Pizza portions

There was an article in the 12^th December issue of New scientist that had some interesting observations on sharing out a pizza between two people.

The essence of the problem is to decide who gets the larger share if the waiter making the cuts across the pizza misses the centre, but manages to get the angles between the cuts all equal and to have the cuts all intersect at a single point.. The two diners eat alternate pieces. Who gets the larger share?

The suprise is that if the number of cuts is even and greater than 2, each diner gets an equal share despite the variability of the sizes of the pizza wedges.

If the number of cuts is 3, 7, 11, etc the diner getting the piece containing the centre eats the most, which is not too surprising. However, if the number of cuts is 5, 9, 13, etc the diner getting the piece containing the centre eats the least, which is not really what might be expected intuititely.

It might be fun to see how much of this we can prove.

4 cuts

When we have 4 cuts there are 8 pieces, with the odd-numbered ones going to diner Oliver, and the even ones going to diner Eric.

The article shows that if any cut goes through the centre, the shares are even.

A strategy:

Slide the horizontal line downwards so that it goes through the centre, as shown by the dashed line.
Slide each of the diagonal lines to go through the intersection of the horizontal and vertical lines.
All cuts now go through the same point, and one cut goes through the center.
Try to prove that there is no nett gain to either diner caused by the relocation of the cuts.

The two triangles are congruent. Let us call the area of each of the triangles t. Let us consider them as still part of their original wedges. Wedge 1 grows by the rectangle with the curved right-hand end, minus the area of the triangle which is still a part of wedge 7. Wedge 8 shrinks by the same amount.

Similarly, wedge 4 grows by the left-hand “rectangle”. Wedge 5 shrinks by the same amount.

Let a be the distance of the vertical line from the centre, and b be the distance of the horizontal line from the centre,

Thus the right-hand “rectangle” is 2a longer than the left-hand “rectangle”.
As the width of the strip is b, Oliver’s share of the pizza has grown by ab, and Eric’s share has shrunk by the same amount.

Let us now consider moving the diagonal cut that divides wedges 7 and 8 and wedges 3 and 4. We slide it until it goes through the intersection of the vertical and horizontal cuts.

The triangle at the apex of wedge 7 now becomes part of wedge 1, but is still part of Oliver’s meal.

The remaining diagonal is now slid so that the new intersection has all four lines.

The six almost-rectangles labelled with capital letters A to F change their ownership, as do the six triangles labelled with lower-case letters p, q, r, s, t and u.

	gains	loses
Wedge 1		A
Wedge 2	A u t
Wedge 3	B p q
Wedge 4	C r	B
Wedge 5	D	C p
Wedge 6		D r q
Wedge 7		E s t
Wedge 8	E	F u

Area	old	new
A	1	2	o->e
B	4	3	e->o
C	5	4	o->e
D	6	5	e->o
E	7	8	o->e
F	8	1	e->o
p	5	3
q	6	3	e->o
r	6	4
s	7	1
t	7	2	o->e
u	8	2

The triangles are all congruent, so there is no nett change due to the triangles.

Eric gains A + C + E from Oliver, and Oliver gains B + D + F from Eric.

Eric’s nett gain is therefore:
      A - D + C - F + E - B

Because all the triangles are the same area:
      C - F = - 2ab
               as 2a is the difference in length between area F and C, and b is the width of the strip.

The width of each of the other two strips is b / √2
We need to find the perpendicular distance of each of the parallel lines form the centre.
For the line that forms an edge of B and E this distance is ( b - a ) / √2
and for the line that forms an edge of A and D this distance is ( b + a ) / √2

So:
      D + r + q + t - A - u = 2 × (b / √2) × ( b - a ) / √2 = b² - ab
Similarly:
      E + s + t + q - B - p = b² + ab

All the triangles are the same area,so if we subtract the penultimate equation from this last one we have:
      E - B + A - D = 2ab
If we now add in our earlier result for C - F we have
      C - F + E - B + A - D = 0

I.e. the original configuration of cuts produces exactly the same share out as the revised configuration of cuts.
The revised configuration has one of the cuts going through the centre, and therefore gives equal shares. Q.E.D.