In an earlier blog post I proposed a puzzle made of five 2 by 2 squares, each of which had been cut along a line from a corner to the midpoint of an opposite side so as to form a right triangle and a trapezoid with two right angles. The challenge of the puzzle was to put these 10 pieces together to form a square.

I mentioned in my previous post, that a colleague and I first saw the puzzle at a NCTM conference. We had thought it might be a relatively new puzzle since neither of us had seen it before, and the people in the booth where we stumbled onto the puzzle didn’t seem to know anything about it–except that it was supposed to be a puzzle.

Ten years or so after finding the puzzle and making hundreds of copies of it for teachers, I discovered the puzzle in an antique shop. It turns out that in the late 1800s and early 1900s, companies would produce and give away put-together puzzles as a way to advertise their wares. This version of the puzzle I found in the antique shop was advertising horse blankets. The instructions on the front of the box told whoever had the puzzle to put the pieces together to form a perfect square.

Several people around the office have accepted the challenge of the puzzle, and I believe one or two may still be working on it. Since I don’t want to ruin the puzzle quite yet for any one still thinking about it, I thought I might use one more post to play around with the puzzle pieces and perhaps give a hint or two for anyone who might be stuck.

So, let me ask a couple of questions. What is the sum of the areas of the five 2 by 2 squares? Of course, the sum is 20 because the area of each of the five squares is four. What would be the length of each side of a square with an area of 20? To help us think about that, let’s look at a square with an area of 16. What is the length of a side? Or how about a square with an area of 25? What is the length of a side?

Now what do you know must be true of the length of a side of a square with an area of 20? Will the length be a whole number? Which side of either the triangle or trapezoid pieces has a length that is not a whole number?

One final thought. Did you notice that a triangle and trapezoid could also be put together to form another triangle? What are the lengths of the legs of this new right triangle? Could those legs be part of the sides of a square with an area of 20?

So much for hints; I’ve likely told you more than you needed to know. I want to pose one other challenge that I see in these puzzle pieces. If you compare corresponding angles of one of the small triangle pieces and the larger triangle you just formed, you will notice that corresponding angles of the two triangles are congruent—they have the same measure. That means that the two triangles are similar. Remember, two triangles are similar if corresponding angles are congruent.

Okay, here’s the new challenge. How many more triangles similar to these two triangles can you form using these puzzle pieces? Does one of those triangles make use of all of the puzzle pieces?

I promise in my next blog post to answer all of these questions. Jesse, our videographer, has spent some time working on these challenges, and we’re planning to create a video in which we have a conversation about how he worked at the problem and about some of the important mathematical ideas involved in the puzzle.

Thanks for the hints. I’ve still been working at the puzzle whenever I wait for my computer to export video.

[…] I will come back with a follow-up post to talk some more about the puzzle and to look at some additional challenges and the math that is there to be explored. Click here for some hints and an extension. […]