Difference between revisions of "Triangle Inequality"

m
m (Problems)
(4 intermediate revisions by 3 users not shown)
Line 1: Line 1:
The '''Triangle Inequality''' says that in a [[nondegenerate]] [[triangle]] <math>\displaystyle ABC</math>:
+
The '''Triangle Inequality''' says that in a [[nondegenerate]] [[triangle]] <math>ABC</math>:
  
<math>\displaystyle AB + BC > AC</math>
+
<math>AB + BC > AC</math>
  
<math>\displaystyle BC + AC > AB</math>
+
<math>BC + AC > AB</math>
  
<math>\displaystyle AC + AB > BC</math>
+
<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.
Line 14: Line 14:
  
  
== Example Problems ==
+
== 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]]
 +
===Intermediate Problems===
 +
*[[2010_AMC_12A_Problems/Problem_25 | 2010 AMC 12A Problem 25]]
 +
=== Olympiad Problems ===
 +
* Belarus 2002 [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59933 Aops Topic]
 +
Given <math>a,b,c,d>0</math>, prove:
  
*[[2006 AIME II Problems/Problem 2 | 2006 AIME II Problem 2]]
+
<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 ==
Line 27: Line 33:
  
 
{{stub}}
 
{{stub}}
 +
 +
[[Category:Geometry]]
 +
 +
[[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

This article is a stub. Help us out by expanding it.

Invalid username
Login to AoPS