Geometric Inequalities
Chongchitnan S. and Pang P.Y.H.1
Department of Mathematics, National University of Singapore
2 Science Drive 2, Singapore 117543.
ABSTRACT
“Given a fixed perimeter, what is the shape that encloses the largest area?”
The Isoperimetric Theorem, which eluded mathematicians from the ancient Greeks
to the Bernoullis to Jacob Steiner of the 19th century, says that it is the circle, which of
course goes with our intuition.
There are also other kinds of isoperimetric problems. For example, what kind of
triangle with a fixed perimeter would enclose the largest area if it is drawn within a
given circle ? or what kind of box would have the largest volume given a fixed
surface area ? I find that these too are interesting problems that fortunately require
proofs that are much less sophisticated than the Isoperimetric Theorem itself.
In this research I investigated some of these problems mostly dealing with
polygons especially the triangle. Figures in three dimensions are also discussed.
Techniques involved are simple geometrical arguments by means of inequalities,
without any use of calculus : simple, self-contained, and very practical.
INEQUALITIES
On the basis of solving geometrical problems using inequalities, as suggested by
the title, we have the inequality that plays the main role in solving problems in this
research.
In words :
Theorem 1 “The product of n positive numbers whose sum is fixed is the greatest
when the numbers are all equal.”
Or mathematically speaking :
“For ai  0 ( i  1,2,3..., n ) , let A be their average. i.e.
a
i
 An
a
i
 nA . Then :
(1)
with equality iff a1  a2  ....  an  A ”
Eq. (1) is the equivalent of the perhaps more popular theorem of geometric and
arithmetic means.
Theorem 2 “ For any set of positive numbers {ai | i  1,2....n} with geometric mean,
G , and arithmetic mean A , given by :
1
Assistant Professor
A =
1
 ai
n
Then :
and
G =
n
a
G A
i
(2)
ALGEBRAIC APPLICATION
One of the interesting examples, although serving as a rather crude estimation, that
can be easily proven using Eq. (2) is this result for all positive integer n :
 n 1
n! 

 2 
n
(3)
GEOMETRIC APPLICATIONS
The Dual Theorem of Isoperimetrics
It was shown that the figure of a certain class with the largest area given a fixed
perimeter is also the one with the least perimeter given a fixed area. This is known as
the Dual Theorem.2 (eg. in the class of all plane figures, the circle is the extremal
figure satisfying the Dual Theorem.) This serves as an extension to theorems to come.
Concerning Triangles
Results on triangles are based algebraically on Eq. (1), the Dual Theorem and the
Heron’s formula for area of triangles given in a less popular form3 as :
16 A2  P( P  2a)( P  2b)( P  2c)
(4)
where A denotes the area and P the perimeter of the triangle with sides a, b and c .
With Eq. (4), and applying Theorem 1 to the factors on the RHS, the following
results can be obtained without any use of calculus.
Theorem 3 “Of all isoperimetric triangles, the equilateral triangle has the greatest
area.” And by the Dual Theorem, we have : “Of all iso-area triangles, the equilateral
triangle has the least perimeter.” (i.e. the equilateral triangle is the extremal figure)
Investigations on triangles circumscribing and inscribing a fixed circle were also
made. (see Fig. 1) Although the results are rather obvious, the proofs are not simple.
They are :
Theorem 4 “Of all triangles circumscribing a given circle, the equilateral triangle
has the least area and least perimeter.”
Theorem 5 “Of all triangles inscribing a given circle, the equilateral triangle has the
greatest area and greatest perimeter.”
2
Strictly speaking, the extremal figure of that certain class must all be similar for the Dual theorem to
work. But this is usually the case anyway.
3 The reader may be more familiar with the expression
A  s(s  a)( s  b)( s  c) where s denotes the
semi-perimeter of the triangle.
Figure 1. Triangle inscribing (left) and circumscribing (right) a given circle.
Concerning Polygons
The first extension of the triangle is the isoperimetric question of quadrilaterals.
The results to follow are based on Theorem 1, the Dual Theorem and the
Brahmagupta’s formula for area of cyclic quad with area A , perimeter P and sides
a, b, c and d , given by :
16 A 2  ( P  2a)( P  2b)( P  2c)( P  2d )
(5)
Firstly, we have the immediate geometrical implication of Theorem 1, which is
“Of all isoperimetric rectangles, the square has the greatest area.” By using Eq. (5), it
can be shown that this result is true for all quadrilaterals, i.e:
Theorem 6 “Of all quadrilaterals, the square is the extremal figure.”
Theorem 6 is actually a consequent of this neat theorem :
Theorem 7 “A quadrilateral of given sides has the greatest area when it can be
inscribed in a circle.”
Having observed that the extremal figures for 3 and 4 sided plane figures are
regular, one should expect that the extremal figure of any polygon should also be
regular. This speculation is correct although there is no elementary proof that does not
resort to the Isoperimetric Theorem itself.
However a close call was made with this theorem:
Theorem 8
perimeter.”
“The circle has greater area than any regular n-gon with the same
The Isoperimetric challenge is however not restricted to only regular polygons, or
polygons of any kinds, but all kinds of possible closed shapes. Because of this
generality, it took mankind more than 2000 years to finally assert that, as intuitive as
it may be, the circle is the extremal figure.
Here is the Isoperimetric Theorem in a more mathematical form :
The Isoperimetric Theorem For any figure with area A and perimeter P :
4A
1
P2
(6)
The LHS is also known as the Isoperimetric Quotient ( IQ ). Using this somewhat
anthropological terminology, we may equivalently state the Isoperimetric Theorem as:
“of all plane figure, the circle has the highest IQ of 1.” (see Table 1)
Table 1. some IQs of plane figures given to 4 sig fig.
Isosceles triangle (1:2:2)
Right triangle (3:4:5)
Equilateral triangle
Rectangle (1:2)
Square
Ellipse (a:b = 1:2)
Pentagon (regular)
Hexagon (regular)
10-gon (regular)
0.4867
0.5236
0.6040
0.6981
0.7854
0.8451
0.8648
0.9069
0.9669
Concerning 3-Dimensional Figures
These results can be obtained by strategically moulding a solid so that Theorem 1
and the Dual Theorem may be applied. Together with results on plane figures, it was
shown that:
Theorem 9 “Of all quadrilateral prisms4 with a given volume, the cube has the least
surface area”
The proof of Theorem 9 is truly eclectic. Here are also other 3D results that may be
easily obtained using the invaluable Theorem 1:
Theorem 10 “Given a fixed surface area, the optimal design for an open box is (not a
cube) but half of a cube, i.e. the box with dimension 1:2:2.”
Theorem 11 “Of all cylinders, the one with equal diameter and height is extremal.”
I cannot stress more the power of Theorem 1 and the Dual Theorem that allow us
to prove some serious 3 dimensional results without the complexity of calculus.
Conclusion.
Isoperimetric problems reflect human’s quest to prove one of nature’s most lucid
statements. Both historically significant and immensely practical, the isoperimetric
problems conjure a mathematical art that is as elegant as it is beautiful.
REFERENCES
Kazarinoff, N.D. (1961), Geometric Inequalities, the L.W Singer Company New
Mathematical Library.
Popple, R. (1992), “Honeybees and Isoperimetrics”, ‘PARABOLA’, University of
New South Wales, Vol 28, No. 2.
Dunham, W. (1994), The Mathematical Universe, J Wiley and Sons Inc.
4
This refers to a solid with parallel ends with a uniform quadrilateral cross section.