Before announcing the thousands (or less!) winners of the Name That Mathematician Challenge, I came across a problem about dissecting a square ABCD with lines PQ and RS which are parallel to the sides of the square. (see diagram).

Naturally, I decided to make it into a deeper investigation. Students and/or readers will be asked to find the maximum value of the product of the areas of either pair of non-adjacent rectangles formed. There are many approaches here, one of which uses the famous Arithmetic Mean-Geometric Inequality. As usual you will work from the particular to the general, beginning with a specific value for the sides of ABCD.

STUDENT/READER INVESTIGATION - PART I

The given conditions about the diagram are given above.

For Part I, we will assume each side of the square has length 4.

(1) (Particular) If AP = 3 and RC = 2, determine the product of the areas of APTS and RTQC. Do the same for the other pair of non-adjacent rectangles formed. Do you believe this product is the maximum possible as we vary the positions of segments PQ and RS?

(2) (General) Show that the product of the areas of either pair of non-adjacent rectangles formed is less than or equal to 16. For example the product of the areas of APTS and RTQC is ≤ 16.

Notes:

(1) Do you think many students would guess what the configuration would be for the maximum product to occur? Is proving the conjecture much more difficult?

(2) The challenge here is to find an effective use of variables to denote the segments. There are many possibilities, some much more efficient than others.

(3) I will add additional parts to this challenge after receiving comments on Part I. How would you generalize this result further? More interestingly, there is a way to prove Part I using the AM-GM Inequality?

## Thursday, November 29, 2007

### Just Another Square Problem? A Means to an End...

Posted by Dave Marain at 5:58 AM

Labels: AM-GM Inequality, geometry, investigations, optimization, squares

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## 2 comments:

Hi Dave,

You could add a 2(a): Show that the product of the areas of each pair of non-adjacent rectangles is equal. I find this result quite interesting, and not at all visually evident, especially if you draw a highly skewed figure wherein the two parallel lines are close to the edges.

TC

Nice, tc!

That is definitely one of the surprises here. Students should experiment with specific values for the segments to discover this!

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