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One Sunday morning, you awake to find yourself completely alone on an infinite, flat plane. You don't remember much about the night before, other than that you may have pissed off a wizard. Next to you, you find a palette with countably infinite colors, and a note, commanding you thus:

You must paint every point on this plane, such that I will never be able to find a triangle with vertices of the same color and rational area.

If you can manage this task, the wizard will let you go free - fail, and you're trapped forever. You don't doubt the wizard's abilities, so no cheap tricks here. Considering the problem, you get to work - and an uncountably infinite amount of time later, the wizard stands beside you, admiring your handiwork.

Does the wizard set you free?


EDIT: To eliminate lateral thinking answers based on the framing of the question, here's a formal mathematical statement of the puzzle:

Does there exist a coloring of $\mathbb{R}^2$ such that it is impossible to find a triangle with vertices of the same color and rational area?

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  • $\begingroup$ Is there a difference between "an uncountably infinite amount of time later" and "never"? $\endgroup$ – Florian F Aug 19 '20 at 20:59
  • $\begingroup$ @FlorianF I'd love to get into a philosophical discussion about ordinals with you, but for the sake of the problem, take it to mean "assume you have successfully colored every point in the plane." $\endgroup$ – Nico A Aug 19 '20 at 21:24
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    $\begingroup$ @EricHirst Not sure what you mean. If you are saying a filled in square, then you can take a triangle of rational area inside that square in infinite different ways. Remember - the triangle doesn’t have to be filled in, only the vertices must be of the same color. $\endgroup$ – Nico A Aug 19 '20 at 23:26
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    $\begingroup$ Are three collinear points considered to form a triangle of area 0, and therefore rational area, or is it considered to not form a triangle? $\endgroup$ – Glen O Aug 21 '20 at 8:41
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    $\begingroup$ What's the wizard's Erdös number? $\endgroup$ – Paul Panzer Aug 23 '20 at 14:25
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Very interesting supertask.

In a 2D plane, any three noncolinear points make a triangle, so only use 2 points of each color. Because you have infinite colors, you will never run out of colors. However, this doesn't save us from our demise, as this task would take an uncountable amount of time, leaving us stuck in the plane. So, we have to approach this like a supertask. Paint the first dot in 1 minute, paint the second dot in half the time, paint the third in half of that of the second, etc. In just two minutes, and however long the wizard needs to check, you will be free from the plane!

Edit:

The above solution runs into the problem that you run out of colors because there is an uncountably infinite number of points on $\mathbb{R}^2$ and there is a countably infinite number of colors. I can get a bit closer by increasing my number colors. Instead of thinking of colors as the discrete paint splotches the wizard has given me, I'll now consider the wavelength of light that the pigment reflects (totally disregarding how paint mixing works here). Now, in each step of the supertask, mix the paints such that you get a new color (e.g. in step 1 you use paint with $700nm$, in step 2 you use paint with $700.\bar01nm$, etc.). Now you have an uncountably infinite number of paint colors. However, I feel as though the plane is full of infinite 2 dimensional points $\mathbb{R}^2$, while I only have paints with $\mathbb{R}>0$, so I still don't have nearly enough colors.

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    $\begingroup$ Love the connection to super tasks, but sadly you will run out of colors! You have countably infinite (i.e 1,2,3....) colors and uncountable infinite points on the plane. $\endgroup$ – Nico A Aug 20 '20 at 4:26
  • $\begingroup$ I hadn't even considered that part. Dang. Even using colinear painting techniques runs into the same problem. $\endgroup$ – rhkoulen Aug 20 '20 at 4:38
  • $\begingroup$ @NicoA I made an edit, but I think I still don't have enough colors... I have an uncountably infinite number of them, but I think $\mathbb{R}^2$ is still too big. I'll think on this overnight. $\endgroup$ – rhkoulen Aug 20 '20 at 4:54
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Surely this is simple

You have an infinity of colours so you use each colour only once. You haven't specified whether the 'points' are true points. If they are true points on a plane then they have no dimension so you can't paint them. Not even 1 molecule of paint can be used.

or

If you can manage this task, the wizard will let you go free - fail, and you're trapped forever.

Given that the task will take forever, you are trapped forever doing the task, no the wizard will not set you free.

or

You paint infinitely-long parallel single-colour straight lines. There will be no triangles having three vertices of the same colour.

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  • $\begingroup$ This is wonderful. +1 $\endgroup$ – Voldemort's Wrath Aug 19 '20 at 23:47
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    $\begingroup$ This is a nice lateral thinking solution, but the points are true points - if we're assuming real physics applies, the "countably infinite time" and "infinite flat plane" would equally present problems, not to mention the wizard! $\endgroup$ – Nico A Aug 20 '20 at 0:58
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    $\begingroup$ Single-color straight lines doesn't work, as you'd have to use uncountably many colors. Unless I'm misunderstanding your solution. $\endgroup$ – Nico A Aug 20 '20 at 1:14
  • $\begingroup$ @Nico A - Maybe you need to edit your question to eliminate any of my solutions. I won't hold this against you. At the moment it seems unclear to me. Maybe you can reframe it without metaphors about wizards and paint? A task that has different infinities and will take infinitely long invites questions about the nature of the reality in which it occurs. Unless of course you have in mind a cheat that enables us to escape without doing the task in its entirety. $\endgroup$ – chasly - supports Monica Aug 20 '20 at 1:18
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    $\begingroup$ @chasly-reinstateMonica Good thinking. I've added a formal statement of the puzzle below the slightly-informal story. I've left the original framing, however, so your lateral thinking solutions still stand. $\endgroup$ – Nico A Aug 20 '20 at 1:22

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