Difference between revisions of "Triangle Inequality"

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The '''Triangle Inequality''' says that in a [[nondegenerate]] [[triangle]] <math>\displaystyle ABC</math>:
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The '''Triangle Inequality''' says that in a [[nondegenerate]] [[triangle]] <math>ABC</math>:
  
<math>\displaystyle AB + BC > AC</math>
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<math>AB + BC > AC</math>
  
<math>\displaystyle BC + AC > AB</math>
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<math>BC + AC > AB</math>
  
<math>\displaystyle AC + AB > BC</math>
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<math>AC + AB > BC</math>
  
 
That is, the sum of the lengths of any two sides is larger than the length of the third side.
 
That is, the sum of the lengths of any two sides is larger than the length of the third side.
 
In [[degenerate]] triangles, the [[strict inequality]] must be replaced by "greater than or equal to."
 
In [[degenerate]] triangles, the [[strict inequality]] must be replaced by "greater than or equal to."
  
This inequality often shows up in contest problems.
 
  
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The Triangle Inequality can also be extended to other [[polygon]]s.  The lengths <math>a_1, a_2, \ldots, a_n</math> can only be the sides of a nondegenerate <math>n</math>-gon if <math>a_i < a_1 + \ldots + a_{i -1} + a_{i + 1} + \ldots + a_n = \left(\sum_{j=1}^n a_j\right) - a_i</math> for <math>i = 1, 2 \ldots, n</math>.  Expressing the inequality in this form leads to <math>2a_i < P</math>, where <math>P</math> is the sum of the <math>a_j</math>, or <math>a_i < \frac{P}{2}</math>.  Stated in another way, it says that in every polygon, each side must be smaller than the [[semiperimeter]].
  
== Example Problems ==
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== Problems ==
 
=== Introductory Problems ===
 
=== Introductory Problems ===
 
* [[2006_AMC_10B_Problems/Problem_10 | 2006 AMC 10B Problem 10]]
 
* [[2006_AMC_10B_Problems/Problem_10 | 2006 AMC 10B Problem 10]]
 
* [[2006 AIME II Problems/Problem 2 | 2006 AIME II Problem 2]]
 
* [[2006 AIME II Problems/Problem 2 | 2006 AIME II Problem 2]]
=== Intermediate Problems ===
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===Intermediate Problems===
* [[2003_AIME_I_Problems/Problem_11 | 2003 AIME I Problem 11]]
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*[[2010_AMC_12A_Problems/Problem_25 | 2010 AMC 12A Problem 25]]
 
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=== Olympiad Problems ===
 
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* Belarus 2002 [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59933 Aops Topic]
== Geometry Resources ==
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Given <math>a,b,c,d>0</math>, prove:
=== Introductory ===
 
* [http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=9 Introduction to Geometry] by [[Richard Rusczyk]]
 
* [http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php#beggeom AoPS Introduction to Geometry course] -- Based on the lessons in the book listed above by the same title.
 
  
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<center><math>\sqrt{(a+c)^2+(b+d)^2}+\frac{2|ad-bc|}{\sqrt{(a+c)^2+(b+d)^2}}\geq \sqrt{a^2+b^2}+\sqrt{c^2+d^2} \geq \sqrt{(a+c)^2+(b+d)^2}</math></center>
  
 
== See Also ==
 
== See Also ==
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{{stub}}
 
{{stub}}
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[[Category:Geometry]]
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[[Category:Theorems]]

Revision as of 17:27, 18 February 2011

The Triangle Inequality says that in a nondegenerate triangle $ABC$:

$AB + BC > AC$

$BC + AC > AB$

$AC + AB > BC$

That is, the sum of the lengths of any two sides is larger than the length of the third side. In degenerate triangles, the strict inequality must be replaced by "greater than or equal to."


The Triangle Inequality can also be extended to other polygons. The lengths $a_1, a_2, \ldots, a_n$ can only be the sides of a nondegenerate $n$-gon if $a_i < a_1 + \ldots + a_{i -1} + a_{i + 1} + \ldots + a_n = \left(\sum_{j=1}^n a_j\right) - a_i$ for $i = 1, 2 \ldots, n$. Expressing the inequality in this form leads to $2a_i < P$, where $P$ is the sum of the $a_j$, or $a_i < \frac{P}{2}$. Stated in another way, it says that in every polygon, each side must be smaller than the semiperimeter.


Problems

Introductory Problems

Intermediate Problems

Olympiad Problems

Given $a,b,c,d>0$, prove:

$\sqrt{(a+c)^2+(b+d)^2}+\frac{2|ad-bc|}{\sqrt{(a+c)^2+(b+d)^2}}\geq \sqrt{a^2+b^2}+\sqrt{c^2+d^2} \geq \sqrt{(a+c)^2+(b+d)^2}$

See Also

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