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Question Number 53066 by ajfour last updated on 16/Jan/19

Commented by ajfour last updated on 17/Jan/19

$$DetermineradiusRoflargercircle$$$$intermsofa,b,c.$$

Commented by ajfour last updated on 17/Jan/19

$$Sir..\stackrel{pleasetrythiseven!}{\equiv \gg }$$

Commented by MJS last updated on 17/Jan/19

$$...{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{busy}\mathrm{trying}...$$

Commented by ajfour last updated on 18/Jan/19

Commented by ajfour last updated on 18/Jan/19

$$R=\frac{\delta }{8(b+c-a)}+\frac{a^2(b+c-a)}{2\delta }$$$$\delta =\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$$$pleasecheckitnowSir.$$

Commented by ajfour last updated on 18/Jan/19

$$BK=R,BM=\frac{a}{2}$$$$\Rightarrow MK=\sqrt{R^2-\frac{a^2}{4}}$$$$JN=r-\sqrt{R^2-\frac{a^2}{4}},KN=R-r$$$$Nowr(\cot \theta +\cot \varphi )=a$$$$r(\cot \theta +\cot \psi )=c$$$$r(\cot \varphi +\cot \psi )=b$$$$\Rightarrow 2r\cot \theta =a+c-b$$$$r\cot \theta =\frac{c-b}{2}+\frac{a}{2}$$$$BP=r\cot \theta$$$$JK=MP=BP-\frac{a}{2}$$$$JK=r\cot \theta -\frac{a}{2}=\frac{c-b}{2}$$$$In△JKN$$$${KN}^2={JK}^2+{JN}^2$$$$\Rightarrow {(R-r)}^2={\left(\frac{c-b}{2}\right)}^2+{(r-\sqrt{R^2-\frac{a^2}{4}})}^2$$$$\Rightarrow R^2-2rR+r^2={\left(\frac{c-b}{2}\right)}^2+r^2+R^2-\frac{a^2}{4}$$$$-2r\sqrt{R^2-\frac{a^2}{4}}$$$$\Rightarrow 2r(R-\sqrt{R^2-\frac{a^2}{4}})=\frac{(a+c-b)(a+b-c)}{4}$$$$letarea(△ABC)=△$$$$weknowr(a+b+c)=2△,and$$$$△=\sqrt{s(s-a)(s-b)(s-c)}$$$$If\delta =\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$$$\Rightarrow 4△=\delta =2r(a+b+c)$$$$\Rightarrow \mathit{r}=\frac{\mathit{\delta }}{2(\mathit{a}+\mathit{b}+\mathit{c})}$$$$2r(R-\sqrt{R^2-\frac{a^2}{4}})=\frac{(a+c-b)(a+b-c)}{4}$$$$gives$$$$2r(R-\sqrt{R^2-\frac{a^2}{4}})=\frac{{\delta }^2}{4(a+b+c)(b+c-a)}$$$$\Rightarrow R-\sqrt{R^2-\frac{a^2}{4}}=\frac{\delta }{4(b+c-a)}$$$$\Rightarrow R^2-\frac{a^2}{4}={[R-\frac{\delta }{4(b+c-a)}]}^2$$$$\Rightarrow \frac{R\delta }{2(b+c-a)}=\frac{a^2}{4}+\frac{{\delta }^2}{16{(b+c-a)}^2}$$$$\Rightarrow R=\frac{a^2(b+c-a)}{2\delta }+\frac{\delta }{8(b+c-a)}.$$

Commented by MJS last updated on 18/Jan/19

$$\mathrm{now}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{same}\mathrm{as}\mathrm{mine}.\mathrm{how}\mathrm{did}\mathrm{you}\mathrm{get}\mathrm{it}?$$

Commented by MJS last updated on 18/Jan/19

$$\mathrm{great}!\mathrm{now}\mathrm{try}\mathrm{the}\mathrm{same}\mathrm{for}\mathrm{the}\mathrm{excircles}$$

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

$$R=\frac{2a^2(p-a)}{8S}+\frac{4S}{16(p-a)}=\frac{a^2+r_a^2}{4r_a}.$$

Answered by MJS last updated on 18/Jan/19

$$\mathrm{coordinate}\mathrm{method}\mathrm{gives}$$$$\delta =\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}$$$$A=\left(\begin{array}{c}-\frac{\delta }{2a}\\ \frac{c^2-b^2}{2a}\end{array}\right)B=\left(\begin{array}{c}0\\ -\frac{a}{2}\end{array}\right)C=\left(\begin{array}{c}0\\ \frac{a}{2}\end{array}\right)$$$$\mathrm{incircle}$$$$\mathrm{center}=\left(\begin{array}{c}-\frac{\delta }{2(a+b+c)}\\ \frac{c-b}{2}\end{array}\right)$$$$\mathrm{radius}=\frac{\delta }{2(a+b+c)}$$$$\mathrm{big}\mathrm{circle}$$$$\mathrm{center}=\left(\begin{array}{c}\frac{5a^3-b^3-c^3-3a^2(b+c)-b^2(a-c)-c^2(a-b)+2abc}{8\delta }\\ 0\end{array}\right)$$$$\mathrm{radius}=-\frac{3a^3+b^3+c^3-5a^2(b+c)+b^2(a-c)+c^2(a-b)-2abc}{8\delta }$$$$\mathrm{I}{\mathrm{couldn}}^{\prime }\mathrm{t}\mathrm{simplify}...\mathrm{but}\mathrm{I}\mathrm{plotted}\mathrm{several}$$$$\mathrm{triangles}\mathrm{with}\mathrm{both}\mathrm{circles}\mathrm{and}\mathrm{it}\mathrm{seems}\mathrm{to}$$$$\mathrm{hold}...$$

Commented by MJS last updated on 18/Jan/19

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}.$$$$\mathrm{I}\mathrm{like}\mathrm{your}\mathrm{way}\mathrm{to}\mathrm{solve}\mathrm{it}!$$

Commented by ajfour last updated on 18/Jan/19

$$thisimissed.thanks.$$

Commented by mr W last updated on 18/Jan/19

$$nicesolutionfromyouboth!$$

Commented by ajfour last updated on 18/Jan/19

$$thanksSir.$$

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