The first Near Equilateral triangles with consecutive integer sides and integer area (sometimes called Brahamagupta triangles) was discovered over 2500 years ago. The discovery of the 3,4,5 right triangle seems lost in antiquity back before 500 BC. All Pythagorean triangles are Heronian, but lots (infinitely many) of other triangles that are not right triangles also are Heronian. The second near equilateral triangle, the 13, 14, 15; was known to Heron of Alexandra as early as 70 AD, almost 2000 years ago. Since then, they've grown in number, and to infinity, and been dissected and diagnosed repeatedly. They've even been generalized to three dimensions in Heronian Tetrahedra. Here is one part of their story.
Heron of Alexandria is known to have developed a method of finding the area of triangles using only the lengths of the three sides. It is known that it was proven in his Metrica around 60 AD. His proof was extended in the 7th century by Brahamagupta extended this property to the sides of inscribable quadrilaterals. Since around 1880, the triangular method of Heron has been known as Heron's formula, or Hero's Formula. It emerged in French, formula d'Heron (1883?) and German, Heronisch formel (1875?) and in George Chrystal's Algebra in 1886 in England.
L E Dickson's History of Number Theory states that Heron stated the 13, 14, 15 triangle and gave its area as 84, the height of 12 being the common side of a 5,12,13 triangle and a 9, 12, 15. The 5 and 9 combining to form the base of length 14. Brahmagupta is cited in the same work for giving an oblique triangle composed of two right triangles with a common leg a, stating that the three sides are \( \frac{1}{2}(\frac{a^2}{b}+ b)\) , \( \frac{1}{2}(\frac{a^2}{c}+ c)\), and \( \frac{1}{2}(\frac{a^2}{b}- b) + ( \frac{1}{2}(\frac{a^2}{c}- c)\)
In 1621 Bachet took two Pythagorean right triangles with a common leg, 12, 35, 37 and 12, 16, 20 and produced a triangle with sides of 37, 20, 51. With an area of 306 if I did my numbers right.
Vieta and Frans van Schooten, both used the same approach of clasping two right triangles with a common leg; and by the first half of the 18th century, the Japanese scholar, Matsunago, realized that any two right triangles would work, by simply multiplying the sides of each by the hypotenuse of the other, he could juxtapose the two resulting triangles.
In the early 1800's through 1825 the problem was alive and hopping on the Ladies Diary and the Gentleman's Math Companion. One method created right triangles in another triangle to be reassembled into a rational triangle, similar in fact, to the problem that would appear in the 1916 American Mathematical Monthly. (Note; any triangle with rational sides and area can be scaled to become a Heronian triangle.)
In a letter of Oct 21, 1847; Gauss to H. C. Schumacher, he stated a method using circumscribed circles, and found lots of others chose the exact same solutions in their response. E. W. Grebe tabulated a set of 46 rational triangles in 1856. W. A. Whitworth noticed that the 13, 14, 15 triangle of antiquity, that had an altitude of 12, was the only one in which the altitude and sides were all consecutive. (1880)
Somehow, among all those, the contributions of a Professor from Scotland was not observed by Dickson.
The first modern western article I can find on the topic of Near Equilateral triangles with integer sides and area is from Edward Sang which appeared in 1864 in the Transactions of the Royal Society of Edinburgh, Volume 23. I find it interesting that this is only a small aside in a much larger article and that he begins with an approach to examining the angles. Then he arrives at the use of a Pell type equation for approximating the square root of three, \(a^2 = 3x^2 + 1 \) and shows that every other convergent in the chain of approximations is a base of a Near Equilateral Triangle, using sides of consecutive integers. The alternate convergents we seek are given by 2/1, 7/4, 26/15, 97/56... each approaching the square root of three more closely, but also each with a numerator that is 1/2 the base of triangle with consecutive integers for sides and integer area. Perhaps it is easier to just use the recurent relation, \(n_i=4n_{i-1} - n_{i-2}\) with \(n_0=2\), and \(n_1=4\) for the actual middle side,2, 4, 14, 52, 194.... The first few such triangles have their even integer base as2x1=2; (1, 2, 3) area 0; 2x2=4; (3, 4, 5); area 12; 2x7=14; (13, 14, 15); area 84; 2x26=52; (51, 52, 53); area 1170... etc. Throughout, he refers to "trigons" rather than triangles, and never invokes the name of Heron throughout.
The next paper using consecutive integer sides was in 1880 by a German mathematician named Reinhold Hoppe, who produced a closed form expression for these almost equilateral Heronian Triangles that was similar to, \( b_n =(2+2\sqrt{3})^n + (2-2\sqrt{3})^n \). His paper calls them "rationales dreieck" (rational triangles) I have not seen the entire paper, and don't know if the term Heronian appeared, or not.
The first American introduction to the phrase "Heronian Triangles", seemed to be an article in the American Mathematical Monthly which posed the introduction as a problem, to divide the triangle whose sides are 52, 56, and 60 into three Heronian Triangles by lines drawn from the vertices to a point within. The problem was posed by Norman Anning, Chillwack, B.C. It then includes a description that suggests it is introducing a new term, "The word Heronian is used in the sense of the German Heronische (with a German citation) to describe a triangle whose sides and area are integral.
In 1621 Bachet took two Pythagorean right triangles with a common leg, 12, 35, 37 and 12, 16, 20 and produced a triangle with sides of 37, 20, 51. With an area of 306 if I did my numbers right.
Vieta and Frans van Schooten, both used the same approach of clasping two right triangles with a common leg; and by the first half of the 18th century, the Japanese scholar, Matsunago, realized that any two right triangles would work, by simply multiplying the sides of each by the hypotenuse of the other, he could juxtapose the two resulting triangles.
In the early 1800's through 1825 the problem was alive and hopping on the Ladies Diary and the Gentleman's Math Companion. One method created right triangles in another triangle to be reassembled into a rational triangle, similar in fact, to the problem that would appear in the 1916 American Mathematical Monthly. (Note; any triangle with rational sides and area can be scaled to become a Heronian triangle.)
In a letter of Oct 21, 1847; Gauss to H. C. Schumacher, he stated a method using circumscribed circles, and found lots of others chose the exact same solutions in their response. E. W. Grebe tabulated a set of 46 rational triangles in 1856. W. A. Whitworth noticed that the 13, 14, 15 triangle of antiquity, that had an altitude of 12, was the only one in which the altitude and sides were all consecutive. (1880)
Somehow, among all those, the contributions of a Professor from Scotland was not observed by Dickson.
The first modern western article I can find on the topic of Near Equilateral triangles with integer sides and area is from Edward Sang which appeared in 1864 in the Transactions of the Royal Society of Edinburgh, Volume 23. I find it interesting that this is only a small aside in a much larger article and that he begins with an approach to examining the angles. Then he arrives at the use of a Pell type equation for approximating the square root of three, \(a^2 = 3x^2 + 1 \) and shows that every other convergent in the chain of approximations is a base of a Near Equilateral Triangle, using sides of consecutive integers. The alternate convergents we seek are given by 2/1, 7/4, 26/15, 97/56... each approaching the square root of three more closely, but also each with a numerator that is 1/2 the base of triangle with consecutive integers for sides and integer area. Perhaps it is easier to just use the recurent relation, \(n_i=4n_{i-1} - n_{i-2}\) with \(n_0=2\), and \(n_1=4\) for the actual middle side,2, 4, 14, 52, 194.... The first few such triangles have their even integer base as2x1=2; (1, 2, 3) area 0; 2x2=4; (3, 4, 5); area 12; 2x7=14; (13, 14, 15); area 84; 2x26=52; (51, 52, 53); area 1170... etc. Throughout, he refers to "trigons" rather than triangles, and never invokes the name of Heron throughout.
The next paper using consecutive integer sides was in 1880 by a German mathematician named Reinhold Hoppe, who produced a closed form expression for these almost equilateral Heronian Triangles that was similar to, \( b_n =(2+2\sqrt{3})^n + (2-2\sqrt{3})^n \). His paper calls them "rationales dreieck" (rational triangles) I have not seen the entire paper, and don't know if the term Heronian appeared, or not.
The first American introduction to the phrase "Heronian Triangles", seemed to be an article in the American Mathematical Monthly which posed the introduction as a problem, to divide the triangle whose sides are 52, 56, and 60 into three Heronian Triangles by lines drawn from the vertices to a point within. The problem was posed by Norman Anning, Chillwack, B.C. It then includes a description that suggests it is introducing a new term, "The word Heronian is used in the sense of the German Heronische (with a German citation) to describe a triangle whose sides and area are integral.
The only other mentions of a Heronian triangle in English in a google search before the midpoint of the 20th century revealed a 1930 article from the Texas Mathematics Teacher's Bulletin. It credits a 1929 talk, it seems, by Dr. Wm. Fitch Cheney Jr. who, "discusses triangles with rational area K and integral sides a, b, c, the g.c.f of the sides 1, under the name Heronian triangles." (Dr Cheney published an article in the American Mathematical Monthly in 1929, The American Mathematical Monthly, Vol. 36, No. 1 (Jan., 1929), pp. 22-28) Since any such rational area can be scaled up to an appropriate integer area with integer sides these address the general Heronian Triangle, but still no Near Equilateral, or at least not revealed in the snippet view.
By the 1980's an article in the Fibonacci Quarterly found a way to produce a Fibonacci like sequence, a second order recursive relation to produce the even bases. Letting \(B_0 = 2, and B_1 = 4\), the recursion was \( U_{n+2} = 4 U{n+1} + U_n\) . This paper by W. H. Gould of West Virginia University addresses the full scope of consecutive sided integer triangles and mentions Hoppe, but not Professor Sang. Gould's paper seems to be his solution to a problem he had posed earlier in the Fibonacci Quarterly, "of finding all triangles having integral area and consecutive integral sides." (H. W. Gould, Problem H-37, Fibonacci Quarterly, Vol. 2 (1964), p. 124. .)
Gould also mentions two other, seemingly earlier posed problems in other journals which I have yet to explore, and given the opportunity, will do so and return to this spot, If you are impatient, they are
7. T. R. Running, Problem 4047, Amer. Math. Monthly, Vol. 49 (1942), p. 479; Solutions by W. B. Clarke and E. P. Starke, ibid. , Vol. 51 (1944), pp. 102-104.
8. W. B. Clarke, Problem 65, National Math. Mag. , Vol. 9 (1934), p. 63
Gould's article is a wonderful read for the geometry of the incircles and Euler lines in such special triangles is well explored.
These are each candidates to be the first American proposal of these consecutive integer sided triangles, but it seems Gould's paper was the first to expand the full scope of the solutions in any detail.
Some of the characteristics of these I think would be found interesting to HS and MS age students I will spell out below.
As mentioned above, the length of the middle (even) side follows a 2nd order recursive relation \(B_n = 4B_{n-1}-B_{n-2}\) so the sequence of these even sides runs 2, 4, 14, 52, 194, 724..... etc. ) is there to represent the degenerate triangle 1,2,3.
Interestingly, the heights follow this same recursive method giving heights of 0, 3, 12, 45, 168....
The height divides the even side into two legs of Pythagorean triangles that make up the whole of the consecutive integer triangle. They are always divide so that one is four greater than the other, or each is b/2 =+/- 2.
Of the two triangles formed by on each side of the altitude, one is a primitive Pythagorean triangle, PPT, and the other is not. The one that is a PPT switches from side to side on each new triangle, alternately with the shorter leg, and then the longer leg. Here are the triangles with the two subdivisions of them with an asterisk Marking the PPT:
Short Base Long small triangle large triangle
3 4 5 *3 4 5
13 14 15 *5, 12, 13 9, 12, 15
51 52 53 24, 45, 51 * 28, 45, 53
193 194 195 *95, 168, 193 99, 168, 195
723 724 725 360, 627, 723 *364, 627, 725
The pattern of the ending digits of 3, 4, 5 repeated twice, and 1,2,3 once by looking at the end number behavior of if the previous two numbers end in 4's or a four followed by a two.
Gould also mentions two other, seemingly earlier posed problems in other journals which I have yet to explore, and given the opportunity, will do so and return to this spot, If you are impatient, they are
7. T. R. Running, Problem 4047, Amer. Math. Monthly, Vol. 49 (1942), p. 479; Solutions by W. B. Clarke and E. P. Starke, ibid. , Vol. 51 (1944), pp. 102-104.
8. W. B. Clarke, Problem 65, National Math. Mag. , Vol. 9 (1934), p. 63
Gould's article is a wonderful read for the geometry of the incircles and Euler lines in such special triangles is well explored.
These are each candidates to be the first American proposal of these consecutive integer sided triangles, but it seems Gould's paper was the first to expand the full scope of the solutions in any detail.
Some of the characteristics of these I think would be found interesting to HS and MS age students I will spell out below.
As mentioned above, the length of the middle (even) side follows a 2nd order recursive relation \(B_n = 4B_{n-1}-B_{n-2}\) so the sequence of these even sides runs 2, 4, 14, 52, 194, 724..... etc. ) is there to represent the degenerate triangle 1,2,3.
Interestingly, the heights follow this same recursive method giving heights of 0, 3, 12, 45, 168....
The height divides the even side into two legs of Pythagorean triangles that make up the whole of the consecutive integer triangle. They are always divide so that one is four greater than the other, or each is b/2 =+/- 2.
Of the two triangles formed by on each side of the altitude, one is a primitive Pythagorean triangle, PPT, and the other is not. The one that is a PPT switches from side to side on each new triangle, alternately with the shorter leg, and then the longer leg. Here are the triangles with the two subdivisions of them with an asterisk Marking the PPT:
Short Base Long small triangle large triangle
3 4 5 *3 4 5
13 14 15 *5, 12, 13 9, 12, 15
51 52 53 24, 45, 51 * 28, 45, 53
193 194 195 *95, 168, 193 99, 168, 195
723 724 725 360, 627, 723 *364, 627, 725
The pattern of the ending digits of 3, 4, 5 repeated twice, and 1,2,3 once by looking at the end number behavior of if the previous two numbers end in 4's or a four followed by a two.
In the 1929 article mentioned above, Dr. Cheney writes that he knows of no examples of Heronian triangles up to that time that were not made up of two right triangles, and then gives an example of one that is not decomposable, 25, 34, 39. He also points out that the altitudes of Heronian triangles are not always integers, and gives the example of 39,58,95 as an example which I calculate to be 4.8.
A paper by Herb Bailey and William Gosnell in Mathematics Magazine, October 2012 demonstrates Heronian triangles in other arithmetic progressions from the near-equilateral ones.
I mentioned that there are also Heronian Tetrahedra, although that use of Heronian seems even later than for triangles, perhaps as late as 2006. The earliest example of an exact rational tetrahedra with all integer edges, surfaces and volume was by Euler. He created a tetrahedron formed by three right triangles parallel to the xyz coordinate axes, and one oblique face connecting them. The triple right angle edges were 153, 104, and 672, and the three edges of the oblique face were 185, 680, and 697. These were each Pythagorean right triangles, the four faces of (104,672,680), (153,680,697), (153,104,185) and (185,672,697)
There are an infinite number of these Eulerian Birectangular tetrahedra, but they seem to get very large very quickly. Euler showed that they can be found by deriving the three axis-parallel sides a, b, and c by using four numbers that are the equal sums of two fourth powers. Euler found an example using , and that's the easy part. Then he constructed the three monster lengths of 386678175, 332273368, and 379083360, Yes, those numbers are each in the hundreds of millions, and each pair had a larger hypotenuse to form a third side.
And as the near end of the Wikipedia discussion of these states, "A complete classification of all Heronian tetrahedra remains unknown."
There are an infinite number of these Eulerian Birectangular tetrahedra, but they seem to get very large very quickly. Euler showed that they can be found by deriving the three axis-parallel sides a, b, and c by using four numbers that are the equal sums of two fourth powers. Euler found an example using , and that's the easy part. Then he constructed the three monster lengths of 386678175, 332273368, and 379083360, Yes, those numbers are each in the hundreds of millions, and each pair had a larger hypotenuse to form a third side.
, and that's the easy part. Then he constructed the three monster lengths of And as the near end of the Wikipedia discussion of these states, "A complete classification of all Heronian tetrahedra remains unknown."
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