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Solutions: 4
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 A2, whence the product is of the form m + n/16 for positive integers m and n, n < 16. 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.
A2 = (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
A2 = wxyzThe inverse relationships between w, x, y, z and the sides of the quadrilateral are:
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(x1,x2,...,xn) over the open domain of nested real variables
Assume the following three conditions hold:
x1 ≤ x2 ≤ ... ≤ xn [6]
Condition 1: The partial derivatives are nested in reverse order over the domain [6]:
Condition 2: There exists a lower bound L such that
∂f/∂xn ≤ ∂f/∂xn-1 ≤ ... ≤ ∂f/∂x1 [7]
Condition 3: The partial derivative
f(L,L,..,L) > 0 [8]
Then it follows that
∂f/∂dxn > 0 for x1 ≥ L [9]
f(x1,x2,...,xn) > 0over the domain L ≤ x1 ≤ x2 ≤ ... ≤ xn.
Proof. By [6], [7] and [9], f is strictly increasing in all variables when x1 ≥ L. The additional condition [8] establishes the conclusion.
Define the multivariate function Ft as follows:
Ft(w,x,y,z) = A2 - t2P2 = wxyz - t2(w + x + y + z)2,where t is the multiplicative factor defined in [1]. From the given constraints1 ≤ 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 Ft.wxy - 2t2(w + x + y + z) ≤ wxz - 2t2(w + x + y + z) ≤ wyz - 2t2(w + x + y + z) ≤ xyz - 2t2(w + x + y + z)
dFt/dz ≤ dFt/dy ≤ dFt/dx ≤ dFt/dw
which holds for all t if
Ft(w,w,w,w) = w4 - t2(4w)2 > 0
w > 4t,[11]
w ≥ 13,so that this lower bound establishes Condition 2 for the function Ft. We have
z ≤ w + x + y - 2 < w + x + ysowxy - 2t2(w + x + y + z) > wxy - 4t2(w + x + y),
dFt/dz > Gt(w,x,y)for
Gt(w,x,y) = wxy - 4t2(w + x + y).From the given constraints [10],
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 Gt. Consider the minimum value of w for which Gt(w,w,w) > 0:wx - 4t2 ≤ wy - 4t2 ≤ xy - 4t2
dGt/dy ≤ dGt/dx ≤ dGt/dw
Gt(w,w,w) = w3 - 4t2(3w) > 0which is already satisfied by [11]. Therefore, Condition 2 is established for the function Gt. From [10] and [11], we can write
wx ≥ w2 > (4t)2 > 4t2which establishes Condition 3 for Gt, and we conclude that Gt(w,x,y) > 0 over the domain 13 ≤ w ≤ x ≤ y. Therefore dFt/dz > 0, which establishes Condition 3 for the function Ft, and it follows that Ft is strictly positive when w ≥ 13. Summarizing, we have the following table:dGt/dy = wx - 4t2 > 0,
w x wx y Solutions to A = tP w ≥ 13 any any any none (Ft > 0 for all t) 1 ≤ w ≤ 12 any any any TBD
Ft(w,x,x,x) = wx3 - t2(3x + w)2 ≥ (1)x3 - (3)2(3x + 12)2 > 0The solution to this cubic inequality is x > 88.4885, orx3 - 81(x + 4)2 > 0
which establishes Condition 2 for the function Ft. As in the preceding section, Condition 3 for Ft is established when
x ≥ 89, [12]
Gt(w,89,89) > 0and
dGt/dy > 0for all w. Condition 1 still holds for Gt on the domain [10]. Again applying the extreme values of w and t, the former inequality is
w(89)2 - 4t2[w + 2(89)] ≥ (1)(89)2 - 4(3)2[12 + 2(89)] = 1081 > 0,which establishes Condition 2 for Gt. From [12], we can write
x ≥ 89 > 36 = 4(3)2 ≥ 4t2 ≥ 4t2/wwhich establishes Condition 3 for Gt and Ft, and it follows that Ft is strictly positive when 1 ≤ w ≤ 12 and x ≥ 89:wx > 4t2
dGt/dy = wx - 4t2 > 0,
w x wx y Solutions to A = tP w ≥ 13 any any any none (Ft > 0 for all t) 1 ≤ w ≤ 12 x ≥ 89 any any none (Ft > 0 for all t) 1 ≤ w ≤ 12 w ≤ x ≤ 88 any any TBD
Ft(w,x,y,z) = wxyz - t2(w + x + y + z)2If wx - 4t2 ≤ 0, then all coefficients of this quadratic in y are zero or negative, and Ft < 0 for all values of y. Based on this result, the last case in the preceding table can be subdivided as follows:< wxy(y + w + x) - t2(2y + w + x)2
= (wx - 4t2)y2 + (wx - 4t2)(w + x)y - t2(w + x)2
w x wx y Solutions to A = tP w ≥ 13 any any any none (Ft > 0 for all t) 1 ≤ w ≤ 12 x ≥ 89 any any none (Ft > 0 for all t) 1 ≤ w ≤ 12 w ≤ x ≤ 88 wx ≤ 4t2 any none (Ft < 0 for all t) 1 ≤ w ≤ 12 w ≤ x ≤ 88 wx > 4t2 any TBD
wx ≥ ⌈4t2⌉ = ⌊4t2⌋ + 1,where ⌈⌉ and ⌊⌋ are the ceiling and floor functions, respectively. Thus, in all cases,
From the prism convention in the Introduction, we have
wx ≥ ⌊4t2⌋ + 1 [13]
1/289 ≤ ⌊4t2⌋ - 4t2 + 1 ≤ 1,or
A minimum for A2 is
A2 = (wx)yz ≥ (⌊4t2⌋ + 1)y2by [13], and a maximum for t2P2 is
t2P2 = t2(w + x + y + z)2Therefore, A2 > t2P2 will occur when the minimum value of A2 exceeds the maximum value of t2P2:≤ t2(w + x + y + w + x + y - 2)2
≤ t2[2(12) + 2(88) + 2y - 2]2
= 4t2y2 + 792t2y + 39,204t2
(⌊4t2⌋ + 1)y2 > 4t2y2 + 792t2y + 39,204t2which holds if
y2 > 289(792y + 39,204)t2,which in turn holds for all t if
y2 - 2,059,992y - 101,969,604 > 0,The solution to this quadratic inequality is
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 (Ft > 0 for all t) 1 ≤ w ≤ 12 x ≥ 89 any any none (Ft > 0 for all t) 1 ≤ w ≤ 12 w ≤ x ≤ 88 wx ≤ 4t2 any none (Ft < 0 for all t) 1 ≤ w ≤ 12 w ≤ x ≤ 88 wx > 4t2 y ≥ 2,060,042 none (Ft > 0 for all t) 1 ≤ w ≤ 12 w ≤ x ≤ 88 wx > 4t2 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 A2 = 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 (A2 = abcd).
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 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 p2 q2 (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 (F1 < 0) occur when wx ≤ 4. Since we also have x ≥ w, this results in exactly five different value pairs for (w,x):
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)
A2 = y2 is a square for all values of yExamples:P = 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 y z P A P-A a b c 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 a1 = 1, b1 = -3, a2 = 1, b2 = 0, so 2b1(p - 1)/a1 = -12, 2b2(p - 1)/a2 = 0, and recursive solutions for the perimeter and area are:A2 = 2y(y + 1)
(P - 3)2 - 2A2 = 1,
Pn = 6Pn-1 - Pn-2 - 12, P0 = 20, P1 = 102Case C0: (w,x,y,z) = (1,3,y,y)An = 6An-1 - An-2, A0 = 12, A1 = 70
A2 = 3y2 has no solution in integersCase C2: (w,x,y,z) = (1,3,y,y+2)
The perimeter and area formulas are
w x y z P A P-A a b c 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 A2 mod 9 = 0, so A/3 is an integer and the preceding is a modified Pell equation with parameters D = 3 and p = 2. The associated coefficients are a1 = 1/2, b1 = -2, a2 = 1/3, b2 = 0, so 2b1(p - 1)/a1 = -8, 2b2(p - 1)/a2 = 0, and recursive solutions for the perimeter and area are:A2 = 3y(y + 2)
(P/2 - 2)2 - 3(A/3)2 = 1
Pn = 4Pn-1 - Pn-2 - 8, P0 = 18, P1 = 56Case D1: (w,x,y,z) = (1,4,y,y+1)An = 4An-1 - An-2, A0 = 12, A1 = 45
A2 = 4y(y + 1) = (2y + 1)2 - 1has no solution in integers.(2y + 1)2 - A2 = 1
Case D3: (w,x,y,z) = (1,4,y,y+3)
A2 = 4y(y + 3) = (2y + 3)2 - 9cannot occur for y ≥ x = 4 (or equivalently, 2y + 3 ≥ 11) since the next lower square will be at least 21 less.(2y + 3)2 - A2 = 9
Case E0: (w,x,y,z) = (2,2,y,y)
A2 = 4y2 is a square for all values of yExamples:P = 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 = 18Case E2: (w,x,y,z) = (2,2,y,y+2)
(w,x,y,z) = (2,2,10,10) or (1,5,8,10), P = 24 & A = 20
A2 = 4y(y + 2) = (2y + 2)2 - 4has no solution in integers since the difference of two positive squares cannot be 4.(2y + 2)2 - A2 = 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 color A0 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 x1,x2,x3,x4 be the real-valued sides of such a quadrilateral, and let s be its semiperimeter. By [A], it suffices to show that the area Ac of a cyclic quadrilateral with these sides is perimeter-dominant. By the generalized polygon inequality, s - xi is positive for all i. It follows from the AM-GM Inequality that
since P < 16, Q.E.D.
A = a2 sin(θ) = 4a = PAs the side length increases, the equable rhombus becomes more elongated in order to maintain equal area and perimeter.θ = 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 A1 be the area of the triangle with sides (a,b,p) and A2 the area of the triangle with sides (p,c,d). By equability, we have
A = A1 + A2 = a + b + c + d = P,or equivalently,
Note that the triangle areas A1 and A2 do not themselves need to be integers, just their sum. Define H1 and H2 as
(16A12 - 16A22)2 - 32(16A12 + 16A22)P2 + 256P4 = 0 [15]
H1 = 16A12Then [15] can be writtenH2 = 16A22
From Heron's area formula, we have
(H1 - H2)2 - 32(H1 + H2)P2 + 256P4 = 0 [16]
H1 = (p + a + b)(-p + a + b)(p - a + b)(p + a - b)and similarly,= -p4 + 2(a2 + b2)p2 - (a2 - b2)2,
H2 = -p4 + 2(c2 + d2)p2 - (c2 - d2)2,so
H1 - H2 = 2(a2 + b2 - c2 - d2)p2 - (a2 - b2)2 + (c2 - d2)2in terms of the following integers:= 2Wp2 - X
W = a2 + b2 - c2 - d2In a similar fashion,X = (a2 - b2)2 - (c2 - d2)2
H1 + H2 = -2p4 + 2Yp2 - Zin terms of the following integers:
Y = a2 + b2 + c2 + d2 > 0We can thus write [16] asZ = (a2 - b2)2 + (c2 - d2)2 > 0
(2Wp2 - X)2 - 32(-2p4 + 2Yp2 - Z)P2 + 256P4 = 0,or
where the integer coefficients of the quadratic in p2 are
c2p4 + c1p2 + c0 = 0, [17]
c2 = 4(W2 + 16P2) > 0By Descartes' Rule of Signs, there can be positive solutions for p2 only when c1 < 0. There are up to two real values of p that satisfy [17] and do not violate the triangle inequalities or the convexity of the resulting quadrilateral; for example, the (5,5,3,4) quadrilateral with descending diagonals 6.28631 and 5.13017:c1 = -4(WX + 16YP2)
c0 = X2 + 32ZP2 + 256P4 > 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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