### Pebbles

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?

### Great Squares

Investigate how this pattern of squares continues. You could measure lengths, areas and angles.

### Square Areas

Can you work out the area of the inner square and give an explanation of how you did it?

# Take Ten

##### Stage: 3 Challenge Level:

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 not change.

I consider the surface area of a small square as a unit of surface.

### First class of solutions

I removed all eight corners, leaving the total area unchanged.

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 (equivalent) solutions.

A drawing of the remained figure is sketched below, using two slightly different points of view. I obtained it using Mathematica.

### Second class of solutions

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:

### Third class of solutions

I removed the central cube of the top surface obtaining an increase of the surface area with four units.

I removed the central cube of the middle surface obtaining an increase of the surface area with eight units (in respect to the initial condition).

I removed the four cubes placed in the middle of each side of the top surface, obtaining the same total surface as at the beginning.

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.

### Fourth class of solutions

I removed the 3 cubes from the centre of the big cube (the central column) obtaining an increase of the surface area with 10 units.

I removed a column (3 cubes) from one (any) corner of the big cube obtaining an increase with altogether 8 units of the surface area.

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: