InterMath | Workshop Support | Write Up Template
Title
Equal Areas
Problem Statement
Construct a circle that has five regions of equal area where none of the perimeters of the
regions can pass through the center of the circle.
Problem setup
For this problem, you had to divide a circle into five equal areas without going through
the center of the circle.
This problem reminded me of The Quartercircle Heritage problem. Once upon a time,
two brothers Pete and Paul inherited a fenced piece of land that had the shape of a quarter
circle. Each brother should own half the area. But Pete had cows and Paul had sheep, so
they needed a separating fence. Pete suggested they build an additional piece of fence
that would divide the land into two equal areas. Paul argued "Every meter of fence is so
expensive nowadays. We should better make sure that we build the shortest
fence possible." What is the shortest fence that divides a quarter circle into two shapes of
equal area?
Even though this problem deals with a quarter of a circle, it helped to give me some ideas
of how I could divide the circle into equal areas without going through the center.
Plans to Solve/Investigate the Problem
My initial thought at the start of this problem was to use a square to help divide the circle.
My reasoning behind this is because if you look in the figure below, we know that the
area of the circle has to lie between the two squares. Also, the area of a square inside the
circle is ½ of the area of the circle. That knowledge gives me something to work with.
Area FGHI = 37.23 cm2
Area JK = 29.24 cm2
Area KLMN = 18.62 cm2
I
F
N
K
J
L
G
M
H
So, I will first draw the circle using GSP, and find out the area. I will then divide that
area to find out what the area of each of my sections will need to be. I will then divide
the circle into 4 equal sections. Then, I will take the midpoints of each of the four
segments, and use them to draw a square. I will see where this gets me and go from there.
Investigation/Exploration of the Problem
The square did not give me what I was looking for, so I had to figure out where to go
from here. I realized that I needed the triangle to be 1/5 of the area of the arc (not ¼) for
the remaining area to be 1/5 of the circle. I had to try a few more different size triangles
before I found the correct size.
C
B
O
A1
H
N
J
P
G
Q
F
E
To get this image, I took the midpoint of each of the radii of the circle, and made the
midpoint to the edge of the circle another line segment. I then took that midpoint and
made a triangle from it. This wasn’t quite right either, but I did notice that the square had
to be between the two I had already drawn. So I then drew intersecting lines from A to O
and N to W, and where they intersected I drew a line segment to create my new triangle,
and got me the correct area.
C
W
O
A1
H
N
J
P
Y
G
Q
X
F
E
As you can see, the area of the triangle is exactly 1/5 of the area of the arc. The area of
each sector is 1/5 of the area of the circle, and the area of each triangle in the square is
1/4 the area of the square.
Area JB = 100.97 cm2
Area ZD1B1C1 = 19.84 cm2
Area ZD1J = 4.96 cm2
Area ABC
Area ABC = 25.24 cm2
4
Area ABC = 25.24 cm2
= 6.31
cm2
ZD1J = 4.96 cm2
Area
Area
C
Area CHG = 25.24 cm2
Area EFG = 25.24 cm2
Area A1 = 25.24 cm2
ZD1J
Area ABC
= 0.20
D1
Area JB
= 20.19 cm2
5
A
Z
J
B1
G
A1
m CJ = 5.67 cm
m JD1 = 3.15 cm
m CJ
= 1.80
m JD1
C1
Area ABC-Area
Area
ZD1J = 20.28 cm2
JB
Area ABC-Area
Area ABC-Area
= 4.98

ZD1J5 = 101.42 cm2
ZD1J
E
The question is why is the
relationship between the
radius of the circle and
any one segment of the
5
square
?
9
As you can see from the image below, it doesn’t matter how big or how small we make
the circle as long as the square changes in direct proportion with it.
Area ZD1B1C1 = 5.01 cm2
Area ZD1J = 1.25 cm2
Area ABC
Area ABC = 6.38 cm2
4
Area CHG = 6.38 cm2
Area EFG = 6.38 cm2
Area A1 = 6.38 cm2
= 1.59 cm2
C
Area JB
= 5.10 cm2
5
S
T
H
G
F
Area ABC = 6.38 cm2
J
B
A
Area
R
A1
U
DE
Area ABC-Area
Area
Area
ZD1J = 1.25 cm2
ZD1J
Area ABC
= 0.20
ZD1J = 5.12 cm2
JB
Area ABC-Area
Area ABC-Area
= 4.98

ZD1J5 = 25.62 cm2
ZD1J
Extensions of the Problem
I realize that there are many different possibilities for this problem, and so I wanted to
figure out at least one other approach. I decided since I originally solved this problem
using a square, that I would now try solving it using a circle. This actually ended up
being easier to construct after having done the problem with the square.
Area JBH = 15.62 cm2
Area IOG = 3.09
Area
M
cm2
D
Area CB
= 12.50 cm2
5
F
A1
Area DKJ = 15.62 cm2
N
E
Area ILE = 3.09 cm2
CB = 62.50 cm2
A
P
Area DMF = 15.62 cm2
L
K
I
Area ENA = 3.09 cm2
Area A1 = 15.62 cm2
Q
C
G
O
J
Area AQG = 3.09 cm2
Area JBH-Area IOG = 12.54 cm2
Area DKJ-Area ILE = 12.54 cm2
Area DMF-Area ENA = 12.54 cm2
Area A1-Area AQG = 12.54 cm2
H
m CA = 1.98 cm
m CF = 4.46 cm
m CF
= 2.25
m CA
B
Area
CA = 12.36 cm2
Area JBH-Area IOG5 = 62.68 cm2
I would like to find some other ways to get 5 equal areas from the circle that don’t
necessarily involve arcs, like perhaps with many circles, or try slicing it from one point
on the exterior of the circle across to many different points.
I feel like this would be a good investigation to bring into the classroom, as long as I had
some of the steps along the way already created. It shows that even though it might not
look like the square has the same area as the four arcs, they actually do all share the same
region. It seems to me to be a great visual for students that are level 1 thinkers because it
should help to guide their thoughts to level 2.
Considering both drawings, when comparing the radius of the circle to that of the square I
got a relationship of 9/5, and when comparing the radius of the larger circle to that of the
smaller circle, I got a relationship of 9/4 in GSP, which is actually an approximation of
the actual ratio, which is sqrt(5)/1.
I now want to consider what happens if you want an n number of equal areas. The smaller
circle one would be easy, because it should be as easy as manipulating the circle in
accordance with 1/nth of the total area and then dividing the rest of the circle into n-1
equal slices. The square one will be much more difficult.
I would venture to say that there are many different ways to cut a circle into 5 equal areas
without any of the perimeters of the regions passing through the center of the circle.
Here is another example just for fun.
Area KJR = 10.03 cm2
Area
KOR = 5.86 cm2
Area
OGR = 5.21 cm2
Area KJR+Area

KOR+Area

OGR - Area OST = 20.18 cm2
Area KDV+Area KOV-Area WLO = 20.14 cm2
Area VWX+Area VFX-Area WQY = 20.16 cm2
Area OST = 0.92 cm2
MV
Area KDV = 7.08 cm2
Area
D
KOV = 17.76 cm2
Area WLO = 4.70 cm2
E
W
Q
C
Y
Area VFX = 10.52 cm2
Area
F
N
K
L
VWX = 10.90 cm2
A
I
Area WQY = 1.27 cm2
X
O
J
Area RHX = 8.92
Area
ZS
G
T
cm2
RYX = 13.31 cm2
B
Area ZTY = 2.09 cm2
Area RHX+Area
H
R


RYX - Area ZTY = 20.15 cm2
Area
AB = 99.94 cm2
Area
AC = 19.99 cm2
Area AB
= 19.99 cm2
5
I will have to further work on Dr. Abney’s way of constructing the circle to have five
equal areas. I got close, but it still needs a little work.
8
Area AB
= 19.60 cm2
5
Area CDEF = 18.09 cm2
6
Area


GDH+ Area GBH = 20.44 cm2
Area
H
GDH = 11.55 cm2
I
4
Area GBH = 8.89 cm2
B
-15
-10
Area HIJ = 8.24
Area
2
F
J
A
-5
5
cm2
HCJ = 10.59 cm2
Area HIJ+Area
C
-2
G
D
E
HCJ = 18.83 cm2
L
-4
M
K
Area GMK = 9.63
Area
-6
cm2
GEK = 11.56 cm2
Area GMK+Area
GEK = 21.20 cm2
Area KLJ = 8.88 cm2
Area
-8
Area KLJ+Area
Author & Contact
Tiffany Graham
Link(s) to resources, references, lesson plans, and/or other materials
http://www.missouri.edu/~tprb62/Flores_Regis.pdf
http://karl.kiwi.gen.nz/prcircle.html
http://karl.kiwi.gen.nz/prcircle.gif
KJF = 10.58 cm2
KJF = 19.46 cm2