At what slope will the best bikes keep a constant speed?
As you might expect, the exact down slope you would need for a full answer will depend on how much drag you and your bicycle produce. If you are very aerodynamically efficient and there are few losses through the bearings and tires, the slope can be shallower. If you create a lot of drag either via aerodynamic inefficiency or mechanical and rolling resistance, the slope will need to be steeper. You can calculate the slope yourself if you know the equations for motion on a bicycle and can make reasonable guesses about the parameter values. The equations are given in one of the answers to this bicycles.stackexchange question. 

In short, you are looking for the slope so that while coasting at equilibrium speed the power will be zero; that is, at equilibrium speed you are neither accelerating or decelerating (so the kinetic energy is zero) and you are coasting so the power is zero. That is, 

Watts = 0 = (Crr + slope) * kg * g * v + 0.5 * rho * CdA * v^3  

where Crr is the coefficient of rolling resistance, kg is the total mass of the rider and bike, g is the gravity constant (9.8 meters/sec^2), rho is the air density in kg/m^3, CdA is the "drag area" (Cd is the coefficient of aerodynamic drag and A is frontal area, so their product Cd*A is also known as the drag area), and v is the speed in meters/sec. 

Slope for a given speed

The equation above can be re-arranged to solve for slope:

slope = (0.5*rhoCdAv^2)/(kg*g) - Crr

In your question, 80 km/h is 22.2 m/s and 50 km/h is 13.9 m/s. 

Air density, rho, at sea level is typically near 1.2 kg/m^3

Crr for a high quality tire on an asphalt road is typically around .0045.

CdA for a rider on a conventional bike in the drops is typically in the range of 0.24 to 0.3 m^2; for a rider on a TT bike in the aerobars, CdA is typically in the range of 0.2 to 0.28 m^2; for a rider in a streamlined fully-faired recumbent bicycle the CdA has been measured at under .02 m^2. 

Let's say you and your bike together weigh a total of 80 kg, rho=1.2, Crr=.0045, and CdA = .24. What is the slope so that you would reach an equilibrium speed of 50 km/h (13.9 m/s) while coasting? That would be  

slope = (0.5*rhoCdAv^2)/(kg*g) - Crr or
slope = (0.5*1.2*0.24*13.9^2)/(80*9.8) - .0045

Thus, the downslope is -0.04, or -4%. 

Suppose you wanted to know how steep the slope would need to be to attain an equilibrium speed of 80 km/h with the same mass, rho, and CdA? That would be -0.095, or -9.5%.

Suppose you wanted to know how steep the slope would need to be to attain an equilibrium speed of 80 km/h in the streamlined Varna Diablo? If one had access to the power and speed data for a speed run, and one also knew the road profile, the air density on the day, the wind speed, and the total mass of vehicle and rider, one can estimate the Crr and CdA given the method described in the linked bicycles.stackexchange answer. I have done that: Crr and CdA appear to be in the range of .0045 and .0175, as noted above. Battle Mountain, NV, where the world human powered vehicle speed runs are attempted, is at an elevation of around 1375 meters (4500 ft) above sea level, but if you could make the run at sea level with a total mass of 90 kg, you could attain an equilibrium coasting speed of 80 km/hr at a down slope of -.011, or -1.1%. 

Terminal Speed

A related question is, given a slope, what would the equilibrium speed be, assuming the slope is long enough to reach maximum speed? This is also known as the "terminal speed." In this case, we can re-arrange the equations of motion to determine the speed in meters per second, v. That equation is

v (in m/s) =  sqrt( -2*(Crr+slope)massg/(rho*CdA))

For example, if Crr = .0045, slope = -3% = -.03, mass=80, g=9.8, rho=1.2, and CdA = 0.24, then v = 11.78 m/s. To convert to km/h, multiply v by 3.6; v = 11.78 m/s is equivalent to 42.4 km/h or 26.4 mph.