### All in the Mind

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?

### Rotating Triangle

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

### Instant Insanity

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.

# Intersecting Circles

##### Stage: 3 Challenge Level:

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.

 Amount of circles intersecting Intersecting points 1 0 2 2 3 6 4 12 5 20

I then discovered a pattern that applied to all of these intersections

1x0=0

2x1=2

3x2=6

4x3=12

5x4=20

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, Oliver.

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 follows:

1 circle can't overlap anything, so number of crossings is 0.

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, giving 20.

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

 Sum = 0 + 2 + 4 + 6 + ... Sum = 198 + 196 + 194 + 196 + ... 2*Sum = 198 + 198 + 198 + 198 + ... = 198 x 100

So the sum = 1/2 x 198 x 100 = 9900, giving the maximum number of crossings for 100 circles.