Analytic geometry

02202019, 09:33 PM
(This post was last modified: 02212019 01:38 PM by Albert Chan.)
Post: #7




RE: Analytic geometry
Again, using assumptions from post #6, prove 3 excircle radiuses are:
r1 =  tan(½(t2t1)) tan(½(t3t1))  r2 =  tan(½(t1t2)) tan(½(t3t2))  r3 =  tan(½(t1t3)) tan(½(t2t3))  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 xvalue = 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 (02192019 10:23 PM)Albert Chan Wrote: s = tan(½(t2t1)) + tan(½(t3t1)) + tan(½(t3t2)) Divide first line by second, with inscribed circle radius of r, not 1: 1/r = 1/r1 + 1/r2 + 1/r3 

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Messages In This Thread 
Analytic geometry  yangyongkang  02182019, 09:27 AM
RE: Analytic geometry  yangyongkang  02182019, 12:09 PM
RE: Analytic geometry  Albert Chan  02182019, 08:48 PM
RE: Analytic geometry  Albert Chan  02182019, 03:16 PM
RE: Analytic geometry  Albert Chan  02182019, 05:51 PM
RE: Analytic geometry  Albert Chan  02192019, 10:23 PM
RE: Analytic geometry  Albert Chan  02202019 09:33 PM

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