Descending without braking: where's the sweet spot?
You are going at constant speed when the driving force from gravity Fg is equal to the drag (air resistance) Fd plus friction (rolling resistance) Ff:

Fg = Fd + Ff

When coasting down a hill, the driving force is the component of the gravitational force parallel to the road:

Fg = m g sin(a)

Here, a is the slope (angle with the horizontal), m is the mass of bike and rider, and g is the gravitational constant. Slopes on roads are often measured in percent, and for small angles sin(a) is equal to the slope in percent. e.g. for a slope of 10% (100m descent on 1km of road) sin(a) = 0.1.

Drag is proportional to the square of the velocity v, the projected frontal area A and a drag coefficient cd, and the density of air rho:

Fd = 1/2 rho v^2 cd A

The drag coefficient cd depends on the shape of the object - streamlined bodies have a low cd, but for a bicycle with rider I find approximately cd = 1 in Wikipedia. I'd estimate the frontal area of a cyclist to about 0.4m^2.

Friction (rolling resistance) is proportional to a coefficient cr and the normal force (the weight component perpendicular to the road) Fn = m g cos(a):

Ff = cr m g cos(a)

The coefficient of rolling resistance cr will depend on the bike (lubrication, tire pressure, surface roughness etc.). For car tires on concrete, Wikipedia gives cr=0.01, so the Ff would be about 10N. That's about a tenth of the driving force on a 10% slope (see above), but on a rough road surface it might well be much more.

So we end up with 

m g sin(a) = Fg = Fd + Ff = 1/2 rho v^2 cd A + cr m g cos(a)

Solving this for v gives:

v^2 = 2 m g (sin(a) - cr cos(a) ) / (rho cd A)

Putting in g=10m/s^2, rho=1.2kg/m^3, cd=1, A=0.4m^2, and m=100kg (nice round number for rider and bike), I get

v = sqrt( sin(a)-cr cos(a) ) * 64 m/s

The rolling resistance cr essentially is like reducing the slope somewhat, so a slope of 10%=0.1 with cr=0.01 is the same as a slope of 9% on a perfectly rolling bike. So, for a slope of 10%, the sqrt becomes 0.3, and I get a speed of 19m/s = 70km/h

It's not unrealistic, but perhaps a bit faster than I would have expected. However, there are simplified assumptions, friction and rolling resistance is probably higher and cd and A have a big effect, they all depend on the bike and the rider.

From the drag equation you can see why the racing position is with the head down. This position decreases the frontal area A considerable, compared to an upright position, and it makes the body somewhat more streamlined (reduces cd). As the drag goes with the square of the velocity, this gets more important at higher speed.