IMPOSSIBLE GREEK MATH PROBLEMS
Back in the good old days, when men were men and nobody watched talk shows(because they didn't want to), there weren't any fancy calculators or some marked rulers. Back then in the ancient Greece we relied soley on compass and a straight edge(an unmarked ruler). These are called the "classical" tools that Euclid(famous geek, eh, I mean Greek mathematician) favoured when he wrote Elements and it was also considered "noble" to be able to solve a geometrical problem without using anything else then these tools and of course a pen(if you really want to get the right feeling; slip away to the nearest beach and draw in the sand).
These restrictions made some problems rather difficult to solve and some got notorious:
1. Trisection of an angle: Find a general method(a method that works on all angles) to split an angle into three smaller angles.
2. Squaring a circle: You start out with a circle and then you have to draw a square that has the same area as the circle.
3. Doubling a cube: From a given cube create another cube with twice the volume.

These problems created quite a hassle for mathematicians since they seemed unsolvable, with the restrictions, but no one could prove that they where unsolvable until 19th century. Then it was proven that they where unsolvable and I will show the reasons briefly further down on this page. However they are based on the fact that when using only a compass and a straight edge there's a limit to the kind of mathematical operations you can do. Translated into "ordinary" math(math where you don't draw but only use numbers and operators such as +, -), these operations are:
Addition, subtraction, multiplication, division and square roots.
I'll now go through some ways of performing these operations:

Addition is pretty straightforward. If you have to lengths 'y' and 'x' you simply put 'y' and 'x' next to each other:
pic1) The red line has the length of 'y' and the blue line has the length of 'x'.

And there's a number of ways of doing that e.g. draw a circle with the radii 'x', mark where the centre of this circle is, and draw another circle with the radii 'y' on the outline of your circle with radii 'x' and mark it's centre also(it's always easier to draw a circle that has a certain length, such as 'x', as it's radii instead of it's diameter since then it's just to accommodate your compass to the length wherever you have it and then the compass will draw a circle with the radii of your length). Then you can erase every part of the 'y' radii circle that is inside the 'x' radii circle, to make it clearer:
pic2a) The red circle has the radii length of 'x' and the blue line has the radii of 'y'. The left dot is the centre of the red circle and the right dot is the centre of the blue circle.
pic2b) Same as pic2a, only now we've erased the part of the blue circle that where in the red circle to make it clearer.

Then draw a line that goes through both of the centres of the circles and intersects the outlines of both circles. Then the space between the centre of the circle with the diameter 'x' and the outline of the circle with the diameter 'y' is 'x' + 'y'.
pic3) Since the red circle had the radii of 'x', means that the part of the purple line that's on the right side of it's centre, and still inside the circles outline, has to have the length of 'x'. The same goes for the blue circle and since we only have the right side of it left, we conclude that everything on the right side of the red circles centre is 'y' + 'x'.

Subtraction is also pretty straightforward. Just place both the lengths you want to subtract on each other and then erase the part of the line that they have incommon, which leaves you with part of the longer line that was left when erasing the smaller line from it:
pic4) The blue is 'x' and the red line is 'y'. We put them on top of each other. The green line shows their mutual length and the red line shows how much longer the red line is from the blue line. If you erase the green, you erase the length of 'x' from the length of 'y'. So the little red line that is left is 'y' - 'x'

Multiplication is a bit more difficult and actually requires us to have some sort of length unit(like metres or inches) despite that we aren't allowed to have any marked ruler to measure them. So how do we do it then?
Well, it's pretty simple: you draw an arbitrary length, I will call this length 1 l.u(length unit), and then all other lines you draw can be said to be measured in this length unit. E.g if you just draw a line and measures it to find out that it's 2cm long and then call this line 1 l.u(that is; you say that all other lines will be measured after this line), then all other lengths have to be measured in 2:s of centimetres. Ex: Using this measuring system, let's say you draw a line that is 8cm(now your using the metre system), in your system this line would be 4 l.u. That is because 1 l.u is 2cm so since there's 2cm for every l.u, you have to divide the amount of centimetres in two to get the corresponding amount of your own measurement system: length units.
Now, before we get on with the actual multiplication there's yet another technicality that we have to go through: Multiplication is communicative, this means that if you have two factors, A and B, that you want to multiply with each other you can write it as: A * B or B * A. It gives the same answer so it doesn't matter which one comes first. And in the same way as it doesn't matter which factor comes on which side of the multiplication operator, it doesn't matter which length('x' or 'y') that you draw in what place. The places I've put them on aren't important and you may interchange their positions.
Finally let's get on with multiplication:
You have two lengths; 'x' and 'y' that you want to multiply. First draw an arbitrarily long line and then make use of the stuff stated above and call this line 1 l.u. Then draw the line 'x' as a continuation of this line:
pic5) The white line is 1 l.u long and the blue line is 'x' long.

Then draw the line 'y' starting from the tip of the 1 l.u. line and continuing to the right. There has to be an angle where 'y' and the 1 l.u. line meets, however this angle is completely arbitrary(note that you wont have to measure all other angles after this angle):
pic6) The red line is 'y' long. The angle labelled with a green A is the arbitrary angle between 'y' and the 1 l.u. line.
Connect the right tip of the 1 l.u. line and the 'y' line, let's call this line 'l1':
pic7) Yellow line connects the tips of the 'y' and the 1 l.u. line.
Next, draw a line that is parallel to 'l1' that touches the tip of the 'x' line, let's call this line 'l2':
pic8) Both the yellow lines are parallel to each other. The left one is 'l1' and the right one is 'l2'.
Draw a line from the right tip of 'y' that continues in the same direction as 'y'. Continue this line until it reaches 'l2', then the line between the 'l1' and 'l2' is as long as 'x' * 'y' using our new measuring system of 1 l.u.
pic9) The purple line is as long as 'y' * 'x' measured in your new l.u system.

So the basic scheme is:
pic10)

Division works in a very similar fashion: You still have to draw an arbitrarily long line and call it 1 l.u. but now we note that division is non communicative. That is 'x'/'y' and 'y'/'x' gives two different answers. So in order for you to carryout the same division as I will below, you have to draw the 'x' and 'y' lines on the same place as I do.

I will do the division 'y'/'x'.
Draw a line with the length 'x'. Then segment this line into two arbitrarily large pieces and call one piece 1 l.u. and the other piece we'll still call 'x'(though it isn't as long as 'x' anymore):
pic11) First the blue line is 'x' long. Then we segment it into two pieces: the white line which we say is 1 l.u. long and the other blue line which isn't as long as 'x' anymore but whom we'll still refer to as 'x'(just to confuse you more, dear reader).
Then we do just as in pic5 and draw the line 'y' from the tip of the line 1 l.u. and at an arbitrary angle at the intersection point:
pic12) Same as pic6.
Now we'll connect the tip of the line 'x' and the line 'y' and we'll call this line 'l1':
pic13) The yellow line is 'l1'.
Then we draw a line that is parallel to 'l1' and that goes through the point where the 1 l.u. line and the 'x' line meets. We'll call this 'l2':
pic14) The left, yellow line is 'l2'. The right one is 'l1'.
Now the segment of 'y' that is between the line 'l2' and the arbitrary angle has a length equal to 'y'/'x', using the l.u. measuring system:
pic15) The purple line equals 'y'/'x' using your own 1 l.u. based measuring system.
So the complete scheme would look like this:
pic)16

Now we come to taking the square root out of something. Once again we need to have 1 l.u. although taking the square root is pretty simple:

Draw an arbitrarily long line and call it 1 l.u. and beside it draw the line 'x', as you did in pic5. Then draw a circle around these two lines, this circle will then have the diameter of 'x' + 1 l.u:
pic17) The white line is 1.l.u long and the blue line is 'x' long. The red circle around them have the diameter of both of the lines combined.

Then Draw a line that starts from the section where the 1 l.u line and the 'x' line meets and goes orthogonal to the 'x' and 1 l.u. lines. The part of this line that is between the 'x' and 1 l.u lines and the outline of the circle is the square root of 'x':
pic18) The yellow line is perpendicular to the 'x' and 1 l.u. lines. The part of the yellow line that is inside the circle equals Ö ('x'), using our own measuring system of course.

So the basic scheme is:
pic19)

And now to why they are impossible to solve:
As said you can only perform addition, subtraction, multiplication, division and square roots using only a compass and a straightedge.
For problem # 1 and 3 this turns out to be the Achilles heel since both of them requires you to solve a cubic equation(meaning an equation involving something raised to the third power e.g: x3) that is taking the cubic root.
It's easy to see why this is required for #3 since it in fact involves cubes and therefore cubic equations simply come in by themselves.
For #1 it might be a bit harder to picture, and it's a bit more involving, but it has to do with trigonometric rules n' stuff. You use cosine when calculating the degree of the three angles and then get a cubic equation(for those interested the equation is: 4cos3(q/3) - 3cos(q/3) - cos(q) = 0 . Where q is the original angle.).
We can solve square roots, as stated above, and this allows us to solve any other roots that are dividable by 2 such as the quadratic root(E.g. take the square root out of 2 and then take the square root out of the answer, and you've got the quadratic root for 2), but for any other roots, such as the cubic(third) root we can't solve the equation with only these tools.

# 2 is a bit trickier but based on the same facts: Using only the above operators we can only get answers that are: rational and irrational.
Rational numbers are numbers that can be expressed by a whole number(1, 2, 3 et cetera) and numbers that can be expressed by division of whole numbers(4/2, 11/9 et cetera). Rational numbers can also be said to be numbers where you can figure out the whole sequence without writing it up. E.g: 11/9 = 1.222222... followed by an infinite number of twos. Now, you don't have to write all the twos to figure out that there in fact are an infinite number of twos, the sequence of numbers follows a specific pattern.
Irrational numbers however, are numbers where you can't figure out the whole sequence of numbers. The square root of 2 is such a number; it is approximately 1,414213... but it doesn't continue like: 1,414213414213414213414213... instead it's something like :1, 414213562373. Irrational numbers can't be expressed as division of whole numbers, but only as a root of a number.

Then there's so called transcendental numbers. These are, just as the irrational numbers, numbers that has got a sequence of numbers that you can't know without writing them up. However transcendental numbers, unlike irrational numbers, can't be written as a root of any number. And since they can't be written as a root of any number(or any of the other operators we could perform with a straightedge and a compass), we can't construct them with a straight edge and a compass.
p is a transcendental number. Since it is involved in the area equation of a circle(p * r2 where r is the radii) we basically have to solve this equation before we can construct a square with the same area as the circle(Though not literarily solve it), and since we are limited by our tool; we can't solve it.

So, here we have three problems that existed for more then a millennia before finally being proven impossible to solve about a hundred years ago. What does that tell you?
That you of course have to try to solve them! So turn up the geek factor, lock yourself in a closet for a holiday or so, and start drawing circles! If not for anything else you might get better at geometry.
The two other documents below in the menu is my own failed attempts to solve two of the problems, for inspiration...

Disclaimer: The Greeks could solve the problems without these restrictions and they did have other tools the the compass and the straight edge, such as a marked ruler.




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