### Building Tetrahedra

Can you make a tetrahedron whose faces all have the same perimeter?

A 1 metre cube has one face on the ground and one face against a wall. A 4 metre ladder leans against the wall and just touches the cube. How high is the top of the ladder above the ground?

Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load.

# Perfectly Square

##### Age 14 to 16Challenge Level

Look at the end for a neat short cut

The solution below is based upon the one submitted by Anna of Parkside school. I liked the explanation of how Anna arrrived at the factorisation.

The youngest person to send in a solution was Sairah of Kings Park School, Lurgan and Barinder sent an excellent solution with lots of clear explanation.

Just three of the large number of solutions to this problem. Well done to you all!

If $x_n = (n+1)^2 +(n+2)^2 + [(n + 1)(n + 2)]^2$ then \begin{eqnarray}x_n &=& n^2 + 2n + 1 +n^2 + 4n + 4 + [n^2 + 3n + 2]^2 \\ &=& 2n^2 + 6n + 5 + n^4 + 3n^3 + 2n^2 + 3n^3 + 9n^2 + 6n +2n^2 + 6n + 4 \\ &=& n^4 + 6n^3 + 15n^2 + 18n + 9 \end{eqnarray} If this can be made into a perfect square the perfect square must have $n^2$ as it's highest term to get $n^4$ when you multiply out. The lowest term must be $3$ to give $9$.

The middle term must be some number of $n$ ($an$).

$an\times n^2 \times 2$ must give $6n^3$

As the only parts of the square which can multiply together to give $n^3$ are $n$ and $n^2$. $an\times n^2 \times 2 = 6n^3$, so $2an=6n$, so $a=3$.

So the perfect square must be: $(n^2+3n+3)^2$.

To check, this multiplies out to: $n^4+3n^3+3n^2+3n^3+9n^2+9n+3n^2+9n+9 =n^4+6n^3+15n^2+18n+9$.

$(n^2+3n+3)^2$ is a perfect square, and it is the solution for $x_n$.

Now you don't actually need to worry about the $n$ value matching the n of the left hand side . . . .

You just need to establish that $(n)^2 +(n+1)^2 + [n(n + 1)]^2$ is always a square of something

Can you follow that through, expanding and simplifying until you get: $n^4 + 2n^3 + 3n^2 + 2n + 1$?

I'll leave you to decide what 'square' that is, but you might think of something inspired just from looking at the algebra, or maybe calculate the first line from the problem (it should come to $49$) and take it from there .

Look out for this kind of 'simpler version' of the algebra when you are problem solving, it isn't always easy to spot but it can save you some effort when you do notice the possibility.