Difference between revisions of "Inradius"
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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>. | 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. | + | 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 == | ||
Revision as of 18:18, 27 July 2021
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)
.
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
- Verify the inequality
. - Verify the identity
(see Carnot's Theorem). - 2007 AIME II Problems/Problem 15
.
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