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Question Number 169539 by Shrinava last updated on 02/May/22

Commented by mr W last updated on 02/May/22

$$\frac{R}{OI}=\sqrt{\frac{3}{2}}?$$

Commented by Shrinava last updated on 02/May/22

$$\mathrm{Yes}\mathrm{dear}\mathrm{sir}$$

Answered by mr W last updated on 02/May/22

$$\sin A=2\sin \frac{A}{2}\cos \frac{A}{2}=\frac{2\sqrt{abcd}}{ad+bc}$$$$\sin B=2\sin \frac{B}{2}\cos \frac{B}{2}=\frac{2\sqrt{abcd}}{ab+cd}$$$$3\sin A\sin B=1$$$$3\times \frac{4abcd}{(ab+cd)(ad+bc)}=1$$$$\frac{abcd}{(ab+cd)(ad+bc)}=\frac{1}{12}$$$$$$$$OI=R\sqrt{1-\frac{2r\mu }{R}}$$$$r=\frac{\sqrt{abcd}}{a+c}$$$$R=\frac{1}{4}\sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{abcd}}$$$$\mu =\sqrt{\frac{(ab+cd)(ad+bc)}{{(a+c)}^2(ac+bd)}}$$$$1-\frac{2r\mu }{R}=1-8\times \frac{\sqrt{abcd}}{a+c}\sqrt{\frac{abcd}{(ab+cd)(ac+bd)(ad+bc)}}\sqrt{\frac{(ab+cd)(ad+bc)}{{(a+c)}^2(ac+bd)}}$$$$=1-\frac{8abcd}{{(a+c)}^2(ac+bd)}$$$$=1-\frac{8abcd}{(a+c)(b+d)(ac+bd)}$$$$=1-\frac{8abcd}{(ab+cd+ad+bc)(ac+bd)}$$$$=1-\frac{4abcd}{(ab+cd)(ad+bc)}$$$$=1-4\times \frac{1}{12}$$$$=\frac{2}{3}$$$$\mathrm{\Omega }=\frac{R}{OI}=\frac{1}{\sqrt{1-\frac{2r\mu }{R}}}=\frac{1}{\sqrt{\frac{2}{3}}}=\sqrt{\frac{3}{2}}✓$$

Commented by Shrinava last updated on 03/May/22

$$\mathrm{Perfect}\mathrm{dear}\mathrm{sir}\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

Commented by mr W last updated on 03/May/22

you may find all the formulas in: https://en.m.wikipedia.org/wiki/Bicentric_quadrilateral

Commented by Shrinava last updated on 03/May/22

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{professor}$$

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