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Question Number 73221 by aliesam last updated on 08/Nov/19

Answered by MJS last updated on 08/Nov/19

$$A=\left(\begin{array}{c}0\\ 0\end{array}\right)B=\left(\begin{array}{c}a\\ 0\end{array}\right)C=\left(\begin{array}{c}a\\ a\end{array}\right)D=\left(\begin{array}{c}0\\ a\end{array}\right)$$$$E=\left(\begin{array}{c}b\\ 0\end{array}\right)F=\left(\begin{array}{c}b\\ b\end{array}\right)G=\left(\begin{array}{c}0\\ b\end{array}\right)$$$$T=\left(\begin{array}{c}\frac{a+b}{2}\\ 0\end{array}\right)I=\left(\begin{array}{c}\frac{a+b}{2}\\ \frac{a-b}{2}\end{array}\right)$$$$M=\left(\begin{array}{c}b\\ \frac{a-b}{2}\end{array}\right)N=\left(\begin{array}{c}\frac{\sqrt{3}}{4}(a+b)+b\\ \frac{3a-b}{4}\end{array}\right)L=\left(\begin{array}{c}b\\ a\end{array}\right)$$$$\mathrm{radius}\mathrm{of}\mathrm{incircle}\mathrm{of}ETIM=\frac{r}{2}=\frac{a-b}{4}$$$$\mathrm{radius}\mathrm{of}\mathrm{incircle}\mathrm{of}MNL=\frac{\sqrt{3}}{12}(a+b)$$$$\frac{a-b}{4}=\frac{\sqrt{3}}{12}(a+b)\Rightarrow a=(2+\sqrt{3})b\Rightarrow r=\frac{1+\sqrt{3}}{2}b$$$$\frac{ab}{a-b}=\frac{(2+\sqrt{3})b^2}{(2+\sqrt{3})b-b}=\frac{1+\sqrt{3}}{2}b$$

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