*Results*

Our hypothesis is that all numbers in the set of perfect squares can be values of N, where N equals the number of sets of tangrams. Examples of numbers in this set are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, etc. One can make a perfect square from four sets of tangrams.

Also, four to the m power sets of tangrams will form a square. (Where m is a positive integer.) All of these numbers appear in the above set. Any number that is not a perfect square cannot be a value for N. We proved this in two ways. One is done algebraically, another is done by using the areas of the figures (rectangles or squares) formed when using N sets of tangrams.

We found that when using two sets of tangrams, one could form what appears to be square, but, in actuality is not a square but a rectangle.

Proof #1

The quadrilateral below appears to be a square. The following proof shows that it is not a square. It is a rectangle.(This picture was drawn in our draw package so we apologize for the inaccuracy of the polygons.) First we found the lengths of all the sides of the seven polygons of one set of tangrams in terms of x.

Let x= a leg of the large right isosceles triangle Last chapter we studied the Pythagorean Theorem to find lengths of sides of triangles. We used the Pythagorean Theorem, a2 + b2 = c2 , where a and b are the legs of a right triangle and c is the hypotenuse, to find the hypotenuse of the large right isosceles triangle.

x2 + x2 = c2 2x2 = c2 Take the square root of both sides. c = x times square root of 2

The hypotenuse of the large triangle is x times the square of 2. We then found that the hypotenuse of the large triangle is also equal to the length of the parallelogram. The sides of the large square and the hypotenuse of the small right isosceles triangle equal x. Since the small triangle is isosceles, a2 = b2. We again used the Pythagorean Theorem to find the legs of the small triangle. a2 + b2 = x a2 and b2 = x divided by the square root of 2 The sides of the small square equal the legs of the small triangle. Finally, we have found all the lengths of the polygons in terms of x. See the picture below for an illustration.

We formed the quadrilateral below from two sets of tangrams. It appeared to be square but we weren't sure. We saw that DC has a length of 3x. AD has a length of 2x + x divided by the square root of 2. Since x divided by the square root of 2 is less than x, AD is less than DC. Therefore, quadrilateral ABCD is a rectangle.

Proof #2

This proof involves the areas of figures formed from N sets of tangrams.

We assigned the length of one side of the square formed from one tangram 2x. Therefore if one had two squares each made from one set of tangrams side by side to form what clearly is a rectangle, the area would be 8x2. If one were to be able to form a square from two sets of tangrams the area would still be 8x2, making the sides of the square 2x times the square root of 2. Looking at the values for the lengths of the sides of all the polygons, it seems mathematically possible for one to form a square from two sets of tangrams, because there are several lengths that have square roots of two in them. It seems impossible, however, to form squares from three, five, or seven (any other number besides two that is not a perfect square) sets of tangrams because there are no square root of those numbers in the polygon side lengths. If a square can be made from two sets of tangrams, other numbers to consider are 8, 18, and 32. These numbers are possible values for N because if there are two sets per square and this pair is multiplied by a perfect square, that number of sets forms a figure that is in fact a square. It took us a bit longer, but we figured out a way to form a square from two sets of tangrams. The picture is shown below.

We decided that two sets was a very special case. This was because not only could you make a rectangle, but a square and another figure which we think is very curious. This figure is made from two sets, has an area of 8x2( the area necessary for the figure to have if it were made from two sets),and has sides that are lengths of 3x. The difference is that there is an empty space in the exact middle of the figure that has an area of x2. This space is equivalent to the large square polygon. If this figure was filled in the area would be 9x2, but minus the area of x2 in the middle, the area is 8x2. We think that this figure fits the definition of a square, but not a square region, as the other figures were. We have just begun to study plane regions in class.

*Conclusion*

The set of numbers that can be values of N are those numbers which are perfect squares. This is true because one can make a square from one set. Since one can make a square from two sets of tangrams, every perfect square doubled can be additional values for N. Our original hypothesis which was any number which is not a perfect square could not be a value for N. We found that we had to add to our original hypothesis because of the two sets case. Two sets is a special case because of the square root of two in the sides of the polygons. This is the reason two sets work. Thanks!

COMMENTS: This group found the right answers, looked at them carefully, and provided great pictures. They show how they expanded on their original hypothesis once they got going, and they said why they think other numbers wouldn't work. Great job of showing the questions they posed and how they attacked them - it's important to include the ones that don't work out (like the rectangle bit in the last paragraph of Proof #1).

Excellent idea to find the lengths of all the sides of the pieces right off the bat, and well done. This helps a lot later on when they talk about working from the possible areas. In the section Proof #2, they mention their square with a hole in the middle, and say that they're going to call this a square, but not a square region, so it should count. A very clear description of the problem, and a solution that many folks wouldn't see.

And it's always good to state a conclusion to wrap things up. Theirs sums up their process, and is absolutely correct. Great job!

First things first. The Tangram set that we used did not look like the one you supplied. However, afterward we tried your set and we got the same answers for both. This is a picture of what we used.

The first thing we found was that any number that was a perfect square worked. This is easy to prove because, the area would be side 1 multiplied by side 2. Since each side of a square has the same length, it is simpler to say the area is side 1 multiplied by side 1, or side 1 squared. If you used a perfect square of tangrams, you would have the same number of tangrams on one side of the square as you would on the other. The following picture is an example.

Therefore, our first solution set for this problem includes <1,4,9,16,25,36,49,64,81,100,.....and all other numbers that are perfect squares>.

However , while we were doing this, we found that there were other numbers that worked, too. The following picture illustrates what I mean.

In that picture, it shows how two tangrams can be used to make a perfect square. The following picture illustrates how eight or eighteen tangrams can be put together (AAP) to form a perfect square.

Therefore, our other solution set includes <2,8,18,32,50,72,98,......and all other numbers that are two multiplied by a perfect square>.

Our conclusion is that if using N sets of Tangrams and trying to make a perfect square, N must either be a perfect square or twice a perfect square.

COMMENTS: Very nice job, and great pictures (some day you will all see them!). Great first paragraph - start things off stating what you used to do the problem - and even better to check and see if the other tangrams worked. Despite the clarity of the pictures, it's often good to provide a little more written explanation about why a certain thing is true. Pictures can lie sometimes! Jerry and Don provided a very clear understanding of the problem, and gave us a good look at the ground they covered in stating a good, correct, conclusion. Their answers exhibits a lot of careful work. However, they never say why other numbers won't work. Why can't you make a square with 3, or 5, or 17 tangrams? It's important to explain why things won't work as much as why they will. Otherwise, excellent job.

Dear Annie,

Our names are Jenn Strong and Kate Rusbasan and we are tenth grade
Honors Geometry students at Shaler Area High School. Here is our Project
of the Month answer.

For the first two methods we have found, each individual tangram is a left as a square and used as a building block for making the larger square. For our purposes we'll call them blocks.

*Method 1*: Any perfect square number of tangrams.
The larger square would have sides of the square root of the perfect square
number of squares. More simply, consider the perfect square 9. The square
root is 3, so the larger square would be 3 rows of three blocks, and the
rest filled in. This would include the perfect square of 1 tangram.

*Method 2*: Any number of tangram perfectly divisible by 4.
In this method, a hollow square would be formed. And example is 12 tangrams.
The larger square would be 4 blocks across and 4 blocks down, going all the
way around to make the outline of a square. (The outline being one block
wide), 20 tangrams would have 6 across and 6 down.

In the following method, each original tangram is divided in half along the diagonal. The two isosceles right triangles formed are then used as building blocks, here called half-blocks. This would require the use of certain types of tangrams, of course. It would only work with tangrams in which there is a definite line of division along one of the diagonals already existing in the tangram. The actual configuration of each half-block is not important here.

*Method 3*: 2 times any perfect square number of tangrams.
This can most easily be seen with 2 tangrams. If the original tangram were 1
unit on each side, the diagonal would be the square root of two units long.
So, we now have 4 isosceles right triangles with sides of 1,1, and the square
root of two. Rearranging them so that the vertexes of the right angles of
each triangle are around on a central point, a new, larger square is make
with diagonals each 2 units long and sides the square root of two units long.

COMMENTS: Very clear descriptions, and the conclusion is right on the nose. Interesting observation in method 2 about the multiples of 4. [Aside: this geometric interpretation of the numbers which are multiples of four may be useful to see the following number theoretic fact: A perfect square is always divisible by 0 or 1 modulo 4....that is, a perfect square is either divisible by four itself, or you can subtract one and get a number divisible by 4. Can you see how to use their method above to prove this? Thus, for example, since 100 is divisible by four, we can immediately say that 834756238703 is not a perfect square! Do you see why?] Good characterization of tangram sets for method 3. But again, they needed to talk about why other number _don't_ work.

Next project || Previous project || Table of Contents || Forum Home Page

2 July 1995