1993 AIME Problems/Problem 10

Revision as of 16:10, 26 March 2009 by Math154 (talk | contribs) (Solution 2)


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 $100P+10T+V$?


Solution 1

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 $100P+10T+V=\boxed{250}$.

Solution 2

As seen above, $E=V+30$. Every vertex $V$, there is a triangle for every $T$ and a pentagon for every $P$ by the given. However, there are three times every triangle will be counted and five times every pentagon will be counted because of their numbers of vertices. From this observation, $\frac{VT}3+\frac{VP}5=32\implies V(5T+3P)=480$. Also, at every vertex $V$, there are $T+P$ edges coming out from that vertex (one way to see this is to imagine the leftmost segment of each triangle and pentagon that is connected to the given vertex, and note that it includes every one of the edges exactly once), so $\frac{V(T+P)}2=E\implies V(T+P)=2E\implies V(5T+5P)=10E$, and subtracting the other equation involving the vertices from this gives $2VP=10E-480\implies VP=5E-240=5(V+30)-240=5V-90$ $\implies V(5-P)=90$. Since $V|480$ from the first vertex-related observation and $P>0\implies5-P<5$, and it quickly follows that $V=30\implies E=60\implies P=2\implies T=2\implies100P+10T+V=\boxed{250}$.

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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