I keep posting problems and you do all the solving. I figured I better walk the walk, so I worked on Lee's problem from last week's post. It was really a nice one. The idea was to give the dimensions of five rectangles such that
1) the sides have unique integer dimensions between 1 and 10
2) they can be arranged into a square.
The first thing I did was narrow the size of the possible squares. I did it originally with an area argument. The minimum area of 5 rectangles that meet criteria 1 would be 1*10+2*9+3*8+4*7+5*6 = 110. The max would be 10*9+8*7+6*5+4*3+2*1= 190. So, 110<11^2, 12^2, 13^2<190. That limits the squares I was considering to only three different dimensions. (Later I found another way to argue for these as the only three possibilities.)
I have to admit I had a little more information than you guys did on the blog. Lee is a puzzle maker and he brought in a wood version of one of the solutions to show me. I was playing around with it on my desk one day and found it surprisingly hard to arrange into a square. It took me around 5 minutes to arrange 5 rectangles into a square.
Seeing the correct configuration was the insight that broke the problem for me. The rectangles looked something like this when assembled.
Note that this isn't drawn to scale, but essentially you have one rectangle in the middle with the others spiraling around it. The dimensions of the middle depend on the dimensions you choose for your outer rectangles.
That's when I started playing with me three possible areas.
1) If the overall square is 11 by 11, then what are my options for the inner rectangle. 10 can't be one of the sides of the inner triangle because I have to add two numbers to it to get 11. 9 also can't be it because I'd have to add two one's to it to get 11, and I'm only allowed one one. 8 is the highest possible side of the inner rectangle, because the outer rectangles could have 2 and 1 as sides.
But then since the sides of the whole rectangle are 11, we can fill in...
Now only 3,4,5,6,7 are left, and the only pairs that add to 11 are (5,6) and (4,7). That means 3 has to be the other dimension of the center rectangle.
What two remaining numbers could you add to 3 to get 11? There aren't any. Conclusion: 8 can't be the largest dimension of the inner rectangle. Try 7. And so the process repeats through all possible cases. I really didn't do it that systemattically, but this makes for a better demonstration.
When I got through with the process I only found two solutions: An 11x11 rectangle like this
and a 13x13 like this
At this point, I freaked out. I thought my logic was impeccable but Lee promised 4 solutions. At some point I thought to go back and look at the solution he gave me and saw that the sides were identical to mine, but his solution had flipped two of my sides giving different dimensions to the surrounding four rectangles. I'll show you with the 13x13,
So that's it, Lee also has a solution page on his website check it out. What I want now is a proof that 5 rectangles with unique sides cannot be arranged into a square in any other way (than the way I described above). There has to be a simple topological proof of this. Help?