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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);
 
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In a triangle, the incenter is where the three angle bisectors meet.
  
 
== A Property ==
 
== 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.
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=Proof=
 
=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>.
 
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>.
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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 [https://artofproblemsolving.com/wiki/index.php/TOTO_SLOT_:_SITUS_TOTO_SLOT_MAXWIN_TERBAIK_DAN_TERPERCAYA TOTO SLOT].
  
 
== Problems ==
 
== Problems ==
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*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]]).  
 
*[[2007 AIME II Problems/Problem 15]]
 
*[[2007 AIME II Problems/Problem 15]]
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[[Category:Geometry]]
 

Latest revision as of 16:57, 19 February 2024

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]

In a triangle, the incenter is where the three angle bisectors meet.

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 $\frac{a+b-c}{2}$ 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 TOTO SLOT.

Problems

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