Guys, would you just confirm that I have understood this properly? In your models, I assume that you are taking the scalar average of the distances to the Earth of (the Sun), Mercury, Venus and Mars over time - and then you get the results that Mercury has the lowest value (with result close to 1AU). I imagine that if you were to take a vector average of the distances, all would come out at 1AU: the average vector distance from the Sun of each planet is zero (by symmetry).
So the reason that any of the planets has an average scalar distance above 1AU is that contributions to distance when the planet is to the side of the Earth-Sun line do not cancel out (contributions to distance when the planet is along the Earth-Sun line do cancel out).
Some of the simplest things do perplex!
I am using standard distance calculation from the X and Y legs of a right triangle and dist = the Hypotenuse:
X^2 + y^2 = dist^2
so, dist = SQR(X^2 + Y^2)
X = difference of x's of Earth and other planet
Y = difference of y's between Earth and another planet
The planets start in alignment in circular orbits around the Sun (origin 0, 0).
They are plotted each day on their orbit according to an Earth day's change in angle calculated by:
2*PI / (Planet's Year in Earth Days)
eg Mercury's change in angle in 1 Earth day = 2*PI / 88 (days) because in 88 Earth days Mercury will have made one complete orbit.
so MercuryX = SunX + MercuryOrbitRadius * (COS(EarthDays * MercuryAngleChangeInDay))
MercuryY = SunY + MercuryOrbitRadius * (SIN(EarthDays * MercuryAngleChangeInDay))
Earth x, y calc'd the same way, SunX, SunY are 0 because origin of Simplified Solar System with Circular Orbits.
DistEarthMerc = SQR( (EarthX - MercX)^2 + (EarthY -MercY)^2)
The distance are accumulated each day, divide by days to get average.
Also each day the distances are compared and the one closest to Earth gets counted as closest that day.
The scalar units are Millions of KM.
x and y coordinates are vectors so "average vector" makes no sense to me.