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Question Number 111926 by ajfour last updated on 05/Sep/20

Commented by ajfour last updated on 05/Sep/20

$$If△ABCisequilateralwithside\mathit{s},$$$$findradiiofthethreecirclesin$$$$termsofs.$$

Answered by mr W last updated on 05/Sep/20

Commented by ajfour last updated on 05/Sep/20

$$from\left(ii\right):$$$${(\beta +\gamma )}^2={\beta }^2+{(1-\gamma )}^2-\beta (1-\gamma )...\left(ii\right)$$$$2\beta \gamma =-2\gamma +1-\beta +\beta \gamma$$$$\mathit{\beta }=\frac{1-2\mathit{\gamma }}{1+\mathit{\gamma }}....\left(I\right)$$$$\mathrm{\&}from\gamma (1-8\beta )=4\beta -1$$$$4\mathit{\beta }=\frac{1+\mathit{\gamma }}{1+2\mathit{\gamma }}...\left(II\right)$$$$\Rightarrow \frac{4(1-2\gamma )}{1+\gamma }=\frac{1+\gamma }{1+2\gamma }$$$$\Rightarrow 4-16{\gamma }^2=1+2\gamma +{\gamma }^2$$$$\Rightarrow 17{\gamma }^2+2\gamma -3=0$$$$\Rightarrow \mathit{\gamma }=\frac{2\sqrt{13}-1}{17}$$$$\beta =\frac{1-2\gamma }{1+\gamma }=\frac{1-\frac{(4\sqrt{13}-2)}{17}}{1+\frac{(2\sqrt{13}-1)}{17}}$$$$=\frac{19-4\sqrt{13}}{16+2\sqrt{13}}=\frac{(19-4\sqrt{13})(16-2\sqrt{13})}{204}$$$$=\frac{304+104-102\sqrt{13}}{204}=\frac{4-\sqrt{13}}{2}$$$$\mathit{\beta }=\frac{4-\sqrt{13}}{2}$$$$\mathit{\alpha }=\frac{1}{2}-\frac{(4-\sqrt{13})}{2}=\frac{\sqrt{13}-3}{2}$$$$\mathit{\gamma }=\frac{2\sqrt{13}-1}{17}$$$$\mathcal{T}hankyouverymuch,mrWSir!$$$$greatsolution!$$$$$$

Commented by mr W last updated on 06/Sep/20

$$wow!i{didn}^{\prime }texpectthatwecanget$$$$theexactsolution.thankssir!$$

Commented by mr W last updated on 05/Sep/20

$$a+b=\frac{s}{2}$$$$\alpha +\beta =\frac{1}{2}...\left(i\right)$$$$$$$$BD=s-c$$$$FD=b+c$$$${FD}^2={BF}^2+{BD}^2-2\times BF\times BD\times \cos \mathrm{\angle }B$$$${(b+c)}^2=b^2+{(s-c)}^2-b(s-c)$$$${(\beta +\gamma )}^2={\beta }^2+{(1-\gamma )}^2-\beta (1-\gamma )...\left(ii\right)$$$$$$$$ED=a+c$$$${(a+c)}^2={(a+2b)}^2+{(s-c)}^2-(a+2b)(s-c)$$$${(\frac{s}{2}-b+c)}^2={(\frac{s}{2}+b)}^2+{(s-c)}^2-(\frac{s}{2}+b)(s-c)$$$${(\frac{1}{2}-\beta +\gamma )}^2={(\frac{1}{2}+\beta )}^2+{(1-\gamma )}^2-(\frac{1}{2}+\beta )(1-\gamma )...\left(iii\right)$$$$$$$$\left(iii\right)-\left(ii\right):$$$$\gamma (1-8\beta )=4\beta -1$$$$\Rightarrow \gamma =\frac{4\beta -1}{1-8\beta }$$$$$$$$\Rightarrow \beta \approx 0.19722=\frac{b}{s}$$$$\Rightarrow \gamma \approx 0.36541=\frac{c}{s}$$$$\Rightarrow \alpha \approx 0.30378=\frac{a}{s}$$

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