A red square. A horizontal rectangle, with three vertical bands (narrow cyan/wide green/narrow cyan), a tall vertical yellow rectangle, a vertical rectangle with three equal horizontal bands (magenta/blue/magenta). An uncoloured hexagon.

Given the colours and banding of the first four rectangles, how should you colour the hexagon, and why?

There's a single, unambiguous solution (ignoring symmetry), which will be obvious when found. Nothing is hidden in the image, you could print it out and still solve it with nothing more than your eyes and some lateral thinking.

Edit: I intended this to be solved by eye, and so my image wasn't drawn 100% accurately. However, since a few people are approaching things somewhat mathematically (which is admittedly a valid approach), I've checked my calculations and updated the image slightly. It shouldn't make much difference in solving it, but I believe the maths is now correct to the nearest pixel.

  • $\begingroup$ Owing to the lack of responses thus far, I think you've stumped a few people :p $\endgroup$
    – nine9
    Commented Jan 20, 2016 at 11:58
  • $\begingroup$ @nine9 - time will tell if I made it too hard. I don't normally make puzzles like this, so it was hard to judge the difficulty. $\endgroup$
    – Alconja
    Commented Jan 20, 2016 at 13:13
  • $\begingroup$ How much do we care about the relative sizes of the side lengths? Just by eye, if we say the cyan rectangles are 1u (1 unit) wide, then I'd say the red square is 6u*6u, cyan rectangles are 1u*6u, green square is 6u*6u, yellow rectangle is 6u*9u, magenta and blue rectangles are 6u*2.5u and the hexagon base is 5u. $\endgroup$ Commented Jan 20, 2016 at 17:20
  • $\begingroup$ Separate observation: I assume we are dealing with additive color mixing due to the black background? $\endgroup$ Commented Jan 20, 2016 at 17:23
  • $\begingroup$ @SpiritFryer - on relative sizes: yes, that's definitely important (though I haven't necessarily got things pixel perfect, just eyeballing should be enough). $\endgroup$
    – Alconja
    Commented Jan 20, 2016 at 20:52

4 Answers 4


These images are all (orthographic) projections of the same cube.

It has the three basic colors, with opposite faces identical. When one is in front of the other their colors are added in the projection.

The final answer is a variation on these two pictures, depending on how you rotate the cube:


A more detailed explanation of the answer:

First: We just see the two red faces superposed.

Second: Now the cube is rotated, with the green faces in place of red. However, it is slightly tilted so the blue sides mix with the green forming the cyan rectangles.

Third: The cube is at 45 degrees with the ground, so red and green faces superpose completely at top, as well as green and red at the bottom, giving an uniform yellow color.

Fourth: Finally, we see a similar arrangement to the second picture except in a different plane of rotation.

Also, we can notice differences in size between the rectangles. That's no mistake: When a cube rotates along a coordinate axis, its visual size increases from $l$ to $l \sqrt2$, then goes back down to $l$.

The yellow is the biggest rectangle because the diagonal of a square is the longest line segment inside it.

Our last projection is the cube viewed vertex-on, which has a hexagonal envelope.

This picture shows how you combine colors of face pairs in the X, Y and Z axis to get the final answer:


This puzzle was actually pretty easy for me to figure out because I love trying to visualize higher-dimensional spaces, so I was very well familiar with projections, envelopes, cross sections and stuff.

It's a fascinating way to improve your spatial reasoning, so I highly recommend anyone interested to take a look. Or you could just try exploring a 4d maze and see if you can find your way out

  • $\begingroup$ Nice solution. If this is correct, though, the colors should be magenta, cyan, and yellow (hex codes #FF00FF, #00FFFF, and #FFFF00.) Your purple and light blue aren't quite right. $\endgroup$ Commented Jan 21, 2016 at 20:32
  • $\begingroup$ Ok. Will fix and add more pictures detailing stuff, wait a minute $\endgroup$
    – MathET
    Commented Jan 21, 2016 at 20:36
  • $\begingroup$ This is a cool answer, but if I'm looking at this right (I'm using a color picker, because I'm the tiniest bit colorblind), the original image uses six colors, and this only uses three - how to account for the other three colors? $\endgroup$ Commented Jan 21, 2016 at 21:30
  • 6
    $\begingroup$ @question_asker I think it is a pun, since lateral means "of, at, towards, or from the side or sides." :-) $\endgroup$ Commented Jan 21, 2016 at 22:09
  • 2
    $\begingroup$ @question_asker - sorry to have frustrated you. I was torn whether to use the tag or not... I guess I felt like making the leap from 2d to 3d required somewhat lateral thinking, and... ok, I admit it, I couldn't resist the pun. $\endgroup$
    – Alconja
    Commented Jan 21, 2016 at 23:12

Either Black or White, depending on how you look at them. If they are basic light colors - then seen together they would be seen as white light, but if they are paint pigments - then black. However, because of the background I would bet on white.

  • $\begingroup$ White is definitely better than black, but neither is right. The solution is multi-coloured and has a specific pattern to it. $\endgroup$
    – Alconja
    Commented Jan 20, 2016 at 21:58

The color of the hexagon should be black because all the other colors are a part of CMYK color model and the black is the only color that is left out.

enter image description here

And there should be no pattern in the color because alternate shapes have been colored.

  • $\begingroup$ He already confirmed (unless I misunderstood his comment) that it's additive color mixing that should be applied, and not subtractive, so this can't be right. (So with your argument, the hexagon should actually be white.) $\endgroup$ Commented Jan 21, 2016 at 4:31
  • $\begingroup$ Sorry, not black. @SpiritFryer is right about the colour mixing being additive. But as I commented on superefka's answer, the solution is multicoloured and has a specific pattern to it. $\endgroup$
    – Alconja
    Commented Jan 21, 2016 at 4:58
  • $\begingroup$ @SpiritFryer Oh yeah sorry. Didn't see that. $\endgroup$ Commented Jan 21, 2016 at 5:30
  • $\begingroup$ @Alconja Oh OK. I will try it again. :) $\endgroup$ Commented Jan 21, 2016 at 5:30
  • $\begingroup$ Isn't black the colour of the background of the image? $\endgroup$ Commented Jan 21, 2016 at 17:32

I spotted something interesting, not sure if it's right, but here's a go:

hexagon with colour stripes

I drew overlapping selections by hand onto a layer mask, so not particularly accurate.


If we look at the relative areas of the four rectangles:
5x5 = 25 (Red)
6x5 = 30 (Cyan/green/cyan)
5x7 = 35 (Yellow)
5x6 = 30 (Magenta/Blue/Magenta)
And the total of these is 120 - the internal angle in degrees of a hexagon. So dividing the hexagon from one vertex using these angles gives us:
25 degrees of red;
30 degrees comprised of 5 cyan + 20 green + 5 cyan;
35 degrees of yellow;
30 degrees of magenta/blue/magenta (sub-divided horizontally, not drawn to scale!).

  • $\begingroup$ I wouldn't call this solution obvious, but it's definitely comprehensive. $\endgroup$
    – nine9
    Commented Jan 21, 2016 at 8:01
  • 1
    $\begingroup$ If you treat areas as angles, then why subareas (magenta/blue/magenta within one rectangle) is treated differently? It is hard to call "a single, unambiguous solution (ignoring symmetry), which will be obvious when found." At least, there are many ways to apply the same idea. $\endgroup$
    – klm123
    Commented Jan 21, 2016 at 9:55
  • 1
    $\begingroup$ And finally, there is author statement "The magenta-blue rectangle should be taller than the cyan-green one is wide". Thereby downvoted, sorry. $\endgroup$
    – klm123
    Commented Jan 21, 2016 at 9:59
  • $\begingroup$ Interesting theory. But unfortunately wrong. If you want to get mathematical about things, I've now updated my original image to be accurate to the nearest pixel. You'll notice the green/cyan is slightly narrower and magenta/blue block is slightly taller than previously. $\endgroup$
    – Alconja
    Commented Jan 21, 2016 at 12:39
  • $\begingroup$ @klm123 fair enough. Although I hadn't seen that comment, since it was posted around the same time I was devising this solution. Regarding the blue/magenta - well the relative areas should still be the same (I drew it by eye, so not to scale as mentioned) and the reason it's different is that they go horizontal. Hey - it's a hypothesis to get feedback and inspire thinking in among not much other activity, I thought that was the whole point. $\endgroup$
    – jhabbott
    Commented Jan 21, 2016 at 13:05

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