Difference between revisions of "1993 AIME Problems/Problem 10"

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== Solution ==
 
== Solution ==
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The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with <math>12_{}^{}</math> equilateral pentagons) in which the <math>20_{}^{}</math> vertices have all been truncated to form <math>20_{}^{}</math> equilateral triangles with common vertices. The resulting solid has then <math>t=20_{}^{}</math> smaller equilateral pentagons and <math>p=12_{}^{}</math> equilateral triangles yielding o total of <math>t+p=F=32_{}^{}</math> faces. In each vertex <math>T=2_{}^{}</math> triangles and <math>P=2_{}^{}</math> pentagons are concurrent. Now, the number of edges <math>E_{}^{}</math> can be obtained if we count the number of sides that each triangle and pentagon contributes: <math>E_{}^{}=\frac{3t+5p}{2}</math>, (the factor <math>2_{}^{}</math> in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus, <math>E_{}^{}=60</math>. Hence, using Euler's formula we have <math>V_{}^{}=E-30=30</math>.
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In summary, the solution to the problem is <math>100P_{}^{}+10+V=240</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1993|num-b=9|num-a=11}}
 
{{AIME box|year=1993|num-b=9|num-a=11}}

Revision as of 19:43, 23 April 2007

Problem

Euler's formula states that for a convex polyhedron with $V\,$ vertices, $E\,$ edges, and $F\,$ faces, $V-E+F=2\,$. A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon. At each of its $V\,$ vertices, $T\,$ triangular faces and $P^{}_{}$ pentagonal faces meet. What is the value of $100P+10+V\,$?

Solution

The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with $12_{}^{}$ equilateral pentagons) in which the $20_{}^{}$ vertices have all been truncated to form $20_{}^{}$ equilateral triangles with common vertices. The resulting solid has then $t=20_{}^{}$ smaller equilateral pentagons and $p=12_{}^{}$ equilateral triangles yielding o total of $t+p=F=32_{}^{}$ faces. In each vertex $T=2_{}^{}$ triangles and $P=2_{}^{}$ pentagons are concurrent. Now, the number of edges $E_{}^{}$ can be obtained if we count the number of sides that each triangle and pentagon contributes: $E_{}^{}=\frac{3t+5p}{2}$, (the factor $2_{}^{}$ in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus, $E_{}^{}=60$. Hence, using Euler's formula we have $V_{}^{}=E-30=30$.

In summary, the solution to the problem is $100P_{}^{}+10+V=240$.

See also

1993 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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