||
+ | |||
+ | Expanding then collecting terms yields a quadratic in <math>\sin{x}:</math> | ||
+ | |||
+ | <cmath>533\sin^2{x}+98\sin{x}-435=0.</cmath> | ||
+ | |||
+ | To make calculations easier, let <math>y=\sin{x}.</math> | ||
+ | |||
+ | <cmath>533y^2+98y-435=0.</cmath> | ||
+ | |||
+ | Upon inspection, <math>y=-1</math> is a root. Dividing by <math>y+1</math>, | ||
+ | |||
+ | <cmath>533y^2+98y-435=(533y-435)(y+1).</cmath> | ||
+ | |||
+ | Substituting <math>y=\sin{x},</math> we see that <math>\sin{x}=-1</math> doesn't work, as <math>\cos{x}=0</math>, leaving <math>\tan{x}</math> undefined. | ||
+ | |||
+ | We conclude that <math>\sin{x}=\frac{435}{533}.</math> | ||
+ | |||
+ | Since <math>\sin^2{x}+\cos^2{x}=1,</math> | ||
+ | |||
+ | <cmath>\cos{x}=\pm \sqrt{\frac{533^2-435^2}{533^2}}.</cmath> | ||
+ | <cmath>=\pm \frac{308}{533}.</cmath> | ||
+ | |||
+ | After checking via the given equation, we know that only the positive solution works. | ||
+ | |||
+ | Therefore, | ||
+ | |||
+ | <cmath>\csc{x}+\cot{x}=\frac{1}{\sin{x}}+\frac{\cos{x}}{\sin{x}}</cmath> | ||
+ | <cmath>=\frac{533}{435}+\frac{308}{435}</cmath> | ||
+ | <cmath>=\frac{29}{15}=\frac{m}{n}.</cmath> | ||
+ | |||
+ | Adding <math>m</math> and <math>n</math>, our answer is <math>\boxed{\textbf{044}}.</math> | ||
+ | |||
+ | -Benedict T (countmath1) | ||
== See also == | == See also == | ||
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[[Category:Intermediate Trigonometry Problems]] | [[Category:Intermediate Trigonometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:04, 20 January 2023
Contents
[hide]Problem
Suppose that and that where is in lowest terms. Find
Solution 1
Use the two trigonometric Pythagorean identities and .
If we square the given , we find that
This yields .
Let . Then squaring,
Substituting yields a quadratic equation: . It turns out that only the positive root will work, so the value of and .
Note: The problem is much easier computed if we consider what is, then find the relationship between and (using , and then computing using and then the reciprocal of .
Solution 2
Recall that , from which we find that . Adding the equations
together and dividing by 2 gives , and subtracting the equations and dividing by 2 gives . Hence, and . Thus, and . Finally,
so .
Solution 3 (least computation)
By the given, and .
Multiplying the two, we have
Subtracting both of the two given equations from this, and simpliyfing with the identity , we get
Solving yields , and
Solution 4
Make the substitution (a substitution commonly used in calculus). By the half-angle identity for tangent, , so . Also, we have Now note the following:
Plugging these into our equality gives:
This simplifies to , and solving for gives , and . Finally, .
Solution 5
We are given that , or equivalently, . Note that what we want is just .
Solution 6
Assign a right triangle with angle , hypotenuse , adjacent side , and opposite side . Then, through the given information above, we have that..
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, ,
Furthermore, by the Pythagorean Theorem, we have that
Solving for in the first equation and plugging in into the second equation...
Hence,
Now, we want
Plugging in, we find the answer is
Hence, the answer is
Solution 7
We know that and that where , , represent the hypotenuse, adjacent, and opposite (respectively) to angle in a right triangle. Thus we have that . We also have that . Set and csc(x)+cot(x) = . Then, notice that ( This is because of the Pythagorean Theorem, recall ). But then notice that . From the information provided in the question, we can substitute for . Thus, . Since, essentially we are asked to find the sum of the numerator and denominator of , we have .
~qwertysri987
Solution 8
Firstly, we write where and . This will allow us to spot factorable expressions later. Now, since , this gives us Adding this to our original expressions gives us or Now since , So we can write Upon simplification, we get We are asked to find so we can write that as Now using the fact that and yields, so
~Chessmaster20000
Solution 9
Rewriting and in terms of and , we know that
Clearing fractions,
Squaring to get an expression in terms of and ,
Substituting
Expanding then collecting terms yields a quadratic in
To make calculations easier, let
Upon inspection, is a root. Dividing by ,
Substituting we see that doesn't work, as , leaving undefined.
We conclude that
Since
After checking via the given equation, we know that only the positive solution works.
Therefore,
Adding and , our answer is
-Benedict T (countmath1)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.