Pizza Slicing Time!

It doesn't matter which slice is which.

Choose the first slice. It can be made in any of 3 ways. 1/4 of the way, 1/2 way, or 3/4 of the way.

If the first was made at the 1/4 mark, there are two choices for slicing the remaining 3/4 into thirds.

If the first was made at 1/2 way, there are two choices for which to slice into the other halves.

If the first was made at the 3/4 mark, there are again two more choices.

Total $3\cdot 3 \cdot 2=\boxed{18}$

Note: some of the choices do lead to identical pieces at the end, but the question is how many ways there are to cut the pizza.

There are a total of $\boxed{18}$ possible arrangements of slices. Eight of these involve two "crust-to-crust" slices, and can be found by simply considering which pairs of "crust-to-crust" slices are possible. The next four involve just one "crust-to-crust" slice, with two more slices from the crust to the cut formed by the first slice. The last six have just one "crust-to-crust" slice, one cut joining the crust to the cut formed by the second slice, and the third cut joining the crust to the cut formed by the second slice. These last ten can all be found by systematically running though the nine possible "crust-to-crust" slices and seeing which permit a dissection that does not contain a second "crust-to-crust" slice. In the diagram below the "crust-to-crust" slices are shown in red.

Let's denote $A_X(n)$ to be the combination of cuts beginning with a cut of $X^*$ enclosing $\frac{n}{4}$ of the triangle on it's left hand side. E.g The examples above lie within ( $A_{90}(1)$ and $A_{60}(3)$ ) and ( $A_{90}(2)$ and $A_{60}(3)$ ) respectively.

Also, for an area of $\frac{3}{4}$ the original, denote $B_X(n)$ to be the combination formed from a cut of $X^*$ enclosing $\frac{n}{3}$ of the shape on it's left hand side, followed by the remaining cut, cutting the final piece in half.

$\lvert A_{90}(1) \rvert$ can be seen to be 2, as $B_{30}(1)$ = $B_{60}(2)$ , $B_{30}(2)$ = $B_{60}(1)$ . $\lvert A_{90}(2) \rvert$ = 2, as you can either assign the $30^*$ cut to the left or right half. $\lvert A_{90}(3) \rvert$ = 4, as $B_{30}(1)$ , $B_{60}(2)$ , $B_{30}(2)$ , $B_{60}(1)$ are all unique. By rotating the triangle $60^*$ , one can see that $\lvert A_{30}(n) \rvert = \lvert A_{90}(4-n) \rvert$ .

Observing the previous cut combinations included. One can note that all the combinations within $A_{60}(2)$ are already included in $A_{30}(3)$ and $A_{90}(3)$ . Also, those in $A_{60}(3)$ are all already present (Combinations where either the $30^*$ or $90^*$ cut touch 2 edges of the original triangle are already counted).

For $A_{60}(1)$ , $B_{30}(2)$ and $B_{90}(2)$ can be seen in $A_{30}(3)$ and $A_{90}(3)$ respectively, but $B_{30}(1)$ and $B_{90}(1)$ are unique.

This sums to 8+8+2=18 total.

• Some maths can be done to work out if cuts meet another cut or not. Mostly it will appear quite obvious, but notably when performing the first cut of $A_{90}(1)$ and $A_{30}(3)$ , these leave a small triangle of area $\frac{1}{4}$ of the original. The edges of this triangle shared with the original are $\frac{L}{\sqrt{2}}$ and $\frac{L}{2\sqrt{2}}$ , and $\frac{3L}{2\sqrt{2}}\approx 1.1L>L$