Red and Blue Points

Topics

Lattice points, reasoning

Overview

We assume that all the points in the plane have been colored either red or blue. It is then required, without knowing anything further about the distribution of the colors, that we find: (a) a line segment whose endpoints and midpoint are the same color, (b) three points of the same color that determine a right angle, (c) an equilateral triangle whose vertices are the same color, and (d) a parallelogram whose vertices are the same color. This takes a considerable amount of logical tracking down, following worst-case scenarios, one after another, until points are found that yield the desired result.

Learning objectives:

• To learn to rigorously examine all possible cases.
• To develop visual inventiveness combined with ingenuity.

The "Hook"

You are aware that there are many, many points on the coordinate plane.  Even on a line, points are "dense," which means that between any two points there's another one, and between those two, there's another one, and on and on, without an end.  This means that no point has an immediate "next door neighbor," that no point has any size to it at all. Yet, oddly, these infinitesimally small points somehow fill up the plane.  We tend not to think about this, and in fact we're not going to think of it much at the moment, either. 

This is because we are going to assume that every point in the coordinate plane is colored either red or blue.  We don't know whether there's any pattern to the distribution of the colors--all we know is that every point is either red or blue--which, as we've said, is impossible, since they're so small.  But never mind. Here is the task: 

The Investigation

Important questions to consider:

 No. Thus you can make some assumptions.  The first task turns out to be the first of four tasks; call the first one Task (a). Worksheet 1 shows points strategically arranged for Task (a). Following is a detailed logical argument for the first task; you might get students started as a class and then see where they can go with it. Link to the first solution sheet to better follow the arguments below for Tasks (a) and (b).

Task (a):  We are looking for a line segment whose endpoints and midpoint are all the same color.

1.  On a line, find a point that's red.  Is there such a point?  If not, the whole line is blue and any line segment we draw will have its endpoints and midpoint blue, and we're done.  So find a point that's red, and call it C (see the diagram). 

2.  Now find another red point.  Is there one?  If not, we're done, because everything else is blue and there are plenty of line segments like the one we need.  So assume there's a second red point, and call it E. 

3.  Draw D, the midpoint of CE.  If D is red, we're done, so assume it's blue. 

4.  Now find point A, such that C is the midpoint of segment AE.  Is A red?  Then we're done: segment AE is what we need, since A, C, and E are all red. 

5. Is A blue?  Then find point G, such that E is the midpoint of CG.  Is G red?  If so we're done: segment CG has red endpoints and a red midpoint (E). 

6.  Is G blue?  If so, we're also done, because segment AG has blue endpoints and a blue midpoint, namely D. Q.E.D.

This is cool.

Here are the three other tasks that will challenge you:

(b) Find three points of the same color that determine a right angle.

(c) Find an equilateral triangle whose vertices are the same color.

(d) Find a parallelogram whose vertices are the same color.

These will take some hard thinking, some real sleuthing. But they can be done. Here are the logical arguments for these tasks:

Task (b):  Now we're looking for three points of the same color that determine a right angle.  Since we worked hard to find that line segment in (a), we'll use it, though we'll rename the points.  Let's assume that line segment AC is red, with a red midpoint B.  Stack some other points underneath it, as in the second drawing on Worksheet 1. Refer to the first solution sheet as we make our argument.

1.  Look at point E.  If it's red, we're done (angle ABE), so assume it's blue. 

2.  Look at F.  If it's red, we're done (angle BCF), so assume it's blue. 

3.  Look at H. If it's red, we're done (angle ABH), but if it's blue we're done too: angle HEF. Q.E.D.

Cool, again.

Tasks (c) and (d) are set up on Worksheet 2. Link to the second solution sheet to better follow the arguments below.

Task (c):  For the equilateral triangle, let us spread out an array of points that can act as vertices of equilateral triangles, as shown in the top drawing.  Let us assume that the segment FH and its midpoint G are all red. 

1.  Look at J.  Is it red?  If so, we're done (triangle FGJ), so assume it's blue. 

2.  Look at K.  Is it red?  If so, we're done (triangle KGH), so assume it's blue. 

3.  Now look at M.  Aha!  If it's red, we're done (triangle FHM), and if it's blue, we're also done (triangle JKM). Q.E.D. 

Task (d):  And now for the parallelogram.  Look at the bottom set of points on Worksheet 2.  Let us assume that equilateral triangle EBC has red vertices. 

1.  Look at A.  Is it red?  If so, we're done (BACE), so assume A is blue.

2.  Look at D.  Is it red?  If so, we're done  (BCED), so assume D is blue. 

3.  Look at F.  If F is red, we're done (BCFE), so assume F is blue. 

4.  Now look up at L.  If L is blue, we're done (AFLD), so assume L is red.  And now look at K.  If K is red, we're done (BCLK, see it?), so assume K is blue. 

5.  And now look at M.  If M is blue, we're done (DFMK), so assume M is red.  But now we're done anyway, because BCML is a parallelogram. Q.E.D.

Relentless logic saves the day!

Teaching Tips

This is a great problem for students because it shows the way visual thinking and sheer logic (the core of geometric reasoning) can support each other. There is a nice "high" involved in tracking down the crucial points that, no matter which color they are, yield the needed results.

The way this problem is received will depend enormously on the nature of your students.  Those who love arguing their parents into tight corners may be delighted with this exploration.  There's a sort of perversity about the logical prowling that can be very satisfying, but finding the layout of points on which to operate can be very difficult, so use your judgment about whether to provide the worksheets. 

If you're doing this with the class as a whole, you may prefer to use the worksheets to make transparencies. Putting dots on the transparency using red and blue markers will speed things along. For Tasks (a) and (b), the points are called lattice points, if we assume that they have integer coordinates. (But note that coordinates are not needed.)  The discussion about density of points is worth having, and is interesting.

An easy extension of the problem, or at least a variant using the same drawings, is to assume that some particular point has coordinates (0, 0), then to specify the distance to a neighboring point and figure out the coordinates of all the remaining points in the drawing.  For instance, in the top drawing on Worksheet 2, let A be at (0, 0) and B at (2, 0).  Then F is at (1, sq rt 3), and so on.  An interesting pattern develops.

I suggest giving students time to cudgel their brains over each task.  Some students will not want to let go of it.

Citation

From the teaching files of Rudd Crawford.