How can i show and prove rigorously that for all integers, i can paint all these segments red?

Question

Three counters, A, B, C are placed at the corners of an equilateral triangle of side n. The triangle is divided into triangles of side length 1. Initially all lines of the figure are painted blue. The counters move along the lines, painting their paths red , according to the following rules:

1. First A moves, then B, then C, then A, and so on in succession.
2. On each turn, each counter moves the full length of a side of one of the short triangles.
3. No counter may retrace a segment already painted red, though it can stop on a red vertex, even if another counter is already there.

Show that for all integers n >0  it is possible to paint all of the segments red in this fashion.

Answer 1

Hint #1: This problem has a symmetry to it. If you rotate the triangle the movements of the counters should look the same. They all follow a similar path.

Hint #2: When you see, "prove this is true for all integers n > 0" that's a hint you need induction.

To do induction properly you need a base case, an inductive hypothesis, and an inductive step.

Here's how I would approach solving it:

For the base case start with a side length n = 2. You can work out the paths for this case by hand (they each move in a path that looks like 3 sides of a trapezoid and end up in a different corner).

Now you need to show that if you can prove it for n = 2, that it must be true for n = 3.

Draw out the triangle for n = 3 and solve it by hand. This is a little harder. Once you do, you should see a pattern. Use this to create your recursive relation.

Hint #3: When n = 3, the drawing will have 18 one-unit segments that need to be painted. So each counter will paint 6 of the segments.

Hint #4: For any size n, look at the small triangles with side = 1 that are in the corners of the big triangle. If the counter follows one of the edges out of the corner, how do can a counter color the other edge leading into the corner? Every counter must end up in one of these corners at the end. So the counter must start in a corner, somehow paints some of the edges in the middle, and heads back to a corner.

If you need more help with this problem or other problems, please feel free to message me!

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