1988 AIME Problems/Problem 10
Problem
A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?
Solution 1
The polyhedron described looks like this, a truncated cuboctahedron.
The number of segments joining the vertices of the polyhedron is . We must now subtract out those segments that lie along an edge or a face.
Since every vertex of the polyhedron lies on exactly one vertex of a square/hexagon/octagon, we have that .
Each vertex is formed by the intersection of 3 edges. Since every edge is counted twice, once at each of its endpoints, the number of edges is .
Each of the segments lying on a face of the polyhedron must be a diagonal of that face. Each square contributes diagonals, each hexagon , and each octagon . The number of diagonals is thus .
Subtracting, we get that the number of space diagonals is .
Solution 2
We first find the number of vertices on the polyhedron: There are 4 corners per square, 6 corners per hexagon, and 8 corners per octagon. Each vertex is where 3 corners coincide, so we count the corners and divide by 3. .
We know that all vertices look the same (from the problem statement), so we should find the number of line segments originating from a vertex, and multiply that by the number of vertices, and divide by 2 (because each space diagonal is counted twice because it has two endpoints).
Counting the vertices that are on the same face as an arbitrary vertex, we find that there are 13 vertices that aren't possible endpoints of a line originating from the vertex in the middle of the diagram. You can draw a diagram to count this better: Since 13 aren't possible endpoints, that means that there are 35 possible endpoints per vertex. The total number of segments joining vertices that aren't on the same face is
Solution 3
Since at each vertex one square, one hexagon, and one octagon meet, then there are a total of vertices. This means that for each segment we have choices of vertices for the first endpoint of the segment.
Since each vertex is the meeting point of a square, octagon, and hexagon, then there are other vertices of the square that are not the first one, and connecting the first point to any of these would result in a segment that lies on a face or edge.
Similarly, there are points on the adjacent hexagon and points on adjoining octagon that, when connected to the first point, would result in a diagonal or edge.
However, the square and hexagon share a vertex, as do the square and octagon, and the hexagon and octagon.
Subtracting these from the vertices we have left to choose from, and adding the that we counted twice, we get
We over-counted, however, as choosing vertex then is the same thing as choosing then , so we must divide .
Alternatively, we could have noted from the diagram:
Our first choice would be the vertex in the middle (there are of these), and our second choice would be any of the remaining 47 points minus the 12 that share a face without chosen vertex. Summing these we get And we divide by as before to get
Solution 4
In the same ways as above, we confined that there are 48 vertices. From each, notice that there are total possible ways to choice two vertices. However, we must remove the cases where the segments do not lie in the interior of the polyhedron. We get
Notice that we remove all the possible edges of the squares, hexagons, and octagons. However, we have undercounted! We must add back the number of edges because when we subtracted the three binomials from we removed each edge twice. This means that we need to add back the number of edges, 72.
Thus, we get .
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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