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.

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.

**A note**about Edward Sang before I continue. In a 2021 article by Dennis Rogel, he calls Professor Sang, " probably the greatest calculator of logarithms of the 19th century" in A guide to Edward Sang’s tables

*(They did not mention the names of the daughters, but in Rogel's paper he cites Flora and Jane.)*These tables went beyond the tables of Henry Briggs, Adriaan Vlacq, and Gaspard de Prony." They add a list of his publications so extensive that they are grouped into five year periods.

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.

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.

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.

I recently (2024) found a very similar formula for the square of the volume of a Tetrahedron in a paper in Letters to the editor of The Mathematical Intellingencer by Martin Lukarevski. The Formula applies to a tetrahedron with all four faces congruent with edges a, b, c. I did a little simple algebra to enhance the similarity to Heron's formula.

And at the near end of the Wikipedia discussion of these states, "A complete classification of all Heronian tetrahedra remains unknown."

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