Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
Investigate how this pattern of squares continues. You could
measure lengths, areas and angles.
Can you work out the area of the inner square and give an
explanation of how you did it?
Congratulations Andrei Lazanu, age: 12 from
School: No. 205, Bucharest, Romania on this very good solution.
I found more solutions, and I shall describe the way I found
them. I'm not sure that these are all.
I observed that if I remove one (or all 8) piece(s) in the
corners, and only them without adjacent ones, the total area does
I consider the surface area of a small square as a unit of
I removed all eight corners, leaving the total area
I removed the central cube of the top surface obtaining an
increase of the surface area with four units.
I removed one cube from the middle of an edge at the top (any of
the four remaining) and I arrived at a figure with ten cubes less
then the original one but with the same surface area.
I observed that the top surface can be replaced with any of the
other five faces of the cube, and for each face in the last move,
there are four possibilities. So, for this class there are 24
A drawing of the remained figure is sketched below, using two
slightly different points of view. I obtained it using
I removed 7 cubes from the top surface so that the remainingd
cubes are in the opposite corners, on the diagonal, obtaining a
decrease of four units.
I removed the central cube, placed in the centre of the original
figure, the centre of the middle surface, obtaining the same total
surface as at the beginning.
I removed any two cubes of the bottom surface placed in the
corners: either on the same side, or on the diagonal, either on the
same diagonal as the corresponding ones of the top surface or on
the other diagonal. In all cases I obtained the same area as before
removing the ten cubes.
I counted 6 distinct configurations, that must be multiplied by
6: the pairs top-bottom, left-right, and front-rear, as well as
bottom-top. right-left and rear front.
I represent below one of these cases:
I removed the central cube of the middle surface obtaining an
increase of the surface area with eight units (in respect to the
I removed the four cubes placed in the middle of each side of
the top surface, obtaining the same total surface as at the
I removed the four cubes placed in the corners of the bottom
surface obtaining the same total area.
For this category there is only one distinct solution, that must
be multiplied, as before, by 6.
I removed the 3 cubes from the centre of the big cube (the
central column) obtaining an increase of the surface area with 10
I removed a column (3 cubes) from one (any) corner of the big
cube obtaining an increase with altogether 8 units of the surface
I removed another column, placed in the middle of a side,
adjacent to the removed corner, obtaining the same surface area as
at the beginning.
I removed a corner cube, not adjacent to those removed.
As example is sketched below: