If R be the circumradius, r the inradius and r 1 , r 2 , r 3 be the radii of the three escribed circles of a triangle, S be the semi-perimeter of the triangle.
(i) The value of r 1 + r 2 + r 3 is –
Text Solution
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Ans.
(i)
Sol. Let r 1 , r 2 , r 3 be the radii of the escribed circles of the Δ ABC, then r 1 , r 2 , r 3 will be the roots of the equation
x 3 – (r 1 + r 2 + r 3 ) x 2 + (r 1 r 2 + r 2 r 3 + r 3 r 1 ) x –r 1 r 2 r 3 = 0
Now, r 1 + r 2 + r 3 = 
+ 
+ 
= 
+ 
+ 
– 
+ 
= Δ 
+ Δ 
+ 
= 
+ 
+ 
= Δ .c 
+ 
= Δ . c 
+ 
= Δ . c. 
+ 
= 
+ 
= 4R + r … 
…(i)
Again, r 1 r 2 + r 2 r 3 + r 3 r 1 =
. 
+ 
. 
+ 
. 
= Δ 2 
= 
= 
= 
= s 2 …(ii)
Finally, r 1 r 2 r 3 = 
= 
= 
= Δ s = (rs) s = rs 2 …(iii)
Thus the required equation using Δ and (i), (ii) and (iii)
is x 3 – (4R + r)x 2 + s 2 x – rs 2 = 0
(ii)
Sol. Let r 1 , r 2 , r 3 be the radii of the escribed circles of the Δ ABC, then r 1 , r 2 , r 3 will be the roots of the equation
x 3 – (r 1 + r 2 + r 3 ) x 2 + (r 1 r 2 + r 2 r 3 + r 3 r 1 ) x –r 1 r 2 r 3 = 0
Now, r 1 + r 2 + r 3 = 
+ 
+ 
= 
+ 
+ 
– 
+ 
= Δ 
+ Δ 
+ 
= 
+ 
+ 
= Δ .c 
+ 
= Δ . c 
+ 
= Δ . c. 
+ 
= 
+ 
= 4R + r … 
…(i)
Again, r 1 r 2 + r 2 r 3 + r 3 r 1 =
. 
+ 
. 
+ 
. 
= Δ 2 
= 
= 
= 
= s 2 …(ii)
Finally, r 1 r 2 r 3 = 
= 
= 
= Δ s = (rs) s = rs 2 …(iii)
Thus the required equation using Δ and (i), (ii) and (iii)
is x 3 – (4R + r)x 2 + s 2 x – rs 2 = 0
(iii)
Sol. Let r 1 , r 2 , r 3 be the radii of the escribed circles of the Δ ABC, then r 1 , r 2 , r 3 will be the roots of the equation
x 3 – (r 1 + r 2 + r 3 ) x 2 + (r 1 r 2 + r 2 r 3 + r 3 r 1 ) x –r 1 r 2 r 3 = 0
Now, r 1 + r 2 + r 3 = 
+ 
+ 
= 
+ 
+ 
– 
+ 
= Δ 
+ Δ 
+ 
= 
+ 
+ 
= Δ .c 
+ 
= Δ . c 
+ 
= Δ . c. 
+ 
= 
+ 
= 4R + r … 
…(i)
Again, r 1 r 2 + r 2 r 3 + r 3 r 1 = 
. 
+ 
. 
+ 
. 
= Δ 2 
= 
= 
= 
= s 2 …(ii)
Finally, r 1 r 2 r 3 = 
= 
= 
= Δ s = (rs) s = rs 2 …(iii)
Thus the required equation using Δ and (i), (ii) and (iii)
is x 3 – (4R + r)x 2 + s 2 x – rs 2 = 0
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