perplexus.info :: Shapes : Polyhedra construction

Polyhedra construction (Posted on 2006-07-17)

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?

Submitted by Tristan

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Solution: (Hide)

12 for the first question
1. 24
2. 8
3. 24
4. 60
5. 120
If you multiply the number of vertices by the number of degrees short from a full circle at each vertex (360 degrees minus the sum of the angles of the intersecting polygons), it seems you will always get 720 degrees. For example, the intersecting angles of five triangles sum to 300 degrees, which is 60 degrees short of a full circle. 60 degrees times 12 vertices equals 720 degrees.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re: A Reference	Dej Mar	2006-07-18 13:56:22
A Reference	Richard	2006-07-18 12:35:02
A solid start....(spoiler)	Dej Mar	2006-07-17 12:24:24
Solution	Bractals	2006-07-17 12:21:44

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