Difference between revisions of "Inradius"

Revision as of 15:56, 22 November 2016

The inradius of a polygon is the radius of its incircle (assuming an incircle exists). It is commonly denoted $r$ .

$[asy] pathpen = linewidth(0.7); pair A=(0,0),B=(4,0),C=(1.5,2),I=incenter(A,B,C),F=foot(I,A,B); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(CR(D(MP("I",I,SW)),inradius(A,B,C))); D(F--I--foot(I,B,C)--I--foot(I,C,A)); D(rightanglemark(I,F,B,5)); MP("r",(F+I)/2,E); [/asy]$

Properties

If $\triangle ABC$ has inradius $r$ and semi-perimeter $s$ , then the area of $\triangle ABC$ is $rs$ . This formula holds true for other polygons if the incircle exists.

Problems

Verify the inequality $R \geq 2r$ .
Verify the identity $\cos{A}+\cos{B}+\cos{C}=\frac{r+R}{R}$ (see Carnot's Theorem).
2007 AIME II Problems/Problem 15

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

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 == Properties ==
 *If <math>\triangle ABC</math> has inradius <math>r</math> and [[semi-perimeter]] <math>s</math>, then the [[area]] of <math>\triangle ABC</math> is <math>rs</math>. This formula holds true for other polygons if the incircle exists.
-*The in radius satisfies the inequality <math>2r \le R</math>, where <math>R</math> is the [[circumradius]] (see below).
-*If <math>\triangle ABC</math> has inradius <math>r</math> and circumradius <math>R</math>, then <math>\cos{A}+\cos{B}+\cos{C}=\frac{r+R}{R}</math>.
 == Problems ==

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Difference between revisions of "Inradius"

Revision as of 15:56, 22 November 2016

Properties

Problems