Difference between revisions of "1993 AIME Problems/Problem 10"
m |
Gabiloncho (talk | contribs) (→Solution) |
||
| Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
| − | {{solution}} | + | The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with <math>12_{}^{}</math> equilateral pentagons) in which the <math>20_{}^{}</math> vertices have all been truncated to form <math>20_{}^{}</math> equilateral triangles with common vertices. The resulting solid has then <math>t=20_{}^{}</math> smaller equilateral pentagons and <math>p=12_{}^{}</math> equilateral triangles yielding o total of <math>t+p=F=32_{}^{}</math> faces. In each vertex <math>T=2_{}^{}</math> triangles and <math>P=2_{}^{}</math> pentagons are concurrent. Now, the number of edges <math>E_{}^{}</math> can be obtained if we count the number of sides that each triangle and pentagon contributes: <math>E_{}^{}=\frac{3t+5p}{2}</math>, (the factor <math>2_{}^{}</math> in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus, <math>E_{}^{}=60</math>. Hence, using Euler's formula we have <math>V_{}^{}=E-30=30</math>. |
| + | |||
| + | In summary, the solution to the problem is <math>100P_{}^{}+10+V=240</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=1993|num-b=9|num-a=11}} | {{AIME box|year=1993|num-b=9|num-a=11}} | ||
Revision as of 19:43, 23 April 2007
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
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 o 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, 
. Hence, using Euler's formula we have 
.
In summary, the solution to the problem 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 | ||
