Let the side lengths of the triangle be a a a b b b c c c
According to the given information:
a + b = x a + b = x a + b = x a b = y ab = y ab = y x 2 − c 2 = y x^2 - c^2 = y x 2 − c 2 = y
Substituting x = a + b x = a + b x = a + b y = a b y = ab y = ab ( a + b ) 2 − c 2 = a b (a + b)^2 - c^2 = ab ( a + b ) 2 − c 2 = ab a 2 + 2 a b + b 2 − c 2 = a b a^2 + 2ab + b^2 - c^2 = ab a 2 + 2 ab + b 2 − c 2 = ab a 2 + b 2 − c 2 = − a b a^2 + b^2 - c^2 = -ab a 2 + b 2 − c 2 = − ab
Using the Law of Cosines to find angle C C C cos C = a 2 + b 2 − c 2 2 a b = − a b 2 a b = − 1 2 \cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{-ab}{2ab} = -\frac{1}{2} cos C = 2 ab a 2 + b 2 − c 2 = 2 ab − ab = − 2 1
Since C ∈ ( 0 , π ) C \in (0, \pi) C ∈ ( 0 , π ) C = 2 π 3 = 120 ∘ C = \frac{2\pi}{3} = 120^\circ C = 3 2 π = 12 0 ∘ sin C = sin ( 2 π 3 ) = 3 2 \sin C = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} sin C = sin ( 3 2 π ) = 2 3
The area of the triangle, Δ \Delta Δ Δ = 1 2 a b sin C = 1 2 y ( 3 2 ) = 3 y 4 \Delta = \frac{1}{2}ab \sin C = \frac{1}{2} y \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}y}{4} Δ = 2 1 ab sin C = 2 1 y ( 2 3 ) = 4 3 y
The semi-perimeter s s s s = a + b + c 2 = x + c 2 s = \frac{a + b + c}{2} = \frac{x + c}{2} s = 2 a + b + c = 2 x + c
Thus, the in-radius r r r r = Δ s = 3 y 4 x + c 2 = 3 y 2 ( x + c ) r = \frac{\Delta}{s} = \frac{\frac{\sqrt{3}y}{4}}{\frac{x + c}{2}} = \frac{\sqrt{3}y}{2(x + c)} r = s Δ = 2 x + c 4 3 y = 2 ( x + c ) 3 y
Using the Sine Rule, the circum-radius R R R 2 R = c sin C ⟹ R = c 2 sin C = c 2 ( 3 2 ) = c 3 2R = \frac{c}{\sin C} \implies R = \frac{c}{2\sin C} = \frac{c}{2\left(\frac{\sqrt{3}}{2}\right)} = \frac{c}{\sqrt{3}} 2 R = s i n C c ⟹ R = 2 s i n C c = 2 ( 2 3 ) c = 3 c
Therefore, the ratio of the in-radius to the circum-radius is:
r R = 3 y 2 ( x + c ) c 3 = 3 y 2 c ( x + c ) \frac{r}{R} = \frac{\frac{\sqrt{3}y}{2(x + c)}}{\frac{c}{\sqrt{3}}} = \frac{3y}{2c(x + c)} R r = 3 c 2 ( x + c ) 3 y = 2 c ( x + c ) 3 y
Hence, the correct option is (B) .