Denote the radius of the sphere to be $r$, and let $O$ be the center of the sphere. Also, let $A$ and $B$ be two vertices of the cube such that $AB$ is a side of the cube. Then, we let $C$ be the foot of the perpendicular drawn from $O$ to $AB$. Notice that right triangle $\triangle AOC$ is a 45-45-90 triangle because $AOC$ is half of a face-diagonal of the cube.
[asy] size(6cm); defaultpen(linewidth(0.4)); import graph; import solids; import geometry; import three; real unit = 1; triple I = (1,0,0), J = (0,1,0), K = (0,0,1); draw(Cube(1)); pair A = draw_circle_baricentre(I+J,K,unit).intersectionpoints(default_circleradius(I+J,K,unit)); pair B1 = draw_circle_baricentre(I,K,unit).intersectionpoints(default_circleradius(I,K,unit)); pair B2 = draw_circle_baricentre(J,K,unit).intersectionpoints(default_circleradius(J,K,unit)); draw(arc(A,K,B1)^^arc(A,K,B2)^^A--B1^^A--B2); draw(I--(I+J)--J^^I--(I+K)--J^^(I+J)--(1,1,1)--(I+K)); draw(Circle(I+J,unit)); D(D((1,1,-1)/3,2/3*unit,fontsize(9pt))); D(D(I+J,fontsize(9pt))); D(D(A,fontsize(9pt))); D(D((I+J)/2,fontsize(9pt))); D(D(B2,fontsize(9pt))); D(anglemark((I+J)/2,(1,1,-1)/3,A,0.6,0.6*size(I+J))); D(anglemark((1,1,-1)/3,(I+J)/2,I+J,0.6,0.6*size(I+J))); draw(I+J--B2^^A--(I+J),dashed); draw_circle_baricentre(I+J,K,unit,linetype("2 3")); [/asy]
It follows that $AO = AC = \frac{a\sqrt{2}}{2}$. Now note that $\triangle AOB$ is a 30-30-120 triangle because $AOB$ is half of a space-diagonal of the cube. Thus, $\angle AOB = 120^\circ$ and $AO = r$. Finally, we have the following equation:
$$\frac{a\sqrt{2}}{2} = r\implies a^3 - \frac{4}{3}\pi \left(\frac{a\sqrt{2}}{2}\right)^3 = \boxed{\frac{a^3\left(3\pi\sqrt{2}-4\right)}{6}}.$$
Each of the twelve edges of a cube of edge 'a' is tangent to a sphere. Find the volume of that portion of the cube which lies outside the sphere.
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1 answer