PROBLEM:
Given a square cake iced on the sides and the top, how can you cut the cake so that
each of five children gets the same amount of cake and the same amount of icing?
E
A
B
K
F
H
D
G
To keep calculations (and cuts) simple, I have
assumed that we will cut the cake into polygonal
pieces like those shown at left. If we assume the
cake’s height is constant (not something we could
assume if I were the one who had baked it!), and that
we will cut straight down, then for each child to get
the same volume of cake, each piece must have the
same area “in plan” (i.e., the area of the base of each
“slice prism” must be equal).
C
When the area in plan is the same, we know each child is getting the same amount of
icing from the top of the cake. So each child must also get the same amount of icing
from the sides of the cake. The amount of icing a child gets from the sides of the cake
is equal to the cake’s height times the length along the outside of the cake from one
slice point to the next. Thus, the lengths along the outside of the cake must be equal
for all children. The sum of these lengths is
(4/5)s
E (1/5)s
A
the perimeter of the square (4s, where s is the
B
length of the cake’s side), and so the distance
(3/5)s
between consecutive points must be 4 s .
K
5
(4/5)s
Let’s say the first cut goes through one corner
of the cake. Then we have a series of cuts that
look something like what is shown at right.
We need to determine where K should be.
H
(2/5)s
(1/5)s
D
We want AreaΔAHK = AreaΔAEK:
AreaΔAHK = AreaΔAEK
⇔
F
(3/5)s
G
(2/5)s
C
( ) ( s ) (AK) (sin ∠HAK) = ( ) ( s ) (AK) (sin ∠EAK)
1
2
4
5
1
2
4
5
⇔ sin ∠HAK = sin ∠EAK ⇔ m∠HAK = m∠EAK
since both are acute angles

Also, ∠HAK and ∠EAK are complementary, so m∠HAK = m∠EAK = 45 .
Thus, K lies on the diagonal AC .
Now consider quadrilateral KFCG.
By SAS, it is made up of two congruent
triangles, so its area is:
1
2
2
5
(3/5)s

2
5
=
E (1/5)s
B
( ) ( s) (CK) (sin 45 )
= ( s ) (CK ) ( )
2⋅
(4/5)s
A
( s) (CK)
(4/5)s
K
2
2
F
2
5
H
(2/5)s
(1/5)s
We want Areaquad KFCG = AreaΔAHK, so:
(3/5)s
D
(2/5)s
G
C
( s) (CK) = 2 ⋅ ( ) ( s) (CK) (sin 45 ) = ( s) (AK) ( ) = ( s) (AK)
2
5
1
2
2
5
2
5

2
2
2
5
⇔ CK = AK
So K is not only on the diagonal AC ; it is the midpoint of AC !
Thus,
Areaquad KFCG = AreaΔAEK = AreaΔAHK
( s) (AK) = ( s)( ) = ( ) s
=
2
5
2
5
s 2
2
1
5
2
s2 = AreaΔAEK + Areaquad EBFK + Areaquad KGDH + Areaquad KFCG + AreaΔAHK
=
=
( )s
( )s
1
5
1
5
2
+ Areaquad EBFK + Areaquad KGDH +
2
+ Areaquad EBFK + Areaquad KGDH
( )s
1
5
2
+
( )s
1
5
2
By reflecting across the diagonal AC , we see that quadrilaterals EBFK and KGDH
are congruent, so Areaquad EBFK = Areaquad KGDH
Thus, s2 =
( )s
3
5
2
+ Areaquad EBFK + Areaquad KGDH =
And so Areaquad EBFK = Areaquad KGDH =
( )s
3
5
2
+ 2∙Areaquad EBFK
( ) s , as desired.
1
5
2
So the cuts as shown (with K the midpoint of the square’s diagonal) give each student
an equal amount of cake and an equal amount of icing.
Maybe next time we should get a round cake, or less demanding kids.