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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.


(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)

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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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