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Difference between revisions of "1986 AIME Problems/Problem 3"

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== Problem ==
 
== Problem ==
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If <math>\tan x+\tan y=25</math> and <math>\cot x + \cot y=30</math>, what is <math>\tan(x+y)</math>?
  
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== Solution ==
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== Solution 1 ==
 +
Since <math>\cot</math> is the reciprocal function of <math>\tan</math>:
 +
 
 +
<math>\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30</math>
 +
 
 +
Thus, <math>\tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}</math>
 +
 
 +
Using the tangent addition formula:
 +
 
 +
<math>\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \cdot \tan y} = \frac{25}{1-\frac{5}{6}} = \boxed{150}</math>.
 +
 
 +
== Solution 2 ==
 +
 
 +
Using the formula for tangent of a sum, <math>\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y} = \frac{25}{1-\tan x \tan y}</math>. We only need to find <math>\tan x \tan y</math>.
 +
 
 +
We know that <math>25 = \tan x + \tan y = \frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}</math>. Cross multiplying, we have <math>\frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} = \frac{\sin(x+y)}{\cos x \cos y} = 25</math>.
 +
 
 +
Similarly, we have <math>30 = \cot x + \cot y = \frac{\cos x}{\sin x} + \frac{\cos y}{\sin y} = \frac{\cos x \sin y + \sin x \cos y}{\sin x \sin y} = \frac{\sin(x+y)}{\sin x \sin y}</math>.
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Dividing:
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<math>\frac{25}{30} = \frac{\frac{\sin(x+y)}{\cos x \cos y}}{\frac{\sin(x+y)}{\sin x \sin y}} = \frac{\sin x \sin y}{\cos x \cos y} = \tan x \tan y = \frac{5}{6}</math>. Plugging in to the earlier formula, we have <math>\tan(x+y) = \frac{25}{1-\frac{5}{6}} = \frac{25}{\frac{1}{6}} = \boxed{150}</math>.
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== Solution 3 (less trig required, use of quadratic formula) ==
 +
 
 +
Let <math>a=\tan x</math> and <math>b=\tan y</math>. This simplifies the equations to:
 +
 
 +
<cmath>a + b = 25</cmath>
 +
 
 +
<cmath>\frac{1}{a} + \frac{1}{b} = 30</cmath>
 +
 
 +
Taking the tangent of a sum formula from Solution 2, we get <math>\tan(x+y) = \frac{25}{1 - ab}</math>.
 +
 
 +
We can use substitution to solve the system of equations from above: <math>b = -a + 25</math>, so <math>\frac{1}{a} + \frac{1}{-a + 25} = 30</math>.
 +
 
 +
Multiplying by <math>-a(a-25)</math>,  we get <math>a + (-a + 25) = -30a(a-25)</math>, which is <math>-30a^2 + 750a = 25</math>. Dividing everything by 5 and shifting everything to one side gives <math>6a^2 - 150a + 5 = 0</math>.
 +
 
 +
Using the quadratic formula gives <math>a = \frac{150 \pm \sqrt {22380}}{12}</math>. Since this looks too hard to simplify, we can solve for <math>b</math> using <math>a + b = 25</math>, which turns out to also be <math>b = \frac{150 \pm \sqrt {22380}}{12}</math>, provided that the sign of the radical in <math>a</math> is opposite the one in <math>b</math>.
 +
 
 +
WLOG, assume <math>a = \frac{150 + \sqrt{22380}}{12}</math> and <math>b = \frac{150 - \sqrt{22380}}{12}</math>. Multiplying them gives <math>ab = \frac{22500 - 22380}{144}</math> which simplifies to <math>\frac{5}{6}</math>.
 +
 
 +
THe denominator of <math>\frac{25}{1 - ab}</math> ends up being <math>\frac{1}{6}</math>, so multiplying both numerator and denominator by 6 gives <math>\boxed{150}</math>.
 +
 
 +
-ThisUsernameIsTaken
 +
 
 +
 
 +
 
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Remark: The quadratic need not be solved. The value of <math>ab</math> can be found through Vieta's.
  
 
== See also ==
 
== See also ==
−
* [[1986 AIME Problems]]
 
−
 
 
{{AIME box|year=1986|num-b=2|num-a=4}}
 
{{AIME box|year=1986|num-b=2|num-a=4}}
 +
* [[AIME Problems and Solutions]]
 +
* [[American Invitational Mathematics Examination]]
 +
* [[Mathematics competition resources]]
 +
{{MAA Notice}}

Latest revision as of 22:44, 12 October 2023

Problem

If $\tan x+\tan y=25$ and $\cot x + \cot y=30$, what is $\tan(x+y)$?

Solution 1

Since $\cot$ is the reciprocal function of $\tan$:

$\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30$

Thus, $\tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}$

Using the tangent addition formula:

$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \cdot \tan y} = \frac{25}{1-\frac{5}{6}} = \boxed{150}$.

Solution 2

Using the formula for tangent of a sum, $\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y} = \frac{25}{1-\tan x \tan y}$. We only need to find $\tan x \tan y$.

We know that $25 = \tan x + \tan y = \frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}$. Cross multiplying, we have $\frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} = \frac{\sin(x+y)}{\cos x \cos y} = 25$.

Similarly, we have $30 = \cot x + \cot y = \frac{\cos x}{\sin x} + \frac{\cos y}{\sin y} = \frac{\cos x \sin y + \sin x \cos y}{\sin x \sin y} = \frac{\sin(x+y)}{\sin x \sin y}$.

Dividing:

$\frac{25}{30} = \frac{\frac{\sin(x+y)}{\cos x \cos y}}{\frac{\sin(x+y)}{\sin x \sin y}} = \frac{\sin x \sin y}{\cos x \cos y} = \tan x \tan y = \frac{5}{6}$. Plugging in to the earlier formula, we have $\tan(x+y) = \frac{25}{1-\frac{5}{6}} = \frac{25}{\frac{1}{6}} = \boxed{150}$.

Solution 3 (less trig required, use of quadratic formula)

Let $a=\tan x$ and $b=\tan y$. This simplifies the equations to:

\[a + b = 25\]

\[\frac{1}{a} + \frac{1}{b} = 30\]

Taking the tangent of a sum formula from Solution 2, we get $\tan(x+y) = \frac{25}{1 - ab}$.

We can use substitution to solve the system of equations from above: $b = -a + 25$, so $\frac{1}{a} + \frac{1}{-a + 25} = 30$.

Multiplying by $-a(a-25)$, we get $a + (-a + 25) = -30a(a-25)$, which is $-30a^2 + 750a = 25$. Dividing everything by 5 and shifting everything to one side gives $6a^2 - 150a + 5 = 0$.

Using the quadratic formula gives $a = \frac{150 \pm \sqrt {22380}}{12}$. Since this looks too hard to simplify, we can solve for $b$ using $a + b = 25$, which turns out to also be $b = \frac{150 \pm \sqrt {22380}}{12}$, provided that the sign of the radical in $a$ is opposite the one in $b$.

WLOG, assume $a = \frac{150 + \sqrt{22380}}{12}$ and $b = \frac{150 - \sqrt{22380}}{12}$. Multiplying them gives $ab = \frac{22500 - 22380}{144}$ which simplifies to $\frac{5}{6}$.

THe denominator of $\frac{25}{1 - ab}$ ends up being $\frac{1}{6}$, so multiplying both numerator and denominator by 6 gives $\boxed{150}$.

-ThisUsernameIsTaken


Remark: The quadratic need not be solved. The value of $ab$ can be found through Vieta's.

See also

1986 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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