You may also like

problem icon

Adding in Rows

List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?

problem icon

Magic Sums and Products

How to build your own magic squares.

problem icon

Weekly Problem 2 - 2013

Weekly Problem 2 -2013

Perimeter Expressions

Stage: 3 Challenge Level: Challenge Level:1

Daniel from Staplehurst Primary School sent us a very clear solution to this problem:

Alison's shape with a perimeter of 8a + 6b combined the largest and smallest rectangles, with the smaller rectangle's side of length a against one of the larger rectangle's sides.

I have made the largest possible perimeter, 12a + 12b, using all the pieces because I have put the shorter side of each shape against the one before it.

Click here to see the examples and all the working out.
 
Rajeev from Haberdashers' Aske's Boys' School managed to improve on Daniel's solution for the largest perimeter. Take a look here at how he combined the rectangles. 

Daniel also considered one of the later questions:

I would like to see Charlie's second shape with a perimeter of 7a + 4b because I think he has made a mistake: it is impossible to find a shape with an odd number of "a"s.

It is impossible to get an odd number of "a" or "b" on the perimeter because every shape has an even number of "a"s or "b"s. When you place the a against an existing shape you are taking away an a from the perimeter and then adding one back again as well as adding 2b and vice versa, this is true with all the rectangles that you add on. Therefore you always get an even number of "a" or "b".
 
Benjamin from Wilson's School agreed:
 
I noticed that whatever way you arrange the two rectangles, the perimeter always equals:
(the longest length + the longest width) x 2