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Question Number 78511 by ajfour last updated on 18/Jan/20

Commented by ajfour last updated on 18/Jan/20

$$Thecylinderhashalfthevolume$$$$ofthecuboid.Ifitscircularfaces$$$$touchtheadjoiningwalls,find$$$$coordinatesofpointsA,B,C$$$$intermsofa,b,c.$$

Commented by ajfour last updated on 19/Jan/20

Answered by mr W last updated on 19/Jan/20

Commented by mr W last updated on 20/Jan/20

$$itistofindtheorientationandsize$$$$ofthecylinder.$$$$P(\alpha ,0,0)$$$$Q(0,\beta ,0)$$$$R(0,0,\gamma )$$$$P^{\prime }(a-\alpha ,0,0)$$$$Q^{\prime }(0,b-\beta ,0)$$$$R^{\prime }(0,0,c-\gamma )$$$$p=QR=\sqrt{{\beta }^2+{\gamma }^2}$$$$q=RP=\sqrt{{\gamma }^2+{\alpha }^2}$$$$r=PQ=\sqrt{{\alpha }^2+{\beta }^2}$$$$\rho =radiusofincircle=radiusofcylinder$$$$\Rightarrow \rho =\frac{1}{2}\sqrt{\frac{(-\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2})(\sqrt{{\alpha }^2+{\beta }^2}-\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2})(\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}-\sqrt{{\gamma }^2+{\alpha }^2})}{\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2}}}$$$$M=centerofincircle=axisofcylinder$$$$\mathit{M}=\frac{p}{p+q+r}\mathit{P}+\frac{q}{p+q+r}\mathit{Q}+\frac{r}{p+q+r}\mathit{R}$$$$x_M=\frac{p}{p+q+r}{x}_P=\frac{\alpha \sqrt{{\beta }^2+{\gamma }^2}}{\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2}}$$$$y_M=\frac{q}{p+q+r}{y}_Q=\frac{\beta \sqrt{{\gamma }^2+{\alpha }^2}}{\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2}}$$$$z_M=\frac{r}{p+q+r}{z}_R=\frac{\gamma \sqrt{{\alpha }^2+{\beta }^2}}{\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2}}$$$$similarly$$$$x_{M^{\prime }}=\frac{p}{p+q+r}{x}_{P^{\prime }}=\frac{(a-\alpha )\sqrt{{\beta }^2+{\gamma }^2}}{\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2}}$$$$y_{M^{\prime }}=\frac{q}{p+q+r}{y}_{Q^{\prime }}=\frac{(b-\beta )\sqrt{{\gamma }^2+{\alpha }^2}}{\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2}}$$$$z_{M^{\prime }}=\frac{r}{p+q+r}{z}_{R^{\prime }}=\frac{(c-\gamma )\sqrt{{\alpha }^2+{\beta }^2}}{\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2}}$$$$\mathrm{\Delta }{x}_{M^{\prime }-M}=\frac{(a-2\alpha )\sqrt{{\beta }^2+{\gamma }^2}}{\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2}}$$$$\mathrm{\Delta }{y}_{M^{\prime }-M}=\frac{(b-2\beta )\sqrt{{\gamma }^2+{\alpha }^2}}{\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2}}$$$$\mathrm{\Delta }{z}_{M^{\prime }-M}=\frac{(c-2\gamma )\sqrt{{\alpha }^2+{\beta }^2}}{\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2}}$$$${MM}^{\prime }=L=\frac{\sqrt{{(a-2\alpha )}^2({\beta }^2+{\gamma }^2)+{(b-2\beta )}^2({\gamma }^2+{\alpha }^2)+{(c-2\gamma )}^2({\alpha }^2+{\beta }^2)}}{\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2}}$$$$L=lengthofcylinder$$$$eqn.ofplanePQR:$$$$\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1$$$${MM}^{\prime }\mathrm{\perp }planePQR:$$$$\alpha (a-2\alpha )\sqrt{{\beta }^2+{\gamma }^2}=\beta (b-2\beta )\sqrt{{\gamma }^2+{\alpha }^2}=\gamma (c-2\gamma )\sqrt{{\alpha }^2+{\beta }^2}$$$$\alpha (a-2\alpha )\sqrt{{\beta }^2+{\gamma }^2}=\gamma (c-2\gamma )\sqrt{{\alpha }^2+{\beta }^2}...\left(i\right)$$$$\beta (b-2\beta )\sqrt{{\gamma }^2+{\alpha }^2}=\gamma (c-2\gamma )\sqrt{{\alpha }^2+{\beta }^2}...\left(ii\right)$$$$volumeofcylinderV=\pi {\rho }^{2}L=\frac{abc}{2}$$$$\frac{(-\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2})(\sqrt{{\alpha }^2+{\beta }^2}-\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2})(\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}-\sqrt{{\gamma }^2+{\alpha }^2})\sqrt{{(a-2\alpha )}^2({\beta }^2+{\gamma }^2)+{(b-2\beta )}^2({\gamma }^2+{\alpha }^2)+{(c-2\gamma )}^2({\alpha }^2+{\beta }^2)}}{{(\sqrt{{\alpha }^2+{\beta }^2}+\sqrt{{\beta }^2+{\gamma }^2}+\sqrt{{\gamma }^2+{\alpha }^2})}^2}=\frac{2abc}{\pi }...\left(iii\right)$$$$from\left(i\right)to\left(iii\right):\alpha ,\beta ,\gamma$$

Commented by mr W last updated on 19/Jan/20

$$thankyoutoosir!$$

Commented by ajfour last updated on 19/Jan/20

$$LooksfabulousSir,ishalltry$$$$tofollow....alreadyihave$$$$learnedsomethingsfromyour$$$$diagrampost,andiamtrying$$$$tosolveeven.Thankyou$$$$immenselySir.$$

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