# COLOR-PUZZLE. Complete the picture

Here is a color-puzzle(Not the best picture but hey :D) If anything is unclear, let me know. But all the information you need should be out there. So... Try complete the picture. Hint1:

It's not {2(1),3,4,6}

• Minor hint, anyone? Dec 22, 2019 at 16:50
• Yes. ,. Please. Dec 22, 2019 at 23:13

I would color it like this The circle is divided into 12 30° sectors.
The red rectangle contains a regular hexagon, each of whose angles is 120° so it takes up 4 sectors. The yellow rectangle contains a square, each of whose angles is 90° or 3 sectors. The grey rectangle contains an equilateral triangle, each of whose angles is 60° or 2 sectors. The blue rectangle contains what appears to be a 30° angle which takes up one sector. The two hatched sectors are unused.

• Haven't thought about this puzzle for a long long time :D But this is the intended answer. Well done! Dec 30, 2021 at 17:52

I believe this to be the answer to your question...(I may be wrong though)

Here, is my explanation:

every colour embedded with the rectangle box have vertices:-

• (R)red triangle 3
• (G)grey hexagon 6
• (B)blue square 4
• (Y)yellow converging lines 1

Every vertices is sum of the vertices to which is diagonal and vertices that is vertically opposite to it, minus the vertices that horizontally opposite to it :-

B=(Y+G)- R = (6 + 1) - 3 = 4
R=(Y+G)- B = (6 + 1) - 4 = 3
Y=(R+B)-G =  (4 + 3) - 6 = 1
G=(R+B)-Y =  (4 + 3) - 1 = 6

[from the placements of rectangle this shows which are vertically opposite, horizontally opposite and diagonal

+--------------------------------------------------------------------------+
| shape-vertices = vertically opposite + diagonally-opp - horizontally-opp |
+--------------------------------------------------------------------------+
| red-tri(R)       grey-hexa(G)          yellow-lines(Y)  Blue-rect(B)     |
| grey-hexa(G)     red-tri(R)            blue-rect(B)     yellow-ines(Y)   |
| yellow-lines(Y)  red-tri(R)            Blue-rect(B)     grey-hexa(G)     |
| blue-rect(B)     grey-hexa(G)          yellow-lines(Y)  red-tri(R)       |
+--------------------------------------------------------------------------+


you may be thinking that how does all of this give you the colours(please bear with me a little :P)

Now the circle has twelve sectors...

the relation between these sectors can be determined with same formula from above. Red(R) Blue(B) Grey(G) Yellow(Y).

B=(Y+G)- R = (8 + 10) - 6 = 12
R=(Y+G)- B = (8 + 10) - 12 = 6
Y=(R+B)-G =  (6 + 12) - 10 = 8
G=(R+B)-Y =  (6 + 12) - 8 = 10


however we can determine the next occurrence of the colour in the sector by subtracting from 12.The next sector with the colour:-

• Blue will be 0 (12-12)
• Red will be 6 (12-6)
• Grey will be 2 (12-10)
• Yellow will be 4(12-8)
• I have a feeling this whole thing should be shorter. Dec 22, 2019 at 20:22
• @FlreCase yes, I agree with you, But I'm afraid that if I shorten my answer, people will have difficulty understanding how I got the answer. Dec 22, 2019 at 20:29
• Maybe you are right FireCase, I too am skeptical about my approach (maybe even its correctness) because the question is tagged as pattern and visual not maths. :) Dec 23, 2019 at 1:44
• Nikhil01: Nice work, but it's not the intended answer. It's much more simple. Also, I can say that the placement of the rounded rectangles outside the circle are just randomly placed. Just focus on what's inside the rectangles and then somehow map it into the circle. Dec 23, 2019 at 5:17
• Prim3numbah: oh I understand, Thank you. :) Dec 23, 2019 at 6:32
+-------------------+---------+-------+
|      Colour       | letters | sides |
+-------------------+---------+-------+
| blue              |       4 |     4 |
| gold              |       4 |     2 |
| grey              |       4 |     6 |
| red               |       3 |     3 |
+-------------------+---------+-------+


Here,

I add a number of edges of each shape with the number of letters in the colour's name, and then divide it by 2. For example, if I want to know how many sectors away from red grey occurs, I add $$(6+4)/2 = 5$$.

From the rounded rectangles you can determine which colour follows which colour. For example, grey follows red, blue follows gold (and vice-versa).

So, I think this could be the answer:

Red $$= (6+4)/2 = 5$$, so red must be 5 sectors away from grey. Similarly, gold $$= (4+4)/2 = 4$$, so gold must be 4 sectors away from blue. As such, my solution is • Hey. I realized this problem wasn't as "elegant" as i hoped. I will apply a similar reasoning to another problem i'll make later on. So for that reason i will not give away the answer. But cudos for the hard work youve done :) Dec 29, 2019 at 12:03
• Thank you, for your kind words. :) I'm looking forward to seeing your question. Dec 31, 2019 at 2:38