I know that's mathematically the right answer, but personally don't consider the middle piece a "slice", since it doesn't have any crust. :)

@Cyclops can't agree more :)

Might be more of a Puzzling.SE or Mathematics.SE question, but I wonder how close you could get to 7 pieces of equal area using this technique? (Never mind the one piece without a crust.)

@DarrelHoffman Would think 100% perfect, just cut an equilateral triangle in the exact center that is 1/7 the total area - then the other bits must be each 1/6 of the remaining 6/7. Of course, subject to ability to actually make the cuts...

@Joe that's straightforward. But you get 3 triangular and 3 "trapezoidal" slices, so I feel Darrel's question was: will their areas be equivalent with this method?

Hmm, I think my drawing skills are not up to par… you’re right, they’re not equal sixths of the remaining.

@pinckerman: You can make the central triangle as large or as small as you like by moving the cuts inward or outward (without changing the angles). Therefore, by the intermediate value theorem, there ought to be a size which is exactly right.

@Kevin The intermediate value theorem in 7 dimensions? That's a lot more subtle than you make it sound!

@Stef: Move all three lines simultaneously and at the same rate. Now you have a one-dimensional function.

@Kevin A one-dimensional function from which space to which space? And what point in the codomain space corresponds to "all sizes are equal"?

@Kevin How does this prove that all 7 slices have the same area? As pinckerman pointed out, if the center piece is a centered equilateral triangle, then the remaining six pieces are 3 "triangular" and 3 "trapezoidal" slices. By symmetry, the 3 triangular pieces have the same area, and the 3 trapezoidal slices have the same area, but there is no reason why the triangular slices should have the same area as the trapezoidal slices.

@Stef: You're right, I oversimplified it. Let me re-explain it. First, find the central triangle of area pi/7. Now, you can move one of the three vertices along a path about the center such that the triangle's area does not change (because there are non-equilateral triangles whose areas are also pi/7). So you can apply intermediate value theorem a second time to that locus, to get one of the outer components to pi/7. Choose to make the curved triangle whose apex is the point you moved the right size. Now do it two more times to the other two triangles. The trapezoids follow by symmetry.

@kevin there's no way you can get the remaining 6 parts equivalent themselves while having an equilateral triangle inside. The only case when you have this is when all 3 cuts pass through the pizza's center, giving us 6 identical triangular slices. Starting from that, you can enlarge the centered triangle, this makes the 3 triangles to shrink and the 3 "trapezoids" to grow. This never leads to a state where you have 7 equal parts, imho.

@pinckerman: Yes, I realized that, that's why I posted an explanation which specifically involves a non-equilateral triangle.

I still can't figure out how a non-equilateral triangle can do this.