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Question-211145

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Question Number 211145 by mr W last updated on 29/Aug/24
Commented by mr W last updated on 30/Aug/24
thanks! but i think people know what i exactly mean.
$${thanks}!\:{but}\:{i}\:{think}\:{people}\:{know}\: \\ $$$${what}\:{i}\:{exactly}\:{mean}. \\ $$
Commented by mr W last updated on 29/Aug/24
a corner of a solid cube with edge length d is chopped off as shown. find the radius R of the largest solid sphere which can be cut from the remaining object. assume 0<a, b, c≤d.
$${a}\:{corner}\:{of}\:{a}\:{solid}\:{cube}\:{with}\:{edge}\: \\ $$$${length}\:\boldsymbol{{d}}\:{is}\:{chopped}\:{off}\:{as}\:{shown}. \\ $$$${find}\:{the}\:{radius}\:\boldsymbol{{R}}\:{of}\:{the}\:{largest}\: \\ $$$${solid}\:{sphere}\:{which}\:{can}\:{be}\:{cut}\:{from}\: \\ $$$${the}\:{remaining}\:{object}. \\ $$$${assume}\:\mathrm{0}<\boldsymbol{{a}},\:\boldsymbol{{b}},\:\boldsymbol{{c}}\leqslant\boldsymbol{{d}}. \\ $$
Commented by MathematicalUser2357 last updated on 30/Aug/24
0<a,b,c≤d is not an equation or logic you should say: “∧_(l∈{a,b,c}) (0<l∧l≤d)”
$$\mathrm{0}<{a},{b},{c}\leqslant{d}\:\mathrm{is}\:\mathrm{not}\:\mathrm{an}\:\mathrm{equation}\:\mathrm{or}\:\mathrm{logic} \\ $$$$\mathrm{you}\:\mathrm{should}\:\mathrm{say}:\:“\underset{\mathrm{l}\in\left\{{a},{b},{c}\right\}} {\wedge}\left(\mathrm{0}<\mathrm{l}\wedge\mathrm{l}\leqslant{d}\right)'' \\ $$
Answered by mr W last updated on 04/Sep/24
say a, b, c are the x, y, x axes. the equation of the cross section is (x/a)+(y/b)+(z/c)=1 if no corner is chopped off, then the largest sphere has center at ((d/2),(d/2),(d/2)) and radius R=R_0 =(d/2) if (((d/(2a))+(d/(2b))+(d/(2c))−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))≥R_0 =(d/2) (1/a)+(1/b)+(1/c)−(√((1/a^2 )+(1/b^2 )+(1/c^2 )))≥(2/d) R=R_0 =(d/2) otherwise case 1: center (R_1 , R_1 , d−R_1 ), radius R_1 (((R_1 /a)+(R_1 /b)+((d−R_1 )/c)−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))=R_1 ⇒(R_1 /d)=(((1/c)−(1/d))/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))−(1/a)−(1/b)+(1/c))) case 2: center (d−R_2 , d−R_2 , R_2 ), radius R_2 ((((d−R_2 )/a)+((d−R_2 )/b)+(R_2 /c)−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))=R_2 ⇒(R_2 /d)=(((1/a)+(1/b)−(1/d))/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))+(1/a)+(1/b)−(1/c))) case 3: center (d−R_3 , d−R_3 , d−R_3 ), radius R_3 ((((d−R_3 )/a)+((d−R_3 )/b)+((d−R_3 )/c)−1)/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))))=R_3 ⇒(R_3 /d)=(((1/a)+(1/b)+(1/c)−(1/d))/( (√((1/a^2 )+(1/b^2 )+(1/c^2 )))+(1/a)+(1/b)+(1/c))) we can show that R_1 <R_3 , R_2 <R_3 that means the radius of maximum sphere from case 1 to case 3 is R_3 .
$${say}\:{a},\:{b},\:{c}\:{are}\:{the}\:{x},\:{y},\:{x}\:{axes}. \\ $$$${the}\:{equation}\:{of}\:{the}\:{cross}\:{section}\:{is} \\ $$$$\frac{{x}}{{a}}+\frac{{y}}{{b}}+\frac{{z}}{{c}}=\mathrm{1} \\ $$$${if}\:{no}\:{corner}\:{is}\:{chopped}\:{off},\:{then} \\ $$$${the}\:{largest}\:{sphere}\:{has} \\ $$$${center}\:{at}\:\left(\frac{{d}}{\mathrm{2}},\frac{{d}}{\mathrm{2}},\frac{{d}}{\mathrm{2}}\right)\:{and} \\ $$$${radius}\:{R}={R}_{\mathrm{0}} =\frac{{d}}{\mathrm{2}} \\ $$$${if} \\ $$$$\frac{\frac{{d}}{\mathrm{2}{a}}+\frac{{d}}{\mathrm{2}{b}}+\frac{{d}}{\mathrm{2}{c}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}\geqslant{R}_{\mathrm{0}} =\frac{{d}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}−\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}\geqslant\frac{\mathrm{2}}{{d}} \\ $$$${R}={R}_{\mathrm{0}} =\frac{{d}}{\mathrm{2}} \\ $$$${otherwise} \\ $$$${case}\:\mathrm{1}:\:{center}\:\left({R}_{\mathrm{1}} ,\:{R}_{\mathrm{1}} ,\:{d}−{R}_{\mathrm{1}} \right),\:{radius}\:{R}_{\mathrm{1}} \\ $$$$\frac{\frac{{R}_{\mathrm{1}} }{{a}}+\frac{{R}_{\mathrm{1}} }{{b}}+\frac{{d}−{R}_{\mathrm{1}} }{{c}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}={R}_{\mathrm{1}} \\ $$$$\Rightarrow\frac{{R}_{\mathrm{1}} }{{d}}=\frac{\frac{\mathrm{1}}{{c}}−\frac{\mathrm{1}}{{d}}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}−\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}} \\ $$$${case}\:\mathrm{2}:\:{center}\:\left({d}−{R}_{\mathrm{2}} ,\:{d}−{R}_{\mathrm{2}} ,\:{R}_{\mathrm{2}} \right),\:{radius}\:{R}_{\mathrm{2}} \\ $$$$\frac{\frac{{d}−{R}_{\mathrm{2}} }{{a}}+\frac{{d}−{R}_{\mathrm{2}} }{{b}}+\frac{{R}_{\mathrm{2}} }{{c}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}={R}_{\mathrm{2}} \\ $$$$\Rightarrow\frac{{R}_{\mathrm{2}} }{{d}}=\frac{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{d}}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{c}}} \\ $$$${case}\:\mathrm{3}:\:{center}\:\left({d}−{R}_{\mathrm{3}} ,\:{d}−{R}_{\mathrm{3}} ,\:{d}−{R}_{\mathrm{3}} \right),\:{radius}\:{R}_{\mathrm{3}} \\ $$$$\frac{\frac{{d}−{R}_{\mathrm{3}} }{{a}}+\frac{{d}−{R}_{\mathrm{3}} }{{b}}+\frac{{d}−{R}_{\mathrm{3}} }{{c}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}}={R}_{\mathrm{3}} \\ $$$$\Rightarrow\frac{{R}_{\mathrm{3}} }{{d}}=\frac{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}−\frac{\mathrm{1}}{{d}}}{\:\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}+\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}} \\ $$$${we}\:{can}\:{show}\:{that}\:{R}_{\mathrm{1}} <{R}_{\mathrm{3}} ,\:{R}_{\mathrm{2}} <{R}_{\mathrm{3}} \\ $$$${that}\:{means}\:{the}\:{radius}\:{of}\:{maximum} \\ $$$${sphere}\:{from}\:{case}\:\mathrm{1}\:{to}\:{case}\:\mathrm{3}\:{is}\:{R}_{\mathrm{3}} . \\ $$

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