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## Inversion of excircles in incircle, and vice versa

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• Dear friends, (1) The A-excircle is inverted in the incircle. The center of the inversive image is the point (a(a^2+3(b-c)^2) : b(3a^2+(b-c)^2) :
Message 1 of 1 , Feb 21, 2013
Dear friends,

(1) The A-excircle is inverted in the incircle. The center of the inversive image is the point

(a(a^2+3(b-c)^2) : b(3a^2+(b-c)^2) : c(3a^2+(b-c)^2)).

The triangle of centers of the inversive images of the excircles in the incircle is perspective with

(i) the intouch triangle at X_{4907} = (a(b+c-a)^2(a^2+3(b-c)^2) : ... : ...),

(ii) the medial triangle at
(a(a^2+3(b-c)^2) : b(b^2+3(c-a)^2) : c(c^2+3(a-b)^2))
with (6-9-13)-search number 0.778919595267....

(2) The incircle is inverted in the A-excircle. The center of the inversive image is the point

(-a(a^2+3(b-c)^2) : b(3a^2+(b-c)^2) : c(3a^2+(b-c)^2)).

The triangle of centers of the inversive images of the incircle in the excircles is perspective with

(i) the intouch triangle at
( a(a^2+3(b-c)^2)/(b+c-a) : b(b^2+3(c-a)^2)/(c+a-b) : c(c^2+3(a-b)^2)/(a+b-c))

with (6-9-13)-search number 0.16447696491...,

(ii) the medial triangle at

(a(b+c-a)^3(a^2+3(b-c)^2) : ... : ...)

with (6-9-13)-search number 3.93642587545...

Best regards
Sincerely
Paul

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