Balmoral Software
12/30/16

## Section 1: Problem definition

A polygon is said to be equable if its area and perimeter are numerically equal. Increasing the sides of a convex polygon by a factor of k > 1 increases the perimeter by the same factor, but the area is increased by the larger factor k2. Therefore, we may intuitively expect equable polygons to be relatively small.

Any non-equable polygon with perimeter P and area A can be made equable by scaling each of its sides by the factor k = P/A; the revised polygon has equal perimeter kP = P2/A and area k2A = P2/A.

This paper is concerned with cyclic integer quadrilaterals, four-sided convex polygons that have integer sides and area, and can be inscribed in a circle. While non-cyclic integer quadrilaterals exist (for example, certain ones created from three congruent right triangles having integer sides), cyclic quadrilaterals have the maximum area amongst all quadrilaterals having the same sides.

Exactly four of these cyclic integer quadrilaterals will be found to be equable:

 Sides Perimeter = Area (4,4,4,4) 16 (6,6,3,3) 18 (8,5,5,2) 20 (14,6,5,5) 30
Following is a proof of these results.

Let the integer sides of a cyclic quadrilateral be a, b, c, d, and denote by s its semiperimeter

Since the sum of any three sides of a realizable quadrilateral must exceed the fourth (the Quadrilateral Inequality), we have
 b + c + d > a [1] -a + b + c + d > 0 [2]
and analogously,
 s - b > 0 [3]

 s - c > 0 [4]

 s - d > 0 [5]

Using Brahmagupta's formula for the area A of a cyclic quadrilateral, we have
A2 = (s - a)(s - b)(s - c)(s - d)
If the perimeter P = 2s = a + b + c + d is odd, then s is of the form k + 1/2 for some integer k, and so is each of the four factors in A2. Therefore, the area and the sides cannot all be integers if P is odd. Henceforth, we will assume that the perimeter is even so that the area is an integer. It follows that the semiperimeter s is also an integer.

Without loss of generality, we can assume that the integer sides of the quadrilateral are ranked alphabetically from longest to shortest:

 a ≥ b ≥ c ≥ d ≥ 1 [6]

## Section 2: Change of variables

To simplify the algebra, we introduce new integers w, x, y, and z:
w = s - a

x = s - b

y = s - c

z = s - d

These integers are all positive by [2]-[5]. Summing both sides,
w + x + y + z = 4s - (a + b + c + d) = 4s - 2s = 2s = a + b + c + d = P
The inverse relationships between w, x, y, z and the sides of the quadrilateral are:
a = s - w

b = s - x

c = s - y

d = s - z

where s = (w + x + y + z)/2.

From [1] and [6], we have

b + c + d > a ≥ b ≥ c ≥ d ≥ 1
Substituting the new variables,
s - x + s - y + s - z > s - w ≥ s - x ≥ s - y ≥ s - z ≥ 1,
which simplifies to
-w < w ≤ x ≤ y ≤ z ≤ s - 1
Since -w < w implies that w ≥ 1, and z ≤ s - 1 is equivalent to
z ≤ (w + x + y + z)/2 - 1

2z ≤ w + x + y + z - 2

z ≤ w + x + y - 2,

we have
 1 ≤ w ≤ x ≤ y ≤ z ≤ w + x + y - 2
This is the fundamental relationship between w, x, y and z that will form the foundation for the analysis below.

## Section 3: Limiting the values of w

Variables in the relationship above are unbounded, so to find all equable integer cyclic quadrilaterals, we must determine limits for w, x and y.

We can define a multivariate function F as follows:

 F(w,x,y,z) = A2 - P2 = wxyz - (w + x + y + z)2, 1 ≤ w ≤ x ≤ y ≤ z ≤ w + x + y - 2 [7]
This leads to a Diophantine polynomial equation for which integer solutions are sought. Solutions (w,x,y,z) for which F(w,x,y,z) = 0 correspond to equable integer cyclic quadrilaterals. We will show that all such quadrilaterals must occur when w ≤ 4 by demonstrating that when w ≥ 5, we have F(w,x,y,z) > 0, or equivalently, area exceeding perimeter.

From the given constraints,

1/z ≤ 1/y ≤ 1/x ≤ 1/w

wxyz/z ≤ wxyz/y ≤ wxyz/x ≤ wxyz/w

wxy - 2(w + x + y + z) ≤ wxz - 2(w + x + y + z) ≤ wyz - 2(w + x + y + z) ≤ xyz - 2(w + x + y + z)

 dF/dz ≤ dF/dy ≤ dF/dx ≤ dF/dw [8]

Let the function G be defined by
G(w,x,y) = wxy - 4(w + x + y)
If w ≥ 5, then
wx ≥ w2 ≥ 25

wx - 4 ≥ 21 > 0

and
1/y ≤ 1/x ≤ 1/w

wxy/y - 4 ≤ wxy/x - 4 ≤ wxy/w - 4

0 < wx - 4 ≤ wy - 4 ≤ xy - 4

0 < dG/dy ≤ dG/dx ≤ dG/dw,

so all partial derivatives of G are strictly positive. It follows that G(5,5,5) = 53 - 4(15) = 65 > 0 is a global minimum of G on the domain 5 ≤ w ≤ x ≤ y, and therefore, G(w,x,y) > 0 over this domain.

We have

G(w,x,y) = wxy - 4(w + x + y) > 0

wxy > 4(w + x + y) > 4(w + x + y) - 4

wxy > 2[w + x + y + (w + x + y - 2)] ≥ 2(w + x + y + z)

wxy - 2(w + x + y + z) > 0,

so dF/dz > 0, and by [8] all other partial derivatives of F are strictly positive on the domain 5 ≤ w ≤ x ≤ y ≤ z ≤ w + x + y - 2. It follows that F(5,5,5,5) = 54 - 202 = 225 > 0 is a global minimum of F on this domain. Thus, F(w,x,y,z) = 0 has no solution for 5 ≤ w ≤ x ≤ y ≤ z ≤ w + x + y - 2, and we therefore must have
1 ≤ w ≤ 4
for equable cyclic integer quadrilaterals. Summarizing, we have the following table:
wxwxyF
w ≥ 5anyanyanyF > 0
1 ≤ w ≤ 4anyanyanyTBD

## Section 4: Limiting the values of x

We will next show that F(w,x,y,z) > 0 for 1 ≤ w ≤ 4 and 10 ≤ x ≤ y ≤ z ≤ w + x + y - 2. We first note that
wx ≥ (1)(10) = 10

 wx - 4 ≥ 6 > 0 [9]

Consider the function
Values of H are:
Since x ≥ 10, and H(w) < 10 for all feasible values of w, we have
 [10]
From the given constraints,
y ≥ x

wy - 4 ≥ wx - 4

We showed in [9] that wx - 4 > 0, so
(wx - 4)(wy - 4) ≥ (wx - 4)2 > 4(w2 + 4) = 4w2 + 16
by [10]. Then
(wx - 4)(wy - 4) - 16 > 4w2

wxy - 4x - 4y > 4w

wxy > 4(w + x + y) > 4(w + x + y) - 4

wxy > 2(2w + 2x + 2y - 2)

wxy > 2[w + x + y + (w + x + y - 2)] ≥ 2(w + x + y + z)

wxy - 2(w + x + y + z) > 0,

so dF/dz > 0, and by [8], all other partial derivatives of F are strictly positive for 1 ≤ w ≤ 4 and 10 ≤ x ≤ y ≤ z ≤ w + x + y - 2. It follows that F(1,10,10,10) = 103 - 312 = 39 > 0 is a global minimum of F on the specified domain. Therefore, F(w,x,y,z) is positive for all values in the specified domain, and all solutions to F(w,x,y,z) = 0 must occur when 1 ≤ w ≤ 4 and w ≤ x ≤ 9:
wxwxyF
w ≥ 5anyanyanyF > 0
1 ≤ w ≤ 4x ≥ 10anyanyF > 0
1 ≤ w ≤ 4w ≤ x ≤ 9anyanyTBD

## Section 5: Limiting the values of wx

We next show that F(w,x,y,z) < 0 always occurs when wx ≤ 4, 1 ≤ w ≤ 4, w ≤ x ≤ 9 and x ≤ y ≤ z ≤ w + x + y - 2. Subtracting y from terms of the last inequality, we have
0 ≤ z - y ≤ w + x - 2
Define the integer t as
t = z - y, where 0 ≤ t ≤ w + x - 2 ≤ 4 + 9 - 2 = 11

z = y + t

Then the function F in [7] can be written in the equivalent form
F(w,x,y,t) = wxy(y + t) - (w + x + 2y + t)2

 = (wx - 4)y2 + [wxt - 4(w + x + t)]y - (w + x + t)2 [11]

The quadratic coefficient of y in F is wx - 4 ≤ 0 by assumption. Then
(wx - 4)t ≤ 0 < 8 = 4(1 + 1) ≤ 4(w + x)

(wx - 4)t + 4t < 4(w + x) + 4t

wxt - 4(w + x + t) < 0

so the linear coefficient of y in F is strictly negative. Finally, the constant coefficient in F is -(w + x + t)2 < 0, so the sum of all terms in [11] is negative, and the inequality F < 0 holds.

Based on this result, the last case in the preceding table can be subdivided as follows:

wxwxyF
w ≥ 5anyanyanyF > 0
1 ≤ w ≤ 4x ≥ 10anyanyF > 0
1 ≤ w ≤ 4w ≤ x ≤ 9wx ≤ 4anyF < 0
1 ≤ w ≤ 4w ≤ x ≤ 9wx ≥ 5anyTBD

## Section 6: Limiting the values of y

Next, we assume that 1 ≤ w ≤ 4, w ≤ x ≤ 9, wx ≥ 5 and x ≤ y ≤ z ≤ w + x + y - 2, and will show that F(w,x,y,z) > 0 when y ≥ 102. A minimum for A2 is
A2 = wxyz ≥ (5)(y)(y) = 5y2
and a maximum for P2 is
Therefore, A2 > P2 always occurs when
5y2 > 4y2 + 96y + 576

y2 - 96y - 576 > 0

This inequality is satisfied when
or
y ≥ 102
(Tighter bounds on y can be found by considering individual cases for values of w.)

Based on this result, the last case in the preceding table can be subdivided as follows:

wxwxyF
w ≥ 5anyanyanyF > 0
1 ≤ w ≤ 4x ≥ 10anyanyF > 0
1 ≤ w ≤ 4w ≤ x ≤ 9wx ≤ 4anyF < 0
1 ≤ w ≤ 4w ≤ x ≤ 9wx ≥ 5y ≥ 102F > 0
1 ≤ w ≤ 4w ≤ x ≤ 9wx ≥ 5x ≤ y ≤ 101TBD

## Section 7: Enumeration of solutions

Now that we have found finite bounds for all variables w, x, y and z, searching for equable integer cyclic quadrilaterals is a manageable problem. There are
outcomes to be considered in the final case in the table above. The following pseudocode describes a simple program to enumerate these outcomes:
```	FOR w = 1 TO 4
FOR x = w TO 9
IF wx ≥ 5 THEN
FOR y = x TO 101
FOR z = y TO y + w + x - 2
P = w + x + y + z
A2 = wxyz
IF A2 is a perfect square and P is even THEN
A = integer square root of A2
IF P < A THEN
Increment count for integer area and P < A
ELSE IF P = A THEN
Increment count for integer area and P = A
PRINT w,x,y,z,P
ELSE
Increment count for integer area and P > A
END IF
ELSE
IF P2 < A2 THEN
Increment count for non-integer area and P < A
ELSE
Increment count for non-integer area and P > A
END IF
END IF
NEXT z
NEXT y
END IF
NEXT x
NEXT w
PRINT table of counts
END
```
The 18,765 outcomes are organized as follows:
1 ≤ w ≤ 4, w ≤ x ≤ 9, wx ≥ 5, x ≤ y ≤ 102
AreaP < AP = AP > ATotal
Integer *479414497
Non-integer17,999026918,268
Total18,478428318,765

*: Requires even perimeter for integer sides

The four equable cases are listed below:
wxyzP=Asabcd
191010301514655
255820108552
33661896633
44441684444

## Section 8: Plots

Since any two adjacent sides of a quadrilateral can be swapped along a diagonal without affecting the area or perimeter, the order of sides is arbitrary. For display purposes, we will follow the convention of orienting an equable quadrilateral with the longest side a along the horizontal x-axis and the origin at its left endpoint. The other sides b, c, d are arranged clockwise around the quadrilateral in decreasing order. The general form of the quadrilateral then is
where e, f, u and v are all positive. Let p be the length of the diagonal joining points (e,f) and (a,0), and let q be the length of the diagonal joining the origin to the point (u,v). By Ptolemy's Theorem, formulas for the diagonals of a cyclic quadrilateral are
If α is the angle between sides a and b, the Law of Cosines states that
and we can evaluate the coordinates of the point (e,f):
Similarly, if β is the angle between sides a and d, then
The coordinates of the point (u,v) are
The circumcircle of this quadrilateral passes through the non-collinear points (0,0), (a,0) and (u,v). The center and radius of the circle defined by these three points can be found by the method of determinants. Let
Then the center of the circle is
Evaluating the formulas for the vertex coordinates and circumcircle parameters of each equable integer cyclic quadrilateral, we have
44443232(4,0)(0,4)(4,4)(2,2)
6633144/545(6,0)(18/5,24/5)(6,3)(3,3/2)
8552412500/41(8,0)(3,4)(310/41,80/41)(4,1/8)
1465510000/109109(14,0)(546/109,360/109)(10,3)(7,-31/6)
The results above can be used to create scale drawings of the four equable integer cyclic quadrilaterals and their corresponding circumcircles:

## Notation

A: Area
a, b, c, d: Integer side lengths
B, C, D, E: Determinants for circumcircle calculations
e, f, u, v: Coordinates of plotted vertices
F: Diophantine polynomial function relating squared area and perimeter
G, H: Subsidiary functions
P: Perimeter
p, q: Lengths of diagonals