1. The document discusses the "pizza problem" of dividing a circle into slices using chords and determining if the black and white areas are equal. 2. It introduces the Tiffany Lemma, which shows that the sum of the squares of the lengths of four chords through a point is a constant equal to 4 times the radius squared. 3. By using polar coordinates and the Tiffany Lemma, the solution shows that the integral of the black and white areas over any interval of length π/2 is π/2, proving the areas are equal.

1. Pizza Problem
Wong Hang Chi
January 19, 2010
1 The Problem
Draw three chords through an arbitrary point inside a circle so that they make
six 60◦ angles at the point and color the resulting “pizza slices” alternately black
and white. Is the black area necessarily equal to the white area? What if four
chords are used, making eight 45◦ angles?
2 Tiffany Lemma
In the figure, AC and BD are two chords of the circle such that AC ⊥ BD. M
and N are the midpoints of AB and CD respectively. Show that 2ON = AB
C
.
. N
−
.
.
.
−
B ... . .. D
. . E .. . . ..
.. . . . ..
= . . . . . . ..
... .
. O ...
. ..
.
M . ...
. .
= .. ...
..
..
..
A
It suffices to show that OM A ∼ CN O.
=
Firstly, the congruence OM A ∼ OM B is established by OA = OB,
=
OM = OM and AM = BM (SSS). Similarly, we have CN O ∼ DN O. Let
=
x = AOM = BOM . Since AOB = 2x is an angle at center and ADB is
an angle at circumference, we have 2x = 2 ADB, and so ADB = x.
Secondly, OM A ∼ DEA because ADB = x = AOM and OM A =
DEA = 90◦ (equiangular). Similarly, we have CN O ∼ DEA, which
1

2. implies OM A ∼ CN O. But OA = CO is the radii, so the similarity is in
fact congruence (AAS), which is the desired result.
We then have the following equality as a corollary.
AE 2 + BE 2 + CE 2 + DE 2 = 4R2
where R is the radius of the circle.
The Pythagorean Theorem implies that AE 2 + BE 2 = AB 2 and CE 2 +
DE 2 = CD2 . Applying the Tiffany Lemma, we have
AE 2 + BE 2 + CE 2 + DE 2 = AB 2 + CD2
= (2ON )2 + (2CN )2
= 4(ON 2 + CN 2 )
= 4OC 2
= 4R2
Therefore, the sum AE 2 + BE 2 + CE 2 + DE 2 is a constant which equals four
times the square of the radius.
3 The Solution
Without loss of generality, suppose that the circle has radius of unit length and
that its interior contains the origin. Using polar coordinates, parametrize the
circle by r(θ), where θ ∈ R. Define the four chords by θ = kπ , where k ∈ N.
4
By the Tiffany Lemma, we have
π 3π
r(θ)2 + r( + θ)2 + r(π + θ)2 + r( + θ)2 = 4
2 2
and so the required area is
π π
1 4 π 3π 1 4
[r(θ)2 + r( + θ)2 + r(π + θ)2 + r( + θ)2 ]dθ = 4dθ
2 0 2 2 2 0
which equals, obviously, π . QED.
2
In general, the integral can be taken over any measurable subset A of the
interval [0, π ] such that m(A) = π .
2 4
2