2009 AMC 12B Problem 20

Attempt Problem 20 of the 2009 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B solutions, or check the answer key.

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20.

A convex polyhedron QQ has vertices V1,V_1, V2,V_2, …,\ldots, Vn,V_n, and 100100 edges. The polyhedron is cut by planes P1,P_1, P2,P_2, …,\ldots, PnP_n in such a way that plane PkP_k cuts only those edges that meet at vertex Vk.V_k. In addition, no two planes intersect inside or on Q.Q. The cuts produce nn pyramids and a new polyhedron R.R. How many edges does RR have?
200200
2n2n
300300
400400
4n4n
Answer: C
Concepts:polyhedrondouble counting
Small hint:
Each original edge is cut once near each of its two endpoints, creating new vertices
Big hint:
Count the new edges (one per edge-endpoint) plus the surviving middle portion of each original edge
Solution:
Each of the 100100 edges is cut once near each endpoint, so RR has 2⋅100=2002 \cdot 100 = 200 vertices.
The cut at vertex VkV_k creates a small polygon whose number of edges equals the degree of VkV_k; summed over all vertices this is 200,200, the total number of edge-endpoints. The middle portion of each original edge also survives, adding 100100 edges. So RR has 200+100=300200 + 100 = 300 edges.
Thus, the correct answer is C.
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