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Question Number 22576 by ajfour last updated on 20/Oct/17

Commented by ajfour last updated on 20/Oct/17

$$Q.22536\left(Alternatesolution\right)$$

Answered by ajfour last updated on 20/Oct/17

$$△PQI\sim △ABC\left(firstdiagram\right)$$$$So\frac{PQ}{AB}=\frac{r}{AC\sin A}$$$$\Rightarrow r(\cot A+\cot B)=\frac{cr}{b\sin A}$$$$or\cot A+\cot B=\frac{c}{b\sin A}$$$$weknow\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2R}$$$$so\cot A+\cot B=\frac{2cR}{ab}$$$$2\mathrm{\Sigma }\cot A=2R(\frac{c}{ab}+\frac{a}{bc}+\frac{b}{ca})$$$$or\mathrm{\Sigma }\cot A=\frac{R(a^2+b^2+c^2)}{abc}...\left(i\right)$$$$Fromseconddiagram:$$$$2\times \frac{1}{2}\times r\times r\mathrm{\Sigma }\cot \frac{A}{2}=△$$$$\Rightarrow r^2\mathrm{\Sigma }\cot \frac{A}{2}=△......\left(ii\right)$$$$and\frac{1}{2}c\times b\sin A=△$$$$or\frac{bc}{2}\times \frac{a}{2R}=△\Rightarrow \frac{abc}{4R}=△..\left(iii\right)$$$$andr\mathrm{\Sigma }\cot \frac{A}{2}=x+y+z$$$$\Rightarrow r\mathrm{\Sigma }\cot \frac{A}{2}=\frac{a+b+c}{2}....\left(iv\right)$$$$squaring\left(iv\right)anddividingby\mathrm{\Sigma }cotA,$$$$\Rightarrow \frac{r^2{\left(\mathrm{\Sigma }\cot \frac{A}{2}\right)}^2}{\mathrm{\Sigma }\cot A}=\frac{{(a+b+c)}^2}{4\mathrm{\Sigma }\cot A}$$$$using\left(ii\right)and\left(i\right)ineqn.above$$$$\frac{\left(\mathrm{\Sigma }\cot \frac{A}{2}\right)△}{\mathrm{\Sigma }\cot A}=\frac{{(a+b+c)}^2}{\frac{4R(a^2+b^2+c^2)}{abc}}$$$$Nowusingeqn.\left(iii\right):abc=4△R$$$$\frac{\mathrm{\Sigma }\cot \frac{A}{2}}{\mathrm{\Sigma }\cot A}=\frac{{(\mathit{a}+\mathit{b}+\mathit{c})}^2}{{\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2}.$$

Commented by NECx last updated on 20/Oct/17

$$thanksboss$$

Commented by math solver last updated on 20/Oct/17

$$+1:)$$

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