Place 4 dots in a plane

[bonanova]

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.]

[Guest]

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.

[bonanova]

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.

[Guest]

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?

[bonanova]

Also, by defining a plane, I assume we are limited to 2 dimensions?

Correct

[bonanova]

For the 2nd 3-3 solution here is a hint:

What if there were 5 points to be placed?

Here are the solutions:

[Guest]

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...

[bonanova]

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.

[Guest]

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?

[bonanova]

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.

[Guest]

For the 2nd 3-3 solution here is a hint:

What if there were 5 points to be placed?

Here are the solutions:

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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.

[bonanova]

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

[Guest]

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?

[Guest]

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.