Difference between revisions of "1993 AIME Problems/Problem 10"
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<math>\implies V(5-P)=90</math>. | <math>\implies V(5-P)=90</math>. | ||
Since <math>V|480</math> from the first vertex-related observation and <math>P>0\implies5-P<5</math>, and it quickly follows that <math>V=30\implies E=60\implies P=2\implies T=2\implies100P+10T+V=\boxed{250}</math>. | Since <math>V|480</math> from the first vertex-related observation and <math>P>0\implies5-P<5</math>, and it quickly follows that <math>V=30\implies E=60\implies P=2\implies T=2\implies100P+10T+V=\boxed{250}</math>. | ||
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+ | === Solution 3 === | ||
+ | Notice that at each vertex, we must have the sum of the angles be less than <math>360</math> degrees or we will not be able to fold the polyhedron. Therefore, we have <math>60T + 108P < 360.</math> Now, let there be <math>t</math> triangles and <math>p</math> pentagons total such that <math>t+p = 32.</math> From the given, we know that <math>E = V + 30.</math> Lastly, we see that <math>E = \frac{3t+5p}{2}</math> and <math>V = \frac{3t}{T}=\frac{5p}{P}.</math> | ||
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+ | Now, we do casework on what <math>P</math> is. | ||
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+ | 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, 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>. | ||
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+ | ~Williamgolly | ||
== See also == | == See also == |
Revision as of 20:32, 15 February 2021
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
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.