Important questions to consider:
No. Thus you can make some assumptions.
Does it matter whether the points turn out to be red or blue?
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).
What would be a good way to start?
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.
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!