Crescent Orange

By realignments of the cutting planes to be parallel, we can preserve the same surface areas, as shown below:

According to Archimedes' Hat-Box Theorem , if the lateral surface areas of the spherical sections (peels) are equal, then their height must be the same for some $h$ .

We know that the blue radius = 12 and red radius = 9. Then let the desired green radius be $R$ . By Pythagorean theorem, we can then obtain the equations:

$R = 2h +\sqrt{R^2 - 12^2}$

$R - h = \sqrt{R^2 - 9^2}$

Hence, $R = 2(R - \sqrt{R^2 - 9^2}) + \sqrt{R^2 - 12^2}$ .

$2\sqrt{R^2 - 9^2} - \sqrt{R^2 - 12^2} = R$

$4(R^2 - 9^2) -4\sqrt{(R^2 - 9^2)(R^2 - 12^2)} + (R^2 -12^2) = R^2$

$(R^2 - 117)^2 = (R^2 - 9^2)(R^2-12^2)$

$9R^2 = 117^2 - 108^2 = 225$

Thus, $R = 15$ .

As a result, the diameter of the whole orange = $\boxed{30}$ .

The particular orientation of the slants (as long as the green remains encompassed by the yellow) doesn't matter so we can let the base of the green slant be parallel to the yellow slant. The solution to the napkin ring problem implies that such slant peel areas, if equal, have the same height (the height is the perpendicular distance between the circular base and the highest point of the slant). Therefore, these two slants have equal height. Let the radius be $r$ and let their heights be $a$ . Then, from Pythagoras' Theorem we have: (for the green slant) $(r-a)^2 + 9^2 = r^2$ and (for the yellow slant) $(r-2a)^2 + 12^2 = r^2$ . Which implies $r^2 = (r-a)^2 + 9^2 = (r-2a)^2 + 12^2 \implies 144-81 = (r-a)^2-(r-2a)^2 \implies 63 = 3\cdot 3 \cdot 7 = a(2r-3a)$ . Checking for integer solutions, we find that $a = 3$ and $r = 15$ satisfy both equations. Therefore the diameter of the orange is $15\cdot 2 = 30$ .