Difference between revisions of "Inradius"
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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); | 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> | </asy></center> | ||
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+ | In a triangle, the incenter is where the three angle bisectors meet. | ||
== A Property == | == A Property == | ||
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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 <math>\frac{a+b-c}{2}</math> 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 | + | Also the inradius of a incircle inscribed in a right triangle is <math>\frac{a+b-c}{2}</math> 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 A = rs |
== Problems == | == Problems == |
Latest revision as of 22:54, 20 June 2024
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)
In a triangle, the incenter is where the three angle bisectors meet.
A Property
- If
has inradius
and semi-perimeter
, then the area of
is
. 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
.
Also the inradius of a incircle inscribed in a right triangle is 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 A = rs
Problems
- Verify the inequality
.
- Verify the identity
(see Carnot's Theorem).
- 2007 AIME II Problems/Problem 15