Difference between revisions of "2009 AMC 12B Problems/Problem 20"

Revision as of 00:31, 11 April 2009

Problem

A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?

$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$

Solution

Solution 1

Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Three edges of $R$ meet at each vertex, so $R$ has $\frac 12 \cdot 3 \cdot 200 = \boxed {300}$ 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 $200$ . A middle portion of each original edge is also present in $R$ , so $R$ has $100 + 200 = \boxed {300}$ edges.

Solution 3

Euler's Polyhedron Formula applied to $Q$ gives $n - 100 + F = 2$ , where F is the number of faces of $Q$ . Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Each cut by a plane $P_k$ creates an additional face on $R$ , so Euler's Polyhedron Formula applied to $R$ gives $200 - E + (F+n) = 2$ , where $E$ is the number of edges of $R$ . Subtracting the first equation from the second gives $300 - E = 0$ , whence $E = \boxed {300}$ . The answer is $\mathrm{(C)}$ .

@@ Line 21: / Line 21: @@
 == See also ==
 {{AMC12 box|year=2009|ab=B|num-b=19|num-a=21}}
+[[Category:Introductory Combinatorics Problems]]

2009 AMC 12B (Problems • Answer Key • Resources)
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