## Neusis constructions (2): trisections

This post, containing a nice slew of methods of trisecting a general angle using neusis methods, can be found at Numbers and Shapes.

## Neusis constructions (1)

You can find this post, introducing geometric constructions which allow the use of a straight-edge with two marks, at my new site Numbers and Shapes.

## Solving a cubic by folding

This post, which is on my new site here:

http://www.numbersandshapes.net/?p=2520

shows how a cubic equation can be solved by origami.  This is not a new result by any means, but it’s hard to find a simple proof of how the construction works.

## MISG day 5

I didn’t explain the problem well enough in my last post: AGL, the energy supply company, are trying (among other things) to create several “reference profiles” of energy use, to which customers can be compared.  This will give them a better idea about supply and demand, and allow them to more accurately forecast energy requirements.

This was a problem more statistical than mathematical, and people were throwing all sorts of clustering analysis techniques (about which I know nothing) at it.  However, an averaging plot showed two parallel “clumps” of data, which a single curve of best fit couldn’t describe.  Since I know nothing about statistics, but quite a lot about imaging, I had the idea of treating each customer plot as an image:

Notice the two clumps at the bottom left – this is an example of a plot to which a curve of best fit could not be usefully applied.

The image has been subdivided into unequal quadrants, of which the upper left and right parts are the energy usage during hot and cold temperatures.  The vertical dividing line is at 20C = 68F.  The horizontal line is one-quarter from the bottom.  This seemed (without any analysis) to be a good place for cutting off “standard” energy usage (below the line) to “high” usage (above the line.  The customers can then be classified by choosing thresholds for “high hot” and “high cold” usage, and clustering them depending on whether they have used more or less than the threshold value.

The nice thing about his approach is that it’s both very easy and very transparent, and also can be adjusted; for example using finer grids on the image, by subdividing the original data into times of day, or into weekdays/weekends (or both), or by using a better approach than “by eye” to determine the cutoff point between standard and high usage.

The workshop ended before I could explore this idea any further… so I might play around with this on my own some more.

## MISG day 3

This week I’m at the Mathematics and Statistics Industry Study Group, a yearly event where companies from Australia and New Zealand provide problems for the assembled mathematicians and statisticians to solve.  I’m in a crowd of people working on a problem provided by the energy company AGL.  They have given us electricity smart meter data for 95 customers (there is a larger data set with 900 customers) of electricity readings taken every 30 minutes over a 199 day period.   That’s 9552 observations for each customer!  You can read the problem here.

So far everybody has (in the words of one participant) “thrown every imaginable package at the data and drawn lots of pretty pictures”.  Other folk are using Excel, Minitab, R and other statistics packages; I’m using Octave.  Today we have decided to break down the data into times of day: night (0000 – 0600), mornings (0600 – 0900), daytime (0900-1700) and evening (1700 – 2400), and to look at electricity usage versus temperature over those times, and over the course of the 199 days, which go from winter to summer.  We can also break down the data into weekdays, and weekends.

Here is an example of one data plot: a customers usage by temperature (which is the horizontal axis and given in celsius, so the upper value, 40, is very hot: 104F).

This customer’s usage, as you can see, is skewed up to the right (where the temperature is hotter), which would seem to indicate the use of an air-conditioner.  Other such plots are skewed to the left (colder temperatures) which would appear to indicate the use of heating in cold weather, but no a/c in hot.  Other plots have humps at both ends.  A few people have been trying to fit quadratic functions to these plots, without a huge amount of success, as the correlation is so low.

If you simple plot the data of a single customer over the entire data period you can see some interesting stuff.  Here for example, is one data set:

Notice the “break” towards the right (this corresponds to early summer).  We can infer that this customer went away at that time, and the residual power usage would have been for refrigerator, maybe security lights etc.

I am not myself any closer to solving the problems proposed by AGL, but I’m getting a better handle on the data.  I still intend to do some data smoothing, probably with a Gaussian filter, and see if it’s possible to classify customers based on the outputs.   Or I might head off to the staffroom in search of tea.

I’m very fond of geometric results which are a bit surprising, but turn out to be easy to prove. Here are two such.

Thébault’s result

The French problemist Victor Thébault published three problems in the late 1930’s. The second problem may be stated thus: on two adjacent sides of a square, construct two equilateral triangles, either both outwards or both inwards. The points at their outer vertices and the square’s opposite vertex form the vertices of another equilateral triangle.

Here’s a picture of the case with outward triangles:

The easiest proof requires a tiny amount of angle chasing. Consider the bottom triangle as pictured here:

The angle at H, being the sum of the internal angles of a square and equilateral triangle is 150°, and since this triangle is isosceles, the outer two angles must be equal and 15°. By symmetry then, the outer two angles at G must be 15° which means that the inner angle is 60°. Since the two edges which form this angle are equal, the triangle is equilateral, as required.

Here’s a sledgehammer proof using vectors. Assume without loss of generality that the square has side length 2, and let ${\bf u}$ be the vector GK, ${\bf v}$ be the vector GL. By construction, we must have

${\bf u} = \langle 2+\sqrt{3},1\rangle$ and ${\bf v} = \langle 1,2+\sqrt{3}\rangle$

Then

$\begin{array}{rcl}\cos\theta &=&\displaystyle{\frac{{\bf u}\cdot{\bf v}}{\|{\bf u}\|\|{\bf v}\|}}\\ &=&\displaystyle{\frac{4+2\sqrt{3}}{1+(2+\sqrt{3})^2}}\\ &=& \displaystyle{\frac{4+2\sqrt{3}}{8+4\sqrt{3}}}\\ &=& \displaystyle{\frac{1}{2}}.\end{array}$

This means $\theta=\pi/3$ as required.

Here is a picture of the case with internal triangles. The proof is very similar to the above case.

Suppose two squares ABCD (vertices given clockwise) and EFGH (likewise) share a common vertex: A and E, say. Then the midpoints of the squares, and the midpoints of the lines BH and DF are the vertices of another square:

First note that this in fact follows from the very general fundamental theorem of directly similar figures: suppose $P=P_0P_1\ldots P_k$ is a polygon (with $P_i$ being its vertices, and suppose $Q=Q_0Q_1\ldots Q_k$ is another polygon obtained from $P$ by translations, rotations, or scalings. The polygons are then said to be directly similar. For any $r$ with $0\le r\le 1$ define the polygon $R$ to have vertices $R_i=rP_i+(1-r)Q_i$. Then $R$ is directly similar to $P$ and $Q$.

The Finsler-Hadwiger theorem follows from this by labelling the vertices $A,B,C,D,E,F,G,H$ as $P_0,P_1,P_1,P_3,Q_2,Q_3,Q_0,Q_2$, and with $r=1/2$. But here’s a direct proof. First observe that vertices $(x_0,y_0)$, $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ are the vertices of a square (given in order around its edges) if and only if

$x_1-x_0=y_1-y_2=x_2-x_3=y_0-y_3$

and

$y_1-y_0=x_2-x_1=y_2-y_3=x_0-x_3$.

Suppose without loss of generality the vertices of the left hand square are $(0,0)$, $(-2,0)$, $(-2,-2)$ and $(0,-2)$, and the vertices of the right hand square be $(0,0)$, $(2a,2b)$, $(2b+2a,2b-2a)$ and $(2b,-2a)$:

The midpoints, in clockwise order, are $(-1,-1)$, $(a-1,b)$, $(b+a,b-a)$ and $(b,-a-1)$. Then the above equalities hold with the top line all being equal to $a$ and the bottom equality all being equal to $b+1$.

Note on figures

The figures were all created using GeoGebra, exported as PNG files, converted to JPG and cropped for uploading into WordPress.

## LaTeX multido

The LaTeX multido files provide a way for providing general loop macros in LaTeX. They can be used within pstricks, for creating complex diagrams with repetition. Here’s a little example which draws a number line from -10 to 10:

\documentclass[fleqn]{article}

\usepackage{pstricks,multido}

\begin{document}

\psset{unit=5mm}
\begin{pspicture}(-11,-1)(11,1)
\psline{<->}(-11,0)(11,0)
\multips(-10,0)(1,0){21}{\psline(0,0.3)(0,-0.3)}
\multido{\n=-10+1}{21}{\rput(\n,-1){\small{\n}}}
\end{pspicture}

\end{document}


In general, multido uses a counter (defined with a backslash), with starting and incremental values separated by a plus sign. So that

\multido{\n=-10+1}{21}{...stuff...}


starts the counter \n at -10, and increments it by 1 for 21 times.

The AIM Network

The Australian Independent Media Network