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**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.

**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.

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?