Friday, March 28, 2008

Classic AMC Contest Square Dissection Problem and more...

Recognize this diagram from a famous math contest problem which I first saw many years ago on an old AHSME contest (now known as AMC)? We'll start with this one and then modify it, creating variations on the basic theme. Finally, we will ask our readers/students to generalize the result algebraically.

In the diagram at the left, nothing is labeled, so we will describe it verbally and hope it will make sense.
We start with a square and dissect it by drawing 4 segments, each connecting a vertex to a midpoint of a side.

Explain why the area of the shaded region is one-fifth of the area of the original square.

(1) This is a wonderful exercise to develop spatial reasoning and to demonstrate a visual approach to a geometry problem when dimensions are not given. Of course, one could use an algebraic or numerical approach if one chooses.
(2) Students who 'see' the jigsaw puzzle approach of rearranging the pieces rarely consider what assumptions are being made. To make the problem even more meaningful, the instructor could ask why the shaded region is, in fact, a square.
(3) Simpler versions of this often appear on the SATs.


This time, both diagonals are drawn. The additional two segments join the midpoint of the bottom side to the midpoints of two other sides.

(a) The red shaded region (does it have to be a square?) is not one-fifth of the original square. What fractional part is it?
(b) The total shaded area is what part of the original square?


This time the smaller segments divide the sides into a 1:2 ratio. The figure is not drawn to scale. The 3 segments on the base are supposed to be equal!

(a) The blue shaded region (is it a square?) is now what fractional part of the original square?
(b) The total shaded area is now what part of the original square?

Use the diagram from Variation 2. Assume the original square has a side length of 1 unit. If the smaller segments divide the sides of the square into an x:(1-x) ratio, do parts (a) and (b) again, expressing your results in terms of x. What restrictions on x make sense here? Make sure your expressions agree with the results above.


Eric Jablow said...

Each of the large triangles is ¼ of the square. So, all 4 large triangles gives you the area of the square. But that area is the full square without the central region, but with the four smaller triangles duplicated. So, the inner region is 4 times the size of the small triangles.

Now, the intermediate triangle is 4 times the size of the small triangle, by similarity. So, the large triangle is 5 times the size of the small triangle. The algebra is clear.

Dave Marain said...

Your approach is my favorite. I would classify it as abstract reasoning using ratios. Some students seem to have an innate ability to do this, others need experiences like this type of problem to develop it.

Instead of working with the 3 types of triangles you mentioned, I thought of it slightly differently. Each large triangle is 1/4 of the whole. The area of each intermediate right triangle is 4/5 of each large triangle or 1/5 of the whole (using ratios of areas of similar figures). The original square can be dissected into 4 of these intermediate triangles WITHOUT overlap. Therefore, the shaded region represents the remaining one-fifth!
Unfortunately, many students do not get enough practice with the fundamental theorem on ratios of areas of similar figures. That's why students generally prefer rearranging and combining pieces without much regard for WHY these pieces fit.

Joshua Zucker said...

OK, Dave, here's your new favorite approach! :)

tessellate the plane with copies of the square you drew. Observe that the various funny-shaped regions meet up to make copies of the small square in the middle. Also observe that there are 5 small squares in the tesselation for each large square. QED.

I don't see quite how to apply this approach to the other variations though.

Dave Marain said...

Nice! Guess that means I have many favorites or none in particular! Actually, the method I suggested to Eric earlier was also arrived at by one of my strong SAT students. She immediately highlighted the 4 non-overlapping right triangle which surrounded the small square. She wasn't quite sure how to finish it from there and I explained the ratio method. 2-3 others saw how to rearrange the pieces to form 5 apparently congruent regions.

Interestingly, I began by posing a simpler version of the diagram, drawing only 2 of the 4 segments and asked the group why they were parallel. Only one student considered proving that a parallelogram was formed by showing that it had 2 pairs of opposite sides congruent. I then asked why the shaded region was a square. One of the stronger students suggested a coordinate approach, first showing that opposite reciprocal slopes were formed. Impressive, although there are several alternate routes.

I should have thought of tessellating that figure with its rotational symmetry. I don't believe this will work for the variations. Try a more traditional approach. However, if anyone can devise an original method, it's you!

Joshua Zucker said...

Thanks for the compliment!

For the parallelogram, I'd use the "if one pair of opposite sides are parallel and congruent, then the quadrilateral is a rectangle" trick.

Then when all four segments are there I would use rotational symmetry to prove it's a square: if it has 90 degree rotational symmetry, it must be a square, right?

Ned Rosen said...

Here are some generalizations of the original problem. They are fairly straightforward but I don't have a really slick argument a la symmetry or tessellation. In each case determine the ratio of the area of the inner quadrilateral to the area of the outer one.

A. Use the x to (1- x) segment ratio instead of 0.5 to 0.5 on an outer 1 x 1 square.

B. Same as (A) except start with an outer rectangle a x b instead of a square. Here x (0< x <1) is the portion of the side at which the inner line hits the side, so. e.g. one of the large right triangles has legs a and xb, and the other has legs b and xa.

C. Same a (B) but start with an outer parallelogram of sides a and b and tilt angle alpha (you can define alpha any way you like) and x again represents the portion of the side at which the inner lines hit the sides.

Of course (C) subsumes the others and the cool thing about the result is ___________.

Dave, I really enjoy your blog. I'm putting together a problem solving course which emphasizes General I. Zation and I thought you might enjoy these, as you seem to be a fan of the General as well.

Dave Marain said...

Nice generalizations. Great problems like the original one will live on in perpetuity in just this way! Yes, like you, I've always had a warm spot in my heart for that "General."

Thanks for the kind words. My goal is to bring these kinds of problems to a wider population of students than might currently exist.

I'd be interested in reading more about this course you're developing. Is this at the secondary or undergraduate level? What type of student will be taking it? Other than your own creations, what kinds of resources will you be using for inspiration. If you'd like you can send me a personal email at dmarain AT GEE-MAIL DOT COM.

Thanks again for visiting.