Question Number 38281 by Tinkutara last updated on 23/Jun/18
Answered by ajfour last updated on 23/Jun/18
$$\frac{{ap}_{{A}\rightarrow{BC}} }{\mathrm{2}}\:=\:\frac{{bp}_{{B}\rightarrow{CA}} }{\mathrm{2}}\:=\:\frac{{cp}_{{C}\rightarrow{AB}} }{\mathrm{2}}\:=\bigtriangleup \\ $$$$\frac{{ap}_{{P}\rightarrow{BC}} }{\mathrm{2}}\:+\:\frac{{bp}_{{P}\rightarrow{CA}} }{\mathrm{2}}\:+\:\frac{{cp}_{{P}\rightarrow{AB}} }{\mathrm{2}}\:=\:\Delta \\ $$$${p}_{{P}\rightarrow{BC}} =\:\frac{{p}_{{A}\rightarrow{BC}} }{{r}}\:\:{and}\:{so}\:{on}.. \\ $$$$\Rightarrow\:\:\frac{{ap}_{{P}\rightarrow{BC}} }{\mathrm{2}}=\:\frac{{ap}_{{A}\rightarrow{BC}} }{\mathrm{2}{r}}\:=\:\frac{\Delta}{{r}} \\ $$$$\frac{{ap}_{{P}\rightarrow{BC}} }{\mathrm{2}}\:=\:\frac{{bp}_{{P}\rightarrow{CA}} }{\mathrm{2}}\:=\frac{{cp}_{{P}\rightarrow{AB}} }{\mathrm{2}}\:=\:\frac{\Delta}{{r}} \\ $$$${So} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{3}\Delta}{{r}}\:=\:\Delta\:\:\:\Rightarrow\:\:\:{r}\:=\:\mathrm{3}\:\:. \\ $$$$\:\:\:\:\:\:\:\: \\ $$
Commented by Tinkutara last updated on 24/Jun/18
Thank you very much Sir! I got the answer. ��������
Very nice solution.