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

• Verify the inequality $R \geq 2r$.
• Verify the identity $\cos{A}+\cos{B}+\cos{C}=\frac{r+R}{R}$ (see Carnot's Theorem).
• 2007 AIME II Problems/Problem 15