Difference between revisions of "1988 AIME Problems/Problem 10"

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== Solution ==
 
== Solution ==
By the [[Euler characteristic]], we have that <math>V - E + F = 2</math>. The number of faces, <math>F</math>, is <math>12 + 8 + 6 = 26</math>. Since every point lies on exactly one vertex of a square/hexagon/octagon, we have that <math>V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48</math>. Substituting gives us <math>E = 72</math>.
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The number of segments joining the vertices of the polyhedron is <math>{48\choose2} = 1128</math>. We must now subtract out those segments that lie along an edge or a face.
  
The number of segments joining the vertices of the polyhedron is <math>{48\choose2} = 1128</math>. Of these segments, <math>72</math> are edges. The number of diagonals of a square is <math>\frac{n(n-3)}{2} = 2</math>, of a hexagon is <math>9</math>, and of an octagon <math>20</math>. Hence the number of face diagonals is <math>2 \cdot 12 + 9 \cdot 8 + 20 \cdot 6 = 216</math>.
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Since every vertex of the polyhedron lies on exactly one vertex of a square/hexagon/octagon, we have that <math>V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48</math>.
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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 <math>E</math> is <math>\frac{3}{2}V = 72</math>.
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Each of the segments lying on a face of the polyhedron must be a diagonal of that face. Each square contributes <math>\frac{n(n-3)}{2} = 2</math> diagonals, each hexagon <math>9</math>, and each octagon <math>20</math>. The number of diagonals is thus <math>2 \cdot 12 + 9 \cdot 8 + 20 \cdot 6 = 216</math>.
  
 
Subtracting, we get that the number of space diagonals is <math>1128 - 72 - 216 = 840</math>.
 
Subtracting, we get that the number of space diagonals is <math>1128 - 72 - 216 = 840</math>.

Revision as of 16:29, 12 December 2010

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

The number of segments joining the vertices of the polyhedron is ${48\choose2} = 1128$. 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 $V = 12 \cdot 4 = 8 \cdot 6 = 6 \cdot 8 = 48$.

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 $E$ is $\frac{3}{2}V = 72$.

Each of the segments lying on a face of the polyhedron must be a diagonal of that face. Each square contributes $\frac{n(n-3)}{2} = 2$ diagonals, each hexagon $9$, and each octagon $20$. The number of diagonals is thus $2 \cdot 12 + 9 \cdot 8 + 20 \cdot 6 = 216$.

Subtracting, we get that the number of space diagonals is $1128 - 72 - 216 = 840$.

See also

1988 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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