Difference between revisions of "Triangle Inequality"
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− | The '''Triangle Inequality''' says that in a [[nondegenerate]] [[triangle]] <math> | + | The '''Triangle Inequality''' says that in a [[nondegenerate]] [[triangle]] <math>ABC</math>: |
− | <math> | + | <math>AB + BC > AC</math> |
− | <math> | + | <math>BC + AC > AB</math> |
− | <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 11: | Line 11: | ||
− | 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> | + | 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]]. |
− | == | + | == Problems == |
=== Introductory Problems === | === Introductory Problems === | ||
+ | * [[2003 AMC 12A Problems/Problem 7]] | ||
* [[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: | ||
+ | |||
+ | <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}} | ||
+ | |||
+ | [[Category:Geometry]] | ||
+ | |||
+ | [[Category:Theorems]] |
Latest revision as of 03:02, 3 January 2021
The Triangle Inequality says that in a nondegenerate triangle :
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 can only be the sides of a nondegenerate
-gon if
for
. Expressing the inequality in this form leads to
, where
is the sum of the
, or
. Stated in another way, it says that in every polygon, each side must be smaller than the semiperimeter.
Contents
Problems
Introductory Problems
Intermediate Problems
Olympiad Problems
- Belarus 2002 Aops Topic
Given , prove:
![$\sqrt{(a+c)^2+(b+d)^2}+\frac{2|ad-bc|}{\sqrt{(a+c)^2+(b+d)^2}}\geq \sqrt{a^2+b^2}+\sqrt{c^2+d^2} \geq \sqrt{(a+c)^2+(b+d)^2}$](http://latex.artofproblemsolving.com/9/a/2/9a2b7ccc620c80325ce20ae5b267d3f8f51a483c.png)
See Also
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