Grade 11Algebra
if a,b,c are in AP then prove that cotA/2,cotB/2,cotC/2 are in AP.
if a,b,c are in AP then prove that cotA/2,cotB/2,cotC/2 are in AP.
1 Answer
Arun
Let a
Then cot(A/2) = (s-a)/r, cot(B/2) = (s-b)/r and cot(C/2) = (s-c)/r where s denotes the semiperimeter (a+b+c)/2 and r denotes the radius of the circle that can be inscribed in triangle ABC.
It follows that cot(A/2) / s-a = cot(B/2) / s-b = cot(C/2) / s-c.
As before, if cot(A/2), cot(B/2), and cot(C/2) are in AP it follows that cot(B/2) = [cot(A/2) + cot(C/2)]/2. But this follows from the previous formulas since
[cot(A/2) + cot(C/2)]/2 = [((s-a)/(s-b)) cot(B/2) + ((s-c)/(s-b)) cot(B/2)]/2
=(1/2) * ((2s - a -c )/(s-b) ) cot (B/2)
=(1/2) * ((a+ 2b +c)/(a+c)) cot(B/2) using 2s=a+b+c
=(1/2) * ((2a+2c)/(a+c)) cot (B/2) using that 2b=a+c by hypothesis
=cot (B/2) as desired.
Then cot(A/2) = (s-a)/r, cot(B/2) = (s-b)/r and cot(C/2) = (s-c)/r where s denotes the semiperimeter (a+b+c)/2 and r denotes the radius of the circle that can be inscribed in triangle ABC.
It follows that cot(A/2) / s-a = cot(B/2) / s-b = cot(C/2) / s-c.
As before, if cot(A/2), cot(B/2), and cot(C/2) are in AP it follows that cot(B/2) = [cot(A/2) + cot(C/2)]/2. But this follows from the previous formulas since
[cot(A/2) + cot(C/2)]/2 = [((s-a)/(s-b)) cot(B/2) + ((s-c)/(s-b)) cot(B/2)]/2
=(1/2) * ((2s - a -c )/(s-b) ) cot (B/2)
=(1/2) * ((a+ 2b +c)/(a+c)) cot(B/2) using 2s=a+b+c
=(1/2) * ((2a+2c)/(a+c)) cot (B/2) using that 2b=a+c by hypothesis
=cot (B/2) as desired.
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