The 21 hexominos below are all those that can be made by joining three dominos together (i.e. they have a perfect matching with respect to the graph of their squares) and are not rectangles. The objective is to tile the three 42-square pears below with these hexominos, seven to a pear; some hexominos have already been placed to give a unique solution.

Bonus: For each of the three sets of hexominos that solve this puzzle, find all ways that set can tile a pear. Up to reflection of the whole set, two sets have two solutions and one (the perfect pear) has only one.

  • 1
    $\begingroup$ I don't understand your last paragraph (bonus). $\endgroup$
    – Florian F
    Sep 24, 2021 at 11:16

1 Answer 1


Here is the finished tiling:


To get started:

We can determine some areas that have to be part of a single hexomino and color them accordingly. In the right pear we can narrow down the left part of the pear to two possible pairings of hexominos.
Step one

Next step:

We can figure out the rest of the right pear. If you try having the piece I have in orange in the upper right have its last square go down, you'll quickly reach a contradiction. Then there is only one combo of pieces that works for the remaining space in the pear. This also lets us determine the two pieces of the middle pear that go along the bottom.
Step two

Looking at the left pear:

Step three
If you look at the available pieces, you can determine that the leftmost piece must include the square directly below it. Also, we have two chokepoints here (in green and blue).

With those chokepoints:

I made the mistake of assuming the chokepoints must belong to different hexominos. It takes a little while, but you'll eventually reach a contradiction based on that. So they actually belong to the same piece, allowing us to finish the right pear as well.
Step four

For the next step I took a guess:

Using some intuition I had from the path I took to the contradiction, I just tried putting these two pieces in the top of the left pear:
Step five

From there, I just looked at the pieces I had left and found something that worked.

  • $\begingroup$ This is in fact the only partition of the given hexominos such that each partition can tile a pear. As for the bonus, exactly one of the pears has only one solution up to pear reflection - I call that The Perfect Pear. (MLPFIM reference there.) $\endgroup$ Sep 25, 2021 at 3:04
  • $\begingroup$ You could have also used the domino property of the hexominos to deduce step 4 (after "with those chokepoints") a little faster. $\endgroup$ Sep 25, 2021 at 3:06

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