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Question: In a triangle ABC, \(\frac { \mathrm { r } _ { 1 } } { \mathrm { bc } } + \frac { \mathrm { r } _ {...

In a triangle ABC, r1bc+r2ca+r3ab\frac { \mathrm { r } _ { 1 } } { \mathrm { bc } } + \frac { \mathrm { r } _ { 2 } } { \mathrm { ca } } + \frac { \mathrm { r } _ { 3 } } { \mathrm { ab } } is equal to

A

B

2R – r

C

r – 2R

D

Answer

Explanation

Solution

r1bc\frac { \mathrm { r } _ { 1 } } { \mathrm { bc } } = 4Rsin⁡(A/2)cos⁡(B/2)cos⁡(C/2)2Rsin⁡B⋅2Rsin⁡C\frac { 4 \mathrm { R } \sin ( \mathrm { A } / 2 ) \cos ( \mathrm { B } / 2 ) \cos ( \mathrm { C } / 2 ) } { 2 \mathrm { R } \sin \mathrm { B } \cdot 2 \mathrm { R } \sin \mathrm { C } }

= sin⁡(A/2)4Rsin⁡(B/2)sin⁡(C/2)\frac { \sin ( \mathrm { A } / 2 ) } { 4 \mathrm { R } \sin ( \mathrm { B } / 2 ) \sin ( \mathrm { C } / 2 ) }

= sin⁡2( A/2)4Rsin⁡(A/2)sin⁡(B/2)sin⁡(C/2)\frac { \sin ^ { 2 } ( \mathrm {~A} / 2 ) } { 4 \mathrm { R } \sin ( \mathrm { A } / 2 ) \sin ( \mathrm { B } / 2 ) \sin ( \mathrm { C } / 2 ) }

=

So that r1bc+r2ca+r3ab\frac { \mathrm { r } _ { 1 } } { \mathrm { bc } } + \frac { \mathrm { r } _ { 2 } } { \mathrm { ca } } + \frac { \mathrm { r } _ { 3 } } { \mathrm { ab } }

=1r= \frac { 1 } { \mathrm { r } } [(sin2 (A/2) + sin2 (B/2)+ sin2 (C/2)]

= 12r\frac { 1 } { 2 r } (1 – cos A + 1 – cos B + 1 – cos C)

= [3 – (cos A + cos B + cos C)]

= [3 – (1 + 4 sin (A/2) sin (B/2) sin (C/2))]

= 12r[2−rR]=1r−12R\frac { 1 } { 2 \mathrm { r } } \left[ 2 - \frac { \mathrm { r } } { \mathrm { R } } \right] = \frac { 1 } { \mathrm { r } } - \frac { 1 } { 2 \mathrm { R } }