Truncated Tetrahedral Numbers

The number associated with my diurnal age today, 28325, has a property that allows it membership of OEIS A100156:

 A100156: structured truncated tetrahedral numbers.

So what is a truncated tetrahedron. It belongs to a class of three-dimensional figurate numbers. They are geometrically constructed by taking a regular tetrahedral number of a given size and removing smaller tetrahedral corners from each of its four vertices.

What does it look like? Figure 1 shows an example.

Figure 1: source

Figure 2 shows the associated net made up from four hexagons and four equilateral triangles.

Figure 2

The details are:

Faces: 8

Edges: 18

Vertices: 12 

A structured truncated tetrahedral number (OEIS A100156) is a three-dimensional figurate number that counts the total number of points (or spheres) in a truncated tetrahedron built outward in nested shells from a single corner vertex, where every polygonal face is populated as a classical vertex-grown figurate polygon rather than a centered polygon.

The initial members of the sequence are: 1, 12, 44, 108, 215, 376, 602, 904, 1293, 1780, 2376, 3092, 3939, 4928, 6070, 7376, 8857, 10524, 12388, 14460, 16751, 19272, 22034, 25048, 28325, 31876, 35712, 39844

 

Figure 3

To quote from Gemini:

The shape first emerged prominently in the 15th and 16th centuries as artists sought to calculate and render perfect three-dimensional forms on flat surfaces:

Leonardo da Vinci & Luca Pacioli (1509): The most celebrated early appearance of the truncated tetrahedron is in Fra Luca Pacioli's treatise, De Divina Proportione. Leonardo da Vinci provided the illustrations, drawing the solid in two revolutionary forms: solidus (a solid form showing surface planes) and vacuus (a skeletal wireframe showing its interior framework). See Figure 3.

Albrecht Dürer (1525): In his foundational work Underweysung der Messung (Instruction in Measurement), German master Albrecht Dürer revolutionized geometric study by publishing the first foldable paper nets. His flat layout of the truncated tetrahedron allowed readers to cut out and construct physical models, making complex perspective easier to perceive in physical space.

Circles Inscribed In Triangles

I watched an interesting video that shows the area of a circle inscribed in a 3 : 4 : 5 right-angled triangle is equal to \( \pi \).

This got me thinking about right-angled triangle in general and I noted in the comments to the video that for a right-angled triangle with sides of \(a, b, c\) where \(c\) is the hypotenuse, the radius \(r\) of the inscribed circle is given by:$$r=\frac{a+b-c}{2}$$One can see that when \(a=3\), \(b=4\) and \(c=5\), we get:$$ r =\frac{3+4-5}{2} = \frac{2}{2}=1$$The area of the inscribed circle is thus \( \pi\) square units. So I investigated the general case, for any triangle with sides of \(a\), \(b\) and \(c\), and this is the explanation and formula that Gemini came up with:

Inscribed Circles in Any General Triangle

The area of the inscribed circle (the incircle) can be expressed purely in terms of the side lengths $a$, $b$, and $c$ for any triangle, whether it is right-angled, acute, or obtuse.

To do this, we use two foundational concepts in geometry: Heron's Formula and the relationship between a triangle's area and its inradius.

Step 1: Define the Semi-perimeter

For any triangle with sides $a$, $b$, and $c$, we first calculate the semi-perimeter (half of the perimeter), usually denoted as $s$:

$$s = \frac{a + b + c}{2}$$

Step 2: The Radius Formula

The radius $r$ of an inscribed circle in any triangle is equal to the area of the triangle divided by its semi-perimeter. Using Heron's Formula for the area of the triangle, this gives us:

$$r = \frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$$

Which simplifies algebraically to:

$$r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$

Step 3: Calculate the Area of the Incircle

Now that we have the radius purely in terms of the sides (via $s$), we just plug it into the circle area formula ($A = \pi \, r^2 \,$). The square root neatly disappears:

$$A = \pi \left( \frac{(s-a)(s-b)(s-c)}{s} \right)$$

Summary of the General Formula

If you want to write the area of the inscribed circle entirely in a single expression using just $a$, $b$, and $c$ (without using the $s$ substitution), it looks like this:

$$A = \pi  \left( \frac{(b+c-a)(a+c-b)(a+b-c)}{4(a+b+c)} \right)$$

Both the formula using $s$ and the expanded algebraic formula above will work perfectly for any valid triangle.

Perimeter Magic Polygons

My previous post was titled Anti-Magic Squares Revisited in which I mentioned heterosquares. Only today however, I came across the concept of a perimeter magic square that is an example of a perimeter magic polygon or PMP to use an acronym. The definition is:

A PMP is defined to be a regular polygon with the consecutive positive integers from 1 to N placed along the perimeter in such a way that the sums of the integers on each side are constant. The order of a polygon refers to the number of integers along each side. The examples in Figures 1, 2 and 3 show a 4th-order triangle, and a 3rd-order square and pentagon. The magic constants are given inside the figures. Source.

Let's look at some perimeter magic triangles to begin with. To quote from Wikipedia:

A magic triangle or perimeter magic triangle is an arrangement of the integers from 1 to \(n\) on the sides of a triangle with the same number of integers on each side, called the order of the triangle, so that the sum of integers on each side is a constant, the magic sum of the triangle. Unlike magic squares, there are different magic sums for magic triangles of the same order. Any magic triangle has a complementary triangle obtained by replacing each integer \(x\) in the triangle with \(1 + n − x\). See Figure 4.

Figure 4

The author of this paper comes up to two sets of formula in which \(C\) is the magic constant,  \(n\) is the order of the polygon and \(k\) is the number of sides. Figures 5 and 6 show these.

Figure 5 shows the formulae for the case of \(n\) even or both \(n\) and \(k\) odd.

Figure 5

Figure 6 shows the formulae for the case of \(n\) is odd and \(k\) is even.

Figure 6

To quote from the article:

With these formulas one can now begin to construct PMPs of any order and number of sides. One word of caution is still in order. There are times when a solution is not possible for certain values of C. The formulas only serve to indicate where solutions may be found, and that there is no need to look elsewhere. However, luck is still with the solver. Based on this author's experience, there are only two cases where solutions cannot be made with values obtained from the formulas. These are 4th-order triangles with \(C \) = 18 and 22, and 3rd-order pentagons with \( C \) = 15 and 18.

By controlling certain of the variables, one can discover some interesting patterns that permit the rapid construction of special PMPs. For example, for 3rd-order PMPs with an odd number of sides, it can be proved that a solution s always possible for the minimum \(C \). And it can be further demonstrated (to the amazement of your friends) that you can produce the solution just as rapidly as it takes to write the \(N\) integers. Rather than present the proof here, two examples will be given, and you can easily note the pattern for yourself.

Figure 7 shows the two examples. 14 is the minimum value of \(C\) for \(n\) = 3 and \(k\) = 5. 19 is the minimum value for \(n\) = 3 and \(k\) = 7.

Figure 7

The pattern in easily perceived once you look at the numbers closely. The article goes further into how to create various PMPs and reference should be made to that for further examples. 

I stumbled upon this topic by way of the number associated with my diurnal age today, 27417. This number is a member of OEIS A135503 for the case where \(n\) = 38.

A135503

\( \text{a} (n) = \dfrac{n \cdot (n^2 - 1)}{2} \)

The initial members of the sequence are:

0, 0, 3, 12, 30, 60, 105, 168, 252, 360, 495, 660, 858, 1092, 1365, 1680, 2040, 2448, 2907, 3420, 3990, 4620, 5313, 6072, 6900, 7800, 8775, 9828, 10962, 12180, 13485, 14880, 16368, 17952, 19635, 21420, 23310, 25308, 27417, 29640, 31980, 34440

The OEIS comments state that for \(n\) > 2, a(\(n\)) is the maximum value of the magic constant in a perimeter-magic \(n\)-gon of order \(n \). For the case of \(n\) = 38, \(n\) is even and so the formula in Figure 5 applies. Substituting in \(n\) = 38 and \(k\) = 38 does indeed give the maximum value of the magic constant as 27417.

The Reuleaux Triangle Revisited

In all my posts since the middle of 2015, I've only mentioned the Reuleaux (pronounciation) Triangle once and there's a good reason for this. Let's look at my original post from the 9th of November 2021 titled Reuleaux Triangle. I wrote (text in blue):

Today, having turned 26518 days old, I found an interesting property of this number that qualifies it for membership in OEIS A340644:

A340644

The number of vertices on a Reuleaux triangle formed by the straight line segments mutually connecting all vertices and all points that divide the sides into n equal parts.

 The initial members of the sequence are:

3, 19, 120, 442, 1332, 2863, 5871, 10171, 17358, 26518, 40590, 57757, 81735, 110209, 148158, 192184, 248772, 313105, 393429, 483283, 593490, 715528, 861660, 1022281, 1211811, 1418515, 1659108, 1919842, 2220204, 2543527, 2912751, 3308305, 3755922, 4233730, 4770150, 5340529, 5977071 

26518 corresponds to the case where \(n\) = 10 and this number stands in splendid isolation from its neighbours: 17358 (\(n\) = 9) and 40590 (\(n \) = 11). The former occurred before I started monitoring my diurnal age and the second will occur long after I'm dead. In my post, I included the image shown in Figure 1.

Figure 1

The post I made back then is a good one, if I do say so myself, and delves into some interesting properties and applications of the Reuleaux triangle. The reason the triangle has popped up again has to do again with my diurnal age (27336) but this time it's regions rather than vertices that are involved. The number is a member of OEIS A340639:

A340639

The number of regions inside a Reuleaux triangle formed by the straight line segments mutually connecting all vertices and all points that divide the sides into n equal parts.

The initial members of the sequence are:

1, 24, 145, 516, 1432, 3084, 6106, 10638, 17764, 27336, 41233, 58902, 82675, 111864, 149497, 194430, 250534, 316020, 395728, 487122, 596434, 720162, 865321, 1027974, 1216291, 1425348, 1664539, 1928022, 2226658, 2553204, 2920378, 3319536, 3764848, 4246638, 4780489, 5355414, 5988973 

As with the number mentioned in the earlier post, 27336 corresponds to the case of \(n\) = 10 and once again its neighbours (17764 and 41233) are far removed. The OEIS comments provide a link to a very colourful image of the regions. See Figure 2.

Figure 2: source

However, the regions can be seen more clearly perhaps in the case of \(n\) = 2. See Figure 3.

Figure 3: source

So it may be some time before the Reuleaux triangle gets a mention again because the only integers associated with it seem to relate to vertices and regions arising from the division of its sides into \(n\) equal parts.

Congruent Numbers

Over the years I've sometimes made note of when a number associated with my diurnal age is congruent. Often I've just ignored the fact. The definition of such number is:

A number is called congruent it is the area of a right triangle with rational sides.

The first congruent number is 5 because it is the area of a right triangle with sides of 20/3, 3/2, and 41/6. The next number, 6, is also congruent being the area of the famous 3, 4, 5 right triangle. See Figure 1.

Figure 1

7 is also a congruent number being the area of a right triangle with sides such that (see Figure 2):$$ \begin{align} \Big ( \frac{35}{12} \Big )^2+ \Big ( \frac{24}{5} \Big )^2 &= \Big ( \frac{337}{60} \Big )^2 \\ \frac{1}{2} \cdot \frac{35}{12} \cdot \frac{24}{5} &= 7 \end{align}$$

Figure 2

With 7, we begin to see the problem of determining whether a number is congruent or not. How do we find those fractions that confirm that 7 is congruent. It's not easy. To quote from Wikipedia:

The question of determining whether a given rational number is a congruent number is called the congruent number problem. This problem has not (as of 2019) been brought to a successful resolution. Tunnell's theorem provides an easily testable criterion for determining whether a number is congruent; but his result relies on the Birch and Swinnerton-Dyer conjecture, which is still unproven.

Without getting too deeply into this topic, it suffices to say that OEIS A006991 provides a list of so-called primitive congruent numbers up to 10,000.  These are congruent numbers that are square-free. If any of these numbers are multiplied by a positive square number, a new congruent number is created.

Let's take the number associated with my diurnal age yesterday as an example. 27260 factorises as follows:$$ \begin{align} 27260 &= 2^2 \times 5 \times 29 \times 47\\ &= 2^2 \times 6815 \end{align}$$Now 6815 is a member of OEIS A006991 and therefore 27260 is thus a congruent number because a primitive congruent number, 6815, has been multiplied by a positive square number (4).

It can be noted that 6814 is also a member of OEIS A006991 and thus 4 x 6814 = 27256 is also a congruent number. It's not uncommon for primitive congruent numbers to occur in pairs or even triples.  For example: (5, 6, 7), (13, 14, 15), (21, 22, 23), (29, 30, 31), 34 is a singleton, (37, 38, 39) etc.

Perimeters and Pythagorean Triples

My diurnal age today is 26800 and one of the properties of this number is that it's a member of OEIS A010814: perimeters of integer-sided right triangles. Such numbers are relatively frequent (almost 22%) as can be seen in the data below:

5824th member is 26780

5825th member is 26784

5826th member is 26790

5827th member is 26796

5828th member is 26800

5829th member is 26808

5830th member is 26810

5831th member is 26814

5832th member is 26820

As I discovered however, it's not easy to find the sides of the triangles with these perimeters. The \(m,n\) method for generating primitive Pythagorean triples is shown in Figure 1:

Figure 1

Thus we have the shortest side \(a=m^2-n^2\) with \(m \gt n\) with \(b=2mn\) and the hypotenuse \(c=m^2+n^2\) where \(m\) and \(n\) are integers. Using these formulae in SageMathCell to generate the triples and associated perimeters, I found that the calculation timed out once the perimeters got even moderately large (far smaller than 26800 for example). Fortunately, there was a web resource that I could call upon. See Figure 2:

Figure 2: https://r-knott.surrey.ac.uk/Pythag/pythag.html

As can be seen, the calculator spits out a plethora of data including the perimeter. It turns out that there are three different right angled triangles with a perimeter of 26800: (6800, 8844, 11156), (5360, 10050, 11390) and (4355, 10800, 11645). If the triangle's dimensions are a multiple of those of a primitive triangle then this is shown as well. This website is certainly the ultimate resource for information about right-angled triangles.

Sometimes there is only one right-angled triangle with a given perimeter. For example, the one triangle with perimeter 26814 has dimensions (3052, 11685, 12077) and it is primitive.  Perimeters that arise from only one triangle form OEIS A098714: only one Pythagorean triangle of this perimeter exists. 26814 is the 2944th member of this sequence and so such numbers make up slightly under 11% of all numbers (at least in the range up to 26814).

The subsite referenced in Figure 2 is maintained by a Dr. Ron Knott and the main site contains links to many other interesting mathematical concepts. Here is some biographical information:

Ph.D (1980, University of Nottingham), M.Sc (1976, University of Nottingham), B.Sc (Pure Maths, University of Wales), C.Math, FIMA, C.Eng, MBCS, CITP

Visiting Fellow, Department of Mathematics,

formerly Lecturer in Mathematics and Computing Science Departments (1979-1998)

Faculty of Electronics and Physical Sciences,

University of Surrey,

Contact me initially by Email: ronknott at mac dot com

I was a lecturer in the Departments of Mathematics and Computing Science at the University of Surrey, Guildford, UK, for 19 years until September 1998 when I left to start working for myself making web pages for maths education sites.

I now give mathematics talks to students at schools and universities as well as to general audiences, teachers' conferences and Science Festivals on topics of the web pages above, especially the Fibonacci Numbers and why they occur so often in plants, Fun with Fractions, As Easy As Pi etc.

I now live in Bolton, near Manchester in NW England.

More on Threes and Fours

This post is a follow up on my previous post so it will make more sense if that post is read first. I made that post yesterday and today I was surprised to find that the threes and fours were still following me. Today I'm 26204 days old today and this number turns out to be divisible by four. In fact:$$26204=4 \times 6551$$So there we have a very clear connection with the number 4 but what about 3? Well, 26204 has a connection to equilateral triangles because it's a member of OEIS A171971 (see Figure 1).

A171971

Integer part of the area of an equilateral triangle with side length \(n\).

Figure 1

So an equilateral triangle with a side of 246 units has an area of 26204 square units when rounded down to the nearest whole number which closer than 99.999% of the exact value. So that's the connection with the number 3. I know that it could be argued that once you start looking for connections to 3 and 4 in the larger numbers, you'll find them but the connections of 26204 to 3 and 4 are not obscure. The connection with 4 is via its factorisation, the most fundamental characteristic of a number, and the connection to 3 is via its very close approximation to the area of an equilateral triangle with integer sides viz. 246. Of course, I could go further and point out that 246 has three digits and an average digit sum of 4 but that might be overkill.

On this same day, I visited the local McDonalds with my granddaughter. While we there, she did some drawing on her iPad and I continued reading the Pauli and Jung book referred to earlier. I had just finished reading a chapter on synchronicity when we decided to return home. Upon exiting the restaurant, I was struck by another flagrant appearance of 3 and 4 in the carpark. Figure 2 shows the sight that confronted me after descending the front stairs of the building. The signs were right in front of me and stopped me in my tracks. I got my phone out and snapped the photo shown.

Figure 2

As I was writing this post, I was reminded of my own date of birth on the 3rd April 1949 that can be written in dd/mm/yy form as 3/4/49 or mm/dd/yy form as 4/3/49. I also got thinking about my year of birth that is a \(4k+1\) prime number and thus expressible as a sum of two squares. It turns out:$$1949 = 1849+100=43^2+10^2$$Thus 3 and 4 turn up even my year of birth. I seem to be on a 3-4 roll at the moment.

The tetrahedron seems to be the perfect fusion of the 3 and 4 numbers, having four triangular faces. See Figure 3 and Figure 4.

Figure 3: source

Figure 4: tetrahedron net (source)