Difference between revisions of "2012 AMC 12A Problems/Problem 22"
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==Solution== | ==Solution== | ||
− | We need | + | We need two different kinds of planes that only intersect <math>Q</math> at the mentioned segments (we call them ''traces'' in this solution). These will be all the possible <math>p_j</math>'s. |
First, there are two kinds of segments joining the midpoints of every pair of edges belonging to the same face of <math>Q</math>: ''long traces'' are those connecting the midpoint of opposite sides of the same face of <math>Q</math>, and ''short traces'' are those connecting the midpoint of adjacent sides of the same face of <math>Q</math>. | First, there are two kinds of segments joining the midpoints of every pair of edges belonging to the same face of <math>Q</math>: ''long traces'' are those connecting the midpoint of opposite sides of the same face of <math>Q</math>, and ''short traces'' are those connecting the midpoint of adjacent sides of the same face of <math>Q</math>. |
Revision as of 19:15, 11 September 2013
Problem
Distinct planes intersect the interior of a cube
. Let
be the union of the faces of
and let
. The intersection of
and
consists of the union of all segments joining the midpoints of every pair of edges belonging to the same face of
. What is the difference between the maximum and minimum possible values of
?
Solution
We need two different kinds of planes that only intersect at the mentioned segments (we call them traces in this solution). These will be all the possible
's.
First, there are two kinds of segments joining the midpoints of every pair of edges belonging to the same face of : long traces are those connecting the midpoint of opposite sides of the same face of
, and short traces are those connecting the midpoint of adjacent sides of the same face of
.
Suppose contains a short trace
of a face of
. Then it must also contain some trace
of an adjacent face of
, where
share a common endpoint with
. So, there are three possibilities for
, each of which determines a plane
containing both
and
.
Case 1: makes an acute angle with
. In this case,
is an equilateral triangle made by three short traces. There are
of them, corresponding to the
vertices.
Case 2: is a long trace.
is a rectangle. Each pair of parallel faces of
contributes
of these rectangles so there are
such rectangles.
Case 3: is the short trace other than the one described in case 1, i.e.
makes an obtuse angle with
. It is possible to prove that
is a regular hexagon (See note #1 for a proof) and there are
of them.
Case 4: contains no short traces. This can only make
be a square enclosed by long traces. There are
such squares.
In total, there are possible planes in
. So the maximum of
is
.
On the other hand, the most economic way to generate these long and short traces is to take all the planes in case 3 and case 4. Overall, they intersect at each trace exactly once (there is a quick way to prove this. See note #2 below.) and also covered all the traces. So the minimum of
is
. The answer to this problem is then
...
.
Note 1: Indeed, let where
, and
be the other endpoint of
that is not
. Draw a line through
parallel to
. This line passes through the center
of the cube and therefore we see that the reflection of
, denoted by
, respectively, lie on the same plane containing
. Thus
is the regular hexagon
. To count the number of these hexagons, just notice that each short trace uniquely determine a hexagon (by drawing the plane through this trace and the center), and that each face has
short traces. Therefore, there are
such hexagons.
Note 2: The quick way to prove the fact that none of the planes described in case 3 and case 4 share the same trace is as follows: each of these plane contains the center and therefore the intersection of each pair of them is a line through the center, which obviously does not contain any traces.
See Also
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.