Geometry Problem
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Friday, March 4, 2011
Problem 591: Triangle, Incenter, Incircle, Inradius, Tangency Point, Midpoint, Altitude
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2b.EM
ReplyDelete= 2b.(AM–AE) = b.(AC–2AE)= b.(b–2cCosA)
= b^2–2bcCosA = b^2–(b^2+c^2–a^2)
= a^2–c^2
2b.DM = b.2DM = b(a–c)from Problem 587
r:EF=DM:EM=b(a – c):(a^2 – c^2)=b:(a + c )
b.EF=r.a + c)
b.BF=b.(BE-EF)=b(cSin A)–r(a+c)
=2Δ–r(a+c)=2rs–r(a+c)=r(2s–a–c)=rb
BF=r follows
Let 2p=a+b+c and h=BE
ReplyDeleteWe have S= Area(ABC)= p.r = ½* b*h
So b=BE= 2.p.r/b = r+r.(a+c)/b (1)
1. Triangle MDO similar to Triangle MEF ( case AA)
So EF=r*(ME/MD) (2)
2. We have a^2-c^2 = CE^2-AE^2=(b/2+ME)^2-(b/2-ME)^2=2.b.ME
So ME=(a^2-c^2)/(2.b)
And MD= (a-c)/2 See Problem 587
Replace value of ME and MD in (2) We have EF=r.(a^2-c^2)/(2.b) *(2/(a-c))= r.(a+c)/b (3)
we have BF=BE-EF= r
Replace value of BE and EF from (1) and (3) to above expression we have BF=r
Peter Tran
(BC+AB).2DM = (BC+AB).(BC-AB)=BC^2–AB^2
ReplyDelete=CE^2–AE^2=(CE+AE).(CE–AE=AC.2EM
EF/r=EF/OD=EM/DM=(BC+AB)/AC=(a+c)/b
So BF=BE–EF=BE–EF
=(2Δ/b)–r(a+c)/b=[2rs-r(a+c)]/b
=r(2s–a–c/b=br/b=r