Difference between revisions of "Inradius"
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The '''inradius''' of a [[polygon]] is the [[radius]] of its [[incircle]] (assuming an incircle exists). It is commonly denoted <math>r</math>. | The '''inradius''' of a [[polygon]] is the [[radius]] of its [[incircle]] (assuming an incircle exists). It is commonly denoted <math>r</math>. | ||
| + | |||
| + | <center><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></center> | ||
== Properties == | == Properties == | ||
| − | *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>. | + | *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 inradius 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 == | ||
| + | *Verify the inequality <math>2r \le R</math>. | ||
| + | *Verify the identity <math>\cos{A}+\cos{B}+\cos{C}=\frac{r+R}{R}</math> (see [[Carnot's Theorem]]). | ||
| + | *[[Special:WhatLinksHere/Inradius]]: [[2007 AIME II Problems/Problem 15]] | ||
| + | |||
| + | {{stub}} | ||
| + | |||
[[Category:Geometry]] | [[Category:Geometry]] | ||
Revision as of 19:16, 15 March 2010
The inradius of a polygon is the radius of its incircle (assuming an incircle exists). It is commonly denoted 
.
![[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]](http://latex.artofproblemsolving.com/3/a/d/3ad1b30f1350fc22f90ce7a79f2567025d4fe4c0.png)
Properties
- If
has inradius 
and semi-perimeter 
, then the area of is 
. This formula holds true for other polygons if the incircle exists. - The inradius satisfies the inequality
, where 
is the circumradius (see below). - If has inradius and circumradius , then
.
has inradius 
, then the 
. This formula holds true for other polygons if the incircle exists.
, where 
is the 
.Problems
- Verify the inequality .
- Verify the identity (see Carnot's Theorem).
- Special:WhatLinksHere/Inradius: 2007 AIME II Problems/Problem 15
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