Difference between revisions of "Inradius"
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| − | The '''inradius''' of a [[polygon]] is the [[radius]] of its [[incircle]] (assuming an incircle exists). | + | 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> | ||
| + | |||
| + | == A Property == | ||
| + | *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. | ||
| + | =Proof= | ||
| + | Add in the incircle and drop the altitudes from the incenter to the sides of the triangle. Also draw the lines <math>\overline{AI}, \overline{BI}</math>, and <math>\overline{CI}</math>. After this AB, AC, and BC are the bases of <math>\triangle{AIB}, {AIC}</math>, and <math>{BIC}</math> respectively. But they all have the same height(the inradius), so <math>[ABC]=\frac{(a+b+c) \times r}{2} =rs</math>. | ||
| + | |||
| + | == Problems == | ||
| + | *Verify the inequality <math>R \geq 2r</math>. | ||
| + | *Verify the identity <math>\cos{A}+\cos{B}+\cos{C}=\frac{r+R}{R}</math> (see [[Carnot's Theorem]]). | ||
| + | *[[2007 AIME II Problems/Problem 15]] | ||
| + | |||
| + | {{stub}} | ||
| + | |||
| + | [[Category:Geometry]] | ||
Revision as of 17:03, 22 November 2016
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)
A Property
- If
has inradius 
and semi-perimeter 
, then the area of is 
. This formula holds true for other polygons if the incircle exists.
has inradius 
, then the 
. This formula holds true for other polygons if the incircle exists.Proof
Add in the incircle and drop the altitudes from the incenter to the sides of the triangle. Also draw the lines 
, and 
. After this AB, AC, and BC are the bases of 
, and 
respectively. But they all have the same height(the inradius), so ![$[ABC]=\frac{(a+b+c) \times r}{2} =rs$](http://latex.artofproblemsolving.com/3/7/2/37201af3741d115348517f124ac85bdb9f605138.png)
.
Problems
- Verify the inequality
. - Verify the identity
(see Carnot's Theorem). - 2007 AIME II Problems/Problem 15
.
(see This article is a stub. Help us out by expanding it.
