1993 AIME Problems/Problem 10

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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 $100P+10T+V$?


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 $p=12_{}^{}$ smaller equilateral pentagons and $t=20_{}^{}$ equilateral triangles yielding a 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, $E_{}^{}=60$. Finally, 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
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All AIME Problems and Solutions
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