Geschosskopf, I still hold that you are wrong. Let me explain it again.
Assuming no irreversible processes happen (like atmospheric braking) then spaceflight is *fully* reversible. There is no time arrow defined -- if time is reversed, we couldn't tell (as we can for example for thermodynamic processes like gas escaping from container into space since reverse process is HIGHLY unlikely), since the dynamics would be exactly described by the same physics.
This means that considering going from different APs to same PE is *exactly* equivalent of going from PE to different APs. Of course this ignores atmosphere, but since you are going to the same predicted PE, it ignores atmosphere anyway. So let's assume that Kerbin has no atmosphere for a moment, and just track how much of the trajectory is "below 70km" or any other set altitude. Then my previous claims hold true. The higher velocity at given fixed PE, the higher the AP, but at the same time less curve and therefore trajectory always above the lower-velocity trajectory.
We have 2 trajectories (both are after any burning you did at AP, so both are ballistic trajectories from that point): A) from AP 12Mm to PE 37km, B) from AP 1.5Mm to Pe 37km
This is exactly equivalent to trajectories: C) from PE 37km to AP 12Mm, D) from PE 37km to AP 1.5Mm. It should be obvious to anyone playing KSP that trajectories A and C have the same velocity at periapsis, and that velocity (Va=Vc) is higher than the periapsis velocity of trajectories B and D (Vb=Vd<Va=Vc) (can't cheat conservation of energy without burning engines, which I assume you don't do except at AP to adjust PE). If your velocity is higher, then the curvature of the trajectory (at a given altitude) MUST be smaller. This is because the gravitational force is proportional only to masses and the distance from the body that you are interacting with. And the mass of the object itself cancels out when calculating acceleration, so it only depends on mass of the planet and distance form the planet center -> altitude. Given higher velocity, the curvature must be smaller = larger curvature radius, as described by r = V^2/a (this is simple relation for circular motion, but if we consider local curvature then it works as well), since a = acceleration from gravity which is the same for both A/C and B/D trajectories.
While it is obviously possible to draw an ellipse of higher eccentrity (and higher AP) with same PE, that is also narrower, such ellipse is not a physical trajectory and will not have Kerbin as one of focal points (which is always true for trajectories in point gravitational field, as calculated in KSP).
In the below picture, the low AP orbit (actually lowest possible, since it's circular) is blue. The high AP orbit is NOT the red one that indeed would spend more time in atmosphere, but the green one, which is always above blue one, except at PE. Black points are supposed focal points of the ellipse. Note that the red one does NOT have focal point at the center of Kerbin (cyan circle), since PE is always *closest* point to the focal point which must be (according to physics) the body with gravity field (so by definition PE is closest to focal point). An even higher AP ellipse would be again wider around kerbin and always above green trajectory.
[IMG]http://i67.tinypic.com/24o9b0o.png[/IMG]