Difference between revisions of "Inradius"

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(Problems)
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== Problems ==
 
== Problems ==
*Verify the inequality <math>2r \le R</math>.  
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*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]]).  
 
*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]]
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*[[2007 AIME II Problems/Problem 15]]
  
 
{{stub}}
 
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[[Category:Geometry]]
 
[[Category:Geometry]]

Revision as of 15:55, 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.
  • The in radius satisfies the inequality $2r \le R$, where $R$ is the circumradius (see below).
  • If $\triangle ABC$ has inradius $r$ and circumradius $R$, then $\cos{A}+\cos{B}+\cos{C}=\frac{r+R}{R}$.

Problems

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