Media used: acrylic on canvas.
Media used: acrylic on canvas.
Pertrigonometric functions are modifications of the three primary trigonometric functions. Unlike the familiar sine, cosine, and tangent functions, the “pertrig” functions include triangle perimeter in their right-triangle-based definitions, which are given in the bulleted list below. The longer form of “pertrigonometric functions” is “perimeter-based trigonometric functions,” and the shorter, informal version is “pertrig functions.”
After defining these terms, I used Geometer’s Sketchpad to construct a right triangle containing a 10º angle, and then used the “measure” and “calculate” functions to find the values of pers(10º), perc(10º), and pert(10º). Since these are ratios, they would have the same values shown for larger or smaller right triangles which contain 10º angles.
An observation: the pertangents of complementary angles are equal. Why? Because complementary angles appear in all right triangles, as pairs of acute angles in the same triangle. For each such complementary angle pair, therefore, the same triangle is used to define pertangent. The hypotenuse/perimeter ratio (which is pertangent) would, it follows, remain unchanged — because both its numerator and denominator remain unchanged. This relationship does not hold for the tangent function; instead, the tangents of complementary acute angles are reciprocals of each other.
Of course, I wanted to know more than just the pers, perc, and pert values for 10º, but I had no desire to repeat the same calculations, many more times, to form a table. Instead, I simply graphed the functions, again using Geometer’s Sketchpad. The units on the x-axis are degrees, not radians.
In the graph above, the dark blue curve is the persine function, with the sine function in light blue, for comparison. Similarly, percosine is shown in red, with cosine shown in pink. Finally, pertangent is shown with a heavy, dark green curve, while tangent is shown as a thinner, light green curve.
Entering the equations for these curves was a little tricky, due to the fact that I wanted this graph to venture beyond 0 and 90 degrees, in both directions, on the x-axis. When that is done, the unit circle must be used (in place of right-triangle based definitions), simply because no right triangles contain angles outside this range. The radius of the unit circle is 1, by definition, and that is the hypotenuse of the right triangle which exists in the zero-to-ninety degree part of the domain of the graph above. As a consequence of setting the length of the hypotenuse of each right triangle at 1, the side opposite the angle in question (used for persine) becomes, simply, the sine of that angle, while the adjacent leg’s length is the angle’s cosine. It then follows that the perimeter (the denominator of the pers, perc, and pert ratios) is equal to sin(x) + cos(x) + 1.
Calculations are shown on the graph above, and you can click on the graph to enlarge it, to make them more readable. In these calculations, one more adjustment had to be made, and that was to the perimeter portion of each pertrigonometric ratio. Using sin(x) + cos(x) + 1 works fine for perimeter, for the zero-to-ninety degree portion of the domain, but, outside that, negative numbers intrude, for values of sin(x) and/or cos(x). It is my contention that triangle perimeter only makes sense as a sum of absolute values of a triangle’s three side lengths. To obtain absolute values for both sin(x) and cos(x) in the perimeter-part of each calculation, then, I simply squared each of these two functions, and then took the square roots of those squares. The result of this can be seen on the graph, in the curve for the pertangent function, which resembles a child’s drawing of waves in the ocean. On the y-axis, it never reaches as low as 0.4, and its maximum value is clearly exactly 0.5 — at the sharp “wave peaks.” At the (smooth) troughs, the actual minimum is equal to the square root of two, minus one, or ~0.414, although I have not yet figured out exactly why that is the case — I simply noticed it on the graph — but, to investigate it further, I know where to look: the 45-45-90 triangle, since these minima are hit when x = (45 ± 90n) degrees, where n is any integer. The pertangent function has the shortest period of all the functions shown above, at a mere 90º. For tangent, by contrast, the period is 180º. All four of the other functions shown have periods of a full 360º.
It is striking that the pertangent and tangent curves bear little resemblance to each other, while marked resemblances do exist between the persine and sine curves, as well as between the percosine and cosine curves. In informal terms, the persine curve is a shorter and spikier (but still recognizable) version of the sine curve (vertically, with the amplitude exactly one-half as great for the shorter persine curve, relative to the sine curve), but, horizontally, the two curves are synchronized. The same relationship holds for the percosine and cosine curves. Also, it is well-known that the cosine curve is simply the sine curve, phase-shifted one-quarter cycle (or 90º, or π/2 radians) to the left. This phase-shift relationship between the cosine and sine curves holds, precisely, for the percosine and persine curves.
There is a simple reason why persine, percosine, and pertangent all peak at exactly y = ½. All three functions generalize, for acute angles, to this ratio — (some side of a right triangle)/(perimeter of that same triangle) — and no side of any triangle can ever exceed, nor even reach, half that same triangle’s perimeter. In all three cases, the maximum y-value is only reached, even in the zero-to-ninety degree portion of the domain, for “degenerate cases” — angles of 0º or 90º, which are, of course, not acute angles at all. Interpreted as triangles, these are cases where either a triangle becomes so short that it collapses to a single segment, or the opposite degenerate situation: two parallel lines, connected by a single segment. If you try to make either (or both) of the acute angles in a right triangle into an additional right angle, after all, that’s what you get.
To my knowledge, no one has described these pertrigonometric functions before, by this or any other name, although I could be wrong. (If I am wrong on this point, please let me know in a comment.) Regardless of whether this is their first appearance, or not, I did not invent them. The reason for this is simple: nothing in mathematics is ever “invented” — only discovered — for mathematics existed long before human beings existed, let alone started writing things down. How do I know this? Simple: there was a universe here before there were people, and all evidence indicates that it operated under the same laws of physics we observe today — and all evidence to date also indicates that those laws are mathematical in nature. Therefore, with the “pertrig” functions, I either discovered them, or, if they have been found before, then I independently rediscovered them.
Finally, I’ll address that question so often asked, about numerous things, in mathematics classes: what are these pertrigonometric functions used for? As far as I know, the answer in this case, so far, is absolutely nothing, other than delighting me by their very existence. It is possible that this may change, for someone might find a way to make a profitable application of these functions — and I won’t get any money if they do, either, for I am not copyrighting any of this. Nothing in mathematics is subject to ownership.
Honestly, though, I hope no one ever finds any practical, “real-world” use, at all, for pers, perc, or pert. Right now, they are pure mathematical ideas, unsullied by tawdry, real-world applications, and, well, I like that. I am far from the only person who ever had such an attitude about a mathematical idea, either — such views are actually fairly common in the mathematical community. Most of those who try to discover previously-unseen things in mathematics do so solely, or primarily, for one reason: the joy of discovery, in its purest form.
As a proposed new “near-miss” to the Johnson solids, I created this polyhedron using Stella 4d, which can be found for purchase, or trial download, here. To make it, I started with a tetrahedron, augmented each face with icosidodecahedra, created the convex hull of the resulting cluster of polyhedra, and then used Stella‘s “try to make faces regular” function, which worked well. What you see is the result.
This polyhedron has no name as of yet (suggestions are welcome), but does have tetrahedral symmetry, and fifty faces. Of those faces, the eight blue triangles are regular, although the four dark blue triangles are ~2.3% larger by edge length, and ~4.6% larger by area, when compared to the four light blue triangles. The twelve yellow triangles are isosceles, with their bases (adjacent to the pink quadrilaterals) ~1.5% longer than their legs, which are each adjacent to one of the twelve red, regular pentagons. These yellow isosceles trapezoids have vertex angles measuring 61.0154º. The six pink quadrilaterals themselves are rectangles, but just barely, with their longer sides only ~0.3% longer than their shorter sides — the shorter sides being those adjacent to the green quadrilaterals.
The twelve green quadrilaterals are trapezoids, and are the most irregular of the faces in this near-miss candidate. These trapezoids have ~90.992º base angles next to the light blue triangles, and ~89.008º angles next to the pink triangles. Their shortest side is the base shared with light blue triangles. The legs of these trapezoids are ~2.3% longer than this short base, and the long base is ~3.5% longer than the short base.
If this has been found before, I don’t know about it — but, if you do, please let me know in a comment.
UPDATE: It turns out that this polyhedron has, in fact, been found before. It’s called the “tetrahedrally expanded tetrated dodecahedron,” and is the second polyhedron shown on this page. I still don’t know who discovered it, but at least I did gather more information about it — the statistics which appear above, as well as a method for constructing it with Stella.
I just had the strange experience of encountering a polyhedral discovery of mine on the Internet from about eight years ago — one that I had completely forgotten, but had shared with others, who posted it online, and were kind enough to give me credit for the discovery. It’s the fifth polyhedron shown on this page: http://www.interocitors.com/polyhedra/Triamonds/ — and is shown with a .stel file, so I was able to use polyhedral-manipulation-and-imaging software, Stella 4d (available at www.software3d.com/Stella.php) to make a rotating image of it:
It’s a member of a class of polyhedra which have, as faces, only regular polygons with unit size, as well as “triamonds.” Triamonds are 1:1:1:2 trapezoids composed of three coplanar, equilateral triangles.
Back in 2006 or earlier, my guess is that I simply made a physical model out of card-stock paper and tape, and then took a photograph of it — something I haven’t done in a very long time, now that making moving pictures of virtual models has become so easy. Another possibility is that I used Zome, an excellent ball-and-stick modelling system available at www.zometool.com. Zome, like Stella, I still use — and I will be using Zome often with students, in class, when school starts next month. Fortunately, I have a lot of Zome!
If one starts with a cuboctahedron, and then creates a zonish polyhedron from it, adding zones (based on the faces) to the faces which already exist, here is the result, below, produced by Stella 4d: Polyhedron Navigator (software you may buy or try at http://www.software3d.com/Stella.php):
The hexagons here, in this second image, are visibly irregular. The four interior hexagon-angles next to the octagons each measure more than 125 degrees, and the other two interior angles of the hexagons each measure less than 110 degrees — too irregular for this to qualify as a near-miss to the Johnson solids. However, Stella includes a “try to make faces regular” function, and applying it to the second polyhedron shown here produces the polyhedron shown in a larger image, at the top of this post.
It is this larger image, at the top, which I am proposing as a new near-miss to the 92 Johnson solids. In it, the twelve hexagons are regular, as are the eight triangles and six octagons. The only irregular faces to be found in it are the near-squares, which are actually isosceles trapezoids with two angles (the ones next to the octagons) measuring ~94.5575 degrees, and two others (next to the triangles) measuring 85.4425 degrees. Three of the edges of these trapezoids have the same length, and this length matches the lengths of the edges of both the hexagons and octagons. The one side of each trapezoid which has a different length is the one it shares with a triangle. These triangle-edges are ~15.9% longer than all the other edges in this proposed near-miss.
My next step is to share this find with others, and ask for their help with these two questions:
Once I learn the answers to these questions, I will update this post to reflect whatever new information is found. If this does qualify as a near-miss, it will be my third such find. The other two are the tetrated dodecahedron (co-discovered, independently, by myself and Alex Doskey) and the zonish truncated icosahedron (a discovery with which I was assisted by Robert Webb, the creator of Stella 4d).
More information about these near-misses, one of my geometrical obsessions, may be found here: https://en.wikipedia.org/wiki/Near-miss_Johnson_solid
Stella’s creator just came out with a new version of Stella 4d, and a discovery of mine made the built-in library that comes with that software. This is my blog, so I get to brag about that, right? My legal name appears in the small print on the right side, at the end of the first long paragraph. I added the red ellipses to make it easier to find.
You can see the earlier posts related to my discovery of this zonish truncated icosahedron here:
If you’d like to try (as a free trial) or buy this software (I recommend Stella 4d over the other available options), here’s the link for that: http://www.software3d.com/Stella.php.
I’ve posted “bowtie” symmetrohedra on this blog, before, which I thought I had discovered before anyone else — only to find, later, that other researchers had found the exact same polyhedra first. Those posts have now been edited to include credit to the original discoverers. With polyhedra, finding something interesting, for the first time ever, is extremely difficult. This time, though, I think I have succeeded — by starting with the idea of using regular pentadecagons as faces.
Software credit: Stella 4d was the tool I used to create this virtual model. You can try a free trial download of this program here: http://www.software3d.com/Stella.php.
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Update: once again, I have been beaten to the punch! A bit of googling revealed that Craig Kaplan and George Hart found this particular symmetrohedron before I did, and you can see it among the many diagrams in this paper: http://archive.bridgesmathart.org/2001/bridges2001-21.pdf.
You’ll also find, in that same paper, a version of this second pentadecagon-based symmetrohedron:
There is a minor difference, though, between the Kaplan-Hart version of this second symmetrohedron, and mine, and it involves the thirty blue faces. I adjusted the distance between the pentadecagons and the polyhedron’s center, repeatedly, until I got these blue faces very close to being perfect squares. They’re actually rectangles, but just barely; the difference in length between the longer and shorter edges of these near-squares is less than 1%. I have verified that, with more work, it would be possible to make these blue faces into true squares, while also keeping the pentadecagons and triangles regular. I may actually do this, someday, but not today. Simply constructing the two symmetrohedra shown in this post took at least two hours, and, right now, I’m simply too tired to continue!
In this diagram, the original triangle is ABC, and is yellow. The brown line, u, is that triangle’s Euler Line, which contains the triangle’s orthocenter (O), circumcenter (K), centroid (R), and nine-point circle center (circle shown, centered at Q). The point I have found on the Euler Line is at W.
To find W, do the following: reflect triangle ABC over the Euler Line to form triangle A’B’C’ (shown uncolored, and with thick black edges). Both triangles, ABC and A’B’C’, have the same circumcircle (shown in green, with uncolored interior). Because of this, a cyclic hexagon may be formed by joining A, A’, B, B’, C, and C’ with segments, linking each in turn as one encounters them on a mental trip once around this circumcircle (the order in which these six points are encountered can change as A, B, and/or C are moved).
A hexagon has 9 diagonals. Of these, six are the sides of ABC and A’B’C’. Here, they are shown in black, and the other three diagonals are shown in red. These red diagonals are not necessarily concurrent, but any two of them do have to intersect, and those three intersections are points T, V, and W. At least one of those points — W, in this case — must be on the Euler Line. To get the other two points on the Euler Line, make the triangle approach regularity. As this is done, K, R, Q, O, and W converge, making the definition of the Euler Line itself problematical.
Point W needs a better definition. Which two of the three hexagon-diagonals which aren’t sides of the original triangle, nor its reflection, intersect on the Euler Line? I haven’t figured that out yet. Also, a formal proof for most of what I have described here is beyond my present abilities.
Why, then, do I believe the statements to be true? Answer: the evidence provided by experiment. This image is a screenshot from Geometer’s Sketchpad — but I don’t know how to post an animation of what happens when A, B, or C are moved. However, I can move them myself, with the program in operation, and observe how everything changes (this is one of the best features of Sketchpad, in my opinion). As these points are moved around, pairs of the heavy red segments (hexagon sides, and three of its diagonals) sometimes “flip” — a side becomes a diagonal, and that diagonal becomes a side. At that point, T, V, and W must be relabeled. Also, some positions of A, B, and C make the area of triangle TVW approaches zero — it collapses to a point on the Euler Line.
Odd things also happen if you make triangle ABC isosceles, because the Euler Line for an isosceles triangle is the perpendicular bisector of the base, which causes triangle A’B’C’, upon reflection of triangle ABC across the Euler Line, to map onto triangle ABC. When this happens, the hexagon becomes a single triangle, making its diagonals vanish — and point W goes and “hides” at the vertex opposite the base of isosceles triangle ABC, by which I mean W approaches that vertex as scalene triangles get closer to being isosceles.
Also, things change a bit if triangle ABC is obtuse:
The Nagel Line (that line which contains the incenter, S, and the centroid, R, where it intersects the Euler Line) has been added, and is shown in purple. As you can see, point W is not on the Nagel Line. With the triangle being obtuse, the earlier all-red convex hexagon is now gone, because two of its sides are black, due to them being sides of triangles ABC and A’B’C’. Point W persists, though, still on the Euler Line, and located in the area between the vertices of these two triangles’ obtuse angles. My hope, in pointing this last fact out, is that it might help define which two hexagon-diagonals’ intersection defines the location of W. It might also be possible to use this to distinguish between the first diagram’s points T, V, and W, for, in the first diagram, W, which is what I am calling the only one of these three points to be on the Euler Line, was the one nearest the largest interior angle of triangle ABC — and the same is also true of triangle A’B’C’, as well. However, the matter of picking W out of the “T, V, and W” set of points may have nothing to do with angle size — it could be, instead, a matter of proximity of A, B, and C, as well as their reflections, to the Euler Line itself. In other words, “Which one is W?” might be answerable simply by examination of which member of the “T, V, and W” set is closest to the member of the set “A, B, and C,” as well as “A’, B’, and C’,” which is, itself, closest to the Euler Line. This matter needs further investigation, with which I would welcome help from anyone.
Also: there are two easier-to-define points on the Euler Line, unlabeled in the diagrams above, which are the two points where triangles ABC and A’B’C’ intersect. The existence of these points on the Euler Line is simply a consequence of the fact that A’B’C’ was formed by reflecting the original triangle over the Euler Line. These two points could use special names, but nothing is immediately springing to my mind which would be appropriate. Another point on the Euler Line, also a consequence of reflection, appears as the midpoint of segment TV in the first diagram, and the midpoint of BB’ in the second — segments which appear analogous. This also seems to apply to the midpoints of AA’ and CC’. At this stage of the discovery process, though, appearances can be misleading.
I want to work out a better definition for this point, W, on the Euler Line, perhaps as an as-yet-undiscovered member of the large collection of triangle centers. I also need to know if it has already been found. If you have information pertinent either of these things, or to any part of this post, please leave it here, in a comment.
No brilliant discovery, nor invention, was ever made by those who compulsively follow rules, demonstrate immense respect for authority, and make a habit of unquestioning obedience.