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

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.

@@ Line 33: / Line 33: @@
 <cmath>\frac{1}{a} + \frac{1}{b} = 30</cmath>
-Taking the tangent of a sum formula from Solution 2, we get <math>\tan x \cdot \tan y = \frac{25}{1 - ab}</math>.
+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>.
@@ Line 46: / Line 46: @@
 -ThisUsernameIsTaken
+Remark: The quadratic need not be solved. The value of <math>ab</math> can be found through Vieta's.
 == See also ==

1986 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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