2009 AMC 12B Problems/Problem 20

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 $\mathrm{(C)}$ .

Solution 4

Each edge connects two points. The plane cuts that edge so it splits into $2$ at each end (like two legs) for a total of $4$ new edges.

$[asy] pair A,B,C,D,E,F; A=(0,50); B=(0,-50); C=(86.602540378443864676372317075294,0); D=(300,0); E=(370.7,70.7); F=(370.7,-70.7); draw(A--C--B); draw(C--D); draw(E--D--F); [/asy]$

But because each new edge is shared by an adjacent original edge cut similarly, the additional edges are overcounted $\times 2$ .

$[asy] pair A,B,C,D,E,F,G,H,I,J,K; A=(0,50); B=(0,-50); C=(86.602540378443864676372317075294,0); D=(300,0); E=(370.7,70.7); F=(370.7,-70.7); G=(-107.5,236.2); H=(-107.5,-236.2); I=(370.7,285); J=(370.7,-285); K=(441.4,0); draw(A--C--B); draw(C--D); draw(E--D--F); draw(G--A, red); draw(H--B, red); draw(I--E, red); draw(J--F, red); draw(A--B, dashed); draw(E--K, dashed); draw(F--K, dashed); [/asy]$

Since there are $100$ edges to start with, $400/2=200$ new edges result. So there are $100+200=\boxed{\textbf{(C) }300}$ edges in the figure.

Solution 5

The question specifies the slices create as many pyramids as there are vertices, implying each vertex owns 4 edge ends. There are twice as many edge-ends as there are edges, and 2 * 100 = 200.

$\frac{200}{4}$ = 50, so there are 50 vertices.

The base of a pyramid has 4 edges, so each sliced vertex would add four edges to $R$ .

100 + 4 * 50 = \boxed{\textbf{(C) } 300}$

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 12 Problems and Solutions

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