1993 AIME Problems/Problem 10


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 $V=30\implies E=60\implies P=2\implies T=2\implies100P+10T+V=\boxed{250}$.

Solution 3

Notice that at each vertex, we must have the sum of the angles be less than $360$ degrees or we will not be able to fold the polyhedron. Therefore, we have $60T + 108P < 360.$ Now, let there be $t$ triangles and $p$ pentagons total such that $t+p = 32.$ From the given, we know that $E = V + 30.$ Lastly, we see that $E = \frac{3t+5p}{2}$ and $V = \frac{3t}{T}=\frac{5p}{P}.$

Now, we do casework on what $P$ is.

Case 1: $P = 2$ Notice that we must have $t$ and $p$ integral. Trying $T = 1, 2$ yields a solution with $t=2.$ Trying other cases of $P$ and $T$ yields no solutions. Therefore, $T=2, P=2$ and after solving for $t, p$ we get $V=30.$ Finally, we have $100P+10T+V = \boxed{250}$.


Solution 4

We know that $V-E = -30 \implies V = E-30$ based off the problem condition. Furthermore, if we draw out a few pentagons as well as triangles on each of side of the pentagons, it's clear that each vertex has 4 edges connected to it, with two triangles and two pentagons for each vertex. However, each edge is used for two vertices and thus counted twice. Therefore, we have that $E = \frac{4V}{2} = 2V$. Plugging this in, we have that $V=30$ and so our answer is $200+20+30 = \boxed{250}$.

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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