Here are two "dissection" problems - cutting up one shape to make another one. The first one is nice, but the second one is pretty hard! I got these out of a neat little book called Fun with Mathematics, by Jerome Meyer. I knew the second one because I have made parts of a quilt that used the answer. These are found in a part labelled "Interesting Problems that Require Some Thought."First, take the shape on the left below and, making just two cuts, turn it into a square. The picture is sparsely labelled, so explain the assumptions you made when you tackled it. (Can you think of more than one way to do it? I think I know two.)
The second problem is to take a square and, using FOUR cuts, make five smaller equal squares whose total area is the same as that of the original square. So cut your square four times so that you could move some pieces and get five smaller squares that won't overlap.
When explaining your answer, you might draw pictures (remember to use a monospaced font and use only spaces when you draw, never tabs). You might also label the existing pictures and then explain your answer based on those labels.
- Annie Fetter
Annie says: Solutions
This was a hard week, partly because I didn't word the question carefully enough, but there were some super solutions! The second part was really tough, so I considered that a bonus question - if there is ever a time when you can't do the whole problem, at least send in what you have.
The part that wasn't worded carefully was how exactly you were supposed to get a square in the first part. The idea of dissections is to cut something up and then put it back together again. Dissections are sort of like tangrams, a puzzle where a square is cut into seven different pieces where you are supposed to rearrange the pieces to make different shapes (roosters, trains, things that don't look like anything in particular). This game has been sold under a number of different names, one of them being Tangoes.
Explaining the answers to this problem was tough without pictures, but a few people managed to do fairly well. We also got a lot of nice pictures. Taking top honors in the picture department this week was Tim Peterson from Rochester, New York. Roger Mong of Zion Heights Junior High provided a superb explanation of how he figured out the second part. That is a tough question, but careful thinking like Roger's will get you a long way.
Terry Haslam of Southern Trinity provided some very nice pictures, and Greg Moore from Shaler Area High School gave a really nice explanation, along with some good pictures. If you were puzzled about how to approach either part of this problem, just read Dave Peterson's solution. I think he covers just about everything.
A list of all the people who got this problem right and most of the solutions are also available; there are a number of good solutions in the full list, so be sure you read them over.
Roger Mong Grade: 8 School: Zion Heights JHS, Richmond Hill, Ontario, Canada 1. The First thing I look at is the area of the figure, then knowing the area (4) I can deduce the dimensions of the square. Here's my solution: * * * * * * * * * * * * * * * * * - - * * * * * * * * * * | * ------> * * * | * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 2. When I first saw the question, I figured out that there will be two pairs of parallel lines (parallel lines are needed to make squares), and the two pairs of lines are also perpendicular to each other, making one square in the middle. For the other four squares to be made, they must be assembled from the other pieces (two each because there will be nine squares and one is used). * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Terry Haslam Grade: 10 School: Southern Trinity High School, Mad River, California My assumption on the first one was that one outside angle was right and two of the others were equal to each other, as shown in my diagrams. I have two answers for the first one.The second took a little time and I'm still not sure if it's right, but I think it works.
I must admit I didn't notice on the question sheet that they had to be the same size. It took me a while and when I finally got it, I kicked myself because it was staring me in the face the whole time. I was really upset.
Tim Peterson Grade: 6 School: Homeschooled, Rochester, New York Answer to Part 1: 8 solutions. For each line that could be made, I went through all of the other lines, figuring out if they could be rearranged to form a square, and if one could, I would add it to my list. Assumptions: (same symbol = same angle):And here are all 8 of my solutions:
P.S. Did you REALLY only find two solutions? There are 8.
Greg Moore Grade: 10 School: Shaler Area High School, Pittsburgh, Pennsylvania Following are two methods of forming a particular shape into a square with two cuts. The distance between two diagonal asterisks is 1/2 * 2^.5. The distance between any other two asterisks is .5. * * * * * * * * *---* * * * * * * * | * * * * * * \ / * * * * * * * * * * * * * * * * * * * * * * * * \ / * * \ * * \ * *-------* * \ * * * * * * * * * Splitting a big square into 5 equal little square making 4 cuts. The distance between two asterisks is 1/5 and the slope of the cuts is 2 or 1/2. * * * * * * * * * * * *\ \ _ - * * | _ + * * \ _ - \ * * _ + | * * \ \_ - * * | _ - | * * \_ - \ * * _ - | | * * - \ \* * * * * * * * * * * * The midpoint of each side of the big square acts as a hinge; flip each small triangle 180 degrees using this hinge. The length of each side of each little square is 2/5 of a cut.
Dave Peterson Homeschool teacher Rochester, NY I found six or seven ways to dissect the first shape into a square. I won't guarantee there aren't more, but others seem to need more than two cuts. I am assuming that all angles are 45 or 90 (or 135) degrees, so that ABF, CDB, FBD, and DEF are right isosceles triangles (where D is the midpoint of CE), and all are the same size. A * * * * * * * 2 2 * * * * * * * * * * C * B * * * * * * * D F * * * * * * * E The area of each of these right triangles is 1, so the area of the square has to be 4 and its side has to be two. By imagining a square of this size and sliding it around over the figure, I found that if it is placed at any of the four "corners" (that is, as far as it can go to the NE, SE, SW, or NW within a circumscribed square), the parts that don't fit under it can be cut to fill in the square, using only two cuts. The last case is the most interesting: the two cuts intersect and make four pieces to assemble, rather than three as in the other cases. I'll show them in the order I discovered them; piece 1 is always where the "sliding square" is. Northwest: * * * * * * * * * * * * * * * * * * * * 2 * * * * * * * 3 * * * * * * * * * * * * * * * * * * * 3 * * * * 1 * * * 1 * * * * * * * * * * * * * * * * * * * * * * * * * * * 2 * * * * Southeast: * * * * * * * * 2 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 1 * * 1 * * * 3 * * * * 2 * * * * * * * * * * * * * 3 * * * * * * * * * * * * * * * * * * * (alternate cuts) * * * * * * 2 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 3 * 1 * * 1 * 2 * * * * * * * * * * * * * * * * * * * * * 3 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 3 * * * * * * * * * * * * * 1 * * 1 * 2 * * * * * * * * * * * * * * 2 * * * * * * * * * * 3 * * * * * * * * * * * * * * * * * Southwest: * * * * * * 2 * * * * * * * * * * * * * * * * * * * * * * * * * 3 * * * * * * * * * 1 * * * 1 * * * * * * * * * * * * * * * * * * * * 2 * * 3 * * * * * * * * * * * Northeast: * * * * * * * * * * * * * * * * * * 2 * * * * * * 3 * * * * * * * * * * * * * * * * * * * * * 2 * * * * * * * 1 * * 1 * * * * * * * 4 * * * * * * * * * * * * * * * * * * * * 3 * 4 * * * * * * * * (alternate - requires flipping piece 2) * * * * * * * * * * * * * * * * * * 1 * * * * * * * * * * * * * * * * * * * * * * 3 * * 2 * * * * * * * 1 * * 2 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * 3 * * * * Now for the second problem: I'll make the side of the square 5 units so the area will be divisible by five to make things easy. Since its area is 25, the five small squares have to have areas = 5, and sides = sqrt(5). To construct that length, I can make a right triangle with legs 1 and 2, since 1^2 + 2^2 = 5. Somehow I expected things to be symmetrical around the center of the original square, so I drew the square with a 5-by-5 grid on it, and noticed that I can draw a square of the desired size in the center. Then I continued its edges out to the edges of the large square, intersecting it at the corners and midpoints, and found that I can cut along those four lines and reassemble to form five identical squares. The grid lines make the proof that all the segments have the right lengths to fit together almost obvious. * * * * * * * * * * * * * * * * * * * * * ** | | * | | * * * * | | * | * * * * | | *| * | * *---*---+-------+-------*-------+-------* * * | | * |* | * * * | * | * | * * *| * | | * | * *-------*-------+-------+---*---+-------* * * |* | | * | * * | * | | * | * * | * | | *| * * *-------+---*---+-------+-------*-------* * | * | | * |* * * | * | * | * * * | *| * | | * * *-------+-------*-------+-------+---*---* * | * |* | | * * * * | * | | * * * * | | * | | ** * * * * * * * * * * * * * * * * * * * * * To construct this dissection, simply draw lines from the midpoint of each side to the first opposite corner in a clockwise direction. To make the five small squares, move the central small square to a new place, and swing each right triangle around until its hypotenuse joins the diagonal side of the trapezoid next to it.
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