For some time now I've been obsessed with the idea that the logo of the Umbrella corporation can be written in math. And I don't mean the version with simple triangles. I mean with the curves. This is high-level stuff so most of you might not understand it.

first we draw a circle with radius r:
k0 -> x^2 + y^2 = r^2

Then we split it with 4 lines
a-> y= x*tan(22.5)
b-> y= x*tan(67.5)
c-> y= -x*tan(22.5)
d-> y= -x*tan(67.5)

to simplify some calculations we add the following variables
g=r*cos(22.5)
i=g*sqrt(2)

next we draw 8 circles with the same radius, but at different starting points:
k1 -> (x-2*g)^2 + y^2 = r^2
k2 -> (x-i)^2 + (y-i)^2 = r^2
k3 -> x^2 + (y-2*g)^2 = r^2
k4 -> (x+i)^2 + (y-i)^2 = r^2
k5 -> (x+2*g)^2 + y^2 = r^2
k6 -> (x+i)^2 + (y+i)^2 = r^2
k7 -> x^2 + (y+2*g)^2 = r^2
k8 -> (x-i)^2 + (y+i)^2 = r^2

and the result is:
u = k0-k1-k2-k3-k4-k5-k6-k7-k8-a-b-c-d

Or otherwise in words: Draw 1 circle with radius ® of your choice, centered in (0,0) on your coordinate system. Draw 8 lines, half of which only outside the circle because you'll have to erase them. The lines that match the X and Y and their bisectors (I'll call them XYb for short) will later later be fully erased, so keep that in mind. Now draw bisectors between all the current lines. These ones will stay. They're the a,b,c and d. Draw the chords only between the nearest a,b,c and d lines crossing the circle. Know the chords will be deleted later. On the lines X,Y and XYb mark the position that is double the length between (0,0) and the chords. Those 8 positions are the centers of 8 circles with radius R as the first circle. Draw the arcs that pass through the main circle. No need to draw the whole 8 circles just to erase most of them. Now clean up all the parts I said would be deleted and voilà! It's ready for coloring.
To achieve the perfect logo you need either a really steady hand, or advanced drawing software. When doing it by hand mistakes of the size of half a millimeter can result in an ugly picture. Best results then are achieved using only a ruler and a wheel-pen.

There is also a wonderful program called Geogebra that can you can do this on. Make sure you have Java installed first. Then just download from www.geogebra.org and install.
when you run it simply set a radius, for example: "r=3"
and then enter the following lines one by one:
CircularArc[(2 Length[(r cos(22.5), 0)], 0), (r; 22.5 °), (r; 337.5 °)]
CircularArc[(r sqrt(2), -r sqrt(2)), (r; 337.5 °), (r; 292.5 °)]
CircularArc[(0, -2 Length[(r cos(22.5), 0)]), (r; 292.5 °), (r; 247.5 °)]
CircularArc[(-r sqrt(2), -r sqrt(2)), (r; 247.5 °), (r; 202.5 °)]
CircularArc[(-2 Length[(r cos(22.5), 0)], 0), (r; 202.5 °), (r; 157.5 °)]
CircularArc[(-r sqrt(2), r sqrt(2)), (r; 157.5 °), (r; 112.5 °)]
CircularArc[(0, 2 Length[(r cos(22.5), 0)]), (r; 112.5 °), (r; 67.5 °)]
CircularArc[(r sqrt(2), r sqrt(2)), (r; 67.5 °), (r; 22.5 °)]
Segment[(r; 337.5 °), (r; 157.5 °)]
Segment[(r; 247.5 °), (r; 67.5 °)]
Segment[(r; 202.5 °), (r; 22.5 °)]
Segment[(r; 292.5 °), (r; 112.5 °)]


So was the lesson useful or just interesting?