Difference between revisions of "Inradius"

m (+ a bit)
m
(9 intermediate revisions by 3 users not shown)
Line 7: Line 7:
 
</asy></center>
 
</asy></center>
  
== Properties ==
+
== 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.
 
*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).
+
=Proof=
*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>.
+
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>.
 +
 
 +
Also the inradius of a incircle inscribed in a right triangle is (a+b-c)/2 as by drawing three inradiuses to the three tangent points, then A to that tangent point is equal to A to the other tangent point (explained in circles) and etc for B and C. After doing it for B and C, C (the hypotenuse) should equal the equivalent tangent of A and the equivalent tangent of B added together, thus our equation.
  
 
== Problems ==
 
== Problems ==
*Verify the inequality <math>2r \le R</math>.  
+
*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]]
+
*[[2007 AIME II Problems/Problem 15]]
 
 
{{stub}}
 
 
 
[[Category:Geometry]]
 

Revision as of 10:03, 30 July 2021

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]

A Property

  • 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.

Proof

Add in the incircle and drop the altitudes from the incenter to the sides of the triangle. Also draw the lines $\overline{AI}, \overline{BI}$, and $\overline{CI}$. After this AB, AC, and BC are the bases of $\triangle{AIB}, {AIC}$, and ${BIC}$ respectively. But they all have the same height(the inradius), so $[ABC]=\frac{(a+b+c) \times r}{2} =rs$.

Also the inradius of a incircle inscribed in a right triangle is (a+b-c)/2 as by drawing three inradiuses to the three tangent points, then A to that tangent point is equal to A to the other tangent point (explained in circles) and etc for B and C. After doing it for B and C, C (the hypotenuse) should equal the equivalent tangent of A and the equivalent tangent of B added together, thus our equation.

Problems