Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

(→‎Solution)
(New solution)
Line 24: Line 24:
  
 
<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>.
 
<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>.
 +
 +
== 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>\frac{25}{1 - ab}</math>.
 +
 +
We can use substitution to solve the system of equations. <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
  
 
== See also ==
 
== See also ==

Revision as of 16:27, 13 March 2021

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 $\frac{25}{1 - ab}$.

We can use substitution to solve the system of equations. $b = -a + 25$, 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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_3&oldid=149420"