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Simple distances (Posted on 2006-05-01) Difficulty: 3 of 5
For this you will need four pennies. Arrange the pennies so that the centers of all four are equidistant from each other.

Example:

(1) (2) (3) (4)

This will not work because penny 1 and penny 4 are farther apart than penny 1 and penny 2.

See The Solution Submitted by Stephen    
Rating: 1.8571 (7 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Pennies of Giza | Comment 3 of 8 |
(In reply to Pennies of Giza by Jyqm)

I believe both Jyqm and TZIDON grasped the concept of what in my mind is a way to solve this. However, Jyqm's pyramid would have 5 vertices for 4 pennies, while an equilateral tetrahedron would do the trick (a vertex for each penny, far from each other by the exact same distance).

I'm sure it was just a matter of speech, but a pyramid wouldn't fit due to it's rectangular or square base (and, therefore, at least two pennies would be separated by this polygon's diagonal).


  Posted by Phil_Osopher on 2006-05-02 11:42:10
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