Author: IMC

It could be argued that the square is the ‘nicest’ rhombus, but the rhombus with angles of 60 and 120 degrees seems nicer still. One of the nice things about the 60/120 rhomb are the plane tilings that can be constructed from it. One of these tilings is the ‘tumbling blocks’ tiling shown at the top of the post, in which at some points you see ‘cubes’ (around the degree 3 vertices), while at others you see ‘flowers’ (around the degree 6 vertices). Because of the two different types of vertices, this is known as a quasi-regular rhombic tiling. (Another tiling that uses the 60/120 rhomb is this one.)

If you want to build a polyhedron that resembles the tumbling block tiling, one method is to reduce the number of petals in your flowers to 5, and then stretch your 60/120 rhombs until they are have angles of 63.435 and 116.565 degrees. Thirty of these rhombs arranged around vertices of degree 3 and 5 produces a quasi-regular polygon known as the rhombic triacontahedron.

The image above is of a model that was built using rhombic units printed onto card stock.

Despite the apparent ugliness of the 63.435 and 116.565 degree angle-measurements, our nice 60/120 rhomb has been, arguably, stretched into an even nicer one – a ‘golden rhombus,’ so called because the ratio of the diagonals is equal to the golden ratio.

The dual of the rhombic triacontahedron is the archemedian polyhedron known as the icosidodecahedron.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

The diagram above illustrates a surprising correspondence – if you take the reciprocal of a particular polynomial whose coefficients match the d+1 row of Pascal’s Triangle, you obtain a formal power series whose coefficients are precisely the elements in the dth column of the triangle.

The case for the third row is shown above (row and column numbering starts at 0 in Pascal’s Triangle), and the case for the fourth row is shown in the diagram below. To actually work out the power series that comes from the reciprocals (well, at least the first few coefficients), the grid method of dividing polynomials works well (I think).

In this way, the rows of the triangle  provide all the information required to obtain the columns – a nice correspondence between a finite sequence and an infinite one.

You’ll see this correspondence if you spend time investigating the “ordinary power series generating functions” (opsgf’s) for the higher dimensional triangular numbers. The row polynomials are just (1-x)^(d+1), while the columns correspond to formal power series whose coefficients are the d-dimensional triangular numbers. A great place to learn about these opsgf’s is H. Wilf’s book generatingfunctionology.

You can uncover the general opsgf for higher dimensional triangular numbers by starting with the (1-x)^(-1) rational function, and applying Wilf’s ‘rule 5’ from chapter 2 of generating functionology.

Applying a few other rules, you can obtain an opsgf for the k-polygonal numbers (k = 4 is squares, k = 5 is pentagonals, k = 6 hexagonals, etc.).

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

Imagine a tetrahedron centered inside a sphere. If you were to project the edges of the tetrahedron out from the center so that they touched the surface of the sphere, the edges would cut the sphere in arcs that lie on the sphere’s ‘great circles’ (the largest possible circles drawn on the surface of the sphere – circles whose radii are the same as the radius of the sphere).

The image at the top of this post shows a model of a sphere with the six great circles formed by the projected edges of a tetrahedron sitting at its center.

The model, described on pages 4 and 5 of Magnus Wennigers’ Polyhedron Models, is easy to construct using the template shown below. Printing copies of the template onto card-stock and gluing them together works well. This pdf file has 10 arcs on a single page for printing (you will need 24 arcs in total).

Each arc is folded and glued to form a spherical triangle, 24 of the spherical triangles glued together will form the sphere.

The diagram below (based on Fig.2, p.5, in Wenniger) shows how to assemble 6 of the triangles into one ‘face’ of the spherical tetrahedron. Four of the faces can be assembled into the ‘spherical tetrahedron.’ When assembling, you should make sure that you align the triangles so that they form great circles.

Spherical triangles, as you can tell from the diagram, do not follow the rule of having their interior angles sum to pi (180 degrees), as plane triangles do.

The arc is divided into angles of 54deg&44min, 70deg&32min, 54deg&44min, as described by Wenniger. The template was created in GSP.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

The triangular numbers (and higher triangular numbers) can be generated using this recurrence relationship:

These will form Pascal’s triangle (if we shift the variables n->n+d-1 and d->r, we get the familiar C(n,r) indexing for the Pascal Triangle). The d=2 case gives the usual “flat” 2d triangular numbers, and other d values provide triangular numbers of different dimensions.

 

It turns out that recurrence relation can be generalized to generate a family of sequences and triangles. Consider this more general relation:

Doing some initial exploring reveals four interesting cases:

The triangular numbers 

With all these additional parameters set to 1, we get our original relation, the familiar triangular numbers, and Pascal’s triangle.

The k-polygonal numbers 

If we set the “zero dimension” to k-2, we end up with the k-polygonal numbers. The triangular numbers arise in the special case where k=3. Except in the k=3 case, the triangles that are generated are not symmetrical.

 

Below is the triangle generated by setting k=5.

The symmetrically shifted k-polygonal numbers

As far as I know, there is not a standard name for these.  Each k value will generate a triangle that is symmetrical about its center and whose edge values are equal to k-2. For a given k value, if you enter sequences generated by particular values of d, you’ll find that some are well known. The codes in the diagrams correspond to the sequence ids from the Encyclopedia.

 

Here is the triangle generated by k=4:

And here is the triangle generated for k=5:

The Eulerian numbers (Euler’s number triangle)

This is a particularly nice way to generate the Eulerian numbers, which have a nice connection to the triangular numbers. There is a little inconsistency in the way the Eulerian numbers are indexed, however. For this formula to work, it should be altered slightly so that d>0. The resulting formula looks like this:

And the triangle looks like this:

It is surprising that so many interesting and well known sequences and triangles can be generated from such a simple formula, and that they can be interpreted as being part of a single family.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

There are plenty of probability problems that have counter-intuitive solutions. Problems like these, and how they can undermine common sense, are among the best reasons for looking at probability theory. One set of humbling questions is the family of Monty Hall problems. Another are those related to conditional probability; nice examples of these are problems that involve medical tests that give ‘false positive’ results.

Simulation is a way of exploring these problems that reveals more than mere theoretical probability calculations do. The structure of the simulation can reflect interesting aspects of the structure of the original problem, and the results reveal a variability that is not apparent when you simply calculate the theoretical probabilities on their own.

This post shows an example of a ‘false positive’ probability problem and a Fathom simulation for it. This problem was adapted from one found in the 4th edition of Ross’s A First Course in Probability (p.75):

A laboratory blood test is 95 percent effective in detecting a certain disease when it is, in fact, present. However, the test also yields a “false positive” result for one percent of healthy persons. (That is, if a healthy person is tested, there is a 0.01 probability that they will test positive for the disease, even though they don’t really have it.) If 0.5 percent of the population actually has the disease, what is the probability that a person who has a positive test result actually has the disease?

Here are the attribute definitions that you could use to build a Fathom simulation for this problem:

The attributes are enough to run the simulation, but it is better to also add the following measures:

To run the simulation you can add new cases (try ~500). Using a measures collection, you can re-run this experiment hundreds of times (collecting a 100 measures re-runs the 500 person experiment 100 times).

If you are calculating the theoretical probability by hand, it helps to write down all of the probabilities (fill in the blanks…):

It also helps to visualize the probabilities in a tree diagram:

The outer tips of the tree are filled in using the multiplicative rule for conditional probabilities:

One nice thing about doing this is that you can see how the tree diagram used to calculate the theoretical probabilities is structured in the same way as the “if” statements in the simulation.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

Students generally learn to divide polynomials using  long division or synthetic division. This post is about another method for dividing polynomials,  the “grid” method. Although this method is used at some schools, I have not found it in any textbook (if you know of any, please let me know).

The grid method makes polynomial division a fun calculation – an almost SUDOKU-like process. The recreational challenge here is to master the method, and to convince yourself that it actually works in all cases (which cases does it work well for, and which are more difficult?).

Before tackling polynomial grid division, you need to be familiar with polynomial grid multiplication. It is the symbolic analog to the familiar “algebra tiles” manipulative, so if you have worked with these, it should be reasonably familiar.

Polynomial Grid Multiplication

In polynomial grid multiplication, the two factors are arranged on the edges of a grid – the number of terms of the factors determine the number of horizontal and vertical lines that make up the grid.

The overall product is found by filling in the cells of the grid with the product of the terms for the row and column, and then summing up all the contents of the interior of the grid.

What polynomial grid multiplication does for us is provide an explicit way to keep track of the distributive property: each term-by-term product gets its own cell.

In the example below, the two factors are placed along the edges of the grid (1). One factor provides the row headings, the other provides the column headings. Then each cell is filled in with the product of the terms from the row and column (2). Finally, the cells are added together (like terms are combined) to find the final product (3). If the terms of the factors are placed according to the same order (descending powers of x, in our example) and there are no missing terms, then like-terms of the product are found along the diagonals of the grid.

Polynomial Grid Division

Polynomial grid division works the same way as polynomial grid multiplication, but in reverse – we start by knowing one of the factors (placed along the edge of the grid), and by knowing what we want the product to be (without knowing exactly how it is ‘split up’ in the grid). Using this knowledge we work backwards, filling in the grid and the top edge one cell at a time until we are done.

Consider the example below. In (1) we create an empty grid with the denominator (divisor) playing the role of one of the factors. Since this question involves a degree 3 polynomial divided by a degree 1 polynomial, we know that the other factor (the quotient) must be degree 2. This allows us to create a grid with the correct size. In step (2) we use the highest power of the dividend (the numerator) to begin to fill in the grid – we know that 27x^3 must be in the top left. This in turn tells us that the first column entry must be 9x^2 (so that the row and column multiply to 27x^3). In step (3) we use this to fill in all of the first column, multiplying 9x^2 by the terms of the row entries.

In step (3) we now have a quadratic term -18x^2. But we know from looking at the dividend (numerator) that in the final answer we actually want 9x^2. Consequently, the other entry on the quadratic diagonal must be 27x^2, so that the whole diagonal sums to 9x^2 .  Filling this in for step (4) tells us what all the entries in the second column should be (step 5). Now that we have a linear entry -18x, we know that we need to add in a 15x so that the overall sum gives a -3x (step 6).

Having a 15x tells us that the top entry must be 5 (the product of 5 and 3x gives us 15x). Filling this in in step 7 allows us to complete the table, and we see that our final constant entry is -10, as hoped for. Now that the grid has been filled in and it matches the dividend, we can read the answer off the top – the factor that we have uncovered is the quotient we were hoping to calculate.

This method is actually easier than it seems at first, and when all steps are carried out on the same grid, is quite compact.

Here is another example for you to try it out on.

In these two examples, the division worked well – there was no remainder. In the case where we are dividing f/g and g is not a factor of f, and the degree of g is less than the degree of f, there is polynomial remainder whose degree is strictly less than that of g. So, for example, when g is a linear function (degree 1), f/g can have a constant remainder (degree 0).

In this case we proceed as above, attempting to fill in the grid with the numerator (dividend). However, if when we are done the grid does not match the numerator, we have a remainder. The remainder is the additional amount that we have to add to the grid in order to arrive at the numerator.

Consider a division question almost identical to the first one that we looked at, except here we change the numerator slightly so that it doesn’t factor well.

Following the same steps as before, we end up with a grid sum that does not match our desired answer: we have -10 instead of -9 for the final constant term. This tells us that we have a remainder of +1, that we choose to write next to the grid. In our final answer, the remainder tells us the “remaining fractional part” that we have to add at the end.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

Here is an outline of how you can go about generating this data. The definition and properties of Farey sequences here are from Hardy & Wright’s An Introduction to the Theory of Numbers (which I am slowly working my way through).
The Farey sequence of order n is the sequence of irreducible fractions between 0 and 1 whose denominator does not exceed n. So, the elements of the sequence are of the form h/k, where h < k < n, and h and k are relatively prime.

The main theorem about Farey numbers provides them with their characteristic property (Theorem 29 in TofN). The characteristic property of Farey sequences is that if h/k, h”/k”, and h’/k’ are successive terms in a Farey sequence, then h”/k” is the mediant of h/k and h’/k’. If h/k and h’/k’ are two reduced fractions, their mediant is given by (h+h’)/(k+k’).

It’s nice when a theorem tells you how to implement an algorithm. This property tells us that Farey sequences can be built iteratively or recursively, beginning with F1={0/1 ,1/1}. The algorithm to do this is a nice one – it’s probably not often used as a textbook exercise in recursion because it helps if you to have some data structure or class to represent the fraction, and a way of telling if integers are relatively prime (you can use the Euclidean algorithm to implement a gcd() function).

Here is an outline of how to calculate the next Farey sequence, given that you have one already.

0) input a Farey sequence oldSequence (initial sequence will be {0/1, 1/1})

1) create a new empty sequence newSequence

2) iterate over oldSequence and find out its level by finding the largest denominator that occurs store this in n

3) set n= n+1

4) iterate over oldSequence, looking at each pair of adjacent elements (left and right)

4.1) add left to newSequence
4.2) if the denominators of left and right sum to n, form their mediant
4.2.1) if the numerator and denominator of the mediant are relatively prime, add mediant to newSequence

5) add the last element of oldSequence to newSequence

Note that you only need to add in new elements where the denominators of existing adjacent elements sum to the n value – when this happens you form the mediant of the two adjacent elements. Furthermore, the mediant is only added if the fraction can’t be reduced.

Below is some Java-ish code corresponding to the above – it assumes that the oldSequence and newSequence are an ArrayList and that you have a class Fraction that has fields num (numerator) and den (denominator).

Here are the first five Farey sequences that you get from the algorithm:

The image at the top of the post was generated by implementing the algorithm in Processing, and using the result to draw the associated Ford circles – you could do something similar in regular Java (or other language). If you draw the Ford Circles associated with the sequence, the circle for a fraction “frac” will be centered at (x,y) and have a radius r where

x = (scale)*frac.num/frac.den

y = r

r = (scale)/(2*(frac.den)^2)

where “scale” is some scaling factor (probably in the 100’s) that increases the size of the image.

Here I decided to draw two copies of each circle, one on top of the other.

 

That it contains only fractions between 0 and 1 and that it contains all reduced fractions for denominators k < n, connects Farey sequences to Euler’s totient function. Euler’s totient function is an arithmetic function that for a given k, counts the integers less than k that are relatively prime to it. This is exactly the number of times that a fraction of the form h/k will appear in the Farey sequence for k>1.

The Farey algorithm, how to draw Ford circles, and the connection to Euler’s totient function are described nicely in J.H. Conway and R.K. Guy’s The Book of Numbers – a great companion to a book like TofN.

 

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

 

Chapter 3 of Hardy & Wright’s An Introduction to the Theory of Numbers involves Farey sequences – which, in addition to showing up in serious number theory books, are an interesting, accessible, and popular topic in recreational mathematics.

Since I am a committed constructivist (in at least a few senses of the word) I thought it would be nice to come up with an activity with Farey sequences that could be carried out without too much advance discussion about what they are. What I came up with is a Fathom activity that starts with some simple but odd looking data that allows you to construct some very interesting plots and displays. The idea is that you will get a feel for what a Farey sequence is by using the data to build the sequence and by looking at the results from different perspectives.

Step 1. Import some data

Here is the data to import into Fathom. It should create 33 cases with two attributes n, and d.

n d

0 1

1 10

1 9

1 8

1 7

1 6

1 5

2 9

1 4

2 7

3 10

1 3

3 8

2 5

3 7

4 9

1 2

5 9

4 7

3 5

5 8

2 3

7 10

5 7

3 4

7 9

4 5

5 6

6 7

7 8

8 9

9 10

1 1

 

Step 2. Add some more attributes

After importing this data, you should create the following attributes

i = caseIndex

q = n/d

dist = q – prev(q)

mediant = (n+prev(n))/(d+prev(d))

d_mediant = mediant – prev(mediant)

disp = concat(n,”/”,d)

The meaning of the “mediant” attributes will become clearer after you read about Farey sequences.

 

Step 3. Explore some plots

Plots you could try creating are:

1. A) n on the y axis and d on the x axis.
2. B) dist on y, i on x (a filter of i>1 makes sense here)
3. C) d_mediant on y, i on x (a filter of i>2 makes sense here)
4. D) mediant on y, q on x (adding a function y=x makes sense here)
5. E) d on y, q on x
6. F) dist on y, q on x
7. H) q on x with no other attributes

While creating these plots, you should be thinking about describing the sequence that the cases represent. What kinds of numbers are they, what values do they have (do they lie in a certain interval?), how close are they to one another?

Step 4. Create a nice display

One of the more visually interesting thing you can do with the Farey sequence is to display its associated Ford circles. This can be done by adding two new sliders and by editing the display settings for the collection.

 

Add the sliders “scale” and “shift,” and give them these initial values:

scale = 400

shift = 150

Now on the collection inspector, click on the Display tab, and edit the formulas for these attributes

x= q*scale +shift

y = scale/(2*d^2)

image = blueCircleIcon

width = scale/(d^2)

height = scale/(d^2)

caption = “”

If you pull open the display, you will see the initial iterations of a fractal pattern known as the Ford Circles that are generated by the Farey sequence. Here is what it should look like:

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

The double recurrence relation that defines the higher triangular numbers is a simple one – it is no surprise that they turn up so often.

The geometric interpretation is stacking: For a given dimension d, you get the n+1 d-triangular number by stacking the nth d-1 triangular number (the gnomon) onto the nth d-triangular number.  The zero dimensional triangular numbers are just the sequence: 1, 1, 1, 1,…, presumably counting stacks of nothing. The one-dimensional triangular numbers are the naturals: 1, 2, 3, 4, …, made by stacking the ones of the one-dimensional case. The two dimensional triangular numbers stack the naturals: 1, 3, 6, 10, …, the three dimensional triangular numbers make pyramids of the triangulars: 1, 4, 10, 20, ….

If you write out a difference table for the higher triangular numbers, you end up with Pascal’s triangle. This suggests a nice formula for the triangulars in terms of binomial coefficients:

From this, you can obtain another recursive formula that you can use when working with higher triangular numbers (this is the “another” formula for this post):

If you vary the defining recurrence relation so that the initial “zero dimensional” value is a number other than 1, you get the other polygonal numbers (square, pentagonal, hexagonal, square-based pyramidal, etc.). In particular, if you let the zero-dimensional value be k-2, you obtain the k-polygonal numbers (k-2 corresponding to the number of triangles in your k-sided polygon).

It turns out there is a nice formula for these in terms of binomial coefficients as well:

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to dan.mackinnon*

As I make my way through Hardy & Wright’s An Introduction to the Theory of Numbers,  I am hoping to work it into my recreational math pursuits – coming up with interesting (but not too heavy) activities that correspond roughly to the material in the text.

The first two chapters are on the sequence of primes. Here’s the activity: obtain a list of primes, import them into Fathom, and construct plots that explore pn and pi(n) and other aspects of the sequence that manifest themselves in the first couple of thousand terms.

In my Fathom experiment, I imported the first 2262 prime numbers.

If you import a sequential list of primes into Fathom (under the attribute prime) and add another attribute n=caseindex, you can create two nice plots. Plot A should have prime as the x axis and n  as the y axis. This shows the function pi(n). To this plot you should add the function y = x/ln(x) and visually compare the two curves. Plot B should have the x and y axis reversed. On this graph, plotting the function y = x*ln(x) shows how closely this approximation for pn (the nth prime) comes to the actual values.

 

You can add further attributes to look at the distance between primes dist=prime-prev(prime), and also the frequency of twin primes is_twin = (dist=2)or(next(dist)=2).

You can also add attributes to keep a running count of twin_primes, and to keep a running average of the twin_primes. The plot above shows how the ratio of tiwn primes diminishes as the number of primes increases. The plot at the top of the post suggests the distribution of primes and twin primes (in blue) in the numbers up to the 2262nd prime.

For more such insights, log into www.international-maths-challenge.com.

Credit for article given to dan.mackinnon*