Difference between revisions of "Triangle Inequality"

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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>
  
This inequality often shows up in contest problems.
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That is, the sum of the lengths of any two sides is larger than the length of the third side.
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In [[degenerate]] triangles, the [[strict inequality]] must be replaced by "greater than or equal to."
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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]].
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== Problems ==
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=== Introductory Problems ===
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* [[2006_AMC_10B_Problems/Problem_10 | 2006 AMC 10B Problem 10]]
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* [[2006 AIME II Problems/Problem 2 | 2006 AIME II Problem 2]]
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===Intermediate Problems===
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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]
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Given <math>a,b,c,d>0</math>, prove:
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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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[[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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