Question-189891

Question Number 189891 by mr W last updated on 23/Mar/23

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

a box of size 1×1×1 is separated by a plate as shown. what is the radius of the largest ball which can be placed inside the box?

$${a}\:{box}\:{of}\:{size}\:\mathrm{1}×\mathrm{1}×\mathrm{1}\:{is}\:{separated} \\ $$$${by}\:{a}\:{plate}\:{as}\:{shown}.\:{what}\:{is}\:{the}\:{radius} \\ $$$${of}\:{the}\:{largest}\:{ball}\:{which}\:{can}\:{be}\:{placed} \\ $$$${inside}\:{the}\:{box}? \\ $$

Answered by manxsol last updated on 24/Mar/23

Commented by manxsol last updated on 24/Mar/23

2(√5)r+2(√2)r=heron(2(√(5,))2(√5),4(√2)) r(2)((√5)+(√2))r=(√((2)((√5)+(√2))(2(√2))^2 (2(√5)−2(√(2))))) 2((√5)+(√2))r=4(√2)(√3) ((2(√6)((√5)−(√(2))))/3) 1.342083/4 0.335521 r=0.33

$$\mathrm{2}\sqrt{\mathrm{5}}{r}+\mathrm{2}\sqrt{\mathrm{2}}{r}={heron}\left(\mathrm{2}\sqrt{\mathrm{5},}\mathrm{2}\sqrt{\mathrm{5}},\mathrm{4}\sqrt{\mathrm{2}}\right) \\ $$$${r}\left(\mathrm{2}\right)\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}\right){r}=\sqrt{\left(\mathrm{2}\right)\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}\right)\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}\sqrt{\left.\mathrm{2}\right)}\right.} \\ $$$$\mathrm{2}\left(\sqrt{\mathrm{5}}+\sqrt{\mathrm{2}}\right){r}=\mathrm{4}\sqrt{\mathrm{2}}\sqrt{\mathrm{3}} \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{6}}\left(\sqrt{\mathrm{5}}−\sqrt{\left.\mathrm{2}\right)}\right.}{\mathrm{3}} \\ $$$$\mathrm{1}.\mathrm{342083}/\mathrm{4} \\ $$$$\mathrm{0}.\mathrm{335521} \\ $$$${r}=\mathrm{0}.\mathrm{33} \\ $$

Answered by mr W last updated on 24/Mar/23

Commented by mr W last updated on 25/Mar/23

$${yes},\:{thanks}\:{sir}! \\ $$

Commented by mr W last updated on 25/Mar/23

OP=OQ=OR=(3/2) eqn. of plate: ((2x)/3)+((2y)/3)+((2z)/3)=1 center of ball (1−r, 1−r, 1−r) r=((∣3×((2(1−r))/3)−1∣)/( (√(3×((2/3))^2 )))) ⇒r=((3−(√3))/4)≈0.317

$${OP}={OQ}={OR}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${eqn}.\:{of}\:{plate}: \\ $$$$\frac{\mathrm{2}{x}}{\mathrm{3}}+\frac{\mathrm{2}{y}}{\mathrm{3}}+\frac{\mathrm{2}{z}}{\mathrm{3}}=\mathrm{1} \\ $$$${center}\:{of}\:{ball}\:\left(\mathrm{1}−{r},\:\mathrm{1}−{r},\:\mathrm{1}−{r}\right) \\ $$$${r}=\frac{\mid\mathrm{3}×\frac{\mathrm{2}\left(\mathrm{1}−{r}\right)}{\mathrm{3}}−\mathrm{1}\mid}{\:\sqrt{\mathrm{3}×\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow{r}=\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{4}}\approx\mathrm{0}.\mathrm{317} \\ $$

Commented by ajfour last updated on 25/Mar/23

if −r=((3×((2(1−r))/3)−1)/( (√(3×((2/3))^2 )))) −((2r)/( (√3)))=1−2r r=(1/(2−(2/( (√3)))))=((√3)/(2(√3)−2))=((3−(√3))/4) ✓

$${if}\:\:\:\:\:\:−{r}=\frac{\mathrm{3}×\frac{\mathrm{2}\left(\mathrm{1}−{r}\right)}{\mathrm{3}}−\mathrm{1}}{\:\sqrt{\mathrm{3}×\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} }} \\ $$$$−\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{3}}}=\mathrm{1}−\mathrm{2}{r} \\ $$$${r}=\frac{\mathrm{1}}{\mathrm{2}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2}}=\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{4}}\:\:\checkmark \\ $$

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