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
![side length of equilateral=1 a^2 =πr^2 ⇒a=(√π)r AG=(r/(√3)) OA=(1/2)−(((√3)r)/3) OD=((√3)/2)−(((√3)a)/2)−a=((√3)/2)−(((√3)/2)+1)(√π)r [((√3)/2)−(((√3)/2)+1)(√π)r−r]^2 +[(1/2)−(((√3)r)/3)−((√π)/2)r]^2 =r^2 ⇒r=0.203 or 0.2501 tan (φ/2)=(a/(2(((√3)/2)−(((√3)a)/2) )))=(((√π)r)/((√3)(1−(√π)r))) ⇒φ=2 tan^(−1) (((√π)r)/((√3)(1−(√π)r)))=35.955° or 49.379°](Q95536.png)
$${side}\:{length}\:{of}\:{equilateral}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} =\pi{r}^{\mathrm{2}} \\ $$$$\Rightarrow{a}=\sqrt{\pi}{r} \\ $$$${AG}=\frac{{r}}{\sqrt{\mathrm{3}}} \\ $$$${OA}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}{r}}{\mathrm{3}} \\ $$$${OD}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}−{a}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{1}\right)\sqrt{\pi}{r} \\ $$$$\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{1}\right)\sqrt{\pi}{r}−{r}\right]^{\mathrm{2}} +\left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}{r}}{\mathrm{3}}−\frac{\sqrt{\pi}}{\mathrm{2}}{r}\right]^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\mathrm{0}.\mathrm{203}\:{or}\:\mathrm{0}.\mathrm{2501} \\ $$$$\mathrm{tan}\:\frac{\phi}{\mathrm{2}}=\frac{{a}}{\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}\:\right)}=\frac{\sqrt{\pi}{r}}{\sqrt{\mathrm{3}}\left(\mathrm{1}−\sqrt{\pi}{r}\right)} \\ $$$$\Rightarrow\phi=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\pi}{r}}{\sqrt{\mathrm{3}}\left(\mathrm{1}−\sqrt{\pi}{r}\right)}=\mathrm{35}.\mathrm{955}°\:{or}\:\mathrm{49}.\mathrm{379}° \\ $$
Commented by MJS last updated on 25/May/20
$$\mathrm{if}\:\mathrm{side}\:\mathrm{length}\:\mathrm{of}\:\mathrm{triangle}\:=\mathrm{1}\:\Rightarrow\:{r}\leqslant\frac{\mathrm{1}}{\mathrm{4}}\:\left(\mathrm{roughly}\right) \\ $$$$\mathrm{or}\:\mathrm{else}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{fit}\:\mathrm{the}\:\mathrm{picture} \\ $$$$\mathrm{if}\:{r}\approx\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{the}\:\mathrm{square}\:\mathrm{is}\:\mathrm{not}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{triangle} \\ $$
Commented by mr W last updated on 25/May/20
$${thank}\:{you}\:{sir}! \\ $$$${i}\:{noticed}\:{the}\:{problem}\:{with}\:{the}\:{second} \\ $$$${case},\:{but}\:{could}\:{not}\:{determine}\:{how} \\ $$$${it}\:{comes}.\:{now}\:{i}\:{think}\:{it}\:{represents} \\ $$$${following}\:{situation}: \\ $$
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
$${anyway},\:{the}\:{second}\:{case}\:{is}\:{not}\:{a} \\ $$$${valid}\:{solution}.\:{thanks}\:{for}\:{reviewing}! \\ $$
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
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate} \\ $$