bonanova 85 Posted August 23, 2007 Report Share Posted August 23, 2007 If you connect all pairs of 4 dots in a plane [geometrical, not air] you create 6 line segments. This puzzle asks how many ways 4 points can be arranged such that the lengths of the 6 connecting lines share no more than 2 values. The points must be distinct. None of the lengths can be zero. Example: the corners of a square. The 4 sides and the 2 diagonals share common lengths: a so-called 4-2 solution. A moment's reflection and some equilateral triangles reveal there are at least 2 other 4-2 solutions, a 5-1 solution and a 3-3 solution. The real stumper is to find one more 3-3 solution - two in all. You can describe these in words, or attach a graphic. [This question was posed to Jr-High school students in a national competition.] Quote Link to post Share on other sites

0 Guest Posted August 24, 2007 Report Share Posted August 24, 2007 You are a touch ambiguous about the "found" shapes, so I don't know if my 3/3 shape is the one you were thinking of, but: If you make one equalateral triangle with a dot in the middle, you have a 3/3. I assume that is the easy one, and you want us to find another. Quote Link to post Share on other sites

0 bonanova 85 Posted August 24, 2007 Author Report Share Posted August 24, 2007 I assume that is the easy one, and you want us to find another. Yup. I didn't want to describe the "found" ones yet - it's fun to find them, as well. Quote Link to post Share on other sites

0 Guest Posted August 24, 2007 Report Share Posted August 24, 2007 So (maybe hints here so don't read if you don't want hints) It's fairly easy to understand that if you have a 3/3 solution, you have 3 "long" and 3 "short" lines, where the long and short must be equal lengths among themselves. If you connect any of the three points with the "long" into a triangle, you MUST have the 4th point inside the trangle and only 1 point will work for all three shorts to connect to it. However, if you conect the three Longs into an open shape (a non-triangle) then I think by necessity you have created a 4/2 or 5/1 configuration. Has this been solved before, or is it an open question? Also, by defining a plane, I assume we are limited to 2 dimensions? Quote Link to post Share on other sites

0 bonanova 85 Posted August 25, 2007 Author Report Share Posted August 25, 2007 Also, by defining a plane, I assume we are limited to 2 dimensions? Correct Quote Link to post Share on other sites

0 bonanova 85 Posted August 26, 2007 Author Report Share Posted August 26, 2007 For the 2nd 3-3 solution here is a hint: What if there were 5 points to be placed? Here are the solutions: Quote Link to post Share on other sites

0 Guest Posted August 26, 2007 Report Share Posted August 26, 2007 Ah, you were too quick with the solutions. I just figured it out and got on to post. A rhombus where top and sides are "short" and bottom and cross pieces are "long". Though I didn't ever see it as a pentagon... Quote Link to post Share on other sites

0 bonanova 85 Posted August 27, 2007 Author Report Share Posted August 27, 2007 Right ... the pentagon is the stumper. The clue, What if there were five points? makes it come to mind, cuz the points of a pentagon create sides and diagonals of only one type, just like as square does. Then you just remove a point to get a 4-dot solution. It's seldom you get one solution to be fundamentally different from all the others. Quote Link to post Share on other sites

0 Guest Posted April 3, 2008 Report Share Posted April 3, 2008 Use the corners of a trapezoid. Assume the height = 2x the longest base. You can also assume the shorter base is 1/3 the other base. None of the 4 "sides" are at all the same length. Neither diagonal is the same length. Since the diagonals do not bisect each other, those lengths won't equal the other lengths either. Did that work like I thought it would? Quote Link to post Share on other sites

0 bonanova 85 Posted April 3, 2008 Author Report Share Posted April 3, 2008 Use the corners of a trapezoid. Assume the height = 2x the longest base. You can also assume the shorter base is 1/3 the other base. None of the 4 "sides" are at all the same length. Neither diagonal is the same length. Since the diagonals do not bisect each other, those lengths won't equal the other lengths either.Did that work like I thought it would? You get too many distinct lengths with your method. The idea was to have all the lengths equal to one of only two different values. There is a trapezoid that works, however. See it here. Quote Link to post Share on other sites

0 Guest Posted April 3, 2008 Report Share Posted April 3, 2008 For the 2nd 3-3 solution here is a hint: What if there were 5 points to be placed? Here are the solutions: does spoiler not work I thought it was interesting that you thought the trapazoidal 3-3 was the hardest... for whatever reason I got stuck on the 4-2 equilateral with interior 4th point. I'm interested in what ppl found to be the hardest configuration to find. Quote Link to post Share on other sites

0 bonanova 85 Posted April 3, 2008 Author Report Share Posted April 3, 2008 I guess it was hardest because I started with a square and looked for ways to morph it. All the others had equilateral triangles or points on a main line of symmetry. The trapezoid - 4/5ths of a pentagon - did not. Thanks for the post, and interesting question for others to answer. - bn Quote Link to post Share on other sites

0 Guest Posted April 13, 2008 Report Share Posted April 13, 2008 For the 2nd 3-3 solution here is a hint: What if there were 5 points to be placed? Here are the solutions: The pentagon-based solution is nice, but unlike the other figures it doesn't seem constructable (using a compass and straight-edge). How would one accurately construct the dot arrangement without computations to many decimal places? Quote Link to post Share on other sites

0 Guest Posted September 5, 2009 Report Share Posted September 5, 2009 The pentagon was the easiest for me, i didnt get the whole long/short line things, i was just trying to get as many diff shapes to create 6 lines. Trapezoid, parrallelagram (rhombus), and square. Those 3 are the easiest to figure out. I didn't get the triangle ones simple triangle with a dot. Quote Link to post Share on other sites

## Question

## bonanova 85

If you connect all pairs of 4 dots in a plane [geometrical, not air] you create 6 line segments.

This puzzle asks how many ways 4 points can be arranged such that the lengths

of the 6 connecting lines share no more than 2 values.

The points must be distinct. None of the lengths can be zero.

Example: the corners of a square.

The

4 sidesand the2 diagonalsshare common lengths: a so-called4-2solution.A moment's reflection and some equilateral triangles reveal there are

at least

2other4-2solutions, a5-1solution and a3-3solution.The real stumper is to find

one more3-3 solution - two in all.You can describe these in words, or attach a graphic.

[This question was posed to Jr-High school students in a national competition.]

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