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Paul Panzer
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UPDATE: I've added a large $N$ solution for multiples of 3 that slightly betters OP's solution at $3\times (\frac N 3 - 1)^2 + 9$, see end of this post.

Just to show that @humn is not the only one capable of wasting eyewatering amounts of pizza here are

15

tiny but equal pieces of pizza made using 6 cuts.

enter image description here
Due to symmetries there are only tree kinds of pieces; equalizing those costs 2 degrees of freedom which we can afford: Let $P$ be the point in the upper center where the blue and orange triangles meet. Then we can adjust the distance of P to the center and the angle between the lines meeting at $P$.

$N = 3n$ solution:

Example $N=12$, 36 slices: enter image description here

Just to show that @humn is not the only one capable of wasting eyewatering amounts of pizza here are

15

tiny but equal pieces of pizza.

enter image description here
Due to symmetries there are only tree kinds of pieces; equalizing those costs 2 degrees of freedom which we can afford: Let $P$ be the point in the upper center where the blue and orange triangles meet. Then we can adjust the distance of P to the center and the angle between the lines meeting at $P$.

UPDATE: I've added a large $N$ solution for multiples of 3 that slightly betters OP's solution at $3\times (\frac N 3 - 1)^2 + 9$, see end of this post.

Just to show that @humn is not the only one capable of wasting eyewatering amounts of pizza here are

15

tiny but equal pieces of pizza made using 6 cuts.

enter image description here
Due to symmetries there are only tree kinds of pieces; equalizing those costs 2 degrees of freedom which we can afford: Let $P$ be the point in the upper center where the blue and orange triangles meet. Then we can adjust the distance of P to the center and the angle between the lines meeting at $P$.

$N = 3n$ solution:

Example $N=12$, 36 slices: enter image description here

Source Link
Paul Panzer
  • 10.4k
  • 19
  • 49

Just to show that @humn is not the only one capable of wasting eyewatering amounts of pizza here are

15

tiny but equal pieces of pizza.

enter image description here
Due to symmetries there are only tree kinds of pieces; equalizing those costs 2 degrees of freedom which we can afford: Let $P$ be the point in the upper center where the blue and orange triangles meet. Then we can adjust the distance of P to the center and the angle between the lines meeting at $P$.