Exradius: Difference between revisions
No edit summary |
there were no latex, so I added, needs rechecking |
||
| Line 1: | Line 1: | ||
Excircle | 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 | The radius of an excircle. Let a triangle have exradius <math>r_A</math> (sometimes denoted <math>\rho_A</math>), opposite side of length <math>a</math> and angle <math>A</math>, area <math>\Delta</math>, and semiperimeter <math>s</math>. Then | ||
r_1 = Delta | <math>r_1 = \frac{\Delta}{(s-a)} | ||
(1) | (1) | ||
= sqrt | = \sqrt{\frac{(s(s-b)(s-c))}{(s-a)}} | ||
(2) | (2) | ||
= | = 4R\sin{\frac{1}{2A}}\cos{\frac{1}{2B}}\cos{\frac{1}{2C}} | ||
(3) | (3) | ||
</math> | |||
(Johnson 1929, p. 189), where <math>R</math> is the circumradius. Let <math>r</math> be the inradius, then | |||
<math>4R=r_1+r_2+r_3-r</math> | |||
4R=r_1+r_2+r_3-r | |||
(4) | (4) | ||
1 | <math>\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=1/r</math> | ||
(5) | (5) | ||
(Casey 1888, p. 65) and | (Casey 1888, p. 65) and | ||
rr_1r_2r_3=Delta^2 | <math>rr_1r_2r_3=\Delta^2</math> | ||
(6) | (6) | ||
Some fascinating formulas due to Feuerbach are | Some fascinating formulas due to Feuerbach are | ||
r(r_2r_3+r_3r_1+r_1r_2)= | <math>r(r_2r_3+r_3r_1+r_1r_2)=s\Delta=r_1r_2r_3</math> | ||
r(r_1+r_2+r_3)=bc+ca+ab-s^2 | <math>r(r_1+r_2+r_3)=bc+ca+ab-s^2 </math> | ||
rr_1+rr_2+rr_3+r_1r_2+r_2r_3+r_3r_1=bc+ca+ab | <math>rr_1+rr_2+rr_3+r_1r_2+r_2r_3+r_3r_1=bc+ca+ab</math> | ||
r_2r_3+r_3r_1+r_1r_2-rr_1-rr_2-rr_3=1 | $r_2r_3+r_3r_1+r_1r_2-rr_1-rr_2-rr_3=\frac{1}{2(a^2+b^2+c^2)} | ||
Revision as of 09:56, 21 May 2020
Excircle
The radius of an excircle. Let a triangle have exradius 
(sometimes denoted 
), opposite side of length 
and angle 
, area 
, and semiperimeter 
. Then
(Johnson 1929, p. 189), where 
is the circumradius. Let 
be the inradius, then
(4)
(5) (Casey 1888, p. 65) and
(6) Some fascinating formulas due to Feuerbach are
$r_2r_3+r_3r_1+r_1r_2-rr_1-rr_2-rr_3=\frac{1}{2(a^2+b^2+c^2)}
