Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Thursday, November 6, 2014
Geometry Problem 1056: Triangle, Exradius, Reciprocals of the Altitudes, Multiplicative Inverse, Perpendicular, Excircle, Circle
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Use subscript 1,2,3 instead of a,b,c.
ReplyDeleteLet BC=a, AC=b, AB=c. Let Δ be the area of ABC.
r₁ = Δ/(s−a), 1/r₁ = (s−a)/Δ = (b+c−a)/(2Δ)
r₂ = Δ/(s−b), 1/r₂ = (s−b)/Δ = (a+c−b)/(2Δ)
r₃ = Δ/(s−c), 1/r₃ = (s−c)/Δ = (a+b−c)/(2Δ)
ah₁ = 2Δ, 1/h₁ = a/(2Δ)
bh₂ = 2Δ, 1/h₂ = b/(2Δ)
ch₃ = 2Δ, 1/h₃ = c/(2Δ)
The rest is obvious.
Let S= area of triangle ABC= Area(ACE)+Area(ABE)-Area(BCE)
ReplyDeleteS= ½(.Ra).b+1/2(Ra).c-1/2(Ra).a= ½.(Ra).(b+c-a)
So 1/(Ra)=1/2(b+c-a)/S ..... (1)
But b/S=2/(Hb) and c/S=2/(Hc) , a/S=2/(Ha)
Replace it in (1) we have 1/(Ra)=1/(Hb)+1/(Hc)-1/(Ha)