b.Theo phần (a) ta có: $a+b=3c \Leftrightarrow p=2c (1)$Mặt khác:$S=pr=\dfrac{abc}{4R}\Rightarrow pabc\dfrac{r}{4R}=S^2=p(p-a)(p-b)(p-c)$Thay $(1)$ vào biểu thức trên ta được:$ab\dfrac{r}{4R}=p^2-(a+b)p+ab\Rightarrow (1-\dfrac{r}{4R})ab=2c^2 (2)$Lại có:$S=\dfrac{1}{2}ab\sin C=pr\Rightarrow r=\dfrac{ab\sin C}{a+b+c}=\dfrac{4R^2\sin A\sin B\sin C}{2R(\sin A+\sin B+\sin C)}$Suy ra: $\dfrac{r}{4R}=\dfrac{\sin A\sin B\sin C}{2(\sin A+\sin B+\sin C)}$Mà ta có:$\sin A+\sin B+\sin C= 2\sin\dfrac{A}{2}\cos\dfrac{A}{2}+2\sin\dfrac{B+C}{2}\cos\dfrac{B-C}{2}$ $=2\cos\dfrac{A}{2}(\cos\dfrac{B+C}{2}+\cos\dfrac{B-C}{2})$ $=4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$$\Rightarrow
\dfrac{r}{4R}=\dfrac{8\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}}{8\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}}=\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}$$\Rightarrow \dfrac{r}{4R}=\dfrac{1}{10}$Thay vào $(2)$ ta được: $ab=\dfrac{20}{9}c^2$$\Rightarrow ab=20(a+b)^2$$\Leftrightarrow 20a^2-41ab+20b^2=0$$\Leftrightarrow \left\{\begin{array}{l}4a=5b\\5a=4b\end{array}\right.$Cả 2 trường hợp đều dẫn tới $\Delta ABc$ vuông. (theo Pytago)
b.Theo phần (a) ta có: $a+b=3c \Leftrightarrow p=2c (1)$Mặt khác:$S=pr=\dfrac{abc}{4R}\Rightarrow pabc\dfrac{r}{4R}=S^2=p(p-a)(p-b)(p-c)$Thay $(1)$ vào biểu thức trên ta được:$ab\dfrac{r}{4R}=p^2-(a+b)p+ab\Rightarrow (1-\dfrac{r}{4R})ab=2c^2 (2)$Lại có:$S=\dfrac{1}{2}ab\sin C=pr\Rightarrow r=\dfrac{ab\sin C}{a+b+c}=\dfrac{4R^2\sin A\sin B\sin C}{2R(\sin A+\sin B+\sin C)}$Suy ra: $\dfrac{r}{4R}=\dfrac{\sin A\sin B\sin C}{\sin A+\sin B+\sin C}$Dễ dàng chứng minh được: $\sin A+\sin B+\sin C=4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$$\Rightarrow \dfrac{r}{4R}=\tan\dfrac{A}{2}\tan\dfrac{B}{2}\tan\dfrac{C}{2}\Rightarrow \dfrac{r}{4R}=\dfrac{1}{10}$Thay vào $(2)$ ta được: $ab=\dfrac{20}{9}c^2$$\Rightarrow ab=20(a+b)^2$$\Leftrightarrow 20a^2-41ab+20b^2=0$$\Leftrightarrow \left\{\begin{array}{l}4a=5b\\5a=4b\end{array}\right.$Cả 2 trường hợp đều dẫn tới $\Delta ABc$ vuông. (theo Pytago)
b.Theo phần (a) ta có: $a+b=3c \Leftrightarrow p=2c (1)$Mặt khác:$S=pr=\dfrac{abc}{4R}\Rightarrow pabc\dfrac{r}{4R}=S^2=p(p-a)(p-b)(p-c)$Thay $(1)$ vào biểu thức trên ta được:$ab\dfrac{r}{4R}=p^2-(a+b)p+ab\Rightarrow (1-\dfrac{r}{4R})ab=2c^2 (2)$Lại có:$S=\dfrac{1}{2}ab\sin C=pr\Rightarrow r=\dfrac{ab\sin C}{a+b+c}=\dfrac{4R^2\sin A\sin B\sin C}{2R(\sin A+\sin B+\sin C)}$Suy ra: $\dfrac{r}{4R}=\dfrac{\sin A\sin B\sin C}{
2(\sin A+\sin B+\sin C
)}$
Mà
ta c
ó:$\sin A+\sin B+\sin C=
2\sin\dfrac{A}{2}\cos\dfrac{A}{2}+2\sin\dfrac{B+C}{2}\cos\dfrac{B-C}{2}$ $=2\cos\dfrac{A}{2}(\cos\dfrac{B+C}{2}+\cos\dfrac{B-C}{2})$ $=4\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}$$\Rightarrow
\dfrac{r}{4R}=\
dfra
c{8\sin\dfrac{A}{2}\
sin\dfra
c{B}{2}\sin\dfrac{
C}{2}\cos\dfrac{A}{2}\cos\dfrac{B}{2}\
cos\dfra
c{C}{2}}{8\cos\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}}=\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}
$$\Rightarrow \dfrac{r}{4R}=\dfrac{1}{10}$Thay vào $(2)$ ta được: $ab=\dfrac{20}{9}c^2$$\Rightarrow ab=20(a+b)^2$$\Leftrightarrow 20a^2-41ab+20b^2=0$$\Leftrightarrow \left\{\begin{array}{l}4a=5b\\5a=4b\end{array}\right.$Cả 2 trường hợp đều dẫn tới $\Delta ABc$ vuông. (theo Pytago)