Difference between revisions of "Exradius"

m (filling in the gap)
Line 5: Line 5:
 
  <math>r_1 = \frac{\Delta}{s-a}
 
  <math>r_1 = \frac{\Delta}{s-a}
 
= \sqrt{\frac{s(s-b)(s-c)}{s-a}}
 
= \sqrt{\frac{s(s-b)(s-c)}{s-a}}
= 4R\sin{\frac{1}{2A}}\cos{\frac{1}{2B}}\cos{\frac{1}{2C}}
+
= 4R\sin{\frac{A}{2}\cos{\frac{B}{2}}\cos{\frac{C}{2}}
 
</math>
 
</math>
 
(Johnson 1929, p. 189), where <math>R</math> is the circumradius. Let <math>r</math> be the inradius, then
 
(Johnson 1929, p. 189), where <math>R</math> is the circumradius. Let <math>r</math> be the inradius, then

Revision as of 12:53, 21 January 2024

Excircle

The radius of an excircle. Let a triangle have exradius $r_A$ (sometimes denoted $\rho_A$), opposite side of length $a$ and angle $A$, area $\Delta$, and semiperimeter $s$. Then

$r_1	=	\frac{\Delta}{s-a}	

= \sqrt{\frac{s(s-b)(s-c)}{s-a}} = 4R\sin{\frac{A}{2}\cos{\frac{B}{2}}\cos{\frac{C}{2}}$ (Error compiling LaTeX. Unknown error_msg) (Johnson 1929, p. 189), where $R$ is the circumradius. Let $r$ be the inradius, then

$4R=r_1+r_2+r_3-r$ 	

and

$\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{r}$ 	


(Casey 1888, p. 65) and

$rr_1r_2r_3=\Delta^2$ 	


Some fascinating formulas due to Feuerbach are

$r(r_2r_3+r_3r_1+r_1r_2)=s\Delta=r_1r_2r_3$ 
$r(r_1+r_2+r_3)=bc+ca+ab-s^2$
$rr_1+rr_2+rr_3+r_1r_2+r_2r_3+r_3r_1=bc+ca+ab$ 
$r_2r_3+r_3r_1+r_1r_2-rr_1-rr_2-rr_3=\frac{1}{2}(a^2+b^2+c^2)$

Reference:

Weisstein, Eric W. "Exradius." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/Exradius.html