One Eighteen Inch Pizza Beats a Twelve and a Ten, by a Corner

One large pizza, eighteen inches across, costs the same as a medium and a small together, twelve inches and ten. All three are round. Which order gives you more pizza, and can you decide it at the counter without arithmetic?

The single large one wins, by about a third. And yes, there is a way to see it that involves no multiplication at all, because the correct comparison turns out to be Pythagoras in disguise.

Widths add and areas do not

The reflex is to add the widths. Twelve and ten make twenty-two, twenty-two is more than eighteen, so two beats one. Every part of that is true except the last clause, and the last clause is the answer.

A pizza of diameter $d$ has area $\pi d^2/4$. The factor $\pi/4$ sits in front of every term on both sides of the comparison, so it cancels and the whole question becomes a comparison of squares:

Which for our three pizzas is not close:

The ratio $324/244 = 1.328$ says the large pizza is about a third more food. In real units the areas are 254.5 and 191.6 square inches, a difference of 62.8 square inches, which is most of another small pizza. Note that 324 and 244 are squared widths, not areas. They are the right numbers to compare and the wrong numbers to attach a unit to.

The comparison is a triangle

Equation (1) is $L^2$ against $M^2 + S^2$, and any time those three quantities appear together there is a triangle nearby. Build one whose sides are the three widths. The law of cosines relates them through the angle $\theta$ at the corner where the $M$ side meets the $S$ side:

Rearranged, the thing you want to know is a single sign:

Since $M$ and $S$ are positive lengths, the sign of the left side is the sign of $-\cos\theta$. That gives a test with no numbers in it.

For our three widths the cosine is a clean fraction:

Wider than 90 degrees, which agrees with equation (2). The picture and the arithmetic are the same statement, and the picture is the one you can perform with the corner of the pizza box.

Where the tie lives

The break-even cases are exactly the Pythagorean triples. Widths 5, 4 and 3 are a dead heat, since $25 = 16 + 9$, and so are 13, 12 and 5, and 25, 24 and 7. A sweep over every integer triple whose largest width runs from 6 to 40 found 15 exact ties, all of them right triangles, as equation (4) requires.

Two equal small pizzas make the rule easy to remember. A tie needs $L^2 = 2M^2$, that is $L = M\sqrt{2} \approx 1.414\,M$. So one pizza beats two of width $M$ as soon as it is more than about 41 per cent wider. Two twelve inch pizzas total $288$ in squared width, so a seventeen inch pizza beats them at $289$, by one unit, and a sixteen inch pizza loses at $256$. The crossover sits at $12\sqrt2 = 16.97$ inches, which is a genuinely narrow window and a good reason to know the rule rather than eyeball it.

More than two, and the honest boundary

Nothing restricts the comparison to two smaller pizzas. Against $n$ of them the test is

and the triangle picture no longer applies, because a triangle has three sides. For $n = 3$ you could reach for a box diagonal instead, comparing $L$ against the space diagonal of a box with edges $d_1, d_2, d_3$, which is the three-dimensional Pythagoras. Beyond that the arithmetic is easier than the geometry.

The triangle test also needs the triangle to exist, which requires $L < M + S$. Here $22 > 18$ so it does. When it fails, the answer is immediate anyway: $L \ge M + S$ gives

so the single large pizza wins outright. The rule therefore has no gap: either the triangle exists and the corner decides, or it does not and the large one has already won. The exhaustive sweep covered both arms, 5,928 real triangles and 5,517 degenerate triples, with no exception in either.

The number to carry around

Once you accept that area follows the square of the width, a useful reflex follows. A pizza 20 per cent wider is $1.2^2 = 1.44$, so 44 per cent more food, and a pizza half again as wide is $1.5^2 = 2.25$, more than double. Width differences that sound modest on a menu are large on a plate, which is the reason the large pizza keeps winning these comparisons.

It also settles fights the triangle cannot reach. One eighteen inch pizza against three ten inch pizzas is $324$ against $3 \times 100 = 300$, so the single one still wins, despite the three being thirty inches of width between them. Against four of them it loses, at $400$. The break-even count for ten inch pizzas against an eighteen is $324/100 = 3.24$, and you cannot buy a quarter of a pizza.

What the model quietly assumes

The comparison is about the top surface, treating a pizza as a disk. Real pizza has crust and depth. The crust is roughly a fixed-width annulus, so the edible interior of a $d$ inch pizza with a $c$ inch crust goes like $(d - 2c)^2$ rather than $d^2$, which favours the large pizza even more strongly. Thickness cuts the other way only if the smaller pizzas are made thicker, which is a question for the kitchen.

The comparison also assumed the prices are equal, which is what makes it a pure area question. If they are not, the quantity to compare is squared width per unit of money, $d^2/p$, and the same cancellation of $\pi/4$ makes that the right ratio to divide in your head.

Sources and further reading

• The formula behind equation (1) — Area of a circle
• The identity in equation (3) — Law of cosines
• The classification the box-corner test is reading off — Acute and obtuse triangles

The areas were checked without using the area formula: four million random points per disk gave 254.45 and 191.60 against the closed forms 254.47 and 191.64, and a ratio of 1.32807 against the exact 1.32787. The equivalence between the box-corner test and the area comparison was checked by exhaustive sweep rather than trusted from equation (4).