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Question Number 188845 by ajfour last updated on 07/Mar/23

Answered by a.lgnaoui last updated on 07/Mar/23

$$△OAB\left(recrangleenO\right)$$$$AB=a+\frac{a\sqrt{2}}{2}\Rightarrow AH=\frac{a}{2}+\frac{a\sqrt{2}}{4}$$$$OA=OB=R$$$$\sin \frac{\pi }{4}=\frac{AH}{R}\Rightarrow R=\frac{2AH}{\sqrt{2}}$$$$\mathit{R}=1+\sqrt{2}$$

Commented by a.lgnaoui last updated on 07/Mar/23

Commented by mr W last updated on 08/Mar/23

$$clearlywrong!$$$$aballwithinthecubecannotbelarger$$$$thanthecubeitself!$$

Answered by mr W last updated on 08/Mar/23

Commented by mr W last updated on 08/Mar/23

$$A(r,r,r)$$$$P(1,0,0)$$$$\overrightarrow{PQ}=(-1,1,1)$$$$\overrightarrow{PA}=(r-1,r,r)$$$$(r-1,r,r)\times (-1,1,1)=(0,2r-1,2r-1)$$$$r=\frac{\sqrt{2{(2r-1)}^2}}{\sqrt{3}}=distancefromAtoPQ$$$$5r^2-8r+2=0$$$$\Rightarrow r=\frac{4-\sqrt{6}}{5}\approx 0.3101<0.5\left(withincube\right)$$

Commented by ajfour last updated on 08/Mar/23

$$(r-1,r,r)\times (-1,1,1)=(0,2r-1,2r-1)$$$$idintunderstandabove..but$$$$PA.\hat{\lambda }=\sqrt{{PA}^2-r^2}$$$${(1-r+r+r)}^2=3\{{(r-1)}^2+2r^2-r^2\}$$$${(1+r)}^2=3r^2+3{(1-r)}^2$$$$\Rightarrow 5r^2-8r+2=0$$$$r=\frac{8\pm \sqrt{64-40}}{10}=\frac{4\pm \sqrt{6}}{5}$$$$herer=\frac{4-\sqrt{6}}{5}$$

Commented by mr W last updated on 08/Mar/23

$$\overrightarrow{PA}\times \overrightarrow{PQ}=(r-1,r,r)\times (-1,1,1)=(0,2r-1,2r-1)$$

Commented by mr W last updated on 08/Mar/23

$$thanksforverifyingsir!$$$${it}^{\prime }sanicequestion!$$

Answered by mr W last updated on 08/Mar/23

Commented by mr W last updated on 08/Mar/23

Commented by mr W last updated on 08/Mar/23

$$forthiscasewegetaslo$$$$r=\frac{4-\sqrt{6}}{5}$$

Commented by ajfour last updated on 08/Mar/23

$$thankyousir,wellexamined$$$$andsolved.$$

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