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Question Number 52881 by ajfour last updated on 14/Jan/19

Commented by ajfour last updated on 14/Jan/19

$$Tofind\mathit{r}intermsofa,b,c.$$

Commented by mr W last updated on 14/Jan/19

$$canitbeconfirmedthat$$$$\mathit{r}=\frac{2\mathit{b}\mathit{c}}{\mathit{a}+\mathit{b}+\mathit{c}}\sqrt{\frac{(\mathit{a}+\mathit{b}-\mathit{c})(\mathit{a}+\mathit{c}-\mathit{b})}{(\mathit{a}+\mathit{b}+\mathit{c})(\mathit{b}+\mathit{c}-\mathit{a})}}$$

Commented by ajfour last updated on 14/Jan/19

$$niceresultSir,how{d}^{\prime }yagetthis?$$

Commented by behi83417@gmail.com last updated on 14/Jan/19

$$r=\frac{bc}{p}tg\frac{A}{2}.itistruesirmrW.$$

Commented by MJS last updated on 14/Jan/19

$$...\mathrm{and}\mathrm{please}\mathrm{re}-\mathrm{check}52061,\mathrm{I}\mathrm{get}\mathrm{a}\mathrm{different}$$$$\mathrm{result}$$

Commented by MJS last updated on 14/Jan/19

$$\mathrm{see}\mathrm{my}\mathrm{answer}\mathrm{to}\mathrm{qu}.52806$$

Commented by mr W last updated on 15/Jan/19

$$MJSsir:yourresultforristhesame.$$$$itcanalsobeformedto$$$$r=\frac{2bc}{a+b+c}\sqrt{\frac{(a+b-c)(a+c-b)}{(a+b+c)(b+c-a)}}$$

Commented by mr W last updated on 15/Jan/19

$$MJSsir:yourresultforQ52061seems$$$$nottobecorrect.$$

Commented by MJS last updated on 15/Jan/19

$$\mathrm{I}\mathrm{will}\mathrm{look}\mathrm{into}52061\mathrm{once}\mathrm{more}$$$$\mathrm{I}\mathrm{saw}\mathrm{that}\mathrm{the}\mathrm{other}\mathrm{result}\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{yours},$$$$\mathrm{I}\mathrm{thought}\mathrm{you}\mathrm{might}\mathrm{have}\mathrm{achieved}\mathrm{it}\mathrm{throuhh}$$$$\mathrm{a}\mathrm{different}\mathrm{path}$$

Answered by ajfour last updated on 15/Jan/19

$$FrompowerofpointQw.r.t.$$$$circumcircle:$$$$(2R\cos \varphi -\frac{r}{\sin \theta })\frac{r}{\sin \theta }=r(2R-r)$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\Rightarrow 2R\cos \varphi =\frac{r}{\sin \theta }+(2R-r)\sin \theta$$$$......\left(i\right)$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$furtherR\cos (\theta +\varphi )=\frac{b}{2}\mathrm{\&}$$$$R\cos (\theta -\varphi )=\frac{c}{2}$$$$Adding:2R\cos \varphi \cos \theta =\frac{b+c}{2}$$$$Sousing\left(i\right)$$$$\frac{r}{\tan \theta }+R\sin 2\theta -\frac{r\sin {}^2\theta }{\tan \theta }=\frac{b+c}{2}$$$$but\frac{\sin 2\theta }{a}=\frac{1}{2R}\Rightarrow$$$$\frac{{\mathit{r}\mathit{c}\mathit{o}\mathit{s}}^2\mathit{\theta }}{\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}=\frac{\mathit{b}+\mathit{c}-\mathit{a}}{2}$$$$\mathit{r}=\frac{(\mathit{b}+\mathit{c}-\mathit{a})}{1+\mathit{c}\mathit{o}\mathit{s}A}\sqrt{\frac{1-\mathit{c}\mathit{o}\mathit{s}A}{1+\mathit{c}\mathit{o}\mathit{s}A}}$$$$=\frac{(b+c-a)}{\left(\frac{2bc+b^2+c^2-a^2}{2bc}\right)}\sqrt{\frac{2bc-b^2-c^2+a^2}{2bc+b^2+c^2-a^2}}$$$$\mathit{r}=\frac{2\mathit{b}\mathit{c}}{(\mathit{a}+\mathit{b}+\mathit{c})}\sqrt{\frac{(\mathit{a}+\mathit{b}-\mathit{c})(\mathit{a}+\mathit{c}-\mathit{b})}{(\mathit{b}+\mathit{c}+\mathit{a})(\mathit{b}+\mathit{c}-\mathit{a})}}.$$

Commented by mr W last updated on 15/Jan/19

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