and related topics

*Balmoral Software*

Solutions: 4

- Problem definition
- Change of variables
- Limiting the values of w
- Limiting the values of x
- Limiting the values of wx
- Limiting the values of y
- Enumeration of solutions
- Plots
- Perimeter-dominant cyclic Heronian quadrilaterals
- Minimum area of a convex quadrilateral
- Equable non-cyclic Heronian quadrilaterals
- References

In general, we will consider cases where the area A of a CHQ is a rational factor t times its perimeter P:

so that t = 1 corresponds to the equable case. If t > 1, it is assumed to be of the form t = h/(h - 2) for some integer h ≥ 3, and will be used to determine equable quadrilateral prisms.

A = tP, 1 ≤ t ≤ 3, [1]

Exactly four CHQs are equable:

Following is a proof of the results for the general relation [1] that includes the four equable quadrilaterals as a subcase.

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

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

By the generalized polygon inequality, the sum of any three sides of a non-degenerate quadrilateral must exceed the fourth, so

and s - a, s - b, s - c and s - d are all strictly positive. Using Brahmagupta's formula for the area of a cyclic quadrilateral, we have

a < b + c + d, [2]

If the perimeter P = 2s = a + b + c + d is odd, then s is of the form k + 1/2 for some positive integer k, and so is each of the four factors in A

A ^{2}= (s - a)(s - b)(s - c)(s - d)[3]

The cyclic area formula [3] does not distinguish between different orderings of the sides, so without loss of generality, we can assume that the integer sides of a CHQ are ranked alphabetically from longest to shortest:

a ≥ b ≥ c ≥ d ≥ 1 [4]

w = s - aThese integers are all strictly positive by the generalized polygon inequality. Summing both sides of the equations above, we havex = s - b

y = s - c

z = s - d

w + x + y + z = 4s - (a + b + c + d) = 4s - 2s = PThe square of the CHQ area can now be written more simply as

AThe inverse relationships between w, x, y, z and the sides of the quadrilateral are:^{2}= wxyz

a = s - wFrom [2] and [4], we haveb = s - x

c = s - y

d = s - z

1 ≤ d ≤ c ≤ b ≤ a < b + c + dwhich can be written in terms of the new variables as

This is the fundamental relationship between w, x, y and z that will form the foundation for the analysis below. Variables in the relationship above are unbounded, so to find all CHQs satisfying [1], we will determine limits for w, x and y. All solutions can then be found by a computer-aided search over a finite and manageable set of possibilities.

1 ≤ w ≤ x ≤ y ≤ z ≤ w + x + y - 2 [5]

We begin by establishing a general technique that will be useful in the
following sections. Consider a differentiable real-valued multivariate function
f(x_{1},x_{2},...,x_{n}) over the open domain of nested
real variables

Assume the following three conditions hold:

x _{1}≤ x_{2}≤ ... ≤ x_{n}[6]

__Condition 1__: The partial derivatives are nested in reverse order over
the domain [6]:

df/dx _{n}≤ df/dx_{n-1}≤ ... ≤ df/dx_{1}[7]

f(L,L,..,L) > 0 [8]

Then it follows that

df/dx _{n}> 0 for x_{1}≥ L[9]

f(xover the domain L ≤ x_{1},x_{2},...,x_{n}) > 0

Proof. By [6], [7] and [9], f is strictly increasing in all variables when
x_{1} ≥ L. The additional condition [8] establishes the conclusion.

Define the multivariate function F_{t} as follows:

Fwhere t is the multiplicative factor defined in [1]. From the given constraints_{t}(w,x,y,z) = A^{2}- t^{2}P^{2}= wxyz - t^{2}(w + x + y + z)^{2},1 ≤ w ≤ x ≤ y ≤ z ≤ w + x + y - 2, 1 ≤ t ≤ 3,

we have

w ≤ x ≤ y ≤ z, [10]

1/z ≤ 1/y ≤ 1/x ≤ 1/wfor all t and all values of w, x, y and z in the domain [10], which establishes Condition 1 for the function Fwxy - 2t

^{2}(w + x + y + z) ≤ wxz - 2t^{2}(w + x + y + z) ≤ wyz - 2t^{2}(w + x + y + z) ≤ xyz - 2t^{2}(w + x + y + z)dF

_{t}/dz ≤ dF_{t}/dy ≤ dF_{t}/dx ≤ dF_{t}/dw

which holds for all t if

F _{t}(w,w,w,w) = w^{4}- t^{2}(4w)^{2}> 0

w > 4t,[11]

w ≥ 13,so that this lower bound establishes Condition 2 for the function F

z ≤ w + x + y - 2 < w + x + ysowxy - 2t

^{2}(w + x + y + z) > wxy - 4t^{2}(w + x + y),

dFfor_{t}/dz > G_{t}(w,x,y)

GFrom the given constraints [10],_{t}(w,x,y) = wxy - 4t^{2}(w + x + y).

1/y ≤ 1/x ≤ 1/wfor all t and all values of w, x and y in the domain [10], which establishes Condition 1 for the function Gwx - 4t

^{2}≤ wy - 4t^{2}≤ xy - 4t^{2}dG

_{t}/dy ≤ dG_{t}/dx ≤ dG_{t}/dw

Gwhich is already satisfied by [11]. Therefore, Condition 2 is established for the function G_{t}(w,w,w) = w^{3}- 4t^{2}(3w) > 0

wx ≥ wwhich establishes Condition 3 for G^{2}> (4t)^{2}> 4t^{2}dG

_{t}/dy = wx - 4t^{2}> 0,

w x wx y Solutions to A = tP w ≥ 13 any any any none (F _{t}> 0 for all t)1 ≤ w ≤ 12 any any any TBD

FThe solution to this cubic inequality is x > 88.4885, or_{t}(w,x,x,x) = wx^{3}- t^{2}(3x + w)^{2}≥ (1)x^{3}- (3)^{2}(3x + 12)^{2}> 0x

^{3}- 81(x + 4)^{2}> 0

which establishes Condition 2 for the function F

x ≥ 89, [12]

Gand_{t}(w,89,89) > 0

dGfor all w. Condition 1 still holds for G_{t}/dy > 0

w(89)which establishes Condition 2 for G^{2}- 4t^{2}[w + 2(89)] ≥ (1)(89)^{2}- 4(3)^{2}[12 + 2(89)] = 1081 > 0,

x ≥ 89 > 36 = 4(3)which establishes Condition 3 for G^{2}≥ 4t^{2}≥ 4t^{2}/wwx > 4t

^{2}dG

_{t}/dy = wx - 4t^{2}> 0,

w x wx y Solutions to A = tP w ≥ 13 any any any none (F _{t}> 0 for all t)1 ≤ w ≤ 12 x ≥ 89 any any none (F _{t}> 0 for all t)1 ≤ w ≤ 12 w ≤ x ≤ 88 any any TBD

FIf wx - 4t_{t}(w,x,y,z) = wxyz - t^{2}(w + x + y + z)^{2}< wxy(y + w + x) - t

^{2}(2y + w + x)^{2}= (wx - 4t

^{2})y^{2}+ (wx - 4t^{2})(w + x)y - t^{2}(w + x)^{2}

w x wx y Solutions to A = tP w ≥ 13 any any any none (F _{t}> 0 for all t)1 ≤ w ≤ 12 x ≥ 89 any any none (F _{t}> 0 for all t)1 ≤ w ≤ 12 w ≤ x ≤ 88 wx ≤ 4t ^{2}any none (F _{t}< 0 for all t)1 ≤ w ≤ 12 w ≤ x ≤ 88 wx > 4t ^{2}any TBD

wx ≥ ⌈4twhere ⌈⌉ and ⌊⌋ are the ceiling and floor functions, respectively. Thus, in all cases,^{2}⌉ = ⌊4t^{2}⌋ + 1,

From the prism convention in the Introduction, we have

wx ≥ ⌊4t ^{2}⌋ + 1[13]

1/289 ≤ ⌊4tor^{2}⌋ - 4t^{2}+ 1 ≤ 1,

A minimum for A

Aby [13], and a maximum for t^{2}= (wx)yz ≥ (⌊4t^{2}⌋ + 1)y^{2}

tTherefore, A^{2}P^{2}= t^{2}(w + x + y + z)^{2}≤ t

^{2}(w + x + y + w + x + y - 2)^{2}≤ t

^{2}[2(12) + 2(88) + 2y - 2]^{2}= 4t

^{2}y^{2}+ 792t^{2}y + 39,204t^{2}

(⌊4twhich holds if^{2}⌋ + 1)y^{2}> 4t^{2}y^{2}+ 792t^{2}y + 39,204t^{2}

ywhich in turn holds for all t if^{2}> 289(792y + 39,204)t^{2},

yThe solution to this quadratic inequality is^{2}- 2,059,992y - 101,969,604 > 0,

or

y ≥ 2,060,042(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 Solutions to A = tP w ≥ 13 any any any none (F _{t}> 0 for all t)1 ≤ w ≤ 12 x ≥ 89 any any none (F _{t}> 0 for all t)1 ≤ w ≤ 12 w ≤ x ≤ 88 wx ≤ 4t ^{2}any none (F _{t}< 0 for all t)1 ≤ w ≤ 12 w ≤ x ≤ 88 wx > 4t ^{2}y ≥ 2,060,042 none (F _{t}> 0 for all t)1 ≤ w ≤ 12 w ≤ x ≤ 88 wx > 4t ^{2}x ≤ y ≤ 2,060,041 TBD

For each of these possibilities, the following steps will be completed:

Verify that P = w + x + y + z is evenThe four equable CHQs are listed below, in order of increasing area:

Verify that A^{2}= wxyz is a square

If A = P, output results for equable quadrilaterals

If A > P and 2A/(A - P) is an integer, output results for equable quadrilateral prisms

If A < P and wx > 4, output preliminary results for perimeter-dominant CHQs

The first three of these are bicentric since each has the square of its area equal to the product of its side lengths (A

a b c d A=P w x y z Possible shapes (depending on order of sides) 4 4 4 4 16 4 4 4 4 Square 6 6 3 3 18 3 3 6 6 Rectangle, right kite 8 5 5 2 20 2 5 5 8 Isosceles trapezoid, trapezium* 14 6 5 5 30 1 9 10 10 Isosceles trapezoid, trapezium*

*: American definition

a < b + c + dmust be satisfied. The general form of the quadrilateral then is:

where y and v are positive. Let p be the length of the descending diagonal joining points (x,y) and (a,0), and let q be the length of the ascending diagonal joining the origin to the point (u,v). By Ptolemy's Theorem, formulas for the diagonals are

If α is the angle between sides a and b, then by the Law of Cosines, we have

Similarly, if β is the angle between sides a and d, then

The circumcircle of the quadrilateral passes through the non-collinear points (0,0), (a,0) and (u,v). As is shown in the results for equable triangles, the center and radius of the circle defined by these three points are:

Evaluating the formulas for the vertex coordinates and circumcircle parameters of each equable CHQ, we have the following table:

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

a b c d p ^{2}q ^{2}(a,0) (x,y) (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)

From the table in a preceding section, all other solutions for perimeter greater than integer area (F

a b c d P A P-A w x y z 3 3 3 3 12 9 3 3 3 3 3 4 4 3 3 14 12 2 3 3 4 4 5 5 3 3 16 15 1 3 3 5 5 7 5 5 1 18 16 2 2 4 4 8 9 5 5 1 20 15 5 1 5 5 9 8 7 4 1 20 18 2 2 3 6 9 10 5 5 2 22 18 4 1 6 6 9 11 7 4 2 24 20 4 1 5 8 10 11 5 5 3 24 21 3 1 7 7 9 12 5 5 4 26 24 2 1 8 8 9 13 5 5 5 28 27 1 1 9 9 9 14 8 7 1 30 28 2 1 7 8 14 15 10 6 1 32 30 2 1 6 10 15 20 16 5 1 42 40 2 1 5 16 20

(w,x) ∈ {(1,1),(1,2),(1,3),(1,4),(2,2)},but the value of y is unbounded. Define the integer k as

k = z - y, where 0 ≤ k ≤ w + x - 2 by [5]The perimeter can then be written

P = 2y + w + x + k,and must be even in order for the quadrilateral area to be an integer, so k and w + x - 2 have the same parity. We then have the following cases where the perimeter may exceed the integer area:

Each subcase can be denoted by the capital letter for the main case followed by a feasible value for k; for example, Case C2 occurs when (w,x) = (1,3) and k = 2. In each subcase, the squared area is wxy(y + k), which may have infinitely-many solutions for y. Some of these infinite sequences of solutions for y have been found in the Online Encyclopedia of Integer Sequences (oeis.org), as shown below:

Case (w,x) w + x - 2 k A (1,1) 0 0 B (1,2) 1 1 C (1,3) 2 0,2 D (1,4) 3 1,3 E (2,2) 2 0,2

__Case A0__: (w,x,y,z) = (1,1,y,y)

AExamples:^{2}= y^{2}is a square for all values of yP = 2y + 2 = 2A + 2 is a line in the P-A plane

w x y z P A P-A a b c d 1 1 1 1 4 1 3 1 1 1 1 1 1 2 2 6 2 4 2 2 1 1 1 1 3 3 8 3 5 3 3 1 1 1 1 4 4 10 4 6 4 4 1 1 1 1 5 5 12 5 7 5 5 1 1 1 1 6 6 14 6 8 6 6 1 1 ...

__Case B1__: (w,x,y,z) = (1,2,y,y+1)

The perimeter and area formulas are

w x yzPAP-Aabc d 1 2 8 9 20 12 8 9 8 2 1 1 2 49 50 102 70 32 50 49 2 1 1 2 288 289 580 408 172 289 288 2 1 1 2 1681 1682 3366 2378 988 1682 1681 2 1 1 2 9800 9801 19604 13860 5744 9801 9800 2 1 1 2 57121 57122 114246 80782 33464 57122 57121 2 1 1 2 332928 332929 665860 470832 195028 332929 332928 2 1 1 2 1940449 1940450 3880902 2744210 1136692 1940450 1940449 2 1 1 2 11309768 11309769 22619540 15994428 6625112 11309769 11309768 2 1 ... OEIS: A001108 A001542 A055997

P = 2y + 4which is a modified Pell equation with parameters D = 2 and p = 3. The associated coefficients are aA

^{2}= 2y(y + 1)(P - 3)

^{2}- 2A^{2}= 1,

P_{n}= 6P_{n-1}- P_{n-2}- 12, P_{0}= 20, P_{1}= 102A

_{n}= 6A_{n-1}- A_{n-2}, A_{0}= 12, A_{1}= 70

A^{2}= 3y^{2}has no solution in integers

The perimeter and area formulas are

w x yzPAP-Aabc d 1 3 6 8 18 12 6 8 6 3 1 1 3 25 27 56 45 11 27 25 3 1 1 3 96 98 198 168 30 98 96 3 1 1 3 361 363 728 627 101 363 361 3 1 1 3 1350 1352 2706 2340 366 1352 1350 3 1 1 3 5041 5043 10088 8733 1355 5043 5041 3 1 1 3 18816 18818 37638 32592 5046 18818 18816 3 1 1 3 70225 70227 140456 121635 18821 70227 70225 3 1 1 3 262086 262088 524178 453948 70230 262088 262086 3 1 ... OEIS: A092184 A102206 A005320 A102206 A092184

P = 2y + 6Since P is even, P/2 - 2 is an integer. The product 3y(y + 2) mod 9 is either 0 or 6 and a square mod 9 is one of {0,1,4,7}. The only common remainder is AA

^{2}= 3y(y + 2)(P/2 - 2)

^{2}- 3(A/3)^{2}= 1

P_{n}= 4P_{n-1}- P_{n-2}- 8, P_{0}= 18, P_{1}= 56A

_{n}= 4A_{n-1}- A_{n-2}, A_{0}= 12, A_{1}= 45

Ahas no solution in integers.^{2}= 4y(y + 1) = (2y + 1)^{2}- 1(2y + 1)

^{2}- A^{2}= 1

__Case D3__: (w,x,y,z) = (1,4,y,y+3)

Acannot occur for y ≥ x = 4 (or equivalently, 2y + 3 ≥ 11) since the next lower square will be at least 21 less.^{2}= 4y(y + 3) = (2y + 3)^{2}- 9(2y + 3)

^{2}- A^{2}= 9

__Case E0__: (w,x,y,z) = (2,2,y,y)

AExamples:^{2}= 4y^{2}is a square for all values of yP = 2y + 4 = A + 4 is a line in the P-A plane

There are exactly two instances where Case E0 produces the same perimeter and integer area as found in the preceding table of results for wx ≥ 5. These duplicates are:

w x y z P A P-A a b c d 2 2 2 2 8 4 4 2 2 2 2 2 2 3 3 10 6 4 3 3 2 2 2 2 4 4 12 8 4 4 4 2 2 2 2 5 5 14 10 4 5 5 2 2 2 2 6 6 16 12 4 6 6 2 2 2 2 7 7 18 14 4 7 7 2 2 ...

(w,x,y,z) = (2,2,9,9) or (1,6,6,9), P = 22 & A = 18

(w,x,y,z) = (2,2,10,10) or (1,5,8,10), P = 24 & A = 20

Ahas no solution in integers since the difference of two positive squares cannot be 4.^{2}= 4y(y + 2) = (2y + 2)^{2}- 4(2y + 2)

^{2}- A^{2}= 4

A scatter diagram shows all the perimeter-dominant CHQs in the P-A plane, with each (P,A) pair represented by a pixel. In this diagram, the perimeter P increases from left to right and the area A from bottom to top, so perimeter-dominant solutions are in the lower right portion of the diagram. For reference, the points P = A are shown as a dimmed diagonal line. A zoomed-in version shows more detail for smaller values of P and A.

The 4 equable cases are shown in purple on the scatter diagrams, and the 14 initial perimeter-dominant cases in the table above are shown in black. Other colors are allocated according to the following summary table:

Case y-sequence A P Diagram colorA0 Any positive integer y 2y + 2 Blue B1 8,49,288,1681,9800,...

(OEIS A001108)

(OEIS A001542)2y + 4 Orange C2 6,25,96,361,1350,...

(OEIS A092184)

(OEIS A005320)2y + 6 Green E0 Any integer ≥ 2 2y 2y + 4 Red

(except for two

duplicates mentioned above)

The minimum area of a convex quadrilateral with specified lengths and order of sides can be determined from the areas of the six triangles formed by combining two of the quadrilateral sides into a straight line segment [B] [C]. To allow a zero minimum area for convex quadrilaterals that can be completely collapsed (such as any rhombus), a non-strict triangle inequality applies:

for a triangle with sides x,y,z. The minimum area of a convex quadrilateral with sides {a,b,c,d} (in any order) is

max{x,y,z} ≤ (x + y + z)/2 [14]

min{T(a+b,c,d),T(a+c,b,d),T(a+d,b,c),T(b+c,a,d),T(b+d,a,c),T(c+d,a,b)}where T(x,y,z) represents the area of triangle {x,y,z}. A triangular area term is excluded from consideration if [14] is not satisfied. For example, the convex quadrilateral {7,3,4,5} has associated triangle areas

T(7,7,5) = 16.346so its minimum area is 8.786. Clearly, all minimum-area convex quadrilaterals are non-cyclic since three of their vertices are collinear and therefore cannot be on a circumcircle.

T(8,7,4) = 13.998

T(9,7,3) = 8.786

The minimum and maximum areas of the 6,328 integer-sided quadrilaterals with perimeters up to 50 are listed here. The order of sides is arbitrary since all arrangements are considered for the area extrema, so we have chosen the convention of listing sides in decreasing order. Somewhat less than half of the quadrilaterals have an area range that includes the perimeter value; these quadrilaterals can be equable and are indicated in the list by an asterisk.

We finish this section with a simple result regarding the minimum perimeter of equable quadrilaterals:

*All quadrilaterals with perimeters less than 16 are perimeter-dominant (P >
A), and therefore cannot be equable.*

Proof. Let x_{1},x_{2},x_{3},x_{4} be the
real-valued sides of such a quadrilateral, and let s be its semiperimeter. By
[A], it suffices to show that the area A_{c}
of a cyclic quadrilateral with these sides is perimeter-dominant. By the
generalized polygon inequality,
s - x_{i} is positive for all i. It follows from the
AM-GM Inequality that

since P < 16, Q.E.D.

A = aAs the side length increases, the equable rhombus becomes more elongated in order to maintain equal area and perimeter.^{2}sin(θ) = 4a = Pθ = Arcsin(4/a)

Unlike their cyclic counterparts, ENCHQs are not required to have an even
perimeter in order to have integer sides and area. Let a,b,c,d be the integer
sides of a convex quadrilateral, arranged clockwise with side a horizontally on
the bottom, such that the quadrilateral
inequality is satisfied. As in the diagram above,
let p be the length of the descending diagonal from the vertex joining sides b
and c to the vertex joining sides d and a. This diagonal divides the
quadrilateral into two triangles; let A_{1} be the area of the triangle
with sides (a,b,p) and A_{2} the area of the triangle with sides
(p,c,d). By equability, we have

A = Aor equivalently,_{1}+ A_{2}= a + b + c + d = P,

Note that the triangle areas A

(16A _{1}^{2}- 16A_{2}^{2})^{2}- 32(16A_{1}^{2}+ 16A_{2}^{2})P^{2}+ 256P^{4}= 0[15]

HThen [15] can be written_{1}= 16A_{1}^{2}H

_{2}= 16A_{2}^{2}

From Heron's area formula, we have

(H _{1}- H_{2})^{2}- 32(H_{1}+ H_{2})P^{2}+ 256P^{4}= 0[16]

Hand similarly,_{1}= (p + a + b)(-p + a + b)(p - a + b)(p + a - b)= -p

^{4}+ 2(a^{2}+ b^{2})p^{2}- (a^{2}- b^{2})^{2},

Hso_{2}= -p^{4}+ 2(c^{2}+ d^{2})p^{2}- (c^{2}- d^{2})^{2},

Hin terms of the following integers:_{1}- H_{2}= 2(a^{2}+ b^{2}- c^{2}- d^{2})p^{2}- (a^{2}- b^{2})^{2}+ (c^{2}- d^{2})^{2}= 2Wp

^{2}- X

W = aIn a similar fashion,^{2}+ b^{2}- c^{2}- d^{2}X = (a

^{2}- b^{2})^{2}- (c^{2}- d^{2})^{2}

Hin terms of the following integers:_{1}+ H_{2}= -2p^{4}+ 2Yp^{2}- Z

Y = aWe can thus write [16] as^{2}+ b^{2}+ c^{2}+ d^{2}> 0Z = (a

^{2}- b^{2})^{2}+ (c^{2}- d^{2})^{2}> 0

(2Wpor^{2}- X)^{2}- 32(-2p^{4}+ 2Yp^{2}- Z)P^{2}+ 256P^{4}= 0,

where the integer coefficients of the quadratic in p

c _{2}p^{4}+ c_{1}p^{2}+ c_{0}= 0,[17]

cBy Descartes' Rule of Signs, there can be positive solutions for p_{2}= 4(W^{2}+ 16P^{2}) > 0c

_{1}= -4(WX + 16YP^{2})c

_{0}= X^{2}+ 32ZP^{2}+ 256P^{4}> 0

**[B]** Garza-Hume et. al., Areas and Shapes of Planar Irregular
Polygons, *Forum Geometricorum*, 18, p. 25.

**[C]** Böröczky, K. et. al., The minimum area of a
simple polygon with given side lengths, *Periodica Mathematica Hungarica*,
39, Numbers 1-3 (2000) 33-49.

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