Parabolas, Tangents, and the Wallace-Simson Line

Re-post from 2012, because of several visitors who ask questions that led me to refer them here.  Thought it worth re-posting.

The oft-called Simson line was attributed to Simson by Poncelet, but is now frequently known as the Wallace-Simson line since it does not actually appear in any work of Simson. (Oh go on, ask your teacher, so WHY do we still call it the Simson line at all?)
The Wallace for whom the line should more probably be named is William Wallace FRSE (23 September 1768, Dysart—28 April 1843, Edinburgh; the Scottish mathematician and astronomer who invented the eidograph, a more complicated version of the pantograph used to make scale images of drawings. He was a protegee of John Playfair, and teacher to Mary Somerville. He wrote about the line in 1799. He is also not credited for his 1807 proof of a result about polygons with an equal area, which has become the Bolyai–Gerwien theorem. He was also one of the first in England/Scotland to promote the calculus as taught on the Continent.  

The theorem says that if a triangle is inscribed in a circle, then if perpendiculars are dropped from a point on this circumcircle to the three sides of the triangle (extended as needed) the feet of these perpendiculars will lie on a straight line. It works the other way too. If you draw a straight line cutting all three sides of the triangle, perpendiculars drawn at these points of intersection will be concurrent at a point on the circumcircle.  (With dynamic geometry software, it is relatively easy for students/teachers to create a single line through three sides of a triangle, then construct perpendiculars to the three intersections and make their intersection a traceable point, the rotate the line about ther middle point of the three to get the circumcircle.)

I mentioned recently in a description of David Well's new book, Games and Mathematics, that I keep finding out new stuff. Well, he pointed out a connection between the Wallace line (he uses Simson, but I believe he knows better) and tangents of a parabola.

If you find three tangents to parabola and construct the circumcircle to the triangle formed by their mutual intersections, the circumcircle will pass through the focus of the parabola.
Tricky and cool, but what does that have to do with the the Wallace line? Well if you drop a perpendicular from the focus to ANY tangent, the foot of the perpendicular will always fall on the line tangent to the parabola at the vertex. The tangent at the vertex is a Wallace line for any triangle formed by three tangents to a parabola.

 

Sang-Heronian Triangles and some History about Near Equilateral Triangles

 

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. 

 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.

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."   

Pythagorean Parabolas

 

I recently came across a note on an Annual Meeting of the Rocky Mountain Section of the MAA in 1923. Among the list of presentations was one by W. J. Hazzard, Professor at the Colorado School of Mines on the topic of "Parabolic Grouping of Pythagorean triangles."
I was a little familiar with Prof. Hazard as I had leapfrogged off one of his old posts in the Mathematics Teacher on methods of solving a quadratic equation to write a little about the history of solving quadratics in twenty or so different ways, which I hope someday to reduce to blog posts, but not today. 
I even had a copy of one of the good Professor's books in my collection of old math books, but I had not read, nor was I aware of the idea he spoke of.  With a few words of guidance from a "very" brief coverage in the article, I was able to extract at least a little that may be of interest to anyone who enjoys Pythagorean relations, and especially if you teach high school math.

If you put one acute vertex of a right triangle at the origin and lay it out so that the shorter leg lies along the positive x-axis, the other vertex will be at the point (a,b) as determined by the two legs of the triangle.  In the graph I have shown the  position of a 3-4-5 triangle and a 5-12-13 triangle to make my meaning clear.

A natural question is, "So What?"  But if we look at several of the points determined by the upper vertex, and select out only some "related" Pythagorean triples, we notice a pattern.  In the image at right the points represent the set of triples 3-4-5; 5-12-13; 7-24-25; and 9-40-41.  (Any teacher or student who is not aware, there is a simple trick to find an infinite number of these triangles with a longer side one less than the hypotenuse.  Just take any odd number to be the short leg, square it, and then divide by two and round up to the next whole for the hypotenuse. For example, 11 is a good odd number, and its square is 121.  If we divide 121 by 2 we get 60.5, which is between 60 and 61, and 11, 60, 61 is a Pythagorean triple.) 
All the points lie on a parabola y= 1/2 x² - 1/2 .  Since the focal length is 1/(4A), with A = 1/2, the focus must be a distance of 1/2 unit above the vertex, making the focal point at the origin. If we think about the definition of a parabola as the set of points equally distant from a focus and directrix, we realize the line of the directrix must be the line y = -1 so that, for instance the point (3,4) which is 5 units from the origin/focus will also be 5 units away from the directrix.  

Admittedly that is a pretty small (although infinity large) sub-set of the Pythagorean triples.  What would happen if we plotted other triangles like 8-15-17?  It turns out they are not on the parabola drawn... they are on another one.  In fact, all the triangles which have a longer leg two less than the hypotenuse will also have a focus at the origin, and the directrix will be ... yeah you knew it would be, y = - 2.  That makes the focus at (0,-1).   You can write the equation with ease for the parabola passing through any of these Pythagorean vertices, and all the ones with a common difference between the longer leg and hypotenuse share a parabola.

All the triples I've picked so far have been primitive triples.  A good question to ask is what would happen if we picked, say, a 6-8-10 triangle. Will it fall on the same parabola as the 3-4-5, or on the ones with a difference of two?

The image below shows parabolas for differences of 1, 2, and 8 between the longer side and hypotenuse, and point D is the 6-8-10 triangle, right there with 8-15-17 and others like it. 

I'm not sure you can swap this information for bread or ale at the local inn, but it's pretty interesting stuff.

Heron's Cube Root Method

 

As is my nature, I love to roam through old math journals, and recently I found a 1920 article on a beautiful hand calculation for cube roots by Heron from his long lost Metrica.  Until nearly the end of the Nineteenth Century, Math Historians were writing that the ancient Greeks could not calculate square roots (they seldom even mentioned the thought of a cube root) because, "We can't find the evidence so they didn't know it." The discovery of Heron's Metrica in 1897 provided the evidence with several examples of square roots, and one single example of a Cube root.

This is a pretty accurate cube root for 2000 years ago, accurate to the third decimal place, and an error in the root of a little less than 0.0013  that's the total error. If you read it carefully, a question that has plagued all who have studied it.  What is the origin of the 100 added to 180 to get 280?  Some early researchers thought it was the original number, 100, and dismissed Heron's method as not very accurate for many numbers.  But this article provides an explanation that is, in the words of the author, closer than seven digit logarithms for large numbers (he didn't define).  

After doing a few by hand, I "cheated" and built a spreadsheet to compute them and give the error.  But if you try a few by hand, you will agree that it is way ahead of that thing they call the cube root algorithm in YouTube videos.  So I'll let You in on the secret, (but don't tell the other kids!).

Heron's method begins with the surrounding roots of the number.  For the 100 example he used, the lower root bound is 4, and the upper is 5, that is, the number 100 is bound by 64 and 125.   Now we need to determine what Heron called the excess, and the deficit.  The excess is 5^3 - 100 or 25.  The deficit is 100 - 4^3 = 36.  

Then we form the numerator of the fractional part of the solution, multiply the deficit by the upper root bound, 5, producing 180.  

Now for the denominator, we take the 180 from the numerator, and add the product of the excess times the lower bound, producing the unexplained 100.  To get our final root estimate, we add the lower root, 4 to this fraction.  My calculator gives 4.642857 but for the actual cube root of 100 it gives  4.6415888..   That's a total error of a smidge more than 00126.  If you cube the approximation you get 100.08199.   

I ran a spreadsheet for numbers from 10 to 330 and notice a few quirks I believe.  The error seems greater when the deficit is more than the excess.  Which actually makes 100 look worse than average.  If you try 90 instead, where the excess and deficit are 35 and 26, almost the opposite of 100 you get much greater accuracy.   For 90, the true value is 4.481404747, and the estimate is 4.481481481, and the total error is 0.0000076734.  The cube of the estimate is 90.0046  (What we country folk call Pretty Dang Good!)

It also seems that as the numbers get bigger, the errors get slightly smaller, and I can believe the 7 digit logarithm quote.  

So if you are driving down the road and factoring the license plate number in front of you is trivial (Oh, Come On, you know you all do that!)  try taking the cube root in you head.  Fun on the Highway .... (I missed what turn off??????)

ADDEDUM, With a Hat Tip to Keith Raskin;

So if you only read really old journals, you may miss something really important.  Case in point, the 2017 Bulletin of Parnas Mathematical Society, which is in Brazil (Brasil)   It

The Bulletin has an extension of Heron's root taking method for any odd root.   I've only just begun playing with it, but I'll add the method here, and my results for 5th roots.  

The system is much the same, but with a power used in the multiplication of the roots and the excess and deficiency, and the first new task is to find out the power k we want to use in those products.  The key is to let the root you seek, n, be equal to 2k+1; so for the fifth root, since 2k+1 = 5 gives us k=2, we want to square the lower and upper bounds of the root when we multiply.  

Example   Find the fifth root of 100 (which your calculator will tell you is 2.51188...  ) .   

The bounding roots are 2 and 3.  The deficit is 100 - 2⁵ = 68, and the excess is 3⁵- 100 = 143.  Now we proceed as before, except the numerator will be 3² times the deficit, 3² x 68 = 612.

 Now for the denominator we find the product we add to the numerator, which is the square of the lower root bound, 2, times the excess, 143, 2² x 143 = 572.  So the fractional part of our estimate will be 612 / (612 + 572) ... which is 153/296, or as a decimal, 0.51689.  Adding our lower root bound we get 2.051689. An error of 0.00500.