All in the Mind

Imagine you are suspending a cube from one vertex and allowing it to hang freely. What shape does the surface of the water make around the cube?

Instant Insanity

Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.

Is There a Theorem?

Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?

Cubes Within Cubes Revisited

Age 11 to 14Challenge Level

Yanqing from Devonport High School for Girls sent us a complete solution to this problem:

We would need to use $26$ red cubes and $98$ blue cubes.

For the $3 \times 3 \times 3$ cube, there is one yellow cube, so there are $27 - 1 = 26$ red cubes.

It is the same for the blue cubes: there are $5 \times 5 \times 5$ cubes in total, and $3 \times 3 \times 3$ of them are not blue.

For the next size up, there would be $7 \times 7 \times 7$ cubes in total, and $5 \times 5 \times 5$ that aren't green.

So there would be $343 - 125 = 218$ green cubes.

The amount you need for each layer added onto a cube with edges of $x$ will be $(x+2)^3 - x^3$.

$(x+2)^3$ is the cube you will get when you add the next layer, and $x^3$ is the number of cubes already there. (This is Martha's method)

With Emma's method, the top and bottom of the layer would both contain $(x+2)^2$ cubes.

Each added face has $x^2$ cubes, and the four vertical columns $4x$ cubes altogether.

So the expression for the total number of cubes needed would be $2(x+2)^2 + 4x^2 + 4x$.

With Charlie's method, the top and bottom of the layer would both contain $x^2$ cubes.

Each column has $x + 2$ cubes, and there would be $4(x + 1)$ columns.

So the expression for the total number of cubes needed would be $2x^2 + 4(x+1)(x+2) = 6x^2 + 12x + 8$.

Martha's method: \begin{eqnarray} (x+2)^3 - x^3 &=& x^3 + 6x^2 + 12x + 8 - x^3\\ &=&6x^2 + 12x + 8 \end{eqnarray} Emma's method: $$2(x+2)^2 + 4x^2 + 4x = 6x^2 + 12x + 8$$ Charlie's method: $$2x^2 + 4(x+1)(x+2) = 6x^2 + 12x + 8$$

These are all equivalent.

Another way of thinking about the problem is:

First you put a square with side of $x$ on each face ($6x^2$ cubes), then a column of $x$ cubes on each edge ($12x$ cubes), and finally one cube on each corner ($8$ cubes).

This would give us our $6x^2 + 12x + 8$ expression, which is what you get when you simplify the other expressions.

Well done Yanqing.