Subject: geometry

Name: Alan

Who is asking: Student

Level: Secondary

Question:

Three spheres of diameter 2 are placed on a level surface so that each sphere
touches the other two. A fourth sphere, also of diameter 2, is placed on top of
the other three so that it touches all of the other spheres. The distance from
the level surface to the highest point of the top sphere is . . .

I have tried attacking this problem by using a tetrahedron with its corners located in the center of each of the four spheres. I do not know where to go from here. Any help would be greatly appreciated. Thanks.

Hi Alan

You have the correct approach by looking at the regular tetrahedron formed by the centers of the four spheres. Now look at the triangle that forms the base of the tetrahedron and bisect the three angles. | |

Since the triangle QBC is a 30-60-90 triangle with BC of length 2, the length of CQ is sqrt(3). Also PRC is a 30-60-90 triangle so the length of PC is twice the length of PR. Since PR and QP have the same length, the length of PC is 2/3(length of QC)=2/3 sqrt(3). | |

Looking again at the tetrahedron and the triangle DPC, CD has length 2 and PC has length 2/3 sqrt(3) so using the Theorem of Pythagoras the height of the tetrahedron is sqrt(8/3). Hence the distance from the level surface to the highest point of the top sphere is 2 + sqrt(8/3) |

Chris and Harley

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