All Squares the Same?

Look at the problem the other way round - if the squares have side $1$ what size of octagon would contain the pattern?

The horizontal width of an octagon holding the green pattern would be $W_G = 3 + \sqrt{2}$ . The horizontal width of an octagon holding the red pattern would be $W_R = \mathrm{cosec}22.5^\circ + 2\cos22.5^\circ$ . Since $W_R = 4.46... > 4.41... = W_G$ , a smaller octagon would hold the green pattern of squares. Thus, if the octagons are to be the same size, the squares in the green pattern will be larger.

Each green square is the largest square in one of the eight isosceles triangle that forms the octagon with the octagon side as its base.

Each red square too is the largest square in one of the eight isosceles triangles with a base as the line joining the mid point of two octagon adjoining sides.
It is similar to the green triangle since in both eight vertex angles of the isosceles triangles add up to 360 degrees.
However as seen from the Figs.
Red base=Sin(3 pi/8) Green base.
So Green base> Red base.
So Green squares are larger than the Red.

My solution is based on comparing the yellow areas of (what I will still call) the green and red configurations, assuming that the squares have unit length.

In the green case the yellow area is made up of

1. The central octagon with unit sides.

2. 8 isosceles triangles, each with two unit sides.

3. 8 little hats sitting on the triangles (and turning them into kites).

In the red case the yellow area is made up of

1. The central octagon with unit sides.

2. 8 isosceles triangles, each with two unit sides.

3. 8 little hats sitting on a unit side of the squares.

The central octagon and the isosceles triangles are identical in both cases, so the whole question can be resolved by deciding if the green-case hats are larger or smaller than the red-case hats.

Well, since the hats are all isosceles triangles, with the octagon's internal angle (135 degrees) at their apex they are certainly similar. So now the question can be settled by deciding whether the green hat base or the red hat base is longer.

The red hat base is one unit long.

Applying the cosine rule to the triangles with sides 1 and included angle $360 - 135 - 90 - 90=45$ degrees gives the green hat base to be $\sqrt{2-\sqrt{2}}=0.765$ (Robert Morewood points out in the comments that you don't even have to do the calculation. The smallest side of a triangle is opposite the smallest angle, so in this case the base, being opposite 45 degrees, is smaller than the other two sides which are each a unit long!)

So for unit squares we see that the green case hats are smaller than the red case hats, and so the yellow area will be less in the green case. And so for unit squares the green configuration is smaller area than the red one.

Finally expand the green configuration so that the boundary is the same size as in the red case to see that the green squares expand to be bigger than the red ones.

Not at all rigorous, but I just imagined rotating the red squares about the centre of the octagon until they were in the same position as the green squares. It seemed clear that they wouldn't quite touch the octagon and were, therefore, smaller than their green equivalents.

Consider a polygon of infinite sides ... a circle. At this point red and green must be equal.

Now consider a 4 sided polygon ... a square.
It is obvious that the green squares would occupy the whole polygon whilst the red ones would occupy half.

By inference, adding more sides to the polygon will tend to zero difference but green will always remain larger.

Solving spatially.

The total green square area is the length of a single side of the inside octagon, squared x 8, simple enough?

The angle between each of the squares starting at the internal octagon on both the red and green is the same as they are at right angles to each octagon line.

As the red squares are closer to the center of the octagon (measured by the height of the triangle at the base of each red square) it is possible that they could be rotated within the octagon. Picture the red squares rotating inside the octagon, the green squares cannot rotate

The only way this is possible is that the inside octagon of the red squares is smaller that the green squares inside octagon and this results in the green squares having a larger area than the red squares as the internal octagon is larger.

So green squares have a larger area

No computer can solve it this way (yet)

JJM

Thought about the math then tried a visual solution. Think about trying to rotate the entire assembly of green squares within the confines of the outer octagon, they cannot be moved. Think about rotating the assembly of red squares to the position held by the green squares. This can be done and in this position it is easy to visualize that the red squares would need to increase in size to match the green squares.

Draw the circumscribed circle of both square sets.

These circles intersect the base of the yellow octagon at two points. For the green configuration, the distance between these intersections is one side of a square. For the red configuration, it is the base of the yellow triangle.

Because the triangle base side is strictly shorter than the square side, as follows from the sum of the 8 inner angles, the intersection points on the red configuration are closer together and the circumscribed circle is smaller.

Hence the red configuration is smaller.

Whilst the green squares fills to the very left of the octagon all the way to the very right and this is not true of the red squares. Although one red square to its opposite red square is close to a 'diagonal' of the octagon, which is a factor of 1/sine(60 degrees) great than the edge of the octagon to the opposite edge which the opposite green squares fill. This is a factor of 2/ squroot 3. The squroot 3 is about 1.73. So 2/squroot 3 = 1.155. The solution depends on which there is sufficient space between the diagonal and where is red square starts. If this seems involved then I think that it is extremely involved. I have calculated that the green squares are bigger by a factor of approximately 1.22 though I think that I will have to do this again tomorrow. Regards, David Ps 'I'll be back'

actually all you have to do is look at the area of the 45 degreed isosceles triangles they form, since total green squares area= (S-(a+8 b+8 d)), where S is the area of large octagon, 8b is area of all the isosceles triangles, and b is the area of the smaller octagon , since both of the smaller octagons is identical, they are also equal in area, so area of red squares = (S-(a+8 c+8 d)), where c is the area of all the isosceles triangles in red that are 135 degree, note that if we do geometry on each of the eight obtuse isosceles of 135 degrees, since both the red and the green case have all 8 of the 45-isoceles triangles, that are identical, both the red and green case have 8 d, is the total area of the 45-isoceles area., which ever one gets the smaller 135 degrees isosceles area, is the one with the larger square are. let s be the side of the square of in the green case, each of the base of the 135-isoceles is 2s^2-2 s^2cos(13), and in the red case, the base of the 135degree-isoceles is s^2, clearly s^2 > s^2*(1-cos(135)), hence the areas of the 135 isosceles red triangles have a larger base, and hence larger area, hence, c>b, so S-(a+8b+8d)> S-(8c+8d), hence green has the larger area in squares

Total angles of octagon is 6(180) = 1080 degrees. Each of the angle of octagon is = 1080/8

= 135 degrees. In the left case if the side of octagon is L and side of each square is x, then

x + 2x tan (pi/8) = L. In the second case, if the side of red squares is y, then y/sin (3pi/8) +

2y/sin (pi/8) = L. Then x = L/[1 + 2 tan(pi/8)] and y = L{sin (pi/8)sin (3pi/8)/[sin (pi/8) + 2

sin (3pi/8)]}. Using a calculator, x is approximately 0.5469….L and y is 0.1585….L. Since

x > y and Green Area is 8x^2 and Red Area is 8y^2, Green Squares is greater.

I thought of the problem by circumscribing the squares with a circle - the squares with the bigger circle will be bigger. The circumscribed circles must be tangent at each corner of each square. The points of tangency also intersect the big octagons along each edge. The further the intersection point is from the center of the edge (equivalently, the closer the intersection point to the vertex), the greater the radius of the circle. The distance between two corners of the different red squares is representative of how far large the circle is, just as the distance between two corners of the same square is for the green. Two corners of green are obviously separated by 1 green side length and 2 corners of different reds are separated by something less than 1 red side length (45 degree vertex of isosceles triangle -> 2sin(22.5 degrees) < 1) At this point I realize I should have explained this differently, but the point is that the cilcumscribed circle is smaller for the reds than for the greens.

A simple solution here; Assuming red and green are same area, i.e. sample edge length, say 1. Then for the green octagon, distance between center and octagon edge is 1+sqrt(2)+0.5==L. However for red octagon L is distance between center and outside edge of small square which is shorter than distance between center and octagon edge. Which means under this assumption red octagon is larger. Now since red and green octagons are same, we conclude that red square edge is shorter.