Fermat's Last Theorem

For anyone that has no clue what Fermat's Last Theorem is, here is a 'quick' overview:

Pierre de Fermat was a French Mathematician. Not what we might call the 'best' mathematician, he is probably only really the one to raise this particular problem to peoples attention.

He owned a copy of "Arithmetica", the famous mathematical text written by Diophantus. This book was composed of 130 algebraic problems. One of which consisted of the following:

We are all aware of the Pythagorean Theorem, a^2 + b^2 = c^2, where c is the length of the hypotenuse and a and b are the lengths of the other two sides in a right angles triangle.

With this equation, you can get numerous whole number solutions,  for example:

3^2 + 4^2 = 5^2
9 + 16 = 25

5^2 + 12^2 = 13^2
25 + 144 = 169

These are called Pythagorean Theorems, and there are infinitely many of these, where a, b and c are all whole numbers.

However, in Arithmetica, Diophantus conjectured this:

for a^n + b^n = c^n when n is bigger than 2, is there any whole number solutions?

In this book, Fermat made a lot of marginal notes. In a lot of these, he claimed he had a proof for the problems, however it seems that we have never found any proofs that he did for the problems. All the problems he says he had proofs to, do have proofs. However, they were all solved by other people later on. It is called Fermat's "Last" theorem, because is was the last of the problems that Fermat claimed to have a proof to, to be proven.

With all honesty, I am unable to see how Fermat could have solved this problem. He was an amateur mathematician. Andrew Wiles, the man who had eventually solved the problem dedicated 7 years of his life towards the problem, going to the best schools, colleges and universities, and an amateur mathematician in the 17th century was able to solve it? In the marginal note, Fermat said that "The proof was to large to put in the margin". Andrew Wiles' proof was over 150 pages long. Oh 'really' Fermat? The proof is 'too big' to put in the margin? Additionally, the mathematics used in the proof for the theorem was not really available to Fermat at the time. And I highly doubt there is an easier way to solve the problem, that we are failing to see, when Wiles spent 7 years on the problem.

Although, credit to Fermat, no one would of really cared about the problem as much as they do if Fermat didn't bring it to peoples awareness by pretending he could solve them.

Area & Perimeter Problem

My former maths teacher gave me this problem to think about - here are my thoughts: 

A rectangle with an area of 48 square units, what is the largest perimeter you can have?

It seems that the bigger the difference between the length and the width, the bigger the perimeter. So if the length and width had to be integers, the largest perimeter you could get would be 98 (48 by 1 rectangle). But you could keep halving one side of the rectangle, and doubling the other, and the perimeter would just keep getting bigger and the area would remain the same. Maybe it's worth noting that the perimeter is getting bigger by a value converging to 2, but I have no idea if that has any significance. 

Having a length and a width that are closer to each others values just makes the perimeter smaller and makes it more "squarish". I'm running into problems when I try and explain why a bigger difference in length and width makes the perimeter bigger with solid maths.

if the area = ab
and perimeter = 2a+2b

then ab=48 and b=48/a

2a+2(48/a)
2a+(96/a)= perimeter 

The smaller the value of a the bigger the value of 96/a, and the bigger the value of a the bigger the value of 2a.

This isn't sufficient as you can plug in any value and the same rules still apply. I'm having difficulty explaining myself. 

The smallest perimeter would be obviously be if the shape was a square, making the length of each side √48, leaving the perimeter as 16√3 ≈ 27.71 

As one of the sides lengths "move away" from that value (the other sides length obviously changing with respect to the other, keeping the area as 48 square units) the perimeter gets bigger and bigger. 

Lets just say that a = 12

2a = 24

96/a = 8

Okay, lets increase a by 4, so a = 16

2a = 32

96/a = 6

Note how 2a increases by a much bigger amount, than the amount 96/a decreases by. In fact, 2a increases by an amount 4 times bigger than the amount 96/a decreased by. If it 2a increases faster than 96/a decreases, then surely the perimeter would be getting bigger.

I'm still working on the problem, I will update when I have anything new to say.

Five Triangles - Circle Area Ratio Solution

This is the solution to the problem:

http://fivetriangles.blogspot.co.uk/2013/08/90-circle-area-ratio.html

To solve the problem above, you have to find the radius of each of the circles. To find the radius of the larger circle inside the triangle...

DG=EG=GF. It is the radius of the circle, so lets just say that they equal X.

AC=16 as shown above.

Therefore AD=16-X

AF=AD 

AD=16-X

CB=12

Therefore EB=12-X

EB=BD

BD=12-X

BD+AD=AB=20

12-X+16-X=20

28-2X=20

2X=8

X=4

The radius of the larger circle is 4 units. So the area of it is 4x4xpi= 16pi

To find the radius of the smaller circle: 

To find the radius of the smaller circle, tangent to the triangle, you have to form a sector.

The radius of the sector is 10, as the diameter (BC) is 20.

 AB=BC=10 (The radius of the whole semi-circle) 

AC=16, therefore EC=8 

Sides EC and BC can form the sides of a right angled triangle, along with BE. 

BE^2+EC^2=BC^2

BE^2+64=100

BE^2=36 

BE=6

BF=10 as it is the radius on the whole sector.

BF-BE=EF (diameter of the small circle)

10-6=EF

EF=4

Therefore, if the diameter is 4, the radius must be 2.

If the radius is 2, Area=4pi

The question asks for the ratio of the two circles areas, not just the area of each.

Ratio = Area of the larger circle/Area of the smaller circle

         = 16pi/4pi 

         =4

So the larger circle inside the triangle, is 4 times bigger than the circle tangent to both AC and the semi-circle.