Difference between revisions of "Triangle Inequality"

Latest revision as of 02:02, 3 January 2021

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

2010 AMC 12A Problem 25

Olympiad Problems

Belarus 2002 Aops Topic

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

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

@@ Line 14: / Line 14: @@
-== Example Problems ==
+== Problems ==
 === Introductory Problems ===
+* [[2003 AMC 12A Problems/Problem 7]]
 * [[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===
+*[[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 ==

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