Analytic geometry

[Albert Chan]

Again, using assumptions from post #6, prove 3 excircle radiuses are:

r1 = | tan(½(t2-t1)) tan(½(t3-t1)) |
r2 = | tan(½(t1-t2)) tan(½(t3-t2)) |
r3 = | tan(½(t1-t3)) tan(½(t2-t3)) |

Due to symmetry, only need to prove r1 is correct.
Let t1=0, then r1 = | tan(t2/2) tan(t3/2) |, calculate actual coordinate:

Post#4 calculated leftmost vertice, with slope relative to (0,0) = tan(½(t2+t3))
Line y = tan(½(t2+t3))*x bisect the angle, thus will hit excircle center too.

To satisfy line x=1, excircle center x-value = 1 + r1 = 1 - tan(t2/2) tan(t3/2)
-> excircle center = [1 - tan(t2/2) tan(t3/2), tan(t2/2) + tan(t3/2)]

Distance from excircle center to line2
= |(1 - tan(t2/2) tan(t3/2)) cos(t2) + (tan(t2/2) + tan(t3/2)) sin(t2) - 1|
= |-tan(t2/2) tan(t3/2) cos(t2) + tan(t3/2) sin(t2)| + (cos(t2) + 2 sin(t2/2)^2 - 1) ; last term=0
= |tan(t3/2) * (sin(t2) cos(t2/2) - cos(t2) sin(t2/2)) / cos(t2)|
= |tan(t2/2) tan(t3/2)| = r1

Due to symmetry, excircle center had same distance to line3 too. QED

(02-19-2019 10:23 PM)Albert Chan Wrote:  s = |tan(½(t2-t1))| + |tan(½(t3-t1))| + |tan(½(t3-t2))|
   = |tan(½(t2-t1)) tan(½(t3-t1)) tan(½(t3-t2))|

Divide first line by second, with inscribed circle radius of r, not 1: 1/r = 1/r1 + 1/r2 + 1/r3