12/30/16

- 1. Problem definition
- 2. Change of variables
- 3. Limiting the values of w
- 4. Limiting the values of x
- 5. Limiting the values of wx
- 6. Limiting the values of y
- 7. Enumeration of solutions
- 8. Plots
- Notation

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 = P^{2}/A and area k^{2}A = P^{2}/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:

Following is a proof of these results.

Sides Perimeter = Area (4,4,4,4) 16 (6,6,3,3) 18 (8,5,5,2) 20 (14,6,5,5) 30

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 haveand analogously,

b + c + d > a

[1]

-a + b + c + d > 0

[2]

Using Brahmagupta's formula for the area A of a cyclic quadrilateral, we have

s - b > 0 [3]

s - c > 0 [4]

s - d > 0 [5]

AIf 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 A^{2}= (s - a)(s - b)(s - c)(s - d)

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]

w = s - aThese integers are all positive by [2]-[5]. Summing both sides,x = s - b

y = s - c

z = s - d

w + x + y + z = 4s - (a + b + c + d) = 4s - 2s = 2s = a + b + c + d = PThe inverse relationships between w, x, y, z and the sides of the quadrilateral are:

a = s - wwhere s = (w + x + y + z)/2.b = s - x

c = s - y

d = s - z

From [1] and [6], we have

b + c + d > a ≥ b ≥ c ≥ d ≥ 1Substituting 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 - 1Since -w < w implies that w ≥ 1, and z ≤ s - 1 is equivalent to

z ≤ (w + x + y + z)/2 - 1we have2z ≤ w + x + y + z - 2

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.

1 ≤ w ≤ x ≤ y ≤ z ≤ w + x + y - 2

We can define a multivariate function F as follows:

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.

F(w,x,y,z) = A ^{2}- P^{2}= wxyz - (w + x + y + z)^{2}, 1 ≤ w ≤ x ≤ y ≤ z ≤ w + x + y - 2[7]

From the given constraints,

1/z ≤ 1/y ≤ 1/x ≤ 1/wLet the function G be defined bywxyz/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]

G(w,x,y) = wxy - 4(w + x + y)If w ≥ 5, then

wx ≥ wand^{2}≥ 25wx - 4 ≥ 21 > 0

1/y ≤ 1/x ≤ 1/wso all partial derivatives of G are strictly positive. It follows that G(5,5,5) = 5wxy/y - 4 ≤ wxy/x - 4 ≤ wxy/w - 4

0 < wx - 4 ≤ wy - 4 ≤ xy - 4

0 < dG/dy ≤ dG/dx ≤ dG/dw,

We have

G(w,x,y) = wxy - 4(w + x + y) > 0so 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) = 5wxy > 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,

1 ≤ w ≤ 4for equable cyclic integer quadrilaterals. Summarizing, we have the following table:

w x wx y F w ≥ 5 any any any F > 0 1 ≤ w ≤ 4 any any any TBD

wx ≥ (1)(10) = 10Consider the function Values of H are: Since x ≥ 10, and H(w) < 10 for all feasible values of w, we have

wx - 4 ≥ 6 > 0 [9]

From the given constraints,

[10]

y ≥ xWe showed in [9] that wx - 4 > 0, sowy - 4 ≥ wx - 4

(wx - 4)(wy - 4) ≥ (wx - 4)by [10]. Then^{2}> 4(w^{2}+ 4) = 4w^{2}+ 16

(wx - 4)(wy - 4) - 16 > 4wso 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) = 10^{2}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,

w x wx y F w ≥ 5 any any any F > 0 1 ≤ w ≤ 4 x ≥ 10 any any F > 0 1 ≤ w ≤ 4 w ≤ x ≤ 9 any any TBD

0 ≤ z - y ≤ w + x - 2Define the integer t as

t = z - y, where 0 ≤ t ≤ w + x - 2 ≤ 4 + 9 - 2 = 11Then the function F in [7] can be written in the equivalent formz = y + t

F(w,x,y,t) = wxy(y + t) - (w + x + 2y + t)The quadratic coefficient of y in F is wx - 4 ≤ 0 by assumption. Then^{2}

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

(wx - 4)t ≤ 0 < 8 = 4(1 + 1) ≤ 4(w + x)so the linear coefficient of y in F is strictly negative. Finally, the constant coefficient in F is -(w + x + t)(wx - 4)t + 4t < 4(w + x) + 4t

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

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

w x wx y F w ≥ 5 any any any F > 0 1 ≤ w ≤ 4 x ≥ 10 any any F > 0 1 ≤ w ≤ 4 w ≤ x ≤ 9 wx ≤ 4 any F < 0 1 ≤ w ≤ 4 w ≤ x ≤ 9 wx ≥ 5 any TBD

Aand a maximum for P^{2}= wxyz ≥ (5)(y)(y) = 5y^{2}

5yThis inequality is satisfied when or^{2}> 4y^{2}+ 96y + 576y

^{2}- 96y - 576 > 0

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:

w x wx y F w ≥ 5 any any any F > 0 1 ≤ w ≤ 4 x ≥ 10 any any F > 0 1 ≤ w ≤ 4 w ≤ x ≤ 9 wx ≤ 4 any F < 0 1 ≤ w ≤ 4 w ≤ x ≤ 9 wx ≥ 5 y ≥ 102 F > 0 1 ≤ w ≤ 4 w ≤ x ≤ 9 wx ≥ 5 x ≤ y ≤ 101 TBD

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 PThe 18,765 outcomes are organized as follows:^{2}< 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

1 ≤ w ≤ 4, w ≤ x ≤ 9, wx ≥ 5, x ≤ y ≤ 102The four equable cases are listed below:

Area P < A P = A P > A Total Integer * 479 4 14 497 Non-integer 17,999 0 269 18,268 Total 18,478 4 283 18,765 *: Requires even perimeter for integer sides

w x y z P=A s a b c d 1 9 10 10 30 15 14 6 5 5 2 5 5 8 20 10 8 5 5 2 3 3 6 6 18 9 6 6 3 3 4 4 4 4 16 8 4 4 4 4

The results above can be used to create scale drawings of the four equable integer cyclic quadrilaterals and their corresponding circumcircles:

a b c d p ^{2}q ^{2}(a,0) (e,f) (u,v) Radius Center 4 4 4 4 32 32 (4,0) (0,4) (4,4) (2,2) 6 6 3 3 144/5 45 (6,0) (18/5,24/5) (6,3) (3,3/2) 8 5 5 2 41 2500/41 (8,0) (3,4) (310/41,80/41) (4,1/8) 14 6 5 5 10000/109 109 (14,0) (546/109,360/109) (10,3) (7,-31/6)

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

R: Radius of circumcircle

s: Semiperimeter

t: Integer variable representing the difference between z and y

w, x, y, z: Integer variables related to sides

α, β, δ: Angles between sides

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