Mock Geometry AIME 2011 Problems/Problem 9
Problem
is a right pyramid with square base
edge length 6, and
The probability that a randomly selected point inside the pyramid is at least
units away from each face can be expressed in the form
where
are relatively prime positive integers. Find
Solution
[I believe solution is wrong - IP' is not sqrt(6)/3 as stated in the second paragraph - The_Turtle]
Let be the set of all points that are at least
units away from each face.
is tetrahedron, and it is similar to
. This can be proved by showing that
is bounded by 5 planes, each of which is parallel to a corresponding plane of
. Let the vertices of
be
such that
is the closest vertex to
and so forth. Consider cross section
. This cross section contains two concentric, similar triangles,
and
. Furthermore, these triangles are equilateral;
is the diagonal of a square with a side length of
and so
.
From symmetry it follows that . Let
intersect
at
and
at
. Then
. We can calculate
, it is the height of an equilateral triangle with a side length of
. Then
. Similarly, let
be the sidelenth of
. Then
is the height of this triangle and so is equal to
. Let
be the foot of the perpendicular from
to
.
bisects
by symmetry, and so
and
. Also
as it just the distance from
to
.
Plugging these values in yields . Solving yields
. Therefore the ratio
to
is
. The ratio of their volumes is then the ratio of their sides cubed, or
. The ratio of the volumes of
to
is equivalent to the probability a point will be in
. Hence
and
.