1980 AHSME Problems/Problem 26
Problem
Four balls of radius are mutually tangent, three resting on the floor and the fourth resting on the others.
A tetrahedron, each of whose edges have length
, is circumscribed around the balls. Then
equals
Solution
Let ,
,
, and
be the centers of the four spheres of radius
. These points must form a regular tetrahedron of side length
. Let
be the center of
. Then
. By the Pythagorean Theorem, the height
of tetrahedron
must be
. Let
be the center of tetrahedron
. Then
,
,
, and
are all congruent tetrahedra, each with
of the volume of
. Since tetrahedra
and
share base
, the height
of tetrahedron
, and therefore the inradius of tetrahedron
, must be
.
Let be the tetrahedron circumscribed around the spheres, with
,
,
, and
intersecting at
. Planes
and
must be parallel and a distance
from each other. So the distance from
to plane
, and therefore the inradius of tetrahedron
, must be
. Therefore, the ratio of the inradius of tetrahedron
to the inradius of tetrahedron
is
. The ratio of the side length of tetrahedron
to the side length of tetrahedron
must also be
, so
.
-j314andrews
See also
1980 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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