A certain polyhedron is constructed such that each vertex is the intersection of five triangles. How many vertices are there?
Was that easy? Try these.
...such that each vertex is the intersection of...
1. three squares and a triangle
2. three triangles and a square
3. four triangles and a square
4. a triangle, square, pentagon, and square in that order
5. a decagon, hexagon, and square
Notice any patterns?
Let
v = number of polyhedron vertices
f = number of polyhedron faces
e = number of polyhedron edges
f_n = number of polyhedron faces
that are n-gons
m_n = number of n-gons at a
polyhedron vertex
Then the following apply to each case:
1) Euler's polyhedron theorem: v + f - e = 2
2) m_n*v = n*f_n for each n
3) f = sum of f_n
4) 2*e = sum of n*f_n
Case 0: five triangles -
3*f_3
v = -------
5
f = f_3
3*f_3
e = -------
2
===> v = 12
Case 1: three squares and a triangle -
3*f_3 4*f_4
v = ------- = -------
1 3
f = f_3 + f_4
3*f_3 + 4*f_4
e = ---------------
2
===> v = 24
Case 2: three triangles and a square -
3*f_3 4*f_4
v = ------- = -------
3 1
f = f_3 + f_4
3*f_3 + 4*f_4
e = ---------------
2
===> v = 8
Case 3: four triangles and a square -
3*f_3 4*f_4
v = ------- = -------
4 1
f = f_3 + f_4
3*f_3 + 4*f_4
e = ---------------
2
===> v = 24
Case 4: a triangle, square, pentagon, and
square in that order -
3*f_3 4*f_4 5*f_5
v = ------- = ------- = -------
1 2 1
f = f_3 + f_4 + f_5
3*f_3 + 4*f_4 + 5*f_5
e = -----------------------
2
===> v = 60
Case 5: a decagon, hexagon, and square -
4*f_4 6*f_6 10*f_10
v = ------- = ------- = ---------
1 1 1
f = f_4 + f_6 + f_10
4*f_4 + 6*f_6 + 10*f_10
e = -------------------------
2
===> v = 120
Edited on July 17, 2006, 12:23 pm
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Posted by Bractals
on 2006-07-17 12:21:44 |
Solution
