The bottom edge must stick in the blue and green sections meaning that if we were to divide the edge in n and see where these points map to, we would find that it would be blue or green and similarly the other edges would match the limitations in q4. If we look at the right corner, we see that the bottom edge maps to green or blue and the right edge maps to green or red, so the bottom corner must be in green. Similarly the other corners match the requirements of q4.
This lets us find a smaller trichromatic triangle. We can repeat this process an arbitrary number of times. Consider the range of possible x and y co-ordinates of elements in these triangles. Each time this will reduce by a particular factor, so we can see that the range of each co-ordinate will approach 0. I’ll leave it to the reader to show that this means that these ranges converge to a point. I’ll also leave it to the reader to show that each trichromatic sub-triangle must contain the center (you may want to look up winding numbers).
I’ve realised that you’ve gotta be careful with this method because when you find a trichromatic subtriangle of the original, it won’t necessarily have the property of only having points of two colours along the edges, and so may not in fact contain a point that maps to the centre.
This isn’t a problem if we just increase the number n by which we divide the whole triangle instead of recursively dividing subtriangles. Unfortunately now we’re not reducing the range of co-ords where this fixed point must be, only finding a triad of arbitrarily close points that map to a triangle surrounding the centre. You can, for example, take the centre point of the first of these triangles (with some method of numbering to make the function definite) for each value of n=1,2,3.. as a sequence in R2. This must have a convergent sequence which should converge to a point that maps to the centre but I can’t prove that last stage.
Here’s the rough idea for 5 (not a full-proof)
The bottom edge must stick in the blue and green sections meaning that if we were to divide the edge in n and see where these points map to, we would find that it would be blue or green and similarly the other edges would match the limitations in q4. If we look at the right corner, we see that the bottom edge maps to green or blue and the right edge maps to green or red, so the bottom corner must be in green. Similarly the other corners match the requirements of q4.
This lets us find a smaller trichromatic triangle. We can repeat this process an arbitrary number of times. Consider the range of possible x and y co-ordinates of elements in these triangles. Each time this will reduce by a particular factor, so we can see that the range of each co-ordinate will approach 0. I’ll leave it to the reader to show that this means that these ranges converge to a point. I’ll also leave it to the reader to show that each trichromatic sub-triangle must contain the center (you may want to look up winding numbers).
I’ve realised that you’ve gotta be careful with this method because when you find a trichromatic subtriangle of the original, it won’t necessarily have the property of only having points of two colours along the edges, and so may not in fact contain a point that maps to the centre.
This isn’t a problem if we just increase the number n by which we divide the whole triangle instead of recursively dividing subtriangles. Unfortunately now we’re not reducing the range of co-ords where this fixed point must be, only finding a triad of arbitrarily close points that map to a triangle surrounding the centre. You can, for example, take the centre point of the first of these triangles (with some method of numbering to make the function definite) for each value of n=1,2,3.. as a sequence in R2. This must have a convergent sequence which should converge to a point that maps to the centre but I can’t prove that last stage.
Yeah, you’re right. That breaks the proof. I don’t know how to deal with it yet.