Difference between revisions of "1993 AIME Problems/Problem 10"
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Case 1: <math>P = 2</math> | Case 1: <math>P = 2</math> | ||
| − | Notice that we must have <math>t</math> and <math>p</math> integral. Trying <math>T = 1, 2</math> yields a solution with <math>t=2.</math> Trying other cases of <math>P</math> and <math>T</math> yields no solutions. Therefore, <math>T=2, | + | Notice that we must have <math>t</math> and <math>p</math> integral. Trying <math>T = 1, 2</math> yields a solution with <math>t=2.</math> Trying other cases of <math>P</math> and <math>T</math> yields no solutions. Therefore, <math>T=2, P=2</math> and after solving for <math>t, p</math> we get <math>V=30.</math> Finally, we have <math>100P+10T+V = \boxed{250}</math>. |
~Williamgolly | ~Williamgolly | ||
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| + | ==Solution 4== | ||
| + | We know that <math>V-E = -30 \implies V = E-30</math> 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 <math>E = \frac{4V}{2} = 2V</math>. Plugging this in, we have that <math>V=30</math> and so our answer is <math>200+20+30 = \boxed{250}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 19:50, 17 April 2022
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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
