### 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?

### Areas and Ratios

What is the area of the quadrilateral APOQ? Working on the building blocks will give you some insights that may help you to work it out.

# Square Product

##### Stage: 4 Short Challenge Level:

We make use of two key facts. First, we have the factorisation $n^2 -1 = (n-1)(n+1)$. Second, when a square number is factorised, each prime factor appears an even number of times.

Now $2^2 -1 = 1\times 3$.
We next get a prime factor $3$ with $4^2-1= 3\times 5$.
We next get a prime factor $5$ with $6^2-1= 5\times 7$.
We next get a prime factor $7$ with $8^2-1= 7\times 9$.

As $9=3^2$, it does not require any further factors. Hence we need $n\geq 8$. Checking the product $n=8$, we get\eqalign{ (2^2 -1)(3^2 -1)(4^2 -1)(5^2 -1)(6^2 -1)(7^2 -1)(8^2 -1) &= 1\times 3\times 2\times4\times3\times5\times4\times6\times5\times7\times6\times8\times7\times9\cr &= 2\times8 \times 3 \times 3\times4\times4\times5\times5\times6\times6\times7\times7\times3\times3\cr &=4\times4\times3\times3\times4\times4\times5\times5\times6\times6\times7\times7\times3\times3\cr &=(4\times3\times4\times5\times6\times7\times3)^2}
So in fact $n=8$ is sufficient, and is thus the minimum.

This problem is taken from the UKMT Mathematical Challenges.