All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
Only Isosceles (Posted on 2005-02-13) Difficulty: 3 of 5
Arrange 8 points in 3-space so that for each of the 56 triples of points that they determine, at least two of the three distances between points in the triple are equal.

See The Solution Submitted by David Shin    
Rating: 4.5000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution | Comment 4 of 6 |

If five points are arranged in a regular pentagon on a given plane, any three of its vertices will form an isosceles triangle.  If the center of the pentagon is added, any three points that include that center point will also form an isosceles triangle, as that point is equidistant from all the vertices.

Then take a line perpendicular to the plane of the pentagon, through the center, and place two points on it, the same distance away from the center of the pentagon as that center is from any vertex of the pentagon.  The center is still equidistant from all the points.  Any set of three points involving only the pentagon's vertices is still forming an isosceles triangle.  If either of the two latest points (the non-co-planar ones) is involved without the center, it involves either two vertices of the pentagon, all of which are the same distance from the non-co-planar point, or it involves the other non-co-planar point, but that's the same distance from any given point on the pentagon as the first non-co-planar point.


  Posted by Charlie on 2005-02-14 04:39:21
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information