Apologies for no accents!
1: Khan Academy 2016 Subtraction to 1000 You don’t have to watch it all! Run it faster if possible.
2: Tom Lehrer (cover version) original from the 60’s The New Math
Following a recent blog post relating a transformation of points on a line to points on another line to the graph of the equation relating the input and output I thought it would be interesting to explore the linear and affine mappings of a plane to itself from a geometrical construction perspective.
It was ! (To me anyway)
These linear mappings (rigid and not so rigid motions) are usually approached in descriptive and manipulative ways, but always very specifically. I wanted to go directly from the transformation as equations directly to the transformation as geometry.
Taking an example, (x,y) maps to (X,Y) with the linear equations
X = x + y + 1 and Y = -0.5x +y
it was necessary to construct a point on the x axis with the value of X, and likewise a point on the y axis with the value of Y. The transformed (x,y) is then the point (X,Y) on the plane.
The construction below shows the points and lines needed to establish the point(X,0), which is G in the picture, starting with the point D as the (x,y)
The corresponding construction was done for Y, and the resulting point (X,Y) is point J. Point D was then forced to lie on a line, the sloping blue line, and as it is moved along the line the transformed point J moves on another line
Now the (x,y) point (B in the picture below, don’t ask why!) is forced to move on the blue circle. What does the transformed point do? It moves on an ellipse, whose size and orientation are determined by the actual transformation. At this point matrix methods become very handy.(though the 2D matrix methods cannot deal with translations)
All this was constructed with my geometrical construction program (APP if you like) called GEOSTRUCT and available as a free web based application from
The program produces a listing of all the actions requested, and these are listed below for this application:
Line bb moved to pass through Point A
New line cc created, through points B and C
New Point D
New line dd created, through Point D, at right angles to Line aa
New line ee created, through Point D, at right angles to Line bb
New line ff created, through Point D, parallel to Line cc
New point E created as the intersection of Line ff and Line aa
New line gg created, through Point E, at right angles to Line aa
New line hh created, through Point B, at right angles to Line bb
New point F created as the intersection of Line hh and Line gg
New line ii created, through Point F, parallel to Line cc
New point G created as the intersection of Line ii and Line aa
G is the X coordinate, from X = x + y + 1 (added by me)
New line jj created, through Point G, at right angles to Line aa
New line kk created, through Point D, at right angles to Line cc
New point H created as the intersection of Line kk and Line bb
New point I created, as midpoint of points H and B
New line ll created, through Point I, at right angles to Line bb
New point J created as the intersection of Line ll and Line jj
J is the Y coordinate, from Y = -x/2 + y (added by me)
and K is the transformed point (X,Y) Point J chosen as the tracking point (added by me)
New Line mm
Point D moved and placed on Line mm
Idly wondering about the tangents from a point to a circle I constructed the figure below. The point A is on a circle and can be moved round the circle. The chord CD then moves around. The interesting thing is to see the envelope of this chord as the point A is moved. ..
Circles and hyperbolas galore, but when the big circle passes through the centre of the smaller circle the envelope is the surprise parabola. The question is “How do I find the equation of such a parabola? Sensible choice of origin and axes is the first thing. the y-axis is best as the line joining the centres of the two circles. Then you need an equation for the line CD, with a suitable parameter. Then a little bit of calculus………..Nice one for calculus students
Constructions done with my web based program. Try it yourself: GEOSTRUCT
And now for January in Puerto Rico. I don’t know the name of the pink and white flowering tree, but I do know that without my machete the whole garden would be infested with babies from the first one.
This one below is a yellow eleconia, also spreading madly.
And this is a cute little tree with probably poisonous berries.
And what the hell have they done to the post editor. It only works in BOLD
Teacher: “Now we’re going to learn about base 10 arithmetic”.
Wise guy: “Is that where 3 + 4 = 12, or is it where 3 x 4 = 12 ?”.
I did a search on the net and found the term “base 10” all over the place. What does it mean?
An apparently annoying question:
“Does the 1 in 10 stand for the number 10’s in 10?”.
The interpretation of 10 in the system described as “Base 10” depends on the base of the system, so what is it? How do I find out?
We have here a logical problem. The term “Base 10” as a definition is self referential. It is more subtle than this definition of a straight line:
“A straight line is a line which is straight”.
The problem arises from the almost universal confusion between the two things:
1: The name of a number, in this case “ten” is supposedly implied
2: The symbols representing a number, in this case 10 in the base ten system”
So the answers to the questions “What is it? How do I find out?” above are “Unknown” and “You can’t”
Writing “Base 10” when you mean “Base ten” is probably the first step in making math meaningless.
Euclid and angle between two lines
Euclid’s definition of angle:
From Euclid’s Elements, Book 1
A plane angle is the inclination to one another of two lines in a plane which meet one another and
do not lie in a straight line.
And when the lines containing the angle are straight, the angle is called rectilinear.
In the diagram we see that angle A can be taken as the inclination, but we can also see that B can be taken as the angle of inclination.
So, which is it?
If the definition is meaningful then the two angles have to be equal in size, regardless of the lack of a measurement system for angles.
My point is that the theorem about vertical angles (Euclid’s Proposition 15) is redundant, and so there is no need to prove it.
This would save students a lot of time and relieve them of the feeling that proof was pointless. This time could be better spent on proving some less obvious things.
Adding angles is a straightforward manipulative activity, but Euclid also uses subtraction of angles, which is not an obvious thing to carry out, and technically requires an additional postulate. See this:
On the formal approach to subtraction
Thanks to intense use of the internet I finally found a simple, understandable way of implementing Save and Fetch operations, enabling the keeping and reusing of any construction.
Here is a reminder of the application (app, program, software, whatever), with the file handling operations:
There is now a not quite finished Spanish option – just click “ESPANOL”
Also a modified “move object” procedure for use with a tablet,or even a smartphone.
The whole application is constructed as a web page, and to run it just click this link: geostruct
I was on the virtually powerless governing body of the local primary school in the UK when the first National Curriculum came out, some time in the early 80’s. Very “New Math”y. Reworked a few years later. Here is some stuff from the UK Dept for Education about the latest rewrite. The old “Back to Basics” brigade are in the ascendant, but at least the UK is not drowning under High Stakes Testing. Have a look:
Key stage 1 and 2 (ages 5 to 10)
Key stage 3 (11 to 13)
key stage 4 (14,15)
ans about assessment
Only the dedicated study math in the last 2 years.
You might find this interesting as well, just look at how little time is spent taking tests, and then only in three of the years.
The main aim is to raise standards, particularly as the UK is slipping down international student assessment league tables. Inspired by what is taught in the world’s most successful school systems, including Hong Kong, Singapore and Finland, as well as in the best UK schools, it’s designed to produce productive, creative and well educated students.
Although the new curriculum is intended to be more challenging, the content is actually slimmer than the current curriculum, focusing on essential core subject knowledge and skills such as essay writing and computer programming.
I found this today. It’s worth a read.
Author: Gary S. Stager
Here’s an extract:
Conrad Wolfram estimates that 20,000 student lifetimes are wasted each year by school children engaged in mechanical (pencil and worksheet) calculations.
Expressed another way, we are spending twelve years educating kids to be a poor facsimile of a $2 calculator.
Forty years after the advent of cheap portable calculators, we are still debating whether children should be allowed to use one.
We are allowing education policy and curriculum to be shaped by the mathematical superstitions of Trump voters.
Educators need to take mathematics back and let Pearson keep “math.”
This is definitely worth a read.
Here is a quote:
“High schools focus on elementary applications of advanced mathematics whereas most people really make more use of sophisticated applications of elementary mathematics. This accounts for much of the disconnect noted above, as well as the common complaint from employers that graduates don’t know any math. Many who master high school mathematics cannot think clearly about percentages or ratios.”
And here is the link:
Link found on f(t)’s blog
This all started with a post by Maya Quinn (mathwater.wordpress.com) on problem solving. An oblong piece has been removed from an oblong cake (why? who did that?).The problem was to cut the remainder of the cake into two equal parts. Lots of solutions, one in particular was very imaginative.
This led me to another problem – what if the cut-out piece was triangular?
So, with Polya at my side (he’s been there since 1962) I decided that an equilateral triangle would be a reasonable starting point, as at least it was obvious that there were a few lines through the centroid doing the job (the medians), so some generality was still around.
Here is the equilateral triangle, nicely resting on the x axis, side length 2, top point on the y axis.
The vertical median bisects the triangle, so I described the general bisector GH by the distances x=DG and y=EH
In order to find out more about the bisector lines I first found a relationship between x and y, which was y=2x/(1+x), based on the area calculations:
Height of equilateral triangle is √3, base is 2, so area is √3
Height of BGH triangle is (2-y)/2 * √3, base is 1+x, so area is (1/2)*(1+x)*(2-y)/2*√3
and this is to be half of √3
So (1/2)*(1+x)*(2-y) must be equal to 1, and this leads to y=2x/(1+x)
This allowed me to find the coordinates of the point H and locate it correctly on the line.
I then joined the points and by moving G the line moved, and I tracked it, shown in green below.
The complete envelope of the bisecting lines consists of two more sections, making a “concave”triangle with the centroid in the middle.
At this point I figured that a change of variable was in order, and looking at the y=2x/(1+x) equation, and at the diagram it looked like the distances of the two points from the left hand vertex would be helpful:
This produced X=1+x and Y=2-y, and to my surprise the resulting equation was Y=2/X, or XY=2.
The significance of this last equation escaped me at this point.
So I found the coordinates of the two points C and K in a coordinate system with origin at point A, in terms of the new variables X and Y, found the line joining them, rewrote in terms of a parameter P, used a bit of calculus to get the envelope (I’ll do a post on this later), and it had the second degree equation
√3/4 = y2 – √3y – √3xy + 3x
which is a hyperbola !!! (see C. Smith “Conic Sections”)
At this point I stopped thinking about halving a triangle, and looked at the full envelope for one of the three sections, and got this:
A complete hyperbola, and, not only that, its asymptotes appear to be sides of the equilateral triangle. (Which when you think about it is not completely unreasonable !).
Then, thinking about doing shear operations on the picture, and with the X, Y variables, and ratios of lengths of segments on the same line being unchanged, I constructed the whole lot on an arbitrary triangle and did the envelope:
The triangle is ADF and the halving line is BG. Yes ! Same result !
At which point I saw a bit of light, and thought that “XY=2” does not involve angles at all.
Ooops, there’s another formula for the area of a triangle: a*b*sin(C)/2
Then straightaway all was revealed. You can draw the picture !
1. Triangle ACB, point X on side AC, point Y on side BC, area=0.5*AC*BC*sin(ACB)
2. Triangle XCY, area 0.5*XC*YC*sin(XCY), but angles ACB and XCY are the same, so if we require the area of triangle XCY to be half the area of triangle ACB we get 0.5*XC*YC*sin(XCY)=0.5*0.5*AC*BC*sin(ACB)
which reduces to (XC/AC)*(YC/BC)=0.5, and this does not involve the angle.
This corresponds to our earlier XY=2 since the equilateral triangle had side length 2
Pure speculation suggests that this result may have some connection with the way that the angle bisector of an angle in a triangle divides the opposite side in the ratio of the two adjacent sides.
The geometric diagrams were all constructed with the web based program