HERONIAN TRIANGLES
If you have ever explored Pythagorean triple triangles (right angled triangles with integer sides), you might be even more amazed at the unique properties of non-Pythagorean, or scalene, integer triangles, better known as Heronian triangles.
A Heronian, or arithmetical, triangle is one whose sides, area, and at least one altitude, are all integers. Some mathematicians refer to Heronian triangles as rational triangles.
Consider the triangle with vertices A, B, and C, opposite side lengths of a, b, and c, a semi-perimeter of s = 1/2(a+b+c), an area of A, and an inscribed circle with the radius r.
Without getting into the proof, the area A of a Heronian triangle can be defined knowing only the 3 sides by using Heron's formula, A = sqrt[s(s-a)(s-b)(s-c)] where a, b, c, are the three sides and s = 1/2(a+b+c). A lesser known expression for the area involves only the sides alone as A = sqrt[(a+b+c)(-a+b+c)(+a-b+c)(a+b-c)]/4 = sqrt[a^2b^2 + 2a^2c^2 + 2b^2c^2 - (a^4 + b^4 + c^4)]/4.
Another surprising characteristic of Heronian triangles is that the area can also be determined from A = rs, where r is the radius of the circle inscribed inside the triangle r = A/s = sqrt[(s-a)(s-b)(s-c)/s]. (It is worth noting that the radii of both the circumscribed and escribed circles are also integers.)
Also, for the angles at A, B, and C, tan(A/2) = r/(s-a), tan(B/2) = r/(s-b), and tan(C/2) = r/(s-c)
Since the sides of a Heronian triangle are integral, the integer altitude from a vertex to the opposite side forms two integer right triangles whose sum, or difference, equal the given triangle, or a smaller Heronian triangle, respectively. Knowing this, it is possible to form any desired Heronian triangle, which is not a right triangle, by combining two integral right triangles of different shapes.
Given two primitive right triangles, we can form similar triangles having either leg of one equal to either leg of the other. We can then form Heronian triangles by bringing the equal legs coincident with one another, with the unequal legs on a common perpendicular to the common leg, either on the same side of it or on opposite sides of the common leg. Since either leg of one right triangle can be used with either leg of the other, four such pairs of Heronian triangles can be so formed.
For example, using the 5-12-13 (1) and 12-16-20 (2) right triangles, with the 12 length sides coincident, and the 5 and 16 sides on the same line. The Heronian triangle thus formed has sides 13-20-21, altitude 12, and area 126. Rotating the smaller 5-12-13 triangle about the common altitude, inside the larger 12-16-20 triangle, we derive another Heronian triangle with sides 11-13-20, altitude 12, and area 66.
Using the 3-4-5 and 5-12-13 primitive right triangles, we can combine the primitive 5-12-13 with the non-primitive derivatives 9-12-15 and 12-16-20, and the non-primitive derivative 5-20-25 with the non-primatives 15-36-39 and 20-48-52, to create eight different Heronian triangles.
Similarly, isosceles Heronian triangles can be formed from either of the given Heronian triangle by rotating the triangle about a common altitude. For example, the 5-12-13 triangle forms 10-13-13 or 24-13-13 isosceles Heronian triangles with areas of 60.
Perhaps more surprising is the fact that all Heronian Triple triangles can be derived from three simple relationships, much the same as it is possible to derive Pythagorean Triples from three simple expressions. It is well known that the altitude from any vertex to the opposite side of a rational triangle is itself rational and divides the opposite side into two rational parts. Denoting the sides by x, y, and z, the altitude to z by h, z = z1 + z2 with z1 adjacent to x and z2 adjacent to y, we can write that h^2 = x^2 - z1^2 = y^2 - z2^2. Through manipulation of these expressions and introducing the variables m, n, and k, every rational integral triangle can be derived from x = n(m^2 + k^2), y = m(n^2 + k^2), and z = (m + n)(mn - k^2) where m, n, and k are positive integers and mn is greater than k^2.
Lets take m = 3, n = 2, and k = 1 for example. Using the three expressions, x = 20, y = 15, and z = 25. From Heron's area formula, the area equals 150 making the altitude h to side z = 12 and z is divided up into 9 and 16 by the altitude. How elegant.
Lets try m = 4, n = 3, and k = 1. Here, we end up with x = 51, y = 40, and z = 77. The area works out to be 924, the altitude to z being 24 and z divided up into lengths of 32 and 45.
Lets try m = 5, n = 3, and k = 2. This yields x = 87, y = 65, and z = 88 for an area of 2640, an altitude of 60 and z broken into lengths of 25 and 63.
A check of all three results will show that each Heronian triangle is made up from two Pythagorean Triple triangles with the common side as the altitude.
Heronian triangles with consecutive sides are possible, the 3-4-5 right triangle being the smallest. Others include 13-14-15, 51-52-53, 193-194-195 and infinitely many others. Any three consecutive sides, a, b and c can be derived from b = (2 + sqrt3)^n + (2 -sqrt3)^n, a = (b - 1) and c = (b + 1), where n = any positive integer from 1 on up. The "b" side can also be derived from b = 2[(2 + sqrt3)^n - 4sqrt3].
Another way of deriving consecutive sided Heronian Triangles is through the well known Pell equation. Consider the three sides to be (m - 1), m and (m + 1). From Heron's triangle area formula, A = sqrt[s(s-a)(s-b)(s-c)], we find the area to be A = [msqrt(3(m^2-4))]/4. By inspection, 3(m^2 - 4) must be a square N^2 or N^2 - 3m^2 = -12, a Pell equation. We can take advantage of one of several unique characteristics of Pell equations where C/D = a square. Solutions to a Pell equation with this characteristic are derivable in a unique three step process described below.
Given x^2 - Dy^2 = -C where C/D = a perfect square.
Define solutions to x^2 - Dy^2 = +1 as follows:
The minimum x/y solution derives from Heron's method for square root estimation; sqrtD = sqrt(a^2 +/- r) = (a +/-r/2a) = p/q.
Subsequent solutions derive from x/y = (p + q\/D)^n.
Then, for x^2 - Dy^2 = -D, the "y" values are the same as the "x" values from x^2 - Dy^2 = +1 and the "x" values follow directly.
Then, the "x" and "y" solutions to x^2 - Dy^2 = -C are sqrt(C/D) times the "x" and "y" solutions to x^2 - Dy^2 = -D.
Example: Given x^2 - 3y^2 = -12.
The minimum solution to x^2 - 3y^2 = +1 derives from sqrtD = x/y = sqrt(a^2 + r) = a + r/2a = 1 + 2/2 = 2/1.
Subsequent solutions follow from (2 + 1\/3)^n.
n......1......2.......3.........4..........5.............6
x/y..2/1...7/4...25/15..97/56..362/209..1351/780.
.
Solutions to x^2 - 3y^2 = -3 follow as
n......1......2.......3.........4..........5...............6
x/y..3/2..12/7..45/26..168/97..627/362..2340/1351 the "y" values being identical to the "x" values from x^2 - 3y^2 = +1.
Solutions to x^2 - 3y^2 = -12 then become sqrt(C/D) = sqrt(4) = 2 times the solutions to x^2 - 3y^2 = -3.
n......1.......2........3..........4............5...............6
x/y..6/4..24/14..90/52..336/194..1254/724..4680/2702
SInce y = b in our analysis, we discover that b = 4, 14, 52, 194, 724, 2702, etc. making the triangles we seek with sides of
3-4-5
13-14-15
51-52-53
193-194-195
723-724-725
2701-2702-2703
10083-10084-10085 and so on.
Heronian triangles whose sides are in arithmetic progression (differing by a constant "d") are also possible. In this case, the middle side "b" is 1, 2, 3, 4, etc., times the "b" side derived for the consecutive sides above resulting in sides differing by 1, 2, 3, 4, ec., respectively. For example, the 3rd consecutive sides solution has sides of 51, 52, 53. A triangle with sides differing by 3 would have sides of (153, 156, 159) and have an area of 2,464,020. By inspection, derived arithmetic progression sides are simply multiples of the initially derived consecutive sided triangles. Triangles with sides having no common factors are derived from a different set of rules.
Some unique recreational problems concerning Heronian Triangles are:
1 What are the sides and area of the unique Heronian triangle whose sides and area are consecutive integers?
2 What are ths sides and area of the unique Heronian triangle whose sides and an altitude are consecutive integers?
3 What are the three Heronian triangles, which are not right-angled, whose perimeters and areas are equal?
4 The area of a Heronian triangle is always a multiple of 6. What is the unique Heronian triangle with area 24?
5 Find an acute Heronian triangle with the smallest area.
6 Find an obtuse Heronian triangle with the smallest area.
7 Find an acute scalene Heronian triangle with the smallest area.
8 Find an obtuse scalene Herronian triangle with the smallest area.
9 Are there any other Heronian triangles with consecutive integer sides other than the 3-4-5 triangle?
You can find additioanal information on Heronian Triangles in many good books on Recreational Mathematics.
Answers:
1 (3-4-5 with area = 6)
2 (13-14-15 with altitude 12 and area 84.)
3 (6-25-29, 7-15-20, and 9-10-17.)
4 (4-13-15 with area = 24)
5 (5-5-6 with area = 12)
6 (5-5-8 with area = 12)
7 (13-14-15 with area = 84.
8 (4-13-15 with area = 24)
9 A few are 13-14-15 with an area of 84, 723-724-725 with an area of 226,974, and 37633-37634-37635 with an area of 613,283,664. There are many others.
