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Question Number 95440 by I want to learn more last updated on 25/May/20

Answered by mr W last updated on 25/May/20

Commented by mr W last updated on 25/May/20

$$sidelengthofequilateral=1$$$$a^2=\pi r^2$$$$\Rightarrow a=\sqrt{\pi }r$$$$AG=\frac{r}{\sqrt{3}}$$$$OA=\frac{1}{2}-\frac{\sqrt{3}r}{3}$$$$OD=\frac{\sqrt{3}}{2}-\frac{\sqrt{3}a}{2}-a=\frac{\sqrt{3}}{2}-(\frac{\sqrt{3}}{2}+1)\sqrt{\pi }r$$$${[\frac{\sqrt{3}}{2}-(\frac{\sqrt{3}}{2}+1)\sqrt{\pi }r-r]}^2+{[\frac{1}{2}-\frac{\sqrt{3}r}{3}-\frac{\sqrt{\pi }}{2}r]}^2=r^2$$$$\Rightarrow r=0.203or0.2501$$$$\tan \frac{\varphi }{2}=\frac{a}{2(\frac{\sqrt{3}}{2}-\frac{\sqrt{3}a}{2})}=\frac{\sqrt{\pi }r}{\sqrt{3}(1-\sqrt{\pi }r)}$$$$\Rightarrow \varphi =2{\tan }^{-1}\frac{\sqrt{\pi }r}{\sqrt{3}(1-\sqrt{\pi }r)}=35.955°or49.379°$$

Commented by MJS last updated on 25/May/20

$$\mathrm{if}\mathrm{side}\mathrm{length}\mathrm{of}\mathrm{triangle}=1\Rightarrow r⩽\frac{1}{4}\left(\mathrm{roughly}\right)$$$$\mathrm{or}\mathrm{else}\mathrm{the}\mathrm{solution}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{fit}\mathrm{the}\mathrm{picture}$$$$\mathrm{if}r\approx \frac{1}{4}\mathrm{the}\mathrm{square}\mathrm{is}\mathrm{not}\mathrm{inside}\mathrm{the}\mathrm{triangle}$$

Commented by mr W last updated on 25/May/20

$$thankyousir!$$$$inoticedtheproblemwiththesecond$$$$case,butcouldnotdeterminehow$$$$itcomes.nowithinkitrepresents$$$$followingsituation:$$

Commented by mr W last updated on 25/May/20

Commented by MJS last updated on 25/May/20

$$\mathrm{you}\mathrm{are}\mathrm{right}.\mathrm{I}\mathrm{only}\mathrm{assumed}\mathrm{the}\mathrm{square}$$$$\mathrm{would}\mathrm{be}\mathrm{outside}\mathrm{without}\mathrm{calculating}\mathrm{nor}$$$$\mathrm{drawing}\mathrm{a}\mathrm{picture}$$

Commented by mr W last updated on 25/May/20

$$anyway,thesecondcaseisnota$$$$validsolution.thanksforreviewing!$$

Commented by I want to learn more last updated on 25/May/20

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}$$

Commented by I want to learn more last updated on 25/May/20

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}$$

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