The Blue Square Enigma

The diagram below (Figure 33.1) was sent to me by someone with an anagrammatic name. Knowing what nasties can emerge from Pandora’s cyber-box, I never open email attachments from unfamiliar sources. However, I know, also, what a rat smells like, so, for once, I felt confident in taking a risk.

Figure 33.1: The pieces fit, with or without the little blue square. How come?

Copyright unidentified. Fair Dealing asserted.

OK, what do we have? Two identical right-angled triangles, each 13 units by 5. ‘A’ contains four constituent shapes: two dissimilar right-angled triangles (red and dark green) and two L-shaped hexagons (orange and light green), all apparently comprising a precise geometric tessellation.

Now look at ‘B’. Within an identical space are the same four shapes along with an additional one, namely the blue, single-unit square. How can it possibly fit? Surely, the total area of any shape is equal to the sum of its parts. This cannot be true in both cases. Therein lies the rub; and it is, I admit, a pretty neat puzzle.

To find the solution, we must remember that the total area of ‘A’ is equal to that of ‘B’.

Area of a right-angled triangle = half the product of its perpendicular sides = ½(13 x 5) = 32½ units.

Next, we must calculate the area of each constituent shape.

Area of red triangle = ½(8 x 3) = 12 units

Area of dark-green triangle = ½(5 x 2) = 5 units

The hexagons are easy: we can just count the unit squares in each. Thus:

Area of orange hexagon = 7 units

Area of light-green hexagon = 8 units

Therefore, the area of the sum of the parts of ‘A’ is 32 (12 + 5 + 7 + 8) units. As for ‘B’, that of the sum of its parts equals 32 plus one, i.e. 33 units, taking into account the blue square.

Hence, neither in the case of ‘A’ nor ‘B’ does the total (32½ units) equal the sum of the parts. For ‘A’, the sum (32 units) amounts to half a unit less than its total, whereas for ‘B’, the sum (33 units) is half a unit more. In other words, the shapes of ‘A’ are slightly too small to fit the outer triangle, those of ‘B’ marginally too large. Neither represents an exact tessellation, despite appearing as such in the diagram. The truth is that a half-unit (1.5%) area discrepancy can be imperceptible to the human eye. It is, of course, an optical illusion.

My eyes light up at the sight of a puzzle (Figure 33.2). Family and close friends know that, which helped me to point the finger. Besides, my anonymous challenger, your anagram was clumsy. I know who you are. Nice try, though!

Figure 33.2: Feliz navidad a todos. Voy a estar en contacto pronto.

Copyright © 2011 Paul Spradbery

Copyright © 2011 Paul Spradbery