Difference between revisions of "2009 AMC 12B Problems/Problem 20"
VelaDabant (talk | contribs) (New page: == Problem == A convex polyhedron <math>Q</math> has vertices <math>V_1,V_2,\ldots,V_n</math>, and <math>100</math> edges. The polyhedron is cut by planes <math>P_1,P_2,\ldots,P_n</math> i...) |
|||
Line 21: | Line 21: | ||
== See also == | == See also == | ||
{{AMC12 box|year=2009|ab=B|num-b=19|num-a=21}} | {{AMC12 box|year=2009|ab=B|num-b=19|num-a=21}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] |
Revision as of 01:31, 11 April 2009
Problem
A convex polyhedron has vertices
, and
edges. The polyhedron is cut by planes
in such a way that plane
cuts only those edges that meet at vertex
. In addition, no two planes intersect inside or on
. The cuts produce
pyramids and a new polyhedron
. How many edges does
have?
Solution
Solution 1
Each edge of is cut by two planes, so
has
vertices. Three edges of
meet at each vertex, so
has
edges.
Solution 2
At each vertex, as many new edges are created by this process as there are original edges meeting at that vertex. Thus the total number of new edges is the total number of endpoints of the original edges, which is . A middle portion of each original edge is also present in
, so
has
edges.
Solution 3
Euler's Polyhedron Formula applied to gives
, where F is the number of faces of
. Each edge of
is cut by two planes, so
has
vertices. Each cut by a plane
creates an additional face on
, so Euler's Polyhedron Formula applied to
gives
, where
is the number of edges of
. Subtracting the first equation from the second gives
, whence
. The answer is
.
See also
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |