Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 765.

## Thursday, June 14, 2012

### Problem 765: Triangle, Exradii, Inradius, Three Exradius, Harmonic Mean

Labels:
exradius,
harmonic mean,
inradius,
triangle

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http://img43.imageshack.us/img43/9397/problem765.png

ReplyDeleteDenote S(XYZ)= area of triangle XYZ

Draw additional lines per attached sketch.

We have ∆AA’F similar to ∆AIE

So IE/A’F=r/rA =AI/AA’= S(IC’B’)/S(A’B’C’)

Similarly r/rB=S(IA’C’)/S(A’B’C’) and r/rC=S(IA’B’)/S(A’B’C’)

But S(A’B’C’)=S(IA’C’)+S(IA’B’)+S(IC’B’)

So r/rA+r/rB+r/rC= 1 or 1/r=1/rA+1/rB+1/rC

Let S be the area of the triangle and X = (1/2)(a+b+c), where a,b,c are the side lengths of the triangle.

ReplyDeleteBy the properties of the exradii, S=(Ra)(X-a) ; S=(Rb)(X-b) ; S=(Rc)(X-c).

Therefore,

S/(Ra)=(X-a)

S/(Rb)=(X-b)

S/(Rc)=(X-c)

By summing up, S[(1/Ra)+(1/Rb)+(1/Rc)]=3X-(a+b+c) = X.

By the properties of the inradii, S=RX => X =S/R

So,

S/R = S[(1/Ra)+(1/Rb)+(1/Rc)]

1/R = (1/Ra)+(1/Rb)+(1/Rc)

q.e.d.

Now,

Thank you very much for that) I searched this answer all day))

DeleteLet BC touch incircle at X and excircle opp to B at Y

ReplyDeleteTriangles BIX and BI₂Y are similar

r/r₂ = AX/AY = (s - b)/s

Similarly r/r₃ = (s - c)/s and r/r₁= (s - a)/s

r/r₁ + r/r₂ + r/r₃

=(s - a)/s + (s - b)/s + (s - c)/s

= [3s - (a + b + c)] = (3s - 2s) /s = 1

Hence 1/r₁ + 1/r₂ + 1/r₃ = 1/r