Difference between revisions of "User:Temperal/The Problem Solver's Resource Tips and Tricks"

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{{User:Temperal/testtemplate|the tips and tricks section}}
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|+ <span style="background:aqua; border:1px solid black; opacity: 0.6;font-size:30px;position:relative;bottom:8px;border-width: 5px;border-color:blue;border-style: groove;position:absolute;top:50px;right:155px;width:820px;height:40px;padding:5px;">The Problem Solver's Resource</span>
 
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==<span style="font-size:20px; color: blue;">Other Tips and Tricks</span>==
 
==<span style="font-size:20px; color: blue;">Other Tips and Tricks</span>==
 
This is a collection of general techniques for solving problems.
 
This is a collection of general techniques for solving problems.
 
*Don't be afraid to use casework! Sometimes it's the only way. (But be VERY afraid to use brute force.)
 
*Don't be afraid to use casework! Sometimes it's the only way. (But be VERY afraid to use brute force.)
 
*Remember that substitution is a useful technique! Example problem:  
 
*Remember that substitution is a useful technique! Example problem:  
===Example Problem Number 1===
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===Example Problem 1===
If <math>\tan x+\tan y=25</math> and <math>\cot x+\ cot y=30</math>, find <math>\tan(x+y)</math>.
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If <math>\tan x+\tan y=25</math> and <math>\cot x+\cot y=30</math>, find <math>\tan(x+y)</math>.
  
 
====Solution====
 
====Solution====
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*Remember the special properties of odd numbers: For any odd number <math>o</math>, <math>o=2n\pm 1</math> for some integer <math>n</math>, and <math>o=a^2-(a-1)^2</math> for some positive integer <math>a</math>.
 
*Remember the special properties of odd numbers: For any odd number <math>o</math>, <math>o=2n\pm 1</math> for some integer <math>n</math>, and <math>o=a^2-(a-1)^2</math> for some positive integer <math>a</math>.
  
===Example Problem Number 2===
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===Example Problem 2===
How many quadruples <math>(a,b,c,d)</math> such that <math>a+b+c+d=98</math> where <math>a,b,c,d</math> are all odd?
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How many quadruples <math>(a,b,c,d)</math> are there such that <math>a+b+c+d=98</math> and <math>a,b,c,d</math> are all odd?
  
 
====Solution====
 
====Solution====
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<math>\Rightarrow 2(n_1+n_2+n_3+n_4)=94</math>
 
<math>\Rightarrow 2(n_1+n_2+n_3+n_4)=94</math>
  
<math>\Rightarrow n_1+n_2+n_3+n_4=47[/latex]
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<math>\Rightarrow n_1+n_2+n_3+n_4=47</math>
Binomial coefficients will yield the answer of </math>\boxed{19600}$.  
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Binomial coefficients will yield the answer of <math>\boxed{19600}</math>.  
  
 
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#Everything on the Combinatorics page.
 
#Everything on the Combinatorics page.
 
#Integrals and derivatives, especially integrals.
 
#Integrals and derivatives, especially integrals.
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Remember, though, don't memorize without understanding!
  
 
*Test your skills on practice [[AIME]]s (<url>resources.php more resources</url>) often!  
 
*Test your skills on practice [[AIME]]s (<url>resources.php more resources</url>) often!  
  
 
[[User:Temperal/The Problem Solver's Resource|Back to Introduction]]
 
[[User:Temperal/The Problem Solver's Resource|Back to Introduction]]
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Latest revision as of 20:06, 10 January 2009

Introduction | Other Tips and Tricks | Methods of Proof | You are currently viewing the tips and tricks section.

Other Tips and Tricks

This is a collection of general techniques for solving problems.

  • Don't be afraid to use casework! Sometimes it's the only way. (But be VERY afraid to use brute force.)
  • Remember that substitution is a useful technique! Example problem:

Example Problem 1

If $\tan x+\tan y=25$ and $\cot x+\cot y=30$, find $\tan(x+y)$.

Solution

Let $X = \tan x$, $Y = \tan y$. Thus, $X + Y = 25$, $\frac{1}{X} + \frac{1}{Y} = 30$, so $XY = \frac{5}{6}$, hence $\tan(x+y)=\frac{X+Y}{1-XY}$, which turns out to be $\boxed{150}$.

This technique can also be used to solve quadratics of high degrees, i.e. $x^{16}+x^4+6=0$; let $y=x^4$, and solve from there.


  • Remember the special properties of odd numbers: For any odd number $o$, $o=2n\pm 1$ for some integer $n$, and $o=a^2-(a-1)^2$ for some positive integer $a$.

Example Problem 2

How many quadruples $(a,b,c,d)$ are there such that $a+b+c+d=98$ and $a,b,c,d$ are all odd?

Solution

Since they're odd, $a, b, c, d$ can each be expressed as $2n+1$ for some positive integer (or zero) $n$. Thus: $2n_1-1+2n_2-1+2n_3+1+2n_4+1=98$

$\Rightarrow 2(n_1+n_2+n_3+n_4)+4=98$

$\Rightarrow 2(n_1+n_2+n_3+n_4)=94$

$\Rightarrow n_1+n_2+n_3+n_4=47$ Binomial coefficients will yield the answer of $\boxed{19600}$.


  • The AM-GM and Trivial inequalities are more useful than you might imagine!
  • Memorize, memorize, memorize the following things:
  1. The trigonometric facts.
  2. Everything on the Combinatorics page.
  3. Integrals and derivatives, especially integrals.

Remember, though, don't memorize without understanding!

  • Test your skills on practice AIMEs (<url>resources.php more resources</url>) often!

Back to Introduction