We have first to prove the Pizza theorem
If we cut a disk (a pizza...) from any interior point by any 4 lines 45 degrees apart, the total area of even numbered pieces equals the total area of odd numbered pieces.
The lengths of segments built by two secant chords at 90° satisfy
a² + b² + c² + d² = 4R²
M midpoint of CD and N midpoint of AB.
In triangle OMD : OD² = OM² + MD²
MD = CD/2 = (PC + PD)/2
OM = NP = NB-BP = (PA + PB)/2 - PB = (PA - PB)/2
Hence R² = OD² = ((PA - PB)/2)² + ((PC + PD)/2)² and expanding
R² = (PA² + PB² + PC² + PD²)/4 + (PC.PD - PA.PB)/2
The power of P is PC.PD = PA.PB
Hence PA² + PB² + PC² + PD² = 4R²
Let's then prove the Pizza theorem.
The total area of blue parts is, without precisely defining functions
a(θ), b(θ), c(θ), d(θ) giving the ray lengths as a function of θ :
S1 = ∫0φ (a²(θ)/2 + b²(θ)/2 + c²(θ)/2 + d²(θ)/2) dθ = ∫0φ4R²/2 dθ = 2R²φ
Similarily, the total area of white pieces is S2 = 2R²(π/2-φ)
When φ = π/4, these two areas are equal. QED.
If two disks overlap, any point P inside the common part then simultaneously shares each of the
two disks into equal areas. This happens with the pizza, the tomato toping and anchovy toping
which are simultaneously equally shared.
The equal sharing is done "automatically" as long as the any number of different areas (topings) have a common part.
Concerning the egg, there is no common part with the anchovy toping, but just draw one of the cuts through the center of the egg will share the egg into two equal parts.
Suppose now there are two concentric disks with radius R and R+dR, the annulus is also equally shared, hence when dR tends toward 0, the circumference (the pizza edge) is also equally shared.
The last problem is because of sweet peppers...
The center of cuts being in the anchovy part, a ray has to go through the egg's center,
and another or the same vector has to go through the sweet peppers area's center.
Of course because of the symetric role of anchovy-egg-pepper, we could as well choose the cutting point inside egg or sweet pepper, the two other centers being on rays.
The locus of cut center P is from where we see the two other centers A and B under a kπ/4 angle, That is the set of AB line, circle with diameter AB and two circles "at 45° " in A and B. If this locus goes through the third area, the problem can be solved. Without any further constraints, we had better choose the cut center P in the largest area !
A proof without words
With a suitable dissection of pieces, the proof becomes obvious by symetries,
each person getting one and only one part of each colour.
M is midpoint of AE, N midpoint of CG, P' symetric of P with respect to M, P" symetric of P w.r.t N, and drawing some parallels to the original cuts.
This construction works if CG<AE and PC<PG, and we always can reach this case by symetry/rotations. Else some parts have "negative" area.
An interactive Java applet.