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

Labels:
altitude,
excircle,
exradius,
perpendicular,
reciprocal,
triangle

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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)