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UPDATE

To refute Sconibulus's answer, I drafted a second puzzle with identical rules.


7 months ago, I released a collection of variations of an original puzzle called Trapezoids, but never mentioned that the cover itself is also a puzzle that resolves to a unique solution if a final rule can be discovered.

Cover Puzzles

Original Puzzle

New Puzzle

Rules

  1. Shade some cells such that they form edge-connected clusters of 3 (trapezoids). A trapezoid may not share an edge with another trapezoid. All remaining unshaded cells are connected edge-to-edge.

  2. Dots on the border between two cells indicate that one cell is shaded and the other is not.

  3. Trapezoids cannot cross interior borders (bold lines). The interior borders create regions, each of which must have at least one shaded cell. Interior borders do not block unshaded cells from connecting.

  4. ???

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  • $\begingroup$ Various thoughts I had: each region has exactly 3 shaded triangles (fails the all unshadeds connected rule). No trapezoid is contained entirely within a region (multiple solutions), each trapezoid goes through exactly two coloured regions (multiple solutions), each region has exactly 2 trapezoid parts in it and each trapezoid goes through exactly 2 coloured regions (no solutions), and one that seems most likely now, but I haven't explored properly, each region has exactly 2 trapezoid parts in it. $\endgroup$ – TheGreatEscaper May 25 '17 at 14:12
  • $\begingroup$ Great puzzles! Why don't you bring your puzzles from your website here?. $\endgroup$ – Jamal Senjaya May 26 '17 at 4:36
  • $\begingroup$ Hi paramesis, you have very interesting puzzles. Can I use some for the logic puzzles section on my website (Puzzle Prime)? Of course, I will give all the credit to you and leave backlinks as well. $\endgroup$ – Puzzle Prime Jun 23 '17 at 13:00
  • 1
    $\begingroup$ Hi @ArturKirkoryan I sent you a message on the contact page of your site. $\endgroup$ – paramesis Jun 23 '17 at 16:12
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The last rule:

At least one cell of each color must be shaded and no two colors may have the same number of shaded cells.

The solutions:

Cover 1

enter image description here

Cover 2

enter image description here

Some notes:

In solving cover 2, I concluded that the bottom-right two purple triangles of the right-most section had to be filled in. I started by drawing a line through the dots to visualize possible paths the connected non-shaded portion could take.
When looking at the possible trapezoid combinations for the top section, it seemed to always block off one side above the dots. I then ruled out filling in the green triangles on the top-left side of the right-most section, leaving the two purple triangles on the bottom-right to be filled. This meant that the purple triangle in the bottom section had to be unfilled, and a path for it to be connected had to go along the bottom below the dots in the bottom section.
The connected non-shaded portion would then have to go through the area between the dots. This allowed me to solve the trapezoids along the dots in the bottom section. There was only one configuration where the path wouldn't be blocked.
This left multiple combinations for the top and right sections, and one of them seemed to have a quirk, which also worked for cover 1.

Coincidentally(?), the number of shaded cells per color are the same for both covers.
Green: 1, Red: 2, Purple: 3, Orange: 4, Blue: 5

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I think the final rule is

Each color is to be treated as it's own 'Unshaded' area, within the larger total Unshaded area.

This leads to a unique solution

enter image description here

This is unique because

The center can't be rotated 60 degrees, because the Purple region would have a separated Triangle, The Red region can't be joined, because it would force abandoned Red triangles, and the only area on the perimeter that doesn't share a side with the interior, and also doesn't result in a separated area of color is the one I have placed on the right side blue squares where they border the purple region.

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  • $\begingroup$ Interesting and apparently valid solution, but not what I had in mind. $\endgroup$ – paramesis May 25 '17 at 14:15

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