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**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 a^{2}+b^{2} = c^{3}+d^{3} = e^{4}+f^{4}.

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:

a^{2}+b^{2} = c^{3}+d^{3} = e^{4}+f^{4}.

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 = 2^{6}, b = 2^{6}, which gives 2^{12}+2^{12}

c = 2^{4}, d = 2^{4}, which gives 2^{12}+2^{12}

e = 2^{3}, f = 2^{3}, which gives 2^{12}+2^{12}

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 2^{12}+2^{12} 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 199^{2} + 199^{2}) is:

18737 = 64^{2}+121^{2} = 17^{3}+24^{3} = 11^{4}+8^{4}.

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 g^{5}+h^{5}?

**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:

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) .

**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): (x_{3}-x_{2})(y_{1}-y_{3})

And the area of triangles will be:

(1): 0.5(x_{3}-x_{2})(y_{2}-y_{3})

(2): 0.5(x_{1}-x_{2})(y_{1}-y_{2})

(3): 0.5(x_{3}-x_{1})(y_{1}-y_{3})

Therefore the area of triangle (4) will be:

Area = (x_{3}-x_{2})(y_{1}-y_{3}) – 0.5(x_{3}-x_{2})(y_{2}-y_{3}) – 0.5(x_{1}-x_{2})(y_{1}-y_{2}) – 0.5(x_{3}-x_{1})(y_{1}-y_{3})

Therefore we have our algorithm! Let’s see if it works with the following coordinates added:

x_{1 } = 2 x_{2 } = 1 x_{3 } = 3

y_{1 } = 3 y_{2 } = 2 y_{3 } = 1

Area = (x_{3}-x_{2})(y_{1}-y_{3}) – 0.5(x_{3}-x_{2})(y_{2}-y_{3}) – 0.5(x_{1}-x_{2})(y_{1}-y_{2}) – 0.5(x_{3}-x_{1})(y_{1}-y_{3})

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 (x_{1}, y_{1}) as (1,1) and then (x_{2}, y_{2}) 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.

**Soap Bubbles and Catenoids**

Soap bubbles form such that they create a shape with the minimum surface area for the given constraints. For a fixed volume the minimum surface area is a sphere, which is why soap bubbles will form spheres where possible. We can also investigate what happens when a soap film is formed between 2 parallel circular lines like in the picture below: [Credit Wikimedia Commons, Blinking spirit]

In this case the shape formed is a catenoid – which provides the minimum surface area (for a fixed volume) for a 3D shape connecting the two circles. The catenoid can be defined in terms of parametric equations:

Where cosh() is the hyperbolic cosine function which can be defined as:

For our parametric equation, t and u are parameters which we vary, and c is a constant that we can change to create different catenoids. We can use Geogebra to plot different catenoids. Below is the code which will plot parametric curves when c =2 and t varies between -20pi and 20 pi.

We then need to create a slider for u, and turn on the trace button – and for every given value of u (between 0 and 2 pi) it will plot a curve. When we trace through all the values of u it will create a 3D shape – our catenoid.

**Individual curve (catenary)**

**Catenoid when c = 0.1**

**Catenoid when c = 0.5**

**Catenoid when c = 1**

**Catenoid when c = 2**

**Wormholes**

For those of you who know your science fiction, the catenoids above may look similar to a wormhole. That’s because the catenoid is a solution to the hypothesized mathematics of wormholes. These can be thought of as a “bridge” either through curved space-time to another part of the universe (potentially therefore allowing for faster than light travel) or a bridge connecting 2 distinct universes.

Above is the Morris-Thorne bridge wormhole [Credit The Image of a Wormhole].

**Further exploration:**

This is a topic with lots of interesting areas to explore – the individual curves (catenary) look similar to, but are distinct from parabola. These curves appear in bridge building and in many other objects with free hanging cables. Proving that catenoids form shapes with minimum surface areas requires some quite complicated undergraduate maths (variational calculus), but it would be interesting to explore some other features of catenoids or indeed to explore why the sphere is a minimum surface area for a given volume.

If you want to explore further you can generate your own Catenoids with the Geogebra animation I’ve made here.

**Plotting the Mandelbrot Set **

The video above gives a fantastic account of how we can use technology to generate the Mandelbrot Set – one of the most impressive mathematical structures you can imagine. The Mandelbrot Set can be thought of as an infinitely large picture – which contains fractal patterns no matter how far you enlarge it. Below you can see a Mandelbrot zoom – which is equivalent to starting with a piece of A4 paper and enlarging it to the size of the universe! Even at this magnification you would still see new patterns emerging.

The way the Mandelbrot set is formed in the first video is by using the following iterative process:

Z_{n+1} = Z_{n}^{2} + c

Here Z is a complex number (of the form a + bi) and c is a constant that we choose. We choose our initial Z value as 0. Z_{1} = 0. We then choose a value of c (which is also a complex number) and see what happens when we follow the iterative process.

Let’s choose c = 2i +1. Z_{1} = 0

Z_{n+1} = Z_{n}^{2} + 2i +1

Z_{2} = (0)^{2} + 2i +1

Z_{2} = 2i + 1

We then repeat this process:

Z_{3} = Z_{2}^{2} + 2i +1

Z_{3} = (2i+1)^{2} + 2i +1

Z_{3} = (2i)(2i) + 2i + 2i + 1 + 2i +1

Z_{3} = 6i-2 (as i.i = -1)

As we continue this process Z_{n} spirals to infinity.

What we are looking for is whether this iterated Z value will diverge to infinity (i.e get larger and larger) or if it will remain bounded. If diverges to infinity we colour the initial point 2i+1 as blue on a complex axis. If it remains bounded we will colour it in black. In this case our initial point 2i+1 will diverge to infinity and so it will be coloured in blue.

So, let’s use Geogebra to see this is action. The Geogrebra online program for this is here.

We choose a value for c. Let’s say c = 0.23 + 0.42i. Z_{1} = 0

Z_{n+1} = Z_{n}^{2} + 0.23 + 0.42i.

Z_{2} = (0)^{2} + 0.23 + 0.42i.

Z_{2} = 0.23 + 0.42i.

Z_{3} = Z_{2}^{2} + 0.23 + 0.42i.

Z_{3} = (0.23 + 0.42i.)^{2} + 0.23 + 0.42i.

Z_{3} = 0.1065 + 0.6132i

Z_{4} = (0.1065 + 0.6132i)^{2} + 0.23 + 0.42i.

Z_{4} = -0.13467199 + 0.5506116i

We carry on with this iterative process and plot the points that we get each time. We can see the (0.23, 0.42), (0.1065, 0.42) and (-0.13467199, 0.5506116) correspond to the first coordinates on the spiral after (0,0). We can see that as this process continues we see a convergence to a point close to (0.05, 0.45).

If we choose another starting value for c: c = 0.17 + 0.56i we get the following diagram:

Again we have a stable spiral which spirals around a geometric shape and does not diverge to infinity.

If we choose another starting value for c: c = -0.25 + 0.64i we get the following diagram:

If we choose another starting value for c: c = 0.11 + 0.59i we get the following diagram:

However, If we choose another starting value for c: c = 0.3 + 0.68i we get the following diagram:

This time we can see that the orbit of points does not converge, but instead it diverges to infinity.

We can then colour in each point – simply categorising whether the value of c leads to an orbit which diverges or remains bounded. Black means it remains bounded, blue that it has escaped to infinity. So, below we can see that when we do the iterative process with c = 0.39+ 0.63i our orbit will escape to infinity (as it is coloured blue)

If we do this exercise in much finer detail we arrive at the following picture:

This is the Mandelbrot Set – and will keep producing fractal patterns as you zoom in to infinity.

**IB Revision**

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

This carries on the previous investigation into Farey sequences, and is again based on the current Nrich task Ford Circles. Below are the Farey sequences for F_{2}, F_{3} and F_{4}. You can read about Farey sequences in the previous post.

This time I’m going to explore the link between Farey sequences and circles. First we need the general equation for a circle:

This has centre (p,q) and radius r. Therefore

**Circle 1:**

has centre:

and radius:

**Circle 2:**

has centre:

and radius:

Now we can plot these circles in Geogebra – and look for the values of a,b,c,d which lead to the circles touching at a point.

**When a = 1, b = 2, c = 2, d = 3:**

Do we notice anything about the numbers a/b and c/d ? a/b = 1/2 and c/d = 2/3 ? These are consecutive terms in the F_{3 }sequence. So do other consecutive terms in the Farey sequence also generate circles touching at a point?

**a = 1, b = 1, c = 2, d = 3**

Again we can see that the fractions 1/1 and 2/3 are consecutive terms in the F_{3 }sequence. So by drawing some more circle we can graphically represent all the fractions in the F_{3 }sequence:

So these four circles represent the four non-zero fractions of in the F_{3 }sequence!

and this is the visual representation of the non-zero fractions of in the F_{4 }sequence. Amazing!

**IB Revision**

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

**The Folium of Descartes**

The folium of Descartes is a famous curve named after the French philosopher and mathematician Rene Descartes (pictured top right). As well as significant contributions to philosophy (“I think therefore I am”) he was also the father of modern geometry through the development of the x,y coordinate system of plotting algebraic curves. As such the Cartesian plane (as we call the x,y coordinate system) is named after him.

**Pascal and Descartes**

Descartes was studying what is now known as the folium of Descartes (folium coming from the Latin for leaf) in the first half of the 1600s. Prior to the invention of calculus, the ability to calculate the gradient at a given point was a real challenge. He placed a wager with Pierre de Fermat, a contemporary French mathematician (of Fermat’s Last Theorem fame) that Fermat would be unable to find the gradient of the curve – a challenge that Fermat took up and succeeded with.

**Calculus – implicit differentiation:**

Today, armed with calculus and the method of implicit differentiation, finding the gradient at a point for the folium of Descartes is more straightforward. The original Cartesian equation is:

which can be differentiated implicitly to give:

Therefore if we take (say) a =1 and the coordinate (1.5, 1.5) then we will have a gradient of -1.

**Parametric equations**

It’s sometimes easier to express a curve in a different way to the usual Cartesian equation. Two alternatives are polar coordinates and parametric coordinates. The parametric equations for the folium are given by:

In order to use parametric equations we simply choose a value of t (say t =1) and put this into both equations in order to arrive at a coordinate pair in the x,y plane. If we choose t = 1 and have set a = 1 as well then this gives:

x(1) = 3/2

y(1) = 3/2

therefore the point (1.5, 1.5) is on the curve.

You can read a lot more about famous curves and explore the maths behind them with the excellent “50 famous curves” from Bloomsburg University.

**IB Revision**

If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

The Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to ready made exams on each topic – again with worked solutions. This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

**Measuring the Distance to the Stars**

This is a very nice example of some very simple mathematics achieving something which for centuries appeared impossible – measuring the distance to the stars. Before we start we need a few definitions:

- 1 Astronomical Unit (AU) is the average distance from the Sun to the Earth. This is around 150,000,000km.
- 1 Light Year is the distance that light travels in one year. This is around 9,500,000,000,000km. We have around 63000AU = 1 Light Year.
- 1 arc second is measurement for very small angles and is 1/3600 of one degree.
- Parallax is the angular difference in measurement when viewing an object from different locations. In astronomy parallax is used to mean the half the angle formed when a star is viewed from opposite sides of the Earth’s solar orbit (marked on the diagram below).

With those definitions it is easy to then find the distance to stars. The parallax method requires that you take a measurement of the angle to a given star, and then wait until 6 months later and take the same measurement. The two angles will be slightly different – divide this difference by 2 and you have the parallax.

Let’s take 61 Cyngi – which Friedrick Bessel first used this method on in the early 1800s. This has a parallax of 287/1000 arc seconds. This is equivalent to 287/1000 x 1/3600 degree or approximately 0.000080 degrees. So now we can simply use trigonometry – we have a right angled triangle with opposite side = 1 AU and angle = 0.0000080. Therefore the distance is given by:

tanΦ = opp/adj

tan(0.000080) = 1/d

d = 1/tan(0.000080)

d = 720000 AU

which is approximately 720000/63000 = 11 light years away.

That’s pretty incredible! Using this method and armed with nothing more than a telescope and knowledge of the Earth’s orbital diameter, astronomers were able to judge the distance of stars in faraway parts of the universe – indeed they used this method to prove that other galaxies apart from our own also existed.

**Orion’s Belt**

The constellation of Orion is one of the most striking in the Northern Hemisphere. It contains the “belt” of 3 stars in a line, along with the brightly shining Rigel and the red super giant Betelgeuse. The following 2 graphics are taken from the great student resource from the Royal Observatory Greenwich:

The angles marked in the picture are in arc seconds – so to convert them into degrees we need to multiply by 1/3600. For example, Betelgeuse the red giant has a parallax of 0.0051 x 1/3600 = 0.0000014 (2sf) degrees. Therefore the distance to Betelgeuse is:

tanΦ = opp/adj

tan(0.0000014) = 1/d

d = 1/tan(0.0000014)

d = 41,000,000 AU

which is approximately 41,000,000/63000 = 651 light years away. If we were more accurate with our rounding we would get 643 light years. That means that when we look into the sky we are seeing Betelgeuse as it was 643 years ago.

**IB Revision**

This post is inspired by the Quora thread on interesting functions to plot.

**The butterfly**

This is a slightly simpler version of the butterfly curve which is plotted using polar coordinates on Desmos as:

Polar coordinates are an alternative way of plotting functions – and are explored a little in HL Maths when looking at complex numbers. The theta value specifies an angle of rotation measured anti-clockwise from the x axis, and the r value specifies the distance from the origin. So for example the polar coordinates (90 degrees, 1) would specify a point 90 degrees ant clockwise from the x axis and a distance 1 from the origin (i.e the point (0,1) in our usual Cartesian plane).

2. **Fermat’s Spiral**

This is plotted by the polar equation:

The next 3 were all created by my students.

3. **Chaotic spiral (by Laura Y9)**

I like how this graph grows ever more tangled as it coils in on itself. This was created by the polar equation:

4. **The flower (by Felix Y9)**

Some nice rotational symmetries on this one. Plotted by:

5. **The heart (by Tiffany Y9)**

Simple but effective! This was plotted using the usual x,y coordinates:

You can also explore how to draw the Superman and Batman logos using Wolfram Alpha here.

**Euler’s 9 Point Circle**

This is a nice introduction to some of the beautiful constructions of geometry. This branch of mathematics goes in and out of favour – back in the days of Euclid, constructions using lines and circles were a cornerstone of mathematical proof, interest was later revived in the 1800s through Poncelot’s projective geometry – later leading to the new field of non Euclidean geometry. It’s once again somewhat out of fashion – but more accessible than ever due to programs like Geogebra (on which the below diagrams were plotted). The 9 point circle (or at least the 6 point circle was discovered by the German Karl Wilhelm von Feuerbach in the 1820s. Unfortunately for Feuerbach it’s often instead called the Euler Circle – after one of the greatest mathematicians of all time, Leonhard Euler.

So, how do you draw Euler’s 9 Point Circle? It’s a bit involved, so don’t give up!

Step 1: Draw a triangle:

Step 2: Draw the perpendicular bisectors of the 3 sides, and mark the point where they all intersect (D).

Step 3: Draw the circle through the point D.

Step 4: From each line of the triangle, draw the perpendicular line through its third angle. For example, for the line AC, draw the perpendicular line that goes through both AC and angle B. (The altitudes of the triangle). Join up the 3 altitudes which will meet at a point (E).

Step 5: Join up the mid points of each side of the triangle with the remaining angle. For example, find the mid point of AC and join this point with angle B. (The median lines of the triangle). Label the point where the 3 lines meet as F.

Step 6: Remove all the construction lines. You can now see we have 3 points in a line. D is the centre of the circle through the points ABC, E is where the altitudes of the triangle meet (the orthoocentre of ABC) and F is where the median lines meet (the centroid of ABC).

Step 7: Join up the 3 points – they are collinear (on the same line).

Step 8: Enlarge the circle through points A B C by a scale factor of -1/2 centered on point F.

Step 9: We now have the 9 point circle. Look at the points where the inner circle intersects the triangle ABC. You can see that the points M N O show the points where the feet of the altitudes (from step 4) meet the triangle.

The points P Q R show the points where the perpendicular bisectors of the lines start (i.e the midpoints of the lines AB, AC, BC)

We also have the points S T U on the circle which show the midpoints of the lines between E and the vertices A, B, C.

Step 10: We can drag the vertices of the triangle and the above relationships will still hold.

In the second case we have both E and D outside the triangle.

In the third case we have E and F at the same point.

In the fourth case we have D and E on opposite sides of the triangle.

So there we go – who says maths isn’t beautiful?

**IB Maths Revision**

I’d strongly recommend starting your revision of topics from Y12 – certainly if you want to target a top grade in Y13. My favourite revision site is Revision Village – which has a huge amount of great resources – questions graded by level, full video solutions, practice tests, and even exam predictions. Standard Level students and Higher Level students have their own revision areas. Have a look!