Difference between revisions of "1993 AIME Problems/Problem 10"
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In summary, the solution to the problem is <math>100P+10T+V=\boxed{250}</math>. | In summary, the solution to the problem is <math>100P+10T+V=\boxed{250}</math>. | ||
=== Solution 2 === | === Solution 2 === | ||
| − | As seen above, <math>E=V+30</math>. Every vertex <math>V</math>, there is a triangle for every <math>T</math> and a pentagon for every <math>P</math> 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, <math>\frac{VT}3+\frac{VP}5=32\implies V(5T+3P)=480</math>. Also, at every vertex <math>V</math>, there are <math>T+P</math> 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 <math>\frac{V(T+P)}2=E\implies V(T+P)=2E\implies V(5T+5P)=10E</math>, and subtracting the other equation involving the vertices from this gives <math>2VP=10E-480\implies VP=5E-240=5(V+30)-240=5V-90\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>. | + | As seen above, <math>E=V+30</math>. Every vertex <math>V</math>, there is a triangle for every <math>T</math> and a pentagon for every <math>P</math> 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, <math>\frac{VT}3+\frac{VP}5=32\implies V(5T+3P)=480</math>. Also, at every vertex <math>V</math>, there are <math>T+P</math> 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 <math>\frac{V(T+P)}2=E\implies V(T+P)=2E\implies V(5T+5P)=10E</math>, and subtracting the other equation involving the vertices from this gives |
| + | <math>2VP=10E-480\implies VP=5E-240=5(V+30)-240=5V-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>. | ||
== See also == | == See also == | ||
Revision as of 15:10, 26 March 2009
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 
.
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 | ||
