Question-185128

Question Number 185128 by Shrinava last updated on 17/Jan/23

Answered by Frix last updated on 17/Jan/23

Formulas for incircle and 3 excircles: r=((2A)/(a+b+c)) r_a =((2A)/(b+c−a)) r_b =((2A)/(a+c−b)) r_c =((2A)/(a+b−c)) A^2 =rr_a r_b r_c ⇒ −a+b+c=((2A)/1)=2A a−b+c=((2A)/4)=(A/2) a+b−c=((2A)/(12))=(A/6) ⇒ a=(A/3)∧b=((13A)/(12))∧c=((5A)/4) ⇒ r=(3/4) ⇒ A=6

$$\mathrm{Formulas}\:\mathrm{for}\:\mathrm{incircle}\:\mathrm{and}\:\mathrm{3}\:\mathrm{excircles}: \\ $$$${r}=\frac{\mathrm{2}{A}}{{a}+{b}+{c}} \\ $$$${r}_{{a}} =\frac{\mathrm{2}{A}}{{b}+{c}−{a}} \\ $$$${r}_{{b}} =\frac{\mathrm{2}{A}}{{a}+{c}−{b}} \\ $$$${r}_{{c}} =\frac{\mathrm{2}{A}}{{a}+{b}−{c}} \\ $$$${A}^{\mathrm{2}} ={rr}_{{a}} {r}_{{b}} {r}_{{c}} \\ $$$$\Rightarrow \\ $$$$−{a}+{b}+{c}=\frac{\mathrm{2}{A}}{\mathrm{1}}=\mathrm{2}{A} \\ $$$${a}−{b}+{c}=\frac{\mathrm{2}{A}}{\mathrm{4}}=\frac{{A}}{\mathrm{2}} \\ $$$${a}+{b}−{c}=\frac{\mathrm{2}{A}}{\mathrm{12}}=\frac{{A}}{\mathrm{6}} \\ $$$$\Rightarrow \\ $$$${a}=\frac{{A}}{\mathrm{3}}\wedge{b}=\frac{\mathrm{13}{A}}{\mathrm{12}}\wedge{c}=\frac{\mathrm{5}{A}}{\mathrm{4}} \\ $$$$\Rightarrow\:{r}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:{A}=\mathrm{6} \\ $$

Commented by Shrinava last updated on 17/Jan/23