# Pythagorean pentagons

To follow up on the theme of so called "pythagorean" dissections, here is one more for you to chew on. I hope you don't get bored.

The pentagons above have sides respectively 3, 4 and 5. Pythagoras says that the sum of the areas of the first two equals the area of the last one. Yes, the Pythagorean theorem works with any shape, not only squares.

Your task is to dissect the first two pentagons in a finite number of pieces and reassemble them to form the third pentagon. Of course, without gaps and without overlaps. You can rotate and flip the pieces as you wish.

These dissection are proven to be always possible. But it can take many pieces and be ugly. So I am asking to minimize the number of pieces. I know at least one elegant solution. But I don't know whether it is optimal.

Just to avoid weird solutions, the pieces must be simple polygons.

• Wow, I never realized that the Pythagorean theorem also works for other shapes than squares. Even circles!
– Ivo
Commented Feb 21 at 10:28
• I would express it as "Pythagoras says that the sum of the areas of the first two equals the area of the last one." For a planar figure, surface would mean perimeter. Commented Feb 21 at 18:13
• It is a better formulation. I changed the question accordingly. Commented Feb 21 at 18:20
• Looks like I unintentionally made @Edward H delete their very cool 6-piece answer. Please consider undeleting it. Commented Feb 25 at 6:16
• FYI, here is an example of a weird solution: puzzling.stackexchange.com/questions/50097/… Commented Feb 25 at 21:24

Here's another possible solution, without flipping:

• Beat you to it by two minutes, I'm afraid. Your shapes are cool, though. Commented Feb 25 at 5:59
• It is one hour 2 minutes actually. But it is still awesome. I wish I could give 2 checkmarks. Commented Feb 25 at 9:28
• Silly me! I looked at the "answered time", not the "edited time". So yes, 2 minutes it was. Commented Feb 25 at 21:46
• Since the solutions are almost simultaneous, I decided to accept this one to give it more visibility. Loopy has already plenty of upvotes. Commented Feb 29 at 23:49
• @FlorianF loopy never has enough upvotes! I'm kidding, of course. This answer is very deserving. Commented Mar 1 at 17:34

6 pieces, no flipping required:

History:

7 pieces, no flipping required:

8 pieces, requires two flips

9 pieces, requires one flip

very similar:

10 pieces, no flipping required

not very elegant, though

• Wow! I managed to do 11. It starts nicely with the large pieces but yeah, the small pieces get a bit messy. But I said it is the count that counts. Commented Feb 20 at 8:43
• The 8 piece solution is stunning! I never would've expected multiple concave pieces. Commented Feb 21 at 18:45
• @FirstNameLastName I use shapely to make the pictures. As for strategy I have found that it helps to work out the characteristic angles (mostly multiples of 36° and rarely 54°) and length units (1 and the golden ratio, phi). Besides that it is quite a bit of incremental work, quite literally moving bits around; you can see in my post the evolution from 10 to 7 pieces. Commented Feb 25 at 3:35
• I just can't wait to see your 5 piece solution :-) Commented Feb 25 at 10:20
• About your 6-piece solution. Not only there is no flipping, but the rotation is minimal. The small pentagon rotation is unavoidable. The length of unmatched edges facing each direction is 3+4 = 7 in the first two pentagons. But you need only 5. Rotating the pentagon "negates" one unit on each face. Turning it 180° makes each edge face the opposite direction and absorb another unit length. That brings the total of unmatched edges to 5 unit per side. I hope you follow my explanations. Commented Feb 25 at 15:35

My efforts to minimize the number of pieces did not improve upon loopy walt's best, but here is another 7-piece solution without flipping:

Sixteen pieces, with some flipping required.

• Nice! A different approach than mine. But let's see if someone finds a better solution ;-). Commented Feb 20 at 0:51
• I'm curious - can this method be extended to any regular polygon? It looks like it probably should work for them but I'm not convinced... Commented Feb 20 at 11:09
• It may not be minimal, but the symmetry is quite pleasing to the eye :) Commented Feb 20 at 15:20
• I believe it works for every regular polygon. So I probably won't post a "pythagorean heptagons" puzzle :-). Commented Feb 20 at 15:54
• @FirstNameLastName geogebra.org/geometry Commented Feb 25 at 3:57

OK, I'll give you my solution for reference.

Elegant, imho, but with 11 pieces not quite optimal.

And just for fun, here is another seven piece solution.

And finally my own 6-piece solution

It looks so broken. Just don't ask how I came up with it!

• +1 Very elegant! Commented Feb 20 at 21:55
• Wow! And you are asking me how I came up with mine?! Btw. if you rotate the top two small pieces in the large pentagon (funny how these two appear in your, Daniel's and my solution) then five out of the seven pieces need translation only, no flipping no rotation. Also, the rotation required of the remaining two can be made 180°. Commented Feb 24 at 16:27
• Yeah, in the meantime I learned how it is done... I actually made it intentionally more convoluted than necessary. The two top pieces, I turned them like this for esthetic reasons. Btw Daniel's and mine are actually related, you can transform mine into his using simple steps. So there are similarities. Commented Feb 24 at 17:43
• I keep coming back to this question, and am impressed anew each time. Commented Mar 1 at 9:35
• Looking at it again, it appears my last 6-piece solution is a variation of loopy's. Both solutions can be dissected into the same 9 pieces. Commented Mar 1 at 9:40