64 = 65

Task 155 ... Years 4 - 10

Summary

64 does not equal 65! Or does it? This jigsaw puzzle suggests that it does The paradox is one of a genre of 'missing square puzzles'. If you begin with an 8 x 8 square and cut the four pieces shown, then their total area must be 64 square units. But when placed in a 5 x 13 frame, they appear to be fit and therefore show an area of 65 square units. Where did the extra square come from?

The explanation of this extra square paradox can be attempted at many year levels. However, noticing that the key numbers involved (5, 8 and 13) are successive terms of the Fibonacci sequence is the starting point for an extended investigation.

 

Materials

  • 4 cut out pieces

Content

  • Area of rectangle
  • Area of parallelogram
  • Visualisation of shapes
  • Pythagoras Theorem
  • Number Patterns (Fibonacci sequence)
  • Gradient
  • Trigonometry
  • Sine Rule
64 = 65

Iceberg

A task is the tip of a learning iceberg. There is always more to a task than is recorded on the card.
   

There are two levels of explanation for the missing square - 'by eye' and by calculation. By eye, if the 'diagonal' of the rectangle is closely examined, there is a long thin parallelogram in the middle ... and you can guess the area of this parallelogram.

If necessary, to help students line up the rectangle a 5 x 13 frame matching the puzzle pieces is supplied. However, be careful when printing from PDF that the Page Scaling option is set to none. An alternative explanation for the extra square calls on mathematician's tools such as Pythagoras' Theorem and Trigonometry. The dissection in the task is based on an 8 x 8 square of unit squares like this:

The diagonals in the dissection are the key. They become four diagonal lengths when cut and one of each makes up the diagonal of the supposed 13 x 5 rectangle. So, if the rectangle actually exists, the sum of the two diagonal lengths above should equal the diagonal of a 13 x 5 rectangle.

Sum of diagonals above = Square Root (2 + 5) + Square Root (3 + 8)
= SQR(29) + SQR(73)
= 5.385 + 8.544
= 13.929

Diagonal of 13 x 5 rectangle = Square Root(13 + 5)
= SQR(194)
= 13.928

Close! Do we conclude therefore that the rectangle can be made because if only two decimal places are used, the lengths would appear to be the same?

We have to turn to trigonometry, or ratio, for more information. There are four pieces in the above dissection, but they come in pairs - 2 trapeziums and 2 triangles. One from each pair makes half of the supposed 13 x 5 rectangle like this:

If the hypotenuse of the shaded area is one straight line, then the slope of the section contributed by the triangle and the slope of the section contributed by the trapezium must be the same.
Triangle
Rise/Run = 3/8 = 0.375
Triangle section of trapezium
Rise/Run = 2/5 = 0.4
Almost the same, but just different enough to prove that the 13 x 5 rectangle can only exist if there is a long thin parallelogram 'up the diagonal'.
At some levels, teachers might expect students to go further and prove that the area of this parallelogram is indeed one square unit.

Extensions

A major extension of the problem, which can turn it into an extended investigation, comes from noticing a link in the key numbers of the problem. The square is 8 x 8 and the rectangle is 13 x 5. The key numbers are 5, 8 and 13. A student might notice that the first two sum to the third. It is not much data, but it might lead to investigating whether 3 other numbers might exist that could generate this 'extra square phenomenon'.
  • One might consider asking whether ?, 5 and 8 could be the three numbers. The ? would then have to be 3 if the sum of the first two is to equal the third.
  • In the original problem it was the middle number that was the square. The rectangle was formed from the other two.
  • If 3, 5 and 8 are to work, the square would be 25 square units and the rectangle 24. Hmmm, one square different. It might be possible to do a construction to create a new 'extra square phenomenon'.
  • If this worked, the next smaller set of numbers would be ?, 3, 5, implying the set 2, 3 and 5.
  • In this case the square would be 9 square units and the rectangle 10. But this time the rectangle is bigger than the square. In the previous set of numbers it was smaller, and in the original problem it was bigger. Hmmm.
  • Continuing the exploration leads to the sets 1, 2 and 3 and 1, 1 and 2. Of course we could also investigate larger sets than the original 5, 8 and 13; such as 8, 13 and 21.
Looking back over the numbers generated in this way, ie: 1, 1, 2, 3, 5, 8, 13, 21..., identifies Fibonacci's sequence ... and therein lies an additional wealth of possibilities.

It is a property of the Fibonacci sequence that for any three consecutive terms of the pattern, the product of the first and third differs from the square of the middle term by 1. For example:

  • 2, 3, 5 ... 5 x 2 = 10 ... 3 x 3 = 9
  • 3, 5, 8 ... 3 x 8 = 24 ... 5 x 5 = 25
  • 5, 8, 13 ... 5 x 13 = 65 ... 8 x 8 = 64
  • 8, 13, 21 ... 8 x 21 = 168 ... 13 x 13 = 169
  • 13, 21, 34 ... 13 x 34= 442 ... 21 x 21 = 441
Students could continue this pattern for several more terms or write a spreadsheet to continue for say the next 100 terms.

An interesting pattern within this is that the square of the middle term alternates between being bigger by 1 and smaller by 1; a connection perhaps with the 'disappearing' unit of area. Could it be therefore that a dissection like that of the 8 x 8 square in the puzzle can be reproduced for any three numbers in the sequence? A hypothesis worthy of testing.

At least, to visually highlight the parallelogram aspect of the puzzle, try the dissection on a 5 x 5 square and 'prove' that 25 = 24. The parallelogram (overlapping in this case) is clearly visible.

Whole Class Investigation

Tasks are an invitation for two students to work like a mathematician. Tasks can also be modified to become whole class investigations which model how a mathematician works.
   

This is easy to state and easy to start as a whole class lesson. Each pair needs graph paper and scissors and you need an image of the dissection for all to see.

All I want you to do to start the lesson today is carefully cut an 8 x 8 square from the graph paper.
...
Now I would like you to draw these lines, again carefully, and cut along them to make four pieces from the square.
Check that the students know where the pieces came from:
...And what was the area of the original square?
and record this fact.
Next I want you to rearrange the four pieces to make a rectangle.
...
And the area of this rectangle is?
Record this fact too ... and wait for the discussion to begin!

For more ideas and discussion about this investigation, open a new browser tab (or page) and visit Maths300 Lesson 132, 64 = 65 which includes using Trigonometry and the Sine Rule to calculate the area of the parallelogram.

Is it in Maths With Attitude?

Maths With Attitude is a set of hands-on learning kits available from Years 3-10 which structure the use of tasks and whole class investigations into a week by week planner.
   

The 64 = 65 task is an integral part of:

  • MWA Chance & Measurement Years 5 & 6
  • MWA Chance & Measurement Years 9 & 10
The 64 = 65 lesson is not in any MWA kit. However it can be used to enrich the Chance & Measurement kits at Years 5/6 and 9/10.

Green Line
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