Tag Archives: circle

Linear transformations, geometrically

 

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)

 

transform of x

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

gif for lin affine trans1

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)

gif for lin affine trans2

All this was constructed with my geometrical construction program (APP if you like) called GEOSTRUCT and available as a free web based application from

http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html

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

 

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Filed under algebra, conics, construction, geometrical, geometry app, geostruct, math, ordered pairs, rigid motion, teaching, transformations, Uncategorized

GEOSTRUCT now with Save and Fetch for your constructions

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:

post example A
The user panel and a simple example of three points on a circle, with the bisectors of the pairs of points.

post example C
The history panel showing the actions that have been carried out

post example D
The Save popup,and, below, the resulting text file.

post example E

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

The full url is http://www.mathcomesalive.com/mathsite/geostruct/geostructforbrowser1.html

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Real problems with conic sections (ellipse, parabola) part two

So suppose we have a parabolic curve and we want to find out stuff about it.

Its equation … Oh, we have no axes.

Its focus … That would be nice, but it is a bit out of reach.

Its axis, in fact its axis of symmetry … Fold it in half? But how?

Try the method of part one, with the ellipse. (previous post)

parabola find the focus1

This looks promising. I even get another axis, for my coordinate system, if I really want the equation.

Now, analysis of the standard equation for a parabola (see later) says that a line at 45 deg to the axis, as shown, cuts the parabola at a point four focal lengths from the axis. In the picture, marked on the “vertical”axis, this is the length DH

parabola find the focus2

So I need a point one quarter of the way from D to H. Easy !

parabola find the focus3

and then the circle center D, with radius DH/4 cuts the axis of the parabola at the focal point (the focus).

Even better, we get the directrix as well …

parabola find the focus4

and now for the mathy bit (well, you do the algebra, I did the picture)

parabola find the focus math

Yes, I know that this one points up and the previous one pointed to the right !

All diagrams were created with my geometrical construction program, GEOSTRUCT

You will find it here:

www.mathcomesalive.com/geostruct/geostructforbrowser1.html

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GEOSTRUCT, a program for investigative geometry

I have been developing this computer software / program / application for some years now, and it is now accessible as a web page, to run in your browser.

It provides basic geometric construction facilities, with lines, points and circles, from which endless possibilities follow.

Just try it out, it’s free.

Click on this or copy and paste for later : www.mathcomesalive.com/geostruct/geostructforbrowser1.html

.Here are some of the basic features, and examples of more advanced constructions, almost all based on straightedge and compass, from “make line pass through a point” to “intersection of two circles”, and dynamic constructions with rolling and rotating circles.

help pic 1
Two lines, with points placed on them
help pic 3
Three random lines with two points of intersection generated
help pic 6
Five free points, three generated circles and a center point
help pic 7
Three free points, connected as point pairs, medians generated
help pic 5
Two free circles and three free points, point pairs and centers generated
gif line and circle
GIF showing points of intersection of a line with a circle
hypocycloid locus
Construction for locus of hypocycloid
circle in a segment
gif002
GIF showing a dilation (stretch) in the horizontal direction
gif piston cylinder
Piston and flywheel
gif touching2circles
Construction for circle touching two circles
gif parabola
Construction for the locus of a parabola, focus-directrix definition.

 

 

 

 

 

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Duality, fundamental and profound, but here’s a starter for you.

Duality, how things are connected in unexpected ways. The simplest case is that of the five regular Platonic solids, the tetrahedron, the cube, the octahedron, the dodecahedron and the icosahedron. They all look rather different, BUT…..

take any one of them and find the mid point of each of the faces, join these points up, and you get one of the five regular Platonic solids. Do it to this new one and you get back to the original one. Calling the operation “Doit” we get

tetrahedron –Doit–> tetrahedron –Doit–> tetrahedron
cube –Doit–> octahedron –Doit–> cube
dodecahedron –Doit–> icosahedron –Doit–> dodecahedron

The sizes may change, but we are only interested in the shapes.

This is called a Duality relationship, in which the tetrahedron is the dual of itself, the cube and octahedron are duals of each other, and the dodecahedron and icosahedron are also duals of each other.

Now we will look at lines and points in the x-y plane.

3x – 2y = 4 and y = (3/2)x + 2 and 3x – 2y – 4 = 0 are different ways of describing the same line, but there are many more. We can multiply every coefficient, including the constant, by any number not 0 and the result describes the same line, for example 6x – 4y = 8, or 0.75x – 0.5y = 1, or -0.75x + 0.5y + 1 = 0

This means that a line can be described entirely by two numbers, the x and the y coefficients found when the line equation is written in the last of the forms given above. Generally this is ax + by + 1 = 0

Now any point in the plane needs two numbers to specify it, the x and the y coordinates, for example (2,3)

So if a line needs two numbers and a point needs two numbers then given two numbers p and q I can choose to use then to describe a point or a line. So the numbers p and q can be the point (p,q) or the line px + qy + 1 = 0

The word “dual” is used in this situation. The point (p,q) is the dual of the line px + qy + 1 = 0, and vice versa.

dual of a rotating line cleaned up1

The line joining the points C and D is dual to the point K, in red.  The line equation is 2x + y = 3, and we rewrite it in the “standard” form as  -0.67x – 0.33y +1 = 0  so we get  (-0.67, -0.33) for the coordinates of the dual point K.

A quick calculation (using the well known formula) shows that the distance of the line from the origin multiplied by the distance of the point from the origin is a constant (in this case 1).

The second picture shows the construction of the dual point.

dual of a rotating line construction1

What happens as we move the line about ? Parallel to itself, the dual point moves out and in.

More interesting is what happens when we rotate the line around a fixed point on the line:

gif duality rotating line

The line passes through the fixed point C.  The dual point traces out a straight line, shown in green.

This can be interpreted as “A point can be seen as a set of concurrent lines”, just as a line can be seen as a set of collinear points (we have fewer problems with the latter).

It gets more interesting when we consider a curve. There are two ways of looking at a curve, one as a (fairly nicely) organized set of points ( a locus), and the other as a set of (fairly nicely) arranged lines (an envelope).

A circle is a set of points equidistant from a central point, but it is also the envelope of a set of lines equidistant from a central point (the tangent lines).

So what happens when we look for the dual of a circle? We can either find the line dual to each point on the circle, or find the point dual to each tangent line to the circle. Here’s both:

dual of a circle4

In this case the circle being dualled is the one with center C, and the result is a hyperbola, shown in green.  The result can be deduced analytically, but it is a pain to do so.

dual of a circle3

The hyperbola again.  It doesn’t look quite perfect, probably due to rounding errors.

The question remains – If I do the dualling operation on the hyperbola, will I get back to the circle ?

Also, why a hyperbola and not an ellipse ? Looking at what is going on suggests that if the circle to be dualled has the origin inside then we will get an ellipse. This argument can be made more believable with a little care !

If you get this far and want more, try this very heavy article:

http://en.wikipedia.org/wiki/Duality_(mathematics)

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Angle between two lines in the plane……Vector product in 3D…….connections???????

So I was in the middle of converting my geometry application Geostruct (we used to call them programs) into javascript

 get it here with the introduction .doc file here

when I decided that the “angle between two lines” routine needed a rewrite. Some surprises ensued !

angle between two lines pic1

Two lines,  ax + by + c  = 0  and  px + qy + r = 0

Their slopes (gradients) are  -a/b = tan(θ)  and  -p/q = tan(φ)

The angle between the lines is  φ – θ,

so it would be nice to know something about  tan(φ – θ)

Back to basics, where  tan(φ – θ) = sin(φ– θ)/cos(φ– θ),

and we have the two expansions

sin(φ– θ) = sin(φ)cos(θ) – cos(φ)sin(θ)   and

cos(φ– θ) = cos(φ)cos(θ) + sin(φ)sin(θ)

So we have  tan(φ – θ) = (sin(φ)cos(θ) – cos(φ)sin(θ))/( cos(φ)cos(θ) + sin(φ)sin(θ))

Dividing top and bottom by  cos(φ)sin(θ)  and skipping some tedious algebra we get

tan(φ – θ)   =  (tan(φ) – tan(θ))/(1 +  tan(φ)tan(θ))

This is where the books stop, which turns out to be a real shame !

Going back to the two lines and their equations, the two lines

ax + by = 0  and  px + qy = 0

have the same angle between them (some things are toooo obvious)

Things are simpler if we look at these two lines through the origin when they both have positive slope.

Take b and q as positive and write the equations as   ax – by = 0  and  px – qy = 0

Then the point whose coordinates are (b,a) lies on the first and (q,p) lies on the second.

angle between two lines pic2

Also, the slopes of the two lines are now  a/b , tan(θ)   and  p/q , tan(φ)

Let us put these into the  tan(φ – θ)   equation above, and once more after tedious algebra

tan(φ – θ)  = (bp – aq)/(ap + bq)

which is a very nice formula for the tan of the angle between two lines.

This is ok if we are interested just in “the angle between the lines”,  but if we are considering rotations, and one of the lines is the “first” one, then the tangent is inadequate. We need both the sine and the cosine of the angle to establish size AND direction (clockwise or anticlockwise).

The formula above can be seen as showing  cos(φ– θ)  as  (ap + bq) divided by something

and  sin(φ– θ)  as  (bp – aq) divided by the same something.

Calling the something  M  it is fairly clear that    (ap + bq)2 + (bp – aq) 2 = M2

and more tedious algebra and some “observation and making use of structure” gives

M= (a2 + b2)(p2 + q2)

and we now have

sin(φ– θ)  = (bp – aq)/M  and  cos(φ– θ) = (bq + ap)/M

and M is the product of the lengths of the two line segments, from the origin to (b,a) and from the origin to (q,p)

It was at this point that I saw M times the sine of the “angle between” as twice the well known formula for the area of a triangle. “half a b sin(C)”, or, if you prefer, the area of the parallelogram defined by the two line segments.

Suddenly I saw all this in 2D vector terms, with bq + ap being the dot product of (b,a) and (q,p) , and bp – aq as being part of the definition of the 3D vector or cross product, in fact the only non zero component (and in the z direction), since in 3D terms our two vectors lie in the xy plane.

Why is the “vector product” not considered in the 2D case ??? It is simpler, and looking at the formula for sine , above, we have a 2D interpretation of the “vector”or cross product as twice the area of the triangle formed by (b,a) and (q,p). (just as in the standard 3D definition, but treated as a scalar).

So “bang goes” the common terms, scalar product for c . d  and vector product for  c X d

Dot product and cross product are much better anyway, and a bit of ingenuity will lead you to the reason for the word “cross”.

This is one of the things implemented using this approach:

gif rolling circle

Anyway, the end result of all this, for rotating points on a circle, was a calculation process which did not require the actual calculation of any angle. No arctan( ) !

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An almost real life geometry problem

I needed to move a point around a circle, in a computer graphics application, using the mouse pointer. It is clearly not sensible to have mouse pointer on the point all the time, so the problem was

“For a point anywhere, where is the point both on the circle and on the radial line?”

point on circle 2

It may help to see the situation without the coordinate grid on show:

point on circle 1

This is a problem with many ways to a solution, some of them incredibly messy !

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The angle bisector theorem and a locus

The theorem states that if we bisect an angle of a triangle then the two parts of the side cut by the bisector are in the same ratio as the two other sides of the triangle. So I thought, what if we move the point, G in the gif below, so that GB always bisects the angle AGC, then where does it go ? Or, more mathematically, what is its locus. Geometry failed me at this point so I did a coordinate geometry version of the ratios and found that the locus was a circle ( a little surprised !). The algebra is fairly simple.

gif bisector locus

What was a real surprise was that if we take the origin of measurements to be the other point where the circle cross the line ABC, then point C is at the harmonic mean of points A and B. There is YOUR problem from this post.

Getting the construction to show all this was tricky. It is shown below.

locus angle bisector point full

I did the construction on my geometry program, Geostruct (see my Software page).

Line bb passes through A. Point D is on line bb. Point E is on this line and also on the line through C parallel to line cc joining B and D, So AD and DE are in the same ratio as AC and CB. When point D is moved the ratios are unchanged (feature of the program). So the point of interest is the intersection of a circle centre A, radius AD, and a circle centre C and radius DE (= FC)

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And now Pythagoras again, with bonus

I was attempting to solve a geometrical problem the other day, a problem which, due to my complete misunderstanding, had no solution, when this popped out. It is probably bog-standard, but new to me, and this time I don’t have the heart to check if it is one of the 100 proofs of The Pythagoras theorem.

sincospythagfor blog1

Now let theta be the angle ACB. Angle ABD is then 2*theta.

Set  r = 1, then  a is  sin(2*theta)  and b is  cos(2*theta),  and so

sin(theta) = (1 – b)/a = 1/a -b/a = cosec(2*theta) – cot(2*theta)

and  cosec(theta) = (1 + b)/a = 1/a + b/a = cosec(2*theta) + cot(2*theta)

I’ve done most of the work,

so now you can show that  sin(2*theta) = 2*sin(theta)*cos(theta)

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