Imagine you are suspending a cube from one vertex (corner) and
allowing it to hang freely. Now imagine you are lowering it into
water until it is exactly half submerged. What shape does the
surface of the water make around the cube?
What happens to the perimeter of triangle ABC as the two smaller
circles change size and roll around inside the bigger circle?
Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.
Many of you tried to solve this problem
and quite a few succeeded.
Congratulations to David, John, Mary and
Kathryn, all from Hethersett High School. Congratulations also to
Oliver and Anna, both from West Flegg GM Middle School, to Alison
of Wymondham High School, all Norfolk schools, and to Arthur of
Sawston Village College, Cambridgeshire.
There are many ways of solving this problem
and two of my favourites were well illustrated to us by Alison and
Arthur. Congratulations and well done to the both of you.
Alison's solution shows good methodology and
observation skills. Unfortunately, I can't reproduce her solution
from the fax. Oliver's solution is similar to Alison's but not
quite as detailed. It illustrates the idea and is as follows:
To start the question I worked out all intersections of circles
up to five.
I then discovered a pattern that applied to all of these
I discovered that if you times the number of circles by one less
than itself you will find the number of intersections.
Therefore, to find the maximum amount of intersections there
were in 100 circles you multiply 100 by 99 which is 9900.
An expression for this pattern is Cx(C-1) = I, where C is the
number of circles and I the number of intersections. Nice solution,
So far we have no proof that this is the correct general
solution. Very good logic and method skills are demonstrated in
Arthur Stratford' s solution, well done Arthur. His proof
1 circle can't overlap anything, so number of crossings is
Maximum crossings for 2 circles is 2
Maximum for 3 is 6 (as shown in puzzle).
Adding another circle, the best you can do is to cross all the 3
circles already there - 2 crossings each - which adds another 6
giving a total of 12.
For 5 circles, you cross the existing 4 twice each, adding 8,
So the sequence is 0, 2, 6, 12, 20...
To get any term you have to add 0+2+4+6+8+10... for as many
terms as you have circles.
For 100 circles, that's 0 + 2 + 4 + ... + 19 8
So the sum = 1/2 x 198 x 100 = 9900, giving the maximum number
of crossings for 100 circles.