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

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

$$aboxofsize1\times 1\times 1isseparated$$$$byaplateasshown.whatistheradius$$$$ofthelargestballwhichcanbeplaced$$$$insidethebox?$$

Answered by manxsol last updated on 24/Mar/23

Commented by manxsol last updated on 24/Mar/23

$$2\sqrt{5}r+2\sqrt{2}r=heron(2\sqrt{5,}2\sqrt{5},4\sqrt{2})$$$$r\left(2\right)(\sqrt{5}+\sqrt{2})r=\sqrt{\left(2\right)(\sqrt{5}+\sqrt{2}){\left(2\sqrt{2}\right)}^2(2\sqrt{5}-2\sqrt{2)}}$$$$2(\sqrt{5}+\sqrt{2})r=4\sqrt{2}\sqrt{3}$$$$\frac{2\sqrt{6}(\sqrt{5}-\sqrt{2)}}{3}$$$$1.342083/4$$$$0.335521$$$$r=0.33$$

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

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

$$yes,thankssir!$$

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

$$OP=OQ=OR=\frac{3}{2}$$$$eqn.ofplate:$$$$\frac{2x}{3}+\frac{2y}{3}+\frac{2z}{3}=1$$$$centerofball(1-r,1-r,1-r)$$$$r=\frac{\mid 3\times \frac{2(1-r)}{3}-1\mid }{\sqrt{3\times {\left(\frac{2}{3}\right)}^2}}$$$$\Rightarrow r=\frac{3-\sqrt{3}}{4}\approx 0.317$$

Commented by ajfour last updated on 25/Mar/23

$$if-r=\frac{3\times \frac{2(1-r)}{3}-1}{\sqrt{3\times {\left(\frac{2}{3}\right)}^2}}$$$$-\frac{2r}{\sqrt{3}}=1-2r$$$$r=\frac{1}{2-\frac{2}{\sqrt{3}}}=\frac{\sqrt{3}}{2\sqrt{3}-2}=\frac{3-\sqrt{3}}{4}✓$$

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