1993 AIME Problems/Problem 10
Contents
[hide]Problem
Euler's formula states that for a convex polyhedron with vertices,
edges, and
faces,
. A particular convex polyhedron has 32 faces, each of which is either a triangle or a pentagon. At each of its
vertices,
triangular faces and
pentagonal faces meet. What is the value of
?
Solution
Solution 1
The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with equilateral pentagons) in which the
vertices have all been truncated to form
equilateral triangles with common vertices. The resulting solid has then
smaller equilateral pentagons and
equilateral triangles yielding a total of
faces. In each vertex,
triangles and
pentagons are concurrent. Now, the number of edges
can be obtained if we count the number of sides that each triangle and pentagon contributes:
, (the factor
in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus,
. Finally, using Euler's formula we have
.
In summary, the solution to the problem is .
Solution 2
As seen above, . Every vertex
, there is a triangle for every
and a pentagon for every
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,
. Also, at every vertex
, there are
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
, and subtracting the other equation involving the vertices from this gives
.
Since
from the first vertex-related observation and
, and it quickly follows that
.
Solution 3
Notice that at each vertex, we must have the sum of the angles be less than degrees or we will not be able to fold the polyhedron. Therefore, we have
Now, let there be
triangles and
pentagons total such that
From the given, we know that
Lastly, we see that
and
Now, we do casework on what is.
Case 1:
Notice that we must have
and
integral. Trying
yields a solution with
Trying other cases of
and
yields no solutions. Therefore,
and after solving for
we get
Finally, we have
.
~Williamgolly
Solution 4
We know that 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
. Plugging this in, we have that
and so our answer is
.
Note
The solid described in this problem is an icosidodecahedron.
See also
1993 AIME (Problems • Answer Key • Resources) | ||
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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