# Maths item of the month

### May 2015

A couple of MEI Conference problems

The following problems appeared in MEI Conference sessions in 2014:

**1×1=2**

You are told that, when rounded to the nearest whole number, *x* and *y* are both 1. What is the probability that, to the nearest whole number, *xy* = 2 ?

**Paper folding a parabola**

Prove that the following steps will produce a parabola:

- Fold a piece of A4 paper in half long-ways (to create the y-axis) and draw this in with a pen.
- Mark a point on the fold a couple of inches from the bottom edge.
- Fold the bottom edge so that it goes through the point and is perpendicular to the vertical fold (to create the x-axis) and mark this with a pen.
- Make repeated folds so that the bottom edge goes through the point – these can be at any angle.

**MEI Conference 2015**

For details of the 2015 MEI Conference see: conference.mei.org.uk/

### April 2015

Consecutive Fibonacci Squares

The first few Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34 ... (each number is the sum of the previous two numbers in the sequence and the first two numbers are both 1).

The sums of the squares of some consecutive Fibonacci numbers are give below:

1^{2} + 1^{2} = 2

3^{2} + 5^{2} = 34

13^{2} + 21^{2} = 610

Is the sum of the squares of consecutive Fibonacci numbers always a Fibonacci number?

### March 2015

4 points, 2 lengths

In how many ways can you arrange 4 distinct points in the plane so there are exactly two different distances amongst the 6 pairs?

e.g. if the four points are at the corners of a square then the four sides are the same length and the two diagonals are the same length.

### February 2015

1, 2, 3, 4

Find two quadratic functions f(*x*), g(*x*) so the equation f(g(*x*)) = 0 has the four roots *x* = 1, 2, 3, 4.

Is it possible to find three quadratic functions f(*x*), g(*x*), h(*x*) so the equation f(g(h(*x*))) = 0 has the eight roots *x* = 1, 2, 3, 4, 5, 6, 7, 8?

### January 2015

Happy 2015: A Triple of Triples

2015 is the product of 3 distinct primes: 5×13×31

2014 and 2013 are also the product 3 distinct primes.

Can you find a smaller triple (*n*, *n*+1, *n*+2) where *n*, *n*+1 and *n*+2 are all the product of 3 distinct primes?

Are there any quadruples (*n*, *n*+1, *n*+2, *n*+3) where *n*, *n*+1, *n*+2 and *n*+3 are all the product of 3 distinct primes?