Slightly OT...four points on a plane. « Next Oldest | Next Newest »

 ▼ Hal Bitton in Boise Senior Member Posts: 291 Threads: 43 Joined: Jun 2007 12-03-2007, 03:54 AM Good evening everyone. Challenge problem: Arrange four points on a plane such that the distance between any 2 points is one of 2 possible lengths. There are supposedly at least 5 arrangements that satisfy this, but alas, I can only find the following 4: 1. One point on each corner of a square. 2. One point on each vertice of an equilateral triangle, and one point in the center of the triangle. 3. One point on each corner of a rhombus which has one diagonal congruent to it's sides. (can also be thought of as 2 equilateral triangles sharing a common side). 4. An equilateral triangle, with a fourth point centered on an arc drawn between two point of the triangle, the arc center (radius point) being the remaining point of the triangle. Any other solutions completely elude me. I'll wait (with baited breath) to see what other solutions the brilliant minds that frequent this forum can come up with. Thanks, and best regards, Hal ▼ Meenzer Member Posts: 150 Threads: 6 Joined: Sep 2007 12-03-2007, 05:15 AM Aren't 1 and 4 special cases of 3? Edited: 3 Dec 2007, 6:21 a.m. Arnaud Amiel Senior Member Posts: 362 Threads: 30 Joined: Jul 2005 12-03-2007, 06:20 AM All 4 points on top of each other? ▼ BruceH Senior Member Posts: 275 Threads: 38 Joined: Jul 2007 12-03-2007, 06:14 PM 4 points on top of each other fails the "one of 2 possible lengths test": since they are all zero distance away from each other there is only one length involved. Two dots on one point and 2 dots on another would satisfy the problem. Bram Member Posts: 182 Threads: 17 Joined: Oct 2005 12-03-2007, 06:51 AM I was thinking of a trapezium, almost immediately. Take a symmetrical trapezium in which the diagonals are as long as the longer parallel side. Imagine the longer side on top then you get a “bucket”. In case of a “low”’/”wide” bucket, the (parallel) base is longer than the (left and right) sides. In case of a “narrow” bucket, the base will be shorter than the sides. So in between you can have a bucket in which the base is exactly the length of the sides, in which case there are only two lengths between four points. I won’t be surprised when the golden ratio appears somewhere in the figure. ▼ Meenzer Member Posts: 150 Threads: 6 Joined: Sep 2007 12-03-2007, 07:39 AM Quote: I was thinking of a trapezium, almost immediately. Take a symmetrical trapezium in which the diagonals are as long as the longer parallel side. Imagine the longer side on top then you get a “bucket”. In case of a “low”’/”wide” bucket, the (parallel) base is longer than the (left and right) sides. In case of a “narrow” bucket, the base will be shorter than the sides. So in between you can have a bucket in which the base is exactly the length of the sides, in which case there are only two lengths between four points. I won’t be surprised when the golden ratio appears somewhere in the figure. When I construct this - not being sure if I get it right - I allways get a third length: (1) base=sides, (2) other parallel, (3) the distance between points A and C diagonally. Edited: 3 Dec 2007, 8:14 a.m. ▼ Bram Member Posts: 182 Threads: 17 Joined: Oct 2005 12-03-2007, 08:22 AM I have a picture to make clear, but unfortunately I cannot post it right now. I will try to describe: When I name de points A, B, C and D (counter clockwise, starting top left) I make sure that AD=AC and AD=BD A is the top of triangle ACD, D is the top of DAB. Both angles the same and hence both bases are of equal length. By varying these top angles (between 0 and 60 degrees), I can adjust the width of the bucket. The length of the base is depending on the chosen angle and will always but in one case be different than the sides AB and CD. Rough sketches show that the angle is somewhere between 30 and 45 degrees. I’ll have to try and compute the exact value. ▼ Dave Shaffer (Arizona) Posting Freak Posts: 776 Threads: 25 Joined: Jun 2007 12-03-2007, 03:58 PM I think that if you make the interior angle between the long top and either of the (shorter) sides to be 72 degrees, you will find that the sides and the short base are all the same length. If alpha is this top interior angle, then the relationship between side and bottom lengths (x) and the top length (t) is cos(alpha) = (x/2)/t Dave Shaffer (Arizona) Posting Freak Posts: 776 Threads: 25 Joined: Jun 2007 12-03-2007, 10:14 AM Quote:(with baited breath) Unless you've been eating fishing worms, you are waiting with "bated" breath! ▼ Hal Bitton in Boise Senior Member Posts: 291 Threads: 43 Joined: Jun 2007 12-03-2007, 03:48 PM Thanks Dave... Got me on that one. Best regards, Hal Hal Bitton in Boise Senior Member Posts: 291 Threads: 43 Joined: Jun 2007 12-03-2007, 05:20 PM Thanks for the responses everyone. The isosceles trapizoid (symmetrical trapizium) most of you described was indeed a solution. I was approaching it upside down when I initially tried to use it...I was trying to make the longer base, both sides, and the diagonals all the same length, with only the top being shorter. This of course caused the top to disappear completely, and my figure to become an equalateral triangle. It's amazing how hindsite can be so clear. Best regards, Hal Werner Member Posts: 163 Threads: 7 Joined: Jul 2007 12-04-2007, 03:36 PM ```There's a sixth (apart from your 4 and the trapezium): Take a unilateral triangle ABC with sides of length S, AB is the base and C the top. Place the fourth point D at a distance S from the top C, so that the line CD is perpendicular to AB. This is actually like your point #4, but the opposite point on the circle. Cheers, Werner ``` ▼ Hal Bitton in Boise Senior Member Posts: 291 Threads: 43 Joined: Jun 2007 12-04-2007, 05:01 PM Of course...if I would have just taken my circle "full circle". Thanks, and best regards, Hal

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