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Question Number 27959 by ajfour last updated on 17/Jan/18

Commented by ajfour last updated on 17/Jan/18

Inspired with Q.#27942

$${Inspired}\:{with}\:{Q}.#\mathrm{27942} \\ $$

Commented by beh.i83417@gmail.com last updated on 17/Jan/18

(r/(a/2))=((Ax)/(xC))⇒((2r)/a)=((Ax)/(xC))⇒((2r+a)/a)=(b/(Cx)) cosC=((a/2)/(xC))⇒xC=(a/(2cosC)) ⇒2r/a+1=(b/(a/(2cosC)))=((2bcosC)/a)⇒ 2r+a=2bcosC⇒r=(1/2)(2bcosC−a) 1)a^2 =b^2 +c^2 −bc 2)cosC=((a^2 +b^2 −c^2 )/(2bc))=((b^2 +c^2 −bc+b^2 −c^2 )/(2bc))= =((2b^2 −bc)/(2bc)) ⇒r=(1/2)(2b.((2b^2 −bc)/(2bc))−a)=((2b^2 −bc−ac)/(2c)) ⇒r=(b^2 /c)−((a+b)/2) .■

$$\frac{\boldsymbol{{r}}}{\boldsymbol{{a}}/\mathrm{2}}=\frac{\boldsymbol{{A}}{x}}{{xC}}\Rightarrow\frac{\mathrm{2}{r}}{{a}}=\frac{\boldsymbol{{A}}{x}}{{xC}}\Rightarrow\frac{\mathrm{2}{r}+{a}}{{a}}=\frac{{b}}{{Cx}} \\ $$$${cosC}=\frac{{a}/\mathrm{2}}{{xC}}\Rightarrow{xC}=\frac{{a}}{\mathrm{2}{cosC}} \\ $$$$\Rightarrow\mathrm{2}{r}/{a}+\mathrm{1}=\frac{{b}}{\frac{{a}}{\mathrm{2}{cosC}}}=\frac{\mathrm{2}{bcosC}}{{a}}\Rightarrow \\ $$$$\mathrm{2}{r}+{a}=\mathrm{2}{bcosC}\Rightarrow{r}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{bcosC}−{a}\right) \\ $$$$\left.\mathrm{1}\right){a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{bc} \\ $$$$\left.\mathrm{2}\right){cosC}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{bc}}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{bc}+{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{bc}}= \\ $$$$=\frac{\mathrm{2}{b}^{\mathrm{2}} −{bc}}{\mathrm{2}{bc}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{b}.\frac{\mathrm{2}{b}^{\mathrm{2}} −{bc}}{\mathrm{2}{bc}}−{a}\right)=\frac{\mathrm{2}{b}^{\mathrm{2}} −{bc}−{ac}}{\mathrm{2}{c}} \\ $$$$\Rightarrow\boldsymbol{{r}}=\frac{\boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{c}}}−\frac{\boldsymbol{{a}}+\boldsymbol{{b}}}{\mathrm{2}}\:.\blacksquare \\ $$

Commented by ajfour last updated on 17/Jan/18

Sir, what does Ax, xC, ..mean ?

$${Sir},\:{what}\:{does}\:{Ax},\:{xC},\:..{mean}\:? \\ $$

Commented by beh.i83417@gmail.com last updated on 17/Jan/18

x, is the intersection of AC with perpendicular line drawn from midpoint of BC.

$${x},\:{is}\:{the}\:{intersection}\:{of}\:\boldsymbol{{AC}}\:{with}\: \\ $$$${perpendicular}\:{line}\:{drawn}\:{from}\:{midpoint} \\ $$$${of}\:\boldsymbol{{BC}}. \\ $$

Commented by beh.i83417@gmail.com last updated on 18/Jan/18

sir Ajfour! what about Q#27942 ?

$${sir}\:{Ajfour}!\:{what}\:{about}\:{Q}#\mathrm{27942}\:? \\ $$

Commented by ajfour last updated on 18/Jan/18

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

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