Triangle Inequality

Revision as of 17:27, 18 February 2011 by Yankeesrule007 (talk | contribs) (Problems)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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