You may also like

problem icon

Just Rolling Round

P is a point on the circumference of a circle radius r which rolls, without slipping, inside a circle of radius 2r. What is the locus of P?

problem icon

Coke Machine

The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design. Coins inserted into the machine slide down a chute into the machine and a drink is duly released. How many more revolutions does the foreign coin make over the 50 pence piece going down the chute? N.B. A 50 pence piece is a 7 sided polygon ABCDEFG with rounded edges, obtained by replacing AB with arc centred at E and radius EA; replacing BC with arc centred at F radius FB ...etc..

problem icon

Rotating Triangle

What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?

Steel Cables

Stage: 4 Challenge Level: Challenge Level:1

If you are a teacher, click here for a version of the problem suitable for classroom use, together with supporting materials. Otherwise, read on...


Cables can be made stronger by compacting them together in a hexagonal formation.

Here is a 'size 5' cable made up of 61 strands:

cable cross section, one cable in the middle with four rings of cables around it

How many strands are needed for a size 10 cable?

How many for a size n cable?

Can you justify your answer?


Once you've had a go at the problem, click below to see how some 15 year old students worked on it.
Can you explain their reasoning?

Group 1

student's picture of hexagon split into three quadrilaterals, two 5*5 rhombuses and a 4*4 rhombus table showing dimensions of the three quadrilaterals and total T for a size 2 up to 10 cable and a size n cable T=n^2+n^2-n+n^2-2n+1 = 3n^2-3n+1 or 3n(n-1) +1

Group 2

student's picture of cable with horizontal arrows showing row lengths n, n+1, n+2 up to 2n-1 in the middle and then decreasing back down to n student's method is adding up pairs of rows to make n pairs of (3n-1), with the 2n-1 row counted twice, giving a total 3n(n-1) +1

Group 3

student's picture of hexagon split into six triangles student's method is to use the formula for the (n-1)th triangular number n(n-1)/2, multiply by six, and then add 1 for the cable in the centre

Group 4

student's picture of hexagon showing four rings and one cable in the centre 1, 6*1, 6*2, 6*3, 6*4 ... 6*(n-1). We noticed that the area of each ring followed this pattern. To find the total we needed to add the area of each ring. 1 + 6*1 + 6*2 + 6*3 ... 6(n-1)=1+6(1+2+3+...+(n-1) 1+6(n(n-1)/2) 1+3n(n-1)


Which of the four approaches makes the most sense to you?
What do you like about your favourite approach?

Can you think of any other approaches?




Notes and Background

Hexagonal packings are often chosen for strength or efficiency. To read more about packings, take a look at the Plus articles Mathematical Mysteries: Kepler's Conjecture and Newton and the Kissing Problem