InterMath
Title
Two squares
Problem Statement
Pick a point P on the line segment AB and make squares: A side of one square is AP and a side
of the other square is PB. Where should the point P be located to satisfy the condition that the
sum of the areas of the two squares is a minimum? Explain.
Problem setup
Start with a line segment and construct two squares using parallel and perpendicular lines after
creating a point anywhere on the line segment. The area of each square will be added to find the
sum.
Plans to Solve/Investigate the Problem
To begin the problem, I will construct a line segment in GSP and from there make perpendicular
lines and parallel lines to construct two squares. From there I will calculate the areas and find the
sum. I will then move point P along line segment AB to show infinite cases and observe the
changes in the area to find the minimum.
Investigation/Exploration of the Problem
In GSP construct two points. Name the points A and B. Connect the points with a segment.
Construct point P anywhere on the line segment.
A
P
B
After the line segment AB is created, make three perpendicular lines through
line segment AB. Position one of the lines going through A, one going
through B, and one through P.
Make a point on the line through point A and name it F. The length of AF
must be the same as that of line segment AP because a square has four
congruent sides. Measure line segment AP and replicate the length. Make a
line parallel to line segment AB going through point F. Where the new line
A
P
B
crosses through the line through P, make point E. Construct a line segment for
EF. Now above line segment EF, construct a parallel line to EF. Before the
parallel line is constructed, line segment BP must be measured. After being
measured, and replicated, translate the new line segment CB so that it is
perpendicular to line segment PB off of the point B. Replicating and translating the line segment
assures there is a square. Then the parallel line can be constructed and be positioned where it
perpendicularly intersects with point A and forms a vertex of a square. Where the line crosses
through the lines through P, make a point at the intersection. Name the intersection point D. Now
after the points are made, finish constructing line segments from A to F, E to F, D to P, C to C,
and from B to C if they have not already been done. The points will be the vertices of two similar
squares.
D
F
A
C
E
B
P
After the two squares are formed, construct the interior of each square. Find the area of square
APEF, and then find the area of square BCDP. The area of a square is A=length x width.
Whenever point P, now a vertex of both squares, is close to point A or point B the sum of the
areas is at a minimum. No matter the size of either square. GSP allows P to be moved all along
line segment AB, so the squares at some time can be all sizes. Line segment AB can be
lengthened so that larger squares can be explored. The squares also can be increased or decreased
in area and show different sums of the total area. Some squares were explored in depth and
measurements show how the squares are proportional.
Area BCDP = 1.68 cm 2
Area BCDP = 0.72 cm 2
cm 2
Area APEF = 0.03 cm 2
Area APEF = 0.40
Area BCDP = 3.33 cm 2
Area BCDP +Area APEF = 2.09 cm 2
Area APEF = 1.30 cm 2
Area BCDP +Area APEF = 4.63 cm 2
D
C
m AF = 1.13 cm
E
m EF = 1.14 cm F
D
m AF = 0.63 cm
m EF = 0.64 cm
F
m DP = 1.30 cm
m DC = 1.29 cm
P
B
Area APEF = 0.01 cm 2
Area BCDP +Area APEF = 0.05 cm 2
A
P
B
m DC = 0.85 cm
m AB = 1.93 cm
m AB = 1.03 cm
m EF / m AB = 0.33
m EF / m AB = 0.18
Area BCDP = 0.03 cm 2
Area APEF = 0.01 cm 2
Area BCDP +Area APEF = 0.03 cm 2
m AF = 0.10 cm
m AF = 0.08 cm
m EF = 0.10 cm
m EF = 0.08 cm
m DP = 0.20 cm
m DP = 0.16 cm
m DC = 0.19 cm
m DC = 0.16 cm
D
m AB = 0.29 cm
m AB = 0.24 cm
m EF / m AB = 0.33
m EF / m AB = 0.33
Size Proportions of Square: Large to small ratio
0.04 / 0.01 = 4
0.03 / 0.01 = 3
C
FE
m DP = 0.85 cm
Size Proportions of Square: Large to small ratio
3.33 / 1.30 = 2.56154
1.68 /0.40 = 4.2
Area BCDP = 0.04 cm 2
m AF = 0.17 cm
m EF = 0.18 cm
E
m DP = 1.83 cm
m DC = 1.82 cm
A
C
Area BCDP +Area APEF = 0.75 cm 2
A P
0.72 / 0.03 = 24
B
The squares are similar squares, and their relation determines if the area will be greater when
vertex P is closer to point A or to point B. The squares are drawn in such a way that square
BCDP is the larger of the two squares. Due to the size difference, it is vital that BCDP has the
smallest possible length and width so the area will be minimized. It will always have a greater
area itself, but having a small length and width will cause the area to be reduced and in turn
reduce the sum of the areas. To acquire the smallest length and width, vertex P must be placed as
close to B as possible. Position vertex P as close to B as possible, without overlapping the
vertices and the area will be small. The other sides of the squares must be reduced each time
vertex P is moved though because a square has to be equal in length and width. Constructing an
accurate square with such small is tedious and problematic. The sides on the square need to
increase at the same rate each time the vertices are moved though to keep the square property.
The fact that P needs to be close to point B is evident though by the calculations, infinite cases
from drawings on GSP, and through formulas.
Looking at the perimeter also shows how the changes in the certain sides produce a change. The
square BCDP will always produce a larger perimeter because it is larger, so the area will also be
larger since the sides produce a greater perimeter. If the area is larger, minimize the area of the
square with the largest perimeter or area until it is as small as possible. To make it smaller and
produce the smallest area possible, vertex P must be in close proximity to vertex B.
0.02+0.02+0.02+0.01+0.01+0.01+0.01 = 0.10
The perimeter, side + side + side +side, of two squares is:
Where the sides of BCDP each measure 0.02 cm and APEF measure 0.01 cm, the perimeter is
0.10 cm. The area of BCDP is .0004 cm2 and APEF is .0002 cm2. The sum is 0.0006 cm2.
Although GSP is unable to show the squares with such a small area, the sum can still be found by
reducing the lengths of the sides. The area is still not minimized as small as it can go though. The
lengths can always be reduced further, but the visual image will start to be microscopic. Vertex P
does not have to stay in the middle of line segment AB and APEF does not have to remain as
half the size of BCDP to get a smaller area. Each square can be resized to any size, but vertex P
must be translated over to vertex A or B, and the stipulation is it must be translated to the vertex
that is apart of the largest square though to achieve the minimum area.
Another way to look at the squares is by translating the squares so they are not side-by-side.
Area HKJP = 3.70 cm 2
Area INOP = 1.03
J
K
cm 2
Area HKJP = 1.09 cm 2
Area INOP = 0.03 cm 2
Area HKJP+Area INOP = 4.73 cm 2
m JK = 1.93 cm
I
Area HKJP+Area INOP = 1.12 cm 2
J
K
m JK = 1.04 cm
P
m JP = 1.92 cm
H
m IP = 1.01 cm
m JP = 1.05 cm
m NI = 1.02 cm
N
O
m NI = 0.17 cm
Area HKJP = 0.12 cm 2
Area HKJP = 0.03 cm 2
cm 2
Area INOP = 0.28 cm 2
Area INOP = 0.10
Area HKJP+Area INOP = 0.22 cm 2
m JK = 0.35 cm
m JP = 0.34 cm
m IP = 0.31 cm
m NI = 0.32 cm
IP
m IP = 0.18 cm
J
IP
N
K
O
H
H
N O
Area HKJP+Area INOP = 0.31 cm 2
J K
I P
m JK = 0.16 cm
H
m JP = 0.17 cm
m IP = 0.53 cm
m NI = 0.54 cm
N
O
Like when the squares are beside each other, the sum of the areas of that are not side-by-side
produces a minimum area when vertex P is closest to the point on the original line segment,
which is now a vertex, of the larger square. Either way the squares are drawn, the area decreases
more when P translates left or right towards the larger square.
Extensions of the Problem
Would the area reach a minimum when there are two rectangles of similar size? What is the
relationship with the perimeter and the rectangles?
Since squares are rectangles, rectangles have the same properties when the sum of two rectangles
area is minimized. Area looks at the length and the width. Unlike squares, rectangles have only
two pairs of congruent sides and not all four. Since the lengths and widths of the rectangle
determine the area, the area of each rectangle needs to be resized so that the length and width are
small. The sum of the areas will be at a minimum if the larger rectangle is positioned so that P is
translated closest towards the vertex I or H respectively.
J
Area HKJP = 2.85 cm 2
Area INOP = 0.23
J K
K
I P
Area HKJP = 0.13 cm 2
H
cm 2
O
N
Area HKJP+Area INOP = 3.07 cm 2
Area INOP = 1.73
cm 2
N
Area HKJP+Area INOP = 1.87 cm 2
Area HKJP = 0.48 cm 2
Area HKJP = 0.87 cm 2
cm 2
Area INOP = 0.05 cm 2
Area INOP = 0.13
I
Area HKJP+Area INOP = 0.61 cm 2
J
K
m JK = 0.46 cm
Area HKJP+Area INOP = 0.92 cm 2
J
K
m JK = 0.83 cm
m JP = 1.05 cm
m JP = 1.05 cm
I
P
m IP = 0.76 cm
m NI = 0.17 cm
O
N
m NI = 0.12 cm
H
O
I P
m IP = 0.39 cm
H
P
N
O
H
Perimeter is calculated by each of the sides added together and can be looked at again with
squares that are not side-by-side. Unlike the perimeter with the squares, the perimeter of the
rectangles will decrease at varying rates because the length and width are different. Depending
on which rectangle is the largest in the beginning determines which vertex that vertex P should
be moved toward. For the perimeter to be minimized the larger rectangle must be reduced in size,
therefore P will be translated on line segment IH to the left or the right towards the larger
rectangle. Perimeter will decrease if the sides measure less.
U
Ci rcumference
Ci rcumference
Ci rcumference
T
S
UV = 9.13 cm
ST = 9.48 cm
ST +Ci rcumference
UV  = 18.61 cm
Looking at the sum of the circumference of two
circles also shows some similar properties to the
squares and rectangles. If center S or U is moved
in a certain direction on the line segment UT, the
circumferences change. To minimize the
circumference of the two circles the center must
be moved where it lays right off the edge of one of
the circles where the radius is a point on the circle.
To maximize the circumference the circles
become one large intertwined circle.
V
Ci rcumference
Ci rcumference
Ci rcumference
S
U
V
S T
U
UV = 17.34 cm
Ci rcumference
ST = 2.83 cm
ST +Ci rcumference
UV  = 20.17 cm
Ci rcumference
Ci rcumference
T
UV = 20.47 cm
ST = 20.47 cm
ST +Ci rcumference
UV  = 40.94 cm
Circles are different though in their properties so they do not function just as the squares or the
rectangles. However, the circumference can be minimized by making one of the circles small.
Author and Contact
Laura Sims
I am a junior at Georgia College and State University. I am majoring in middle grades education
with a concentration in language arts and math.
Click here to send me an Email