Two Problems About Squares, or Maybe Just One

It started with this problem, a decade ago when I student walked into class and said, "My Dad got this from a friend in the states and wondered if you could solve it, since  he got stuck."  It was lunch and, as often would happen, other students stopped in and joined in our exploration.  I don't even remember how the problem was posed, but by the end of the lunch, the students and I had found a good half dozen solutions to the problem and extensions of the problem.  I'll only tell you here that we began by discovering that the square of the distances connecting opposite vertices of the two squares were equal.  Then in forty + minutes of student led thinking that would make any teacher proud, we discovered much more, only some of which I remember.

So flash forward a decade, and I'm sliding down my twitter feed and there is a problem similar to the one below.  A random point in a square  and segments connecting it to the midpoints of the four sides of the square.  They gave areas of three of the four quadrilaterals shown below, and asked for the fourth.  I instinctively knew the answer, but I wasn't sure why I knew it.  I had never seen the problem, or had I.

The solution boiled down to realizing that the two opposing sections (red and blue or the two whites) would each add up to 1/2 the total area.  It's some pretty simple geometry, add dotted lines from the point to each corner of the square and think on it a while.  The triangle formed in the red area with it's base on the upper leg of the square and the triangle in the blue area with its base on the bottom side of the square are treated as corresponding, and together their heights add up to the length of a side of the square,s.  Sine each has a base of s/2 the formula for their combined area would be \( \frac{1}{2} ( \frac{s}{2}) s \)   = \( \frac{s^2}{4} \)   doing the same thing with the two corresponding triangles with bases on the left and right side gives you the rest to prove that these two opposite sections contain 1/2 the total area, and therefore, are equal to the remaining area.  

 My mind kept searching for why I knew the solution instantly, and all I could think about was the problem above with two squares. One about sums of squares, one about areas, but were they connected.  

I set my head to wondering, and within a few minutes I realized that if you draw the square connecting the midpoints of the sides of the square in the area problem, they also form a square.  And one of the quick deductions my students jumped to from the first problem above, was that the size of the square didn't matter, and so if you let it degenerate to a single point, the sum of the squares of segments connecting opposite vertices were equal.   So in this area problem, the sum of the squares of the segments connecting opposite midpoints were also equal.  But would the areas work in the original problem with two squares?

I created an area version of the four areas with a square cut out of a square.  Now I wondered if, in our rushed lunch, we had discovered a slightly altered version of this diagram and discovered that the four outside areas were in equal pairs.  Our version would have had the corners of the inner square connecting the corners of the outer square.  Using the square formed by the midpoints would reduce it to the former problem, and if that problem dividing the areas equally, adding back the four isosceles right triangles cut off would not create an imbalance in the areas since all four triangles outside the midpoints were congruent.  

I set about convincing myself that all the area properties true of the single point problem about areas were true in the two square part, and all the length properties of the two square problem were true about the area problem.   

I will add that the single point, or the inner square, do not have to be "inside" the square, as long as appropriate allowances are made for negative lengths and areas.  In fact, they don't have to be parallel to the sides of the original square, or even in the same plane as the original square, (although I did constrain my problems to offsets in parallel planes.)  

So the proof of the inner square area problem.  By drawing a diagonal line from the corner of the inner square to the corner of the outer square in two opposite pentagons, you divide each pentagon into a trapezoid and a triangle.  I used S for the length of the larger square, and 1 for the inner, (any other variable would have worked),  and so the area to be divided by the four pentagons was s² -1.  We need to prove that the sum of two opposite pentagons if half this amount.  

Once more the triangles in each opposite region can be summed and their area will be \( \frac {(s)(s-1)}{4}\)  .  The trapezoids on opposite corners would have a combined area of \( \frac{1}{2} ( \frac{s}{2}+1)(s-1) \) .  A little algebra adding those together gives one-half of s²-1.