Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of the larger sphere?
The answer is 3+ root 69/3.
I really can't think of an approach, yet...
Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is the distance from the plane to the top of the larger sphere?
The answer is 3+ root 69/3.
I really can't think of an approach, yet...
1. The centers $\displaystyle (C_1, C_2, C_3)$of the smaller spheres form a isosceles triangle with the side-length 2.
2. The center M of the larger sphere is vertically above the centroid F of the isosceles triangle. The larger sphere is tangent to all three smaller spheres.
3. $\displaystyle MFC_1$ is a right triangle with
hypotenuse $\displaystyle |\overline{MC_1}| = 3$
horizontal leg $\displaystyle |\overline{FC_1}| = \frac 23 \cdot \sqrt{2^2-1^2}=\frac 23 \cdot \sqrt{3}$
vertical leg $\displaystyle |\overline{MF}| =\sqrt{3^2-\left(\frac 23 \cdot \sqrt{3} \right)^2} = \sqrt{\frac{23}{3}}$
4. You have to add one small radius and one large radius to the distance $\displaystyle |\overline{MF}|$ and you'll get the given result.
5. For additional information have a look here: Close-packing of spheres - Wikipedia, the free encyclopedia