The Aha!" problems 4. Six integral points

[bonanova]

Determine the coordinates of six points on the plane with the following properties.

1. No three points are collinear.
2. Every pairwise distance is an integer.

You may use sketchpad, compass, ruler, straight edge, whatever you think may be useful.

The answer will be six pairs of coordinates: (x_i , y_i).

[k-man]

[Perhaps check it again]

There is an example with a set of seven coordinates on the plane such that no three points are
collinear and every pairwise distance is an integer. So, if you want a set of six coordinates, just
omit one of the seven. The example is at the top of the second page:

www.wm.uni-bayreuth.de/fileadmin/Sascha/Publikationen2/7-cluster_arxiv.pdf

[bonanova]

Some of the considerations discussed in that paper bear on the problem at hand.

But in the spirit of the Aha! theme, let's rule out exhaustive computer constructions and go with compass and straightedge.

[bonanova]

begin with a 3 4 5 triangle and derive three more points with integer separations.

Then show that all the nodal separations are rational.

[plasmid]

I don't think you could start with the points of a 3 4 5 triangle and add three more points to make this work.

If you call the points of that triangle A, B, and C such that AB=3, AC=4, BC=5, then add a point D, the lengths AB, AD, and BD must form a Pythagorean triple. There are no Pythagorean triples with 3 as a side length aside from 3 4 5 (this can be inferred from Euclid's formula).

In general, if you are given two points and told that they must form the length=3 side of a 345 triangle, then there are four other points which could be the other vertex of that triangle. In this case, point C is already included in the set of points, so there are only three other possible points. All of them must be included since the solution to this problem must have six points, but one can easily see that if you include all of those points then there will be sets of three colinear points and there will be edges with non-integer length, violating the conditions of a solution.

So there cannot be a 345 triangle in the solution. But perhaps there could be a triangle with edge lengths that are a multiple of 345.

[plainglazed]

think bn was saying find other rational points which can then be extrapolated to make all six integers. also, don't think they necessarily have to be Pythagorean triples. Non Pythagorean, Heronian triangles might also be needed. That said Ive been working on this since first posted and am still stumped

[DeGe]

Place two 345 triangles A1B1C1 and A2B2C such that B1 and B2 are offset by 4 and 3 units on x and y axis respectively

Edited May 7, 2014 by DeGe

[bonanova]

Place two 345 triangles A1B1C1 and A2B2C such that B1 and B2 are offset by 4 and 3 units on x and y axis respectively

DeGe, sorry for the delay in answering.

I think your right triangle A₁B₁C₂ will have sides of 4, 6 and sqrt(52).

That is, A₁C₂ will be irrational. But did I get your construction right?

[bonanova]

Here is the solution I had in mind, and k-man found it.

Center

AC, the hypotenuse of 3 4 5 triangle ABC, on the origin: (Black points)
A: (-2.5, 0); C: (2.5, 0); B: (-.7, 2.4)

Drop a perpendicular from B to a point D (-.7, 0) on the x-axis.

Reflect on the x-axis. (Green point)

Reflect on the y-axis. (Red points)

Scale by 10x to get integer coordinates (+/-25, 0), (+/-7, +/-24).

Pairwise distances are AB = 30; BC = 40; AC = 50.

Red rectangle has sides and diagonals of 14, 48 and 50.

Since points A, B and C lie on a circle, they all do.