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Question Number 19610 by ajfour last updated on 13/Aug/17

$$\mathrm{Prove}\mathrm{that}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{a}\mathrm{circle}$$$$\mathrm{passing}\mathrm{through}\mathrm{the}\mathrm{midpoints}$$$$\mathrm{of}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{ABC}\mathrm{is}$$$$\mathrm{half}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{circum}-$$$$\mathrm{scribed}\mathrm{about}\mathrm{the}\mathrm{triangle}.$$

Commented by ajfour last updated on 13/Aug/17

Commented by ajfour last updated on 13/Aug/17

$$\mathrm{Let}\mathrm{B}(0,0);\mathrm{A}(2\mathrm{b},2\mathrm{c});\mathrm{C}(2\mathrm{a},0)$$$${\mathrm{R}}^2={\mathrm{h}}^2+{\mathrm{k}}^2....\left(1\right)$$$$={(2\mathrm{a}-\mathrm{h})}^2+{\mathrm{k}}^2...\left(2\right)$$$$={(\mathrm{h}-2\mathrm{b})}^2+{(2\mathrm{c}-\mathrm{k})}^2...\left(3\right)$$$$\mathrm{from}\left(1\right)\mathrm{and}\left(2\right):$$$$\mathrm{h}=\mathrm{a}\mathrm{and}\mathrm{from}\left(1\right)\mathrm{and}\left(3\right):$$$${(\mathrm{a}-2\mathrm{b})}^2+{(2\mathrm{c}-\mathrm{k})}^2={\mathrm{a}}^2+{\mathrm{k}}^2$$$$\Rightarrow {4\mathrm{b}}^2-4\mathrm{ab}=(2\mathrm{k}-2\mathrm{c})\left(2\mathrm{c}\right)$$$$\mathrm{or}\mathrm{k}=\mathrm{c}-\frac{\mathrm{b}(\mathrm{a}-\mathrm{b})}{\mathrm{c}}$$$$\mathrm{hence}{\mathrm{R}}^2={\mathrm{a}}^2+{\left[\frac{{\mathrm{b}}^2+{\mathrm{c}}^2-\mathrm{ab}}{\mathrm{c}}\right]}^2$$$${\mathrm{r}}^2={({\mathrm{x}}_0-\mathrm{a})}^2+{\mathrm{y}}_0^2...\left(\mathrm{i}\right)$$$$={({\mathrm{x}}_0-\mathrm{b})}^2+{({\mathrm{y}}_0-\mathrm{c})}^2....\left(\mathrm{ii}\right)$$$$={({\mathrm{x}}_0-\mathrm{a}-\mathrm{b})}^2+{({\mathrm{y}}_0-\mathrm{c})}^2...\left(\mathrm{iii}\right)$$$$\mathrm{using}\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right)$$$$\Rightarrow {\mathrm{x}}_0-\mathrm{a}-\mathrm{b}=\mathrm{b}-{\mathrm{x}}_0$$$$\mathrm{or}{\mathrm{x}}_0=\mathrm{b}+\frac{\mathrm{a}}{2};$$$$\mathrm{Also}\mathrm{using}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right):$$$${\mathrm{r}}^2={(\mathrm{b}+\frac{\mathrm{a}}{2}-\mathrm{a})}^2+{\mathrm{y}}_0^2={(\mathrm{b}+\frac{\mathrm{a}}{2}-\mathrm{b})}^2+{({\mathrm{y}}_0-\mathrm{c})}^2$$$$\Rightarrow \mathrm{c}({2\mathrm{y}}_0-\mathrm{c})=\frac{{\mathrm{a}}^2}{4}-{(\mathrm{b}-\frac{\mathrm{a}}{2})}^2$$$$\Rightarrow \mathrm{c}({2\mathrm{y}}_0-\mathrm{c})=\mathrm{ab}-{\mathrm{b}}^2$$$$\Rightarrow {\mathrm{y}}_0=\frac{\mathrm{c}}{2}+\frac{\mathrm{b}(\mathrm{a}-\mathrm{b})}{2\mathrm{c}}$$$$\Rightarrow {({\mathrm{y}}_0-\mathrm{c})}^2={[\frac{\mathrm{b}(\mathrm{a}-\mathrm{b})}{2\mathrm{c}}-\frac{\mathrm{c}}{2}]}^2$$$$\mathrm{using}\left(\mathrm{ii}\right):$$$${\mathrm{r}}^2=\frac{{\mathrm{a}}^2}{4}+{\left[\frac{\mathrm{ab}-{\mathrm{b}}^2-{\mathrm{c}}^2}{2\mathrm{c}}\right]}^2$$$$\Rightarrow {4\mathrm{r}}^2={\mathrm{a}}^2+{\left[\frac{{\mathrm{b}}^2+{\mathrm{c}}^2-\mathrm{ab}}{\mathrm{c}}\right]}^2={\mathrm{R}}^2$$$$\Rightarrow \mathbf{r}=\frac{\mathbf{R}}{2}.$$

Answered by Tinkutara last updated on 13/Aug/17

$$\mathrm{Let}\mathrm{circumradius}\mathrm{of}\mathrm{original}\mathrm{triangle}\mathrm{is}$$$$R=\frac{a_1{b}_1{c}_1}{4{\mathrm{\Delta }}_1}$$$$\mathrm{New}\mathrm{sides}\mathrm{are}\frac{a_1}{2},\frac{b_1}{2}\mathrm{and}\frac{c_1}{2}\mathrm{and}{\mathrm{\Delta }}_2$$$$=\frac{{\mathrm{\Delta }}_1}{4}$$$$\mathrm{So}{R}^{\prime }=\frac{\frac{a_1{b}_1{c}_1}{8}}{4\frac{{\mathrm{\Delta }}_1}{4}}=\frac{1}{2}\left(\frac{a_1{b}_1{c}_1}{4{\mathrm{\Delta }}_1}\right)=\frac{R}{2}$$

Commented by ajfour last updated on 13/Aug/17

$$\mathrm{Excellent}!$$

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