| [2] |
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Solutions: 5
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 triangular prisms. In a later section, other results are presented for cases where the perimeter of an HT exceeds its area (t < 1).
A = tP, 1 ≤ t ≤ 3, [1]
As was proven over a century ago, exactly five HTs are equable [A]:
Following is a proof of the results for the general relation [1] that includes the five equable triangles as a particular case.
Sides Perimeter = Area (10,8,6) * 24 (13,12,5) * 30 (17,10,9) 36 (20,15,7) 42 (29,25,6) 60 *: Right triangle
Let the integer sides of an HT be a, b, c, and denote by s its semiperimeter
By the Triangle Inequality,
[2]
and analogously, s - b > 0 and s - c > 0. Using Heron's area formula,
If the perimeter P = 2s 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, 1 < 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(s - a)(s - b)(s - c) [3] 16A2 = P(P - 2a)(P - 2b)(P - 2c)
Without loss of generality, we can assume that the integer sides of an HT are ranked alphabetically from longest to shortest:
1 ≤ c ≤ b ≤ a < b + c [4]
These integers are all strictly positive by the previous results. Summing both sides of the equations above, we have
x = s - a [5] y = s - b [6] z = s - c [7]
The square of the HT area can now be written more simply as
x + y + z = 3s - (a + b + c) = 3s - 2s = s [8]
The inverse relationships between x, y, z and the sides of the triangle are:
A2 = sxyz [9]
Substituting variables in [4], we have
a = s - x = y + z b = s - y = x + z c = s - z = x + y [10]
The variables in the preceding inequality are unbounded, so to find all HTs satisfying [1], we will determine limits for x, y and z. All solutions can then be found by a computer-aided search over a finite and manageable set of possibilities.
1 ≤ x ≤ y ≤ z [11]
First, assume x ≥ 36. Then by [11], we have
xyz = 4t2(x + y + z) [12] (xy - 4t2)z - 4t2(x + y) = 0 [13]
yz ≥ x2 ≥ 362 > 36 ≥ 4t2Since x ≥ 4t2, we can write [12] asyz - 4t2 > 0
4t2(y + z) = x(yz - 4t2) ≥ 4t2(yz - 4t2)It follows that
But
(y - 1)(z - 1) ≤ 4t2 + 1 ≤ 37 [14]
(y - 1)(z - 1) ≥ (x - 1)2 ≥ 352 = 1225,which contradicts [14]. Therefore, x ≤ 35:
Next, assume x ≤ 35 and y ≥ 87. Then
x y xy z Solutions to A = tP x ≥ 36 any any any none 1 ≤ x ≤ 35 any any any TBD
(y - 36)2 > 2556Since z ≥ y, we have
(y - 36)(z - 36) > 2556Since 1 ≤ x ≤ 35 and 4t2 ≤ 36, it follows thatyz > 36(y + z + 35)
xyz > 4t2(x + y + z),so there are no solutions to [12] when x ≤ 35 and y ≥ 87:
Note that if xy - 4t2 ≤ 0, then
x y xy z Solutions to A = tP x ≥ 36 any any any none 1 ≤ x ≤ 35 y ≥ 87 any any none 1 ≤ x ≤ 35 x ≤ y ≤ 86 any any TBD
(xy - 4t2)z - 4t2(x + y)is negative, and there is no solution to [13]:
Henceforth, assume x ≤ 35, x ≤ y ≤ 86 and xy > 4t2. If 4t2 ∈ ℤ, then xy ≥ 4t2 + 1, but if 4t2 ∉ ℤ, then
x y xy z Solutions to A = tP x ≥ 36 any any any none 1 ≤ x ≤ 35 y ≥ 87 any any none 1 ≤ x ≤ 35 x ≤ y ≤ 86 xy ≤ 4t2 any none 1 ≤ x ≤ 35 x ≤ y ≤ 86 xy > 4t2 any TBD
xy ≥ ⌈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
xy ≥ ⌊4t2⌋ + 1 [15]
1/289 ≤ ⌊4t2⌋ - 4t2 + 1 ≤ 1or
Let z > 1,258,884. Then
by [15], so there is no solution to [12]:
(⌊4t2⌋ - 4t2 + 1)z > 4t2(x + y)
(xy)z ≥ (⌊4t2⌋ + 1)z > 4t2(x + y + z)
x y xy z Solutions to A = tP x ≥ 36 any any any none 1 ≤ x ≤ 35 y ≥ 87 any any none 1 ≤ x ≤ 35 x ≤ y ≤ 86 xy ≤ 4t2 any none 1 ≤ x ≤ 35 x ≤ y ≤ 86 xy > 4t2 z > 1,258,884 none 1 ≤ x ≤ 35 x ≤ y ≤ 86 xy > 4t2 y ≤ z ≤ 1,258,884 TBD
For each of these possibilities, the following steps will be completed:
Compute s = x + y + z and P = 2sThe five equable HTs (t = 1) are listed below, in order of increasing area:
Verify that A2 = sxyz is a square
If A = P, output results for equable triangles
If A > P and 2A/(A - P) is an integer, output results for equable triangular prisms
a b c A=P x y z Triangle type 10 8 6 24 2 4 6 Right 13 12 5 30 2 3 10 Right 17 10 9 36 1 8 9 Obtuse 20 15 7 42 1 6 14 Obtuse 29 25 6 60 1 5 24 Obtuse
By the distance formula,
Subtracting both sides,
u2 + v2 = b2 [16] (u - a)2 + v2 = c2
From [16] and A = P, we have
The inradius of the triangle is
r = A/s = P/s = 2Since the base of the triangle lies on the x-axis and the incircle is tangent to it, the y-coordinate of the incenter is also 2. Therefore, the incenter is
The circumcircle of this triangle passes through the non-collinear vertices (0,0), (a,0) and (u,v). Its circumcenter can be found by the method of determinants. Let
Then the circumcenter is
and the circumradius is
Evaluating the formulas for the vertex coordinates and the inscribed and circumscribed circles of each equable HT, we have
The results above can be used to create scale drawings of the five equable HTs and their corresponding incircles and circumcircles:
a b c P=A (a,0) (u,v) Inradius Incenter Circumradius Circumcenter 10 8 6 24 (10,0) (32/5,24/5) 2 (6,2) 5 (5,0) 13 12 5 30 (13,0) (144/13,60/13) 2 (10,2) 13/2 (13/2,0) 17 10 9 36 (17,0) (154/17,72/17) 2 (9,2) 85/8 (17/2,-51/8) 20 15 7 42 (20,0) (72/5,21/5) 2 (14,2) 25/2 (10,-15/2) 29 25 6 60 (29,0) (715/29,120/29) 2 (24,2) 145/8 (29/2,-87/8)
a < b + 1 and b < a + 1or
b - 1 < a < b + 1This is satisfied only when a = b, in which case the perimeter 2a + 1 is odd, and therefore the triangle cannot be Heronian.
Next, let (a,b,2) be the integer sides of an HT. Again applying the triangle inequalities,
b - 2 < a < b + 2or
b - 1 ≤ a ≤ b + 1If a is b ± 1, then the perimeter (b ± 1) + b + 2 is again odd.
The remaining possibility is the isosceles triangle (b,b,2), in which case the semiperimeter s is b + 1 and by [3], the square of the area is
A2 = (b + 1)(1)(1)(b - 1)which has no solution in integers since the difference of two squares cannot be 1.b2 - A2 = 1,
Case 1
Case Shortest
side cxy Number of perimeter-dominant
integer-sided triangles1 c ≥ 17 any None 2 1 ≤ c ≤ 16 xy > 4 317 3 1 ≤ c ≤ 16 xy ≤ 4 Infinitely many
In [4], apply a lower bound to the integer triangle sides:
We'll use the same variables s, x, y and z defined in [2], [5], [6] and [7], but won't restrict the analysis to triangles with even perimeters, so s may be an integer or an odd integer multiple of 1/2. The sum of any two of {x,y,z} must be an integer since it is equal to one of the triangle sides, so {x,y,z} are either all integers or all odd multiples of 1/2. Equivalently, the doubles {2x,2y,2z} all have the same parity. The constraints in [17] may be written
17 ≤ c ≤ b ≤ a < b + c, [17]
1/2 ≤ x ≤ y ≤ z and x + y ≥ 17We can write
Perimeter dominance requires that
17 - y ≤ x ≤ y z ≥ y ≥ 17/2 [18]
Note that for x < 3, we have
P2 = (2s)2 > s(s - a)(s - b)(s - c) = A2 4(x + y + z) > (s - a)(s - b)(s - c) = xyz [19] (xy - 4)z < 4(x + y) [20]
so that
x 17 - x > 0.5 16.5 > 16.246 1 16 > 8.472 1.5 15.5 > 6 2 15 > 4.828 2.5 14.5 > 4.161
which contradicts [19]. We henceforth take x ≥ 3, so that xy ≥ 9. Then
(y - 4/x)(z - 4/x) ≥ (y - 4/x)2 > 4 + 16/x2
xyz > 4y + 4z + 4x
(x - 8/17)2 > 1220/289Dividing by xy - 4 ≥ (3)(3) - 4 > 0 and using [18], we have(x - 8/17)(y - 8/17) > 4 + 64/289
17xy - 8x - 8y > 68
4x + 4y < (17/2)xy - 4(17/2)
which contradicts [20]. We conclude that there are no perimeter-dominant integer-sided triangles when c ≥ 17.
(xy - 4)z > 4x + 4y
For 1 ≤ c ≤ 16, first assume xy > 4 so that we can perform division in [20]:
which by [10] reduces to the cubic
[21]
Since 1/2 ≤ x ≤ y = c - x, it follows that
y3 - cy2 + 4y + 4c > 0 [22]
To determine all the triangle solutions, we simply iterate over integers c such that 1 ≤ c ≤ 16, and half-integers y satisfying [23]. For each of the 136 resulting combinations of c and y, we set x = c - y, determine if xy > 4 and [22] are satisfied, and if so, apply the limits for z in [21] such that the double of z has the same parity as the doubles of x and y. There are a total of 317 solutions when 5 ≤ c ≤ 16 and xy > 4, listed here and summarized below:
c/2 ≤ y ≤ c - 1/2 [23]
For example, if (x,y,z) = (0.5,12.5,17.5), the corresponding integer triangle (a,b,c) = (30,18,13) has perimeter 61 that is greater than its area 57.758. There aren't any solutions for 1 ≤ c ≤ 4 since that implies xy ≤ 4.
Shortest
side cNumber of perimeter-dominant
integer-sided triangles (xy > 4)5 27 6 30 7 10 8 4 9 137 10 44 11 25 12 16 13 11 14 7 15 4 16 2 Total 317
Now assume 1 ≤ c ≤ 16 and xy ≤ 4. We have
1/2 ≤ x ≤ y ≤ 4/x,so x has only the four values
1/2 ≤ x ≤ 2In [20], we have
(xy - 4)z < 4(x + y) = 4cSince xy ≤ 4 and the right side of the inequality above is strictly positive, the two sides bracket 0 regardless of the value of z, and so there are infinitely-many perimeter-dominant integer triangles. We have the following possibilities for x and y since their doubles must have the same parity:
For example, if x = 1 and y = 3, then c = x + y = 4, xy - 4 = -1 and inequality [20] is written
x 4/x y ≤ 4/x Number of (x,y)
combinationsc = x + y 0.5 8 0.5,1.5,...,7.5 8 1,2,3,4,5,6,7,8 1 4 1,2,3,4 4 2,3,4,5 1.5 8/3 1.5,2.5 2 3,4 2 2 2 1 4 Total 15
(-1)z < 4(4),which is true for any value of z. Since b = a - (y - x), we have the following equivalent representations:
One conclusion that can be drawn from these results is that any isosceles triangle with a base of 4 or less has a perimeter greater than its area (this is true whether the triangle sides are integers or not).*: Isosceles
c (x,y) Triangle forms (a,b,c) 1 (0.5,0.5) (a,a,1)* 2 (0.5,1.5),(1,1) (a,a-1,2), (a,a,2)* 3 (0.5,2.5),(1,2),(1.5,1.5) (a,a-2,3), (a,a-1,3), (a,a,3)* 4 (0.5,3.5),(1,3),(1.5,2.5),(2,2) (a,a-3,4), (a,a-2,4), (a,a-1,4), (a,a,4)* 5 (0.5,4.5),(1,4) (a,a-4,5), (a,a-3,5) 6 (0.5,5.5) (a,a-5,6) 7 (0.5,6.5) (a,a-6,7) 8 (0.5,7.5) (a,a-7,8)
First, we can scan the results from Case 2 above to find that there is only one perimeter-dominant HT for which xy > 4: the (6,5,5) isosceles triangle with perimeter 16 and area 12. This triangle consists of two back-to-back copies of the smallest primitive Pythagorean triangle (3,4,5).
For Case 3, we remove the odd-perimeter cases from the preceding table to find the following possibilities for a perimeter-dominant HT:
(a,a,2) triangles
Triangle forms (a,b,c) (a,a,2) (a,a-1,3) (a,a-2,4) (a,a,4) (a,a-3,5)
We have from [3],
A2 = (a + 1)(1)2(a - 1)which has no solution since the difference of two squares cannot be 1.a2 - A2 = 1,
(a,a,4) triangles
The area formula is
A2 = (a + 2)(2)2(a - 2)The possible factors are(2a + A)(2a - A) = 16,
Since neither of these results in an integer value for a, there is no solution.
2a + A > 2a - A sum = 4a 16 > 1 17 8 > 2 10
(a,a-3,5) triangles
We have
A2 = (a + 1)(1)(4)(a - 4)The possible factors are(2a - 3 + A)(2a - 3 - A) = 25,
The single integer solution a = 8 produces one of the two isosceles perimeter-dominant HTs: (8,5,5) with perimeter 18 and area 12. This triangle also consists of two back-to-back copies of the smallest primitive Pythagorean triangle (3,4,5).
2a - 3 + A > 2a - 3 - A sum = 4a - 6 a 25 > 1 26 8
The remaining two triangle forms each have infinitely-many solutions. For consistency, triangle sides will continue to be denoted by (a,b,c), but for the remainder of this section the order sequence a ≥ b ≥ c is not required.
The perimeter and area formulas are
a b c P A P-A 5 4 3 12 6 6 26 25 3 54 36 18 149 148 3 300 210 90 866 865 3 1734 1224 510 5045 5044 3 10092 7134 2958 29402 29401 3 58806 41580 17226 171365 171364 3 342732 242346 100386 998786 998785 3 1997574 1412496 585078 5821349 5821348 3 11642700 8232630 3410070 ... OEIS: A072221 A075848
In [24], (2A2 + 9) mod 9 ∈ {0,2,5,8} and (P2 - 6P + 9) mod 9 ∈ {0,1,4,7}, so the only common remainder is {0}, and it follows that P and A are both divisible by 3. Therefore, [25] is a modified Pell equation with parameters D = 2 and p = 3. The associated coefficients are a1 = 1/3, b1 = -1, a2 = 1/3, b2 = 0, so 2b1(p - 1)/a1 = -12, 2b2(p - 1)/a2 = 0, and recursive solutions for the perimeter and area are:
P = 2a + 2 A2 = 2a2 - 2a - 4, a ≥ 3 2A2 + 9 = (P - 3)2 [24] (P/3 - 1)2 - 2(A/3)2 = 1 [25]
Pn = 6Pn-1 - Pn-2 - 12, P0 = 12, P1 = 54An = 6An-1 - An-2, A0 = 6, A1 = 36
Case 3b: (a,a-2,4) triangles
The perimeter and area formulas are
a b c P A P-A 5 3 4 12 6 6 15 13 4 32 24 8 53 51 4 108 90 18 195 193 4 392 336 56 725 723 4 1452 1254 198 2703 2701 4 5408 4680 728 10085 10083 4 20172 17466 2706 37635 37633 4 75272 65184 10088 140453 140451 4 280908 243270 37638 ... OEIS: A016064 A102091 A001352
In [26], (4A2 + 48) mod 48 ∈ {0,4,16,36} and (3P2 - 24P + 48) mod 48 ∈ {0,3,12,27}, so the only common remainder is {0}, and it follows that A is divisible by 6 and P is divisible by 4. Therefore, [27] is a modified Pell equation with parameters D = 3 and p = 2. The associated coefficients are a1 = 1/4, b1 = -1, a2 = 1/6, b2 = 0, so 2b1(p - 1)/a1 = -8, 2b2(p - 1)/a2 = 0, and recursive solutions for the perimeter and area are:
P = 2a + 2 A = 3a2 - 6a - 9, a ≥ 4 4A2 + 48 = 3(P - 4)2 [26] (P/4 - 1)2 - 3(A/6)2 = 1 [27]
Pn = 4Pn-1 - Pn-2 - 8, P0 = 12, P1 = 32Due to the different coefficients in their respective difference equations, the Case 3a and Case 3b solution sequences do not always alternate; for example, the two Case 3a perimeter values 392 and 1452 lie between the consecutive Case 3b perimeters 300 and 1734. Similar remarks apply for the areas.An = 4An-1 - An-2, A0 = 6, A1 = 24
Obtuse and right Heronian perimeter-dominant triangles
The only Heronian perimeter-dominant triangle that is also a right triangle is the first one listed in the Case 3a and Case 3b tables; namely, the smallest primitive Pythagorean triangle (5,4,3) (or (5,3,4)). For an obtuse or right triangle, we require
First consider Case 3a (a,a-1,3). Since a ≥ 3, it is the longest side of the triangle and the preceding inequality is
a2 ≥ b2 + c2, with equality in the case of a right triangle [28]
a2 ≥ (a - 1)2 + 32, or a ≥ 5,so (5,4,3) is the only possible right triangle, and all other triangles in the the Case 3a results table above are obtuse. Similarly, in Case 3b (a,a-2,4), we have a ≥ 4, so a is again the longest side of the triangle. The inequality [28] is
a2 ≥ (a - 2)2 + 42, or a ≥ 5,so again the only right triangle possibility is (5,3,4), and all other triangles in the Case 3b results table above are obtuse.
Summary
Perimeter-dominant Heronian triangles:Perimeter-dominant HTs are relatively rare; there are only 19 of them with their longest side length under a million. A merged table of the smaller perimeter-dominant HTs can be found here.
Triangle sides Perimeter > Area Triangle type (6,5,5) 16 > 12 Acute (8,5,5) 18 > 12 Obtuse (5,4,3) 12 > 6 Right (a,a-1,3), a ≥ 6 2a + 2 > A075848 Obtuse (a,a-2,4), a ≥ 6 2a + 2 > A001352 Obtuse
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