2009 AMC 12B Problems/Problem 20
Contents
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 .
Solution 4
Each edge connects two points. The plane cuts that edge so it splits into at each end (like two legs) for a total of new edges.
But because each new edge is shared by an adjacent original edge cut similarly, the additional edges are overcounted .
Since there are edges to start with, new edges result. So there are edges in the figure.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.