Art of Problem Solving

Exradius: Difference between revisions

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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 r_A (sometimes denoted  rho_A), opposite side of length a and angle A, area Delta, and semiperimeter s. Then
<math>
<math>
r_1 = Delta/(s-a)
r_1 = \Delta/(s-a)
(1)
(1)
= sqrt((s(s-b)(s-c))/(s-a))
= sqrt((s(s-b)(s-c))/(s-a))
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</math>
</math>
(Johnson 1929, p. 189), where R is the circumradius. Let r be the inradius, then
(Johnson 1929, p. 189), where R is the circumradius. Let r be the inradius, then
 
<math>
  4R=r_1+r_2+r_3-r
  4R=r_1+r_2+r_3-r
</math>
(4)
(4)
  1/(r_1)+1/(r_2)+1/(r_3)=1/r
  <math>
1/(r_1)+1/(r_2)+1/(r_3)=1/r
</math>
(5)
(5)
(Casey 1888, p. 65) and
(Casey 1888, p. 65) and
 
<math>
  rr_1r_2r_3=Delta^2.
  rr_1r_2r_3=Delta^2.
</math>
(6)
(6)
Some fascinating formulas due to Feuerbach are
Some fascinating formulas due to Feuerbach are
 
<math>
  r(r_2r_3+r_3r_1+r_1r_2)=sDelta=r_1r_2r_3  
  r(r_2r_3+r_3r_1+r_1r_2)=sDelta=r_1r_2r_3  
r(r_1+r_2+r_3)=bc+ca+ab-s^2  
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  
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=1/2(a^2+b^2+c^2)
r_2r_3+r_3r_1+r_1r_2-rr_1-rr_2-rr_3=1/2(a^2+b^2+c^2)
</math>

Revision as of 21:18, 26 June 2019

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 $r_1	=	\Delta/(s-a)	 (1) 	=	sqrt((s(s-b)(s-c))/(s-a))	 (2) 	=	4Rsin(1/2A)cos(1/2B)cos(1/2C)	 (3)$ (Johnson 1929, p. 189), where R is the circumradius. Let r be the inradius, then $4R=r_1+r_2+r_3-r$ (4)

$1/(r_1)+1/(r_2)+1/(r_3)=1/r$

(5) (Casey 1888, p. 65) and $rr_1r_2r_3=Delta^2.$ (6) Some fascinating formulas due to Feuerbach are $r(r_2r_3+r_3r_1+r_1r_2)=sDelta=r_1r_2r_3  r(r_1+r_2+r_3)=bc+ca+ab-s^2  rr_1+rr_2+rr_3+r_1r_2+r_2r_3+r_3r_1=bc+ca+ab  r_2r_3+r_3r_1+r_1r_2-rr_1-rr_2-rr_3=1/2(a^2+b^2+c^2)$