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The Perfect Rugby Kick
This was inspired by the ever excellent Numberphile video which looked at this problem from the perspective of Geogebra. I thought I would look at the algebra behind this.
In rugby we have the situation that when a try is scored, there is an additional kick (conversion kick) which can be taken. This must be in a perpendicular line with where the try was scored, but can be as far back as required.
We can represent this in the diagram above. The line AB represents the rugby goals (5.6 metres across). For a try scored at point D, a rugby player can then take the kick anywhere along the line DC.
Let’s imagine a situation where a player has scored a try at point D – which is a metres from the rugby post at B. For this problem we want to find the distance, x for this value of a such that this maximises the value of θ . The larger the value of θ, the more of the rugby goal the player can aim at and so we are assuming that this is the perfect angle to achieve.
Making an equation:
We can use the diagram to achieve the following equation linking θ and x:
We can use Desmos to plot this graph for different values of a:
We can then find the maximum points from Desmos and record these
This then allows us to find the exponential regression line of the coordinates of the maximum points:
This regression line is given by the equation:
This graph is shown below:
We can also plot the x values of the maximum points against a to give the following linear regression line:
This graph is shown below:
This means that if we know the value of a we can now very easily calculate the value of x which will provide the optimum angle.
Bring in the calculus!
We can then see how close our approximations are by doing some calculus:
We can find the x coordinate of the maximum point in terms of a by differentiating, setting equal to 0 and then solving. This gives:
When we plot this (green) versus our earlier linear approximation we can see a very close fit:
And if we want to find an equation for optimum theta in terms of x we can also achieve this as follow:
When we plot this (green) we can also see a good fit for the domain required:
Conclusion
A really nice investigation – could be developed quite easily to score very highly as an HL IA investigation as it has a nice combination of modelling, trigonometry, calculus and generalised functions. We can see that our approximations are pretty accurate – and so we can say that a rugby player who scores a try a metres from the goal should then take the resultant conversion kick about a+2 metres perpendicular distance from the try line in order to maximise the angle to the goal.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Prime Spirals – Patterns in Primes
One of the fundamental goals of pure mathematicians is gaining a deeper understanding of the distribution of prime numbers – hence why the Riemann Hypothesis is one of the great unsolved problems in number theory and has a $1 million prize for anyone who can solve it. Prime numbers are the the building blocks of our number system and are essential to our current encryption methods such as RSA encryption. Hence finding patterns in the primes is one of the great mathematical pursuits.
Polar coordinates
The beautiful prime spiral was generated above on Desmos using polar coordinates. We can see a clear spiral pattern – so let’s see how to create this. Polar coordinates (r, θ) need a length (r) from the origin and an angle of anti-clockwise rotation from the origin (θ). So for example in polar coordinates (2,2) means a length of 2 from the origin and a rotation of 2 radians. By considering trigonometry and the unit circle we can say that the polar coordinates (r, θ) are equivalent to the Cartesian coordinate (r.cosθ, r.sinθ).
Plotting prime pairs
So we plot the first few prime pairs:
Polar: (2,2). Cartesian: (2cos2, 2sin2).
Polar: (3,3). Cartesian: (3cos3, 3sin3).
Polar: (5,5). Cartesian: (5cos5, 5sin5).
In Desmos (making sure we are in radians) we input:
We can then change the Desmos graph view to polar (first click on the spanner on the right of the screen). This gives the first 3 points of our spirals. Note I have labeled the points as polar coordinates.
I then downloaded the first 1000 prime numbers from here. I then copied this list of comma separated values and pasted it into an empty part of square brackets M = [ ] in Desmos to create a list.
I can then plot every point in the list as a prime pair by doing the following:
We can then generate our prime spiral for the first 1000 prime pairs:
Just to see how powerful Desmos really is, I then downloaded all the prime numbers less than or equal to 100,000 from here. This time we see the following graph:
We can see that we lose the clear definition of the spiral – though there are still circular spirals with higher densities of primes than others. Also we can see that there are higher densities of the primes on some of the radial lines out from the origin – and other radial lines where no primes appear.
Prime Number Theorem
We can also use our Desmos result to investigate another (more fundamental) result about the distribution of prime numbers. The prime number theorem states:
Here pi(N) is the number of prime numbers less than or equal to N. The little squiggle means that as N gets large pi(N) becomes better and better approximated by the function on the RHS.
For our purple “spiral” above we downloaded all the primes less than or equal to 100,000 – and Desmos tells us that there were 9,592 of them. So let’s see how close the prime number theorem gets us:
We can see that we are off by an error of around 9.46% – not too bad, though still a bit out. As we make N larger we will find that we get a better and better approximation.
Let’s look at what would happen if we took N as 1,000,000,000. From Wikipedia we can see that there are 50,847,534 primes less than or equal to 1,000,000,000. Therefore:
This time we are off by an error of only 5.10%. Have a look at the table of values in Wikipedia to find how large N has to be to be within 1% accuracy.
So this is a nice introduction to looking for patterns in the primes – and a good chance to explore some of the nice graphical capabilities of Desmos. See if you can find any more patterns of your own!
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
The Barnsley Fern: Mathematical Art
This pattern of a fern pictured above was generated by a simple iterative program designed by mathematician Michael Barnsely. I downloaded the Python code from the excellent Tutorialspoint and then modified it slightly to run on repl.it. What we are seeing is the result of 40,000 individual points – each plotted according to a simple algorithm. The algorithm is as follows:
Transformation 1: (0.85 probability of occurrence)
xi+1 = 0.85xi +0.04yi
yi+1= -0.04xi+0.85yi+1.6
Transformation 2: (0.07 probability of occurrence)
xi+1 = 0.2xi -0.26yi
yi+1= 0.23xi+0.22yi+1.6
Transformation 3: (0.07 probability of occurrence)
xi+1 = -0.15xi -0.28yi
yi+1= 0.26xi+0.24yi+0.44
Transformation 4: (0.01 probability of occurrence)
xi+1 = 0
yi+1= 0.16yi
So, I start with (0,0) and then use a random number generator to decide which transformation to use. I can run a generator from 1-100 and assign 1-85 for transformation 1, 86-92 to transformation 2, 93-99 for transformation 3 and 100 for transformation 4. Say I generate the number 36 – therefore I will apply transformation 1.
xi+1 = 0.85(0)+0.04(0)
yi+1= -0.04(0)+0.85(0)+1.6
and my new coordinate is (0,1.6). I mark this on my graph.
I then repeat this process – say this time I generate the number 90. This tells me to do transformation 2. So:
xi+1 = 0.2(0) -0.26(1.6)
yi+1= 0.23(0)+0.22(1.6)+1.6
and my new coordinate is (-0.416, 1.952). I mark this on my graph and carry on again. The graph above was generated with 40,000 iterations – let’s see how it develops over time:
1000 iterations:
10,000 iterations:
100,000 iterations:
500,000 iterations:
If we want to understand what is happening here we can think of each transformation as responsible for a different part of our fern. Transformation 1 is most likely and therefore this fills in the smaller leaflets. Transformations 2 and 3 fill in the bottom left and right leaflet (respectively) and transformation 4 fills in the stem.
It’s quite amazing to think that a simple computer program can create what looks like art – or indeed that is can replicate what we see in nature so well. This fern is an example of a self-similar pattern – i.e one which will look the same at different scales. You could zoom into a detailed picture and see the same patterns repeating. You might want to explore the idea of fractals in delving into this topic in more detail.
Changing the iterations
We can explore what happens when we change the iterations very slightly.
Christmas tree
Crazy spiral
Modern art
You can modify the code to run this here. Have a go!
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Sphere packing problem: Pyramid design
Sphere packing problems are a maths problems which have been considered over many centuries – they concern the optimal way of packing spheres so that the wasted space is minimised. You can achieve an average packing density of around 74% when you stack many spheres together, but today I want to explore the packing density of 4 spheres (pictured above) enclosed in a pyramid.
Considering 2 dimensions
First I’m going to consider the 2D cross section of the base 3 spheres. Each sphere will have a radius of 1. I will choose A so that it is at the origin. Using some basic Pythagoras this will give the following coordinates:
Finding the centre
Next I will stack my single sphere on top of these 3, with the centre of this sphere directly in the middle. Therefore I need to find the coordinate of D. I can use the fact that ABC is an equilateral triangle and so:
3D coordinates
Next I can convert my 2D coordinates into 3D coordinates. I define the centre of the 3 base circles to have 0 height, therefore I can add z coordinates of 0. E will be the coordinate point with the same x and y coordinates as D, but with a height, a, which I don’t yet know:
In order to find a I do a quick sketch, seen below:
Here I can see that I can find the length AD using trig, and then the height DE (which is my a value) using Pythagoras:
Drawing spheres
The general equation for spheres with centre coordinate (a,b,c) and radius 1 is:
Therefore the equation of my spheres are:
Plotting these on Geogebra gives:
Drawing a pyramid
Next I want to try to draw a pyramid such that it encloses the spheres. This is quite difficult to do algebraically – so I’ll use some technology and a bit of trial and error.
First I look at creating a base for my pyramid. I’ll try and construct an equilateral triangle which is a tangent to the spheres:
This gives me an equilateral triangle with lengths 5.54. I can then find the coordinate points of F,G,H and plot them in 3D. I’ll choose point E so that it remains in the middle of the shape, and also has a height of 5.54 from the base. This gives the following:
As we can see, this pyramid does not enclose the spheres fully. So, let’s try again, this time making the base a little bit larger than the 3 spheres:
This gives me an equilateral triangle with lengths 6.6. Taking the height of the pyramid to also be 6.6 gives the following shape:
This time we can see that it fully encloses the spheres. So, let’s find the density of this packing. We have:
Therefore this gives:
and we also have:
Therefore the density of our packaging is:
Given our diagram this looks about right – we are only filling less than half of the available volume with our spheres.
Comparison with real data
[Source: Minimizing the object dimensions in circle and sphere packing problems]
We can see that this task has been attempted before using computational power – the table above shows the average density for a variety of 2D and 3D shapes. The pyramid here was found to have a density of 46% – so our result of 44% looks pretty close to what we should be able to achieve. We could tweak our measurements to see if we could improve this density.
So, a nice mixture of geometry, graphical software, and trial and error gives us a nice result. You could explore the densities for other 2D and 3D shapes and see how close you get to the results in the table.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Projectile Motion III: Varying gravity
We can also do some interesting things with projectile motion if we vary the gravitational pull when we look at projectile motion. The following graphs are all plotted in parametric form.
Here t is the parameter, v is the initial velocity which we will keep constant, theta is the angle of launch which we will vary, and g is the gravitational constant which on Earth we will take as 9.81 m/s2.
Earth
Say we take a projectile and launch it with a velocity of 10 m/s. When we vary the angle of launch we get the folowing graphs:
On the y axis we have the vertical height, and on the x axis the horizontal distance. Therefore we can see that the maximum height that we achieve is around 5m and the maximum horizontal distance is around 10m.
Other planets and universal objects
We have the following values for the gravitational pull of various objects:
Enceladus (Moon of Saturn): 0.111 m/s2, The Moon: 1.62 m/s2, Jupiter: 24.8 m/s2, The Sun: 274 m/s2, White dwarf black hole surface gravity: 7×1012m/s2.
So for each one let’s see what would happen if we launched a projectile with a velocity of 10 m/s. Note that the mass of the projectile is not relevant (though it would require more force to achieve the required velocity).
Enceladus:
The Moon:
Jupiter:
The Sun:
Black hole surface gravity:
This causes some issues graphically! I’ll use the equations derived in the last post to find the coordinates of the maximum point for a given launch angle theta:
Here we have v = 10 and g = 7×1012m/s2. For example if we take our launch angle (theta) as 45 degrees this will give the coordinates of the maximum point at:
(7.14×10-12, 3.57×10-12).
Summary:
We can see how dramatically life would be on each surface! Whilst on Earth you may be able to throw to a height of around 5m with a launch velocity of 10 m/s., Enceladus would see you achieve an incredible 450m. If you were on the surface of the Sun then probably the least of your worries would be how hight to throw an object, nevertheless you’d be struggling to throw it 20cm high. And as for the gravity at the surface of a black hole you wouldn’t get anywhere close to throwing it a nanometer high (1 billionth of a meter).
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
Projectile Motion Investigation II
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Another example for investigating projectile motion has been provided by fellow IB teacher Ferenc Beleznay. Here we fix the velocity and then vary the angle, then to plot the maximum points of the parabolas. He has created a Geogebra app to show this (shown above). The locus of these maximum points then form an ellipse.
We can see that the maximum points of the projectiles all go through the dotted elliptical line. So let’s see if we can derive this equation.
Let’s start with the equations for projectile motion, usually given in parametric form:
Here v is the initial velocity which we will keep constant, theta is the angle of launch which we will vary, and g is the gravitational constant which we will take as 9.81.
We can plot these curves parametrically, and for each given value of theta (the angle of launch) we will create a projectile motion graph. If we plot lots of these graphs for different thetas together we get something like this:
We now want to see if the maximum points are in any sort of pattern. In order to find the maximum point we want to find when the gradient of dy/dx is 0. It’s going to be easier to keep things in parametric form, and use partial differentiation. We have:
Therefore we find the partial differentiation of both x and y with respect to t. (This simply means we can pretend theta is a constant).
We can then say that:
We then find when this has a gradient of 0:
We can then substitute this value of t back into the original parametric equations for x:
and also for y:
We now have the parametric equations in terms of theta for the locus of points of the maximum points. For example, with g= 9.81 and v =1 we have the following parametric equations:
This generates an ellipse (dotted line), which shows the maximum points generated by the parametric equations below (as we vary theta):
And here is the graph:
We can vary the velocity to create a new ellipse. For example the ellipse generated when v =4 creates the following graph:
So, there we go, we have shown that different ellipses will be created by different velocities. If you feel like a challenge, see if you can algebraically manipulate the parametric equations for the ellipse into the Cartesian form!
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Classical Geometry Puzzle: Finding the Radius
This is another look at a puzzle from Mind Your Decisions. The problem is to find the radius of the following circle:
We are told that line AD and BC are perpendicular and the lengths of some parts of chords, but not much more! First I’ll look at my attempt to solve this. It’s not quite as “nice” as the solution in the video as it requires the use of a calculator, but it still does the job.
Method 1, extra construction lines:
These are the extra construction lines required to solve this problem. Here is the step by step thought process:
- Find the hypotenuse of triangle AGC.
- Use the circle theorem angles in the same segment are equal to show that angle CBD = angle CAG.
- Therefore triangle AGC and GBD are similar, so length BG = 4. We can now use Pythagoras to find length BD.
- We can find length CD by Pythagoras.
- Now we have 3 sides of a triangle, CDB. This allows use to find angle BDC using the cosine rule.
- Now we the circle theorem angles in the same segment are equal to show that angle BDC = angle BEC.
- Now we use the circle theorem angles in a semi circle are 90 degrees to show ECB = 90.
- Now we have a right angled triangle BCE where we know both an angle and a side, so can use trigonometry to find the length of BE.
- Therefore the radius is approximately 4.03.
Method 2, creating a coordinate system
This is a really beautiful solution – which does not require a calculator (and which is discussed in the video above). We start by creating a coordinate system based around point G at (0,0). Because we have perpendicular lines we can therefore create coordinates for A, B and C. We also mark the centre of the circle as (p,q).
First we start with the equation of a circle centre (p.q):
Next we create 3 equations by substituting in our coordinates:
Next we can do equation (3) – equation (1) to give:
Next we can substitute this value for p into equations (1) and (3) and equate to get:
Lastly we can substitute both values for p and q into equation (1) to find r:
We get the same answer as before – though this definitely feels like a “cleaner” solution. There are other ways to solve this – but some of these require the use of equations you may not already know (such as the law of sines in a circumcircle, or the equation for perpendicular chords and radius). Perhaps explore any other methods for solving this – what are the relative merits of each approach?
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
When do 2 squares equal 2 cubes?
Following on from the hollow square investigation this time I will investigate what numbers can be written as both the sum of 2 squares, 2 cubes and 2 powers of 4. i.e a2+b2 = c3+d3 = e4+f4.
Geometrically we can think of this as trying to find an array of balls such that we can arrange them into 2 squares, or we can rearrange them and stack them to form 2 cubes, or indeed we can arrange them into 2 4-dimensional cubes. I’ll add the constraints that all of a,b,c,d,e,f should be greater than 1 and that the pair of squares or cubes (etc) must be distinct. Therefore we can’t for example have 2 squares the same size.
Infinite solutions
Let’s look at why we can easily find infinite solutions if the squares or cubes (etc) can be the same size.
We want to find solutions to:
a2+b2 = c3+d3 = e4+f4.
so we look at the powers 2,3,4 which have LCM of 12. Therefore if we choose powers with the same base we can find a solution. For example we chose to work with base 2. Therefore we choose
a = 26, b = 26, which gives 212+212
c = 24, d = 24, which gives 212+212
e = 23, f = 23, which gives 212+212
Clearly these will be the same. So we can choose any base we wish, and make the powers into the same multiples of 12 to find infinite solutions.
Writing some code
Here is some code that will find some other solutions:
list1=[]
for a in range(2, 200):
for b in range(2,200):
list1.append(a**2+b**2)
list2=[]
for j in list1:
for c in range(2,200):
for d in range(2,200):
if c**3+d**3 == j:
list2.append(c**3+d**3)
print(list2)
for k in list2:
for e in range(2,200):
for f in range(2,200):
if k == e**4+f**4:
print(k,e,f)
This returns the following solutions: 8192, 18737, 76832. Of these we reject the first as this is the solution 212+212 which we found earlier and which uses repeated values for the squares, cubes and powers of 4. The 3rd solution we also reject as this is formed by 14 4 + 14 4. Therefore the only solution up to 79202 (we checked every value up to and including 1992 + 1992) is:
18737 = 642+1212 = 173+243 = 114+84.
Therefore if we had 18,737 balls we could arrange them into 2 squares, a 64×64 square and a 121×121 square. Alternatively we could rearrange them into 2 cubes, one 17x17x17 and one 24x24x24. Or we could enter a higher dimensional space and create 2 tesseracts one with sides 11x11x11x11 and the other with 14x14x14x14.
With only 1 solution for around the first 80,000 numbers it looks like these numbers are quite rare – could you find another one? And could you find one that also satisfies g5+h5?
Essential resources for IB students:
1) Exploration Guides and Paper 3 Resources
I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
Ramanujan’s Taxi Cabs and the Sum of 2 Cubes
The Indian mathematician Ramanujan (picture cite: Wikipedia) is renowned as one of great self-taught mathematical prodigies. His correspondence with the renowned mathematician G. H Hardy led him to being invited to study in England, though whilst there he fell sick. Visiting him in hospital, Hardy remarked that the taxi that had brought him to the hospital had a very “rather dull number” – number 1729. Ramanujan remarked in reply, ” No Hardy, it’s a very interesting number! It’s the smallest number expressible as the sum of 2 cubes in 2 different ways!”
Ramanujan was profoundly interested in number theory – the study of integers and patterns inherent within them. The general problem referenced above is finding integer solutions to the below equation for given values of A:
In the case that A = 1729, we have 2 possible ways of finding distinct integer solutions:
The smallest number which can be formed through 3 distinct (positive) integer solutions to the equation is A = 87, 539, 319.
Although this began as a number theory problem it has close links with both graphs and group theory – and it is from these fields that mathematicians have gained a deeper understanding as to the nature of its solutions. The modern field of elliptical curve cryptography is closely related to the ideas below and provides a very secure method of encrypting data.
We start by sketching the graph of:
For some given integer value of A. We will notice that the graph has a line of symmetry around y = x and also an asymptote at y = -x. If we plot:
We can see that both our integer solutions to this problem (1,12) and (9,10) lie on the curve:
Group theory
Groups can be considered as sets which follow a set number of rules with regards to operations like multiplication, addition etc. Establishing that a set is a group then allows certain properties to be inferred. If we can establish the following rules hold then we can create an Abelian group. If we start with a set A and and operation Θ.
1) Identity. For an element e in A, we have a Θ e = a for all a in A.
(for example 0 is the identity element for the addition operation for the set of integers numbers. a+0 = a for all a in the real numbers).
2) Closure. For all elements a,b in A, a Θ b = c, where c is also in A.
(For example with the addition operation, the addition of 2 integers numbers is still an integer)
3) Associativity. For all elements a,b,c in A, (a Θ b) Θ c = a Θ (b Θ c)
(For example with the addition operation, (1+2) + 3 = 1 + (2+3) )
4) Inverse. For each a in A there exists a b in A such that a Θ b = b Θ a = e. Where e is the identity.
(For example with the addition operation, 4+-4 = -4+4 = 0. 0 is the identity element for addition)
5) Commutativity. For all elements a,b in A, a Θ b = b Θ a
(For example with the addition operation 1+2 = 2+1).
As we have seen, the set of integers under the operation addition forms an abelian group.
Establishing a group
So, let’s see if we can establish a Abelian group based around the rational coordinates on our graph. We can demonstrate with the graph:
We then take 2 coordinate points with rational coordinates (i.e coordinates that can be written as a fraction of integers). In this case A (1,12) and B (9,10).
We then draw the line through A and B. This will intersect the graph in a 3rd point, C (except in a special case to be looked at in a minute).
We then reflect this new point C in the line y = x, giving us C’.
In this case C’ is the point (46/3, -37/3)
We therefore define addition (our operation Θ) in this group as:
A + B = C’.
(1,12) + (9,10) = (46/3, -37/3).
We now need to deal with the special case when a line joining 2 points on the curve does not intersect the curve again. This will happen whenever the gradient of this line is -1, which will make it parallel to the graph’s asymptote y = -x.
In this case we affix a special point at infinity to the Cartesian (x,y) plane. We define this point as the point through which all lines with gradient -1 intersect. Therefore in our expanded geometry, the line through AB will intersect the curve at this point at infinity. Let’s call our special point Φ. Now we have a new geometry, the (x,y) plane affixed with Φ.
We can now create an Abelian group. For any 2 rational points P(x,y), Q(x,y) we will have:
1) Identity. P + Φ = Φ + P = P
2) Closure. P + Q = R. (Where R(x,y) is also a rational point on the curve)
3) Associativity. (P+Q) + R = P+(Q+R)
4) Inverse. P + (-P) = Φ
5) Commutativity. P+Q = Q+P
Understanding the identity
Let’s see if we can understand some of these. For the identity, if we have a point A on the line and the point at infinity then this will contain the line with gradient -1. Therefore the line between the point at infinity and A will intersect the curve again at B. Our new point, B’ will be created by reflecting this point in the line y = x. This gets us back to point A. Therefore P + Φ = P as required.
Understanding the inverse
With the inverse of our point P(x,y) given as -P = (-x,-y) we can see that this is the reflection in the line y = x. We can see that we we join up the 2 points reflected in the line y = x we will have a line with slope -1, which will intersect with the curve at our point at infinity. Therefore P + (-P) = Φ.
Through our graphical understanding the commutativity rule also follows immediately, It doesn’t matter which of the 2 points come first when we draw a line that connects them, therefore P+Q = Q+P.
Understanding associativity and closure
Neither associativity nor closure are obvious from our graph. We could check individual points to show that (P+Q) + R = P+(Q+R), but it would be harder to explain why this always held. Equally whilst it’s clear that P+Q will always create a point on the curve it’s not obvious that this will be a rational point.
In fact we do have both associativity and closure for our group as we have the following algebraic definition for our addition operation:
For a given point:
On a given curve of the form:
The addition of 2 points is given by:
In the case of our curve:
If we take P = (1,12). P + P will be given by:
We can check this result graphically. If P and Q are the same point, then the line that passes through both P and Q has to be the tangent to the curve at that point. Therefore we would have:
Here the tangent at A does indeed meet the curve again – at point C, which does reflect in y = x to give us the coordinates above.
We could also find this intersection point algebraically. If we differentiate the original curve to find the gradient when x = 1 we can find the equation of the tangent when x=1 and then substitute this back into the equation of the curve to find the intersection point. This would give us:
We would then reverse the x and y coordinates to reflect in the line y = x. This also gives us the same coordinates.
More generally if we have the 2 rational coordinates on the curve:
We have the algebraic formula for addition as:
If P = (1,12) and Q = (9,10), P + Q would give (after much tedious substitution!):
This agrees with the coordinates we found earlier using the much easier geometrical approach. As we can see from this formula, both coordinate points will always be rational – as they will be composed of combinations of our original rational coordinates. For any given curve there will be a generator set of coordinates through which we can generate all other rational coordinates on the curve through our addition operation.
So, we seem to have come a long way from our original goal – finding integer solutions to an algebraic equation. Instead we seem to have got sidetracked into studying graphs and establishing groups. However by reinterpreting this problem as one in group theory then this then opens up many new mathematical techniques to help us understand the solutions to this problem.
A fuller introduction to this topic is the very readable, “Taxicabs and the Sum of Two Cubes” by Joseph Silverman (from which the 2 general equations were taken) .
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!
The Shoelace Algorithm to find areas of polygons
This is a nice algorithm, formally known as Gauss’s Area formula, which allows you to work out the area of any polygon as long as you know the Cartesian coordinates of the vertices. The case can be shown to work for all triangles, and then can be extended to all polygons by first splitting them into triangles and following the same approach.
Let’s see if we can work out the algorithm ourselves using the construction at the top of the page. We want the area of the triangle (4), and we can see that this will be equivalent to the area of the rectangle minus the area of the 3 triangles (1) (2) (3).
Let’s start by adding some other coordinate points for the rectangle:
Therefore the area of the rectangle will be:
(1) + (2) +(3) +(4): (x3-x2)(y1-y3)
And the area of triangles will be:
(1): 0.5(x3-x2)(y2-y3)
(2): 0.5(x1-x2)(y1-y2)
(3): 0.5(x3-x1)(y1-y3)
Therefore the area of triangle (4) will be:
Area = (x3-x2)(y1-y3) – 0.5(x3-x2)(y2-y3) – 0.5(x1-x2)(y1-y2) – 0.5(x3-x1)(y1-y3)
Therefore we have our algorithm! Let’s see if it works with the following coordinates added:
x1 = 2 x2 = 1 x3 = 3
y1 = 3 y2 = 2 y3 = 1
Area = (x3-x2)(y1-y3) – 0.5(x3-x2)(y2-y3) – 0.5(x1-x2)(y1-y2) – 0.5(x3-x1)(y1-y3)
Area = (3-1)(3-1) – 0.5(3-1)(2-1) – 0.5(2-1)(3-2) – 0.5(3-2)(3-1)
Area = 4 – 1 – 0.5 – 1 = 1.5 units squared
We could check this using Pythagoras to find all 3 sides of the triangle, followed by the Cosine rule to find an angle, followed by the Sine area of triangle formula, but let’s take an easier route and ask Wolfram Alpha (simply type “area of a triangle with coordinates (1,2) (2,3) (3,1)). This does indeed confirm an area of 1.5 units squared. Our algorithm works. We can of course simplify the area formula by expanding brackets and simplifying. If we were to do this we would get the commonly used version of the area formula for triangles.
The general case for finding areas of polygons
The general formula for the area of an n-sided polygon is given above.
For a triangle this gives:
For a quadrilateral this gives:
For a pentagon this gives:
You might notice a nice shoelace like pattern (hence the name) where x coordinates criss cross with the next y coordinate along. To finish off let’s see if it works for an irregular pentagon.
If we arbitrarily assign our (x1, y1) as (1,1) and then (x2, y2) as (3,2), and continue in a clockwise direction we will get the following:
area = absolute of 0.5( 1×2 + 3×4 + 3×1 + 4×0 + 2×1 – 3×1 – 3×2 – 4×4 – 2×1 – 1×0)
area = 4.
Let’s check again with Wolfram Alpha – and yes it does indeed have an area of 4.
It could be a nice exploration task to take this further and to explore how many different methods there are to find the area of polygons – and compare their ease of use, level of mathematics required and aesthetic appeal.
Essential Resources for IB Teachers
If you are a teacher then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over 2000 pages of pdf content for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:
- Original pdf worksheets (with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.
- Original Paper 3 investigations (with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.
- Over 150 pages of Coursework Guides to introduce students to the essentials behind getting an excellent mark on their exploration coursework.
- A large number of enrichment activities such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.
There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!
Essential Resources for both IB teachers and IB students
1) Exploration Guides and Paper 3 Resources
I’ve put together a 168 page Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made Paper 3 packs for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.
IB Maths Resources from Intermathematics 
