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| Exradii and Circumradius (Posted on 2008-06-10) |
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If ra, rb, rc are exradii of triangle ABC with sides a,b,c and circumradius R, then show that:
(a/ra)+(b/rb)+(c/rc) ≤ 2R(1/a+1/b+1/c)
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Submitted by Praneeth
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Rating: 4.0000 (1 votes)
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Solution:
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Let s be the semi-perimeter of the triangle, then
Area(A) = ra(s-a) = rb(s-b) = rc(s-c)=abc/4R
s=(a+b+c)/2
LHS=1/A*(a(s-a)+b(s-b)+c(s-c))
=1/A*((a+b+c)2/2-a2-b2-c2)
=1/A*(ab+bc+ca-1/2(a2+b2+c2))
=1/A*(1/2*(ab+bc+ca)-1/4{(a-b)2+(b-c)2+(c-a)2})
≤1/2A*(ab+bc+ca)
≤2R(ab+bc+ca)/abc
≤2R(1/a+1/b+1/c)
Hence Proved.
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