Tag Archives: geometry

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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Complex Numbers via Rigid Motions

https://howardat58.files.wordpress.com/2015/02/complex-numbers-by-rotations.doc

Complex numbers via rigid motions
Just a bit mathematical !

I wrote this in response to a post by Michael Pershan:
http://rationalexpressions.blogspot.com/2015/02/could-this-introduce-kids-to-complex.html?

The way I have presented it is showing how mathematicians think. Get an idea, try it out, if it appears to work then attempt to produce a logical and mathematically sound derivation.
(This last part I have not included)
The idea is that wherever you have operations on things, and one operation can be followed by another of the same type, then you can consider the combinations of the operations separately from the things being operated on. The result is a new type of algebra, in this case the algebra of rotations.
Read on . . .

Rotations around the origin.
angle 180 deg or pi
Y = -y, and X = -x —> coordinate transformation
so (1,0) goes to (-1,0) and (-1,0) goes to (1,0)
Let us call this transformation H (for a half turn)

angle 90 deg or pi/2
Y = x, and X = -y
so (1,0) goes to (0,1) and (-1,0) goes to (0,-1)
and (0,1) goes to (-1,0) and (0,-1) goes to (1,0)
Let us call this transformation Q (for a quarter turn)

Then H(x,y) = (-x,-y)
and Q(x,y) = (-y,x)

Applying H twice we have H(H(x,y)) = (x,y) and if we are bold we can write HH(x,y) = (x,y)
and then HH = I, where I is the identity or do nothing transformation.
In the same way we find QQ = H

Now I is like multiplying the coodinates by 1
and H is like multiplying the coordinates by -1
This is not too outrageous, as a dilation can be seen as a multiplication of the coordinates by a number <> 1

So, continuing into uncharted territory,
we have H squared = 1 (fits with (-1)*(-1) = 1
and Q squared = -1 (fits with QQ = H, at least)

So what is Q ?
Let us suppose that it is some sort of a number, definitely not a normal one,
and let its value be called k.
What we can be fairly sure of is that k does not multiply each of the coordinates.
This appears to be meaningful only for the normal numbers.

Now the “number” k describes a rotation of 90, so we would expect that the square root of k to describe a rotation of 45

At this point it helps if the reader is familiar with extending the rational numbers by the introduction of the square root of 2 (a surd, although this jargon seems to have disappeared).

Let us assume that sqrt(k) is a simple combination of a normal number and a multiple of k:
sqrt(k) = a + bk
Then k = sqr(a) + sqr(b)*sqr(k) + 2abk, and sqr(k) = -1
which gives k = sqr(a)-sqr(b) + 2abk and then (2ab-1)k = sqr(a) – sqr(b)

From this, since k is not a normal number, 2ab = 1 and sqr(a) = sqr(b)
which gives a = b and then a = b = 1/root(2)

Now we have a “number” representing a 45 degree rotation. namely
(1/root(2)*(1 + k)

If we plot this and the other rotation numbers as points on a coordinate axis grid with ordinary numbers horizontally and k numbers vertically we see that all the points are on the unit circle, at positions corresponding to the rotation angles they describe.

OMG there must be something in this ! ! !

The continuation is left to the reader (as in some Victorian novels)

ps. root() and sqrt() are square root functions, and sqr() is the squaring function .

pps. Diagrams may be drawn at your leisure !

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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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Completing the (four sided) square

completing the square

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Quadrilaterals – a Christmas journey – part 3

So what else do stretches and shears do?
A stretch will turn a square into a rectangle if it acts in the direction of one of the sides.
A stretch will turn a square into a rhombus if it acts in the direction of a diagonal.
A shear will turn a square or a rectangle into a parallelogram.
Try it out !

Now to continue the journey – the mathematician now thinks “Is that it ? Are these the only transformations of the plane that map straight lines to straight lines ?”. His answer, with a Eureka moment, is “No ! What about those artists, with their perspective drawings ? Not only do straight lines in reality go to straight lines in their pictures, but when the lines are parallel in reality they go to convergent straight lines in the picture. “This is projection !”, cries the mathematician, and pursues the matter further and further…..

old master picArchitect_drawing_conceptual_2012

If I fix a sheet of glass or acrylic, sit still with marker pen in hand, and copy onto the glass exactly what I see through the glass I get a point projection of reality on the glass. Doing this when reality is a flat wall creates a projection from the reality plane to my glass plane. This is a new type of line preserving transformation of the points on a plane.

What is really nice about projection is that all the transformations we have seen so far can be described by projections.

First we have to classify projections, from a source plane to a target plane:
1. Point projection. The projection lines all pass through a fixed point (that’s the point where your eye was earlier). then each projection line passes through a point on the source plane, and where it hits the target plane is the transformed or projected point.
2. Line projection. The projection lines all start on a fixed line and are at right angles to that line. Imagine a spout brush or a very hairy caterpillar.
3. Parallel projection, in which the projection lines are all parallel to each other, as in rainfall or a bundle of spaghetti.

Here are translations, reflections, stretches and shears as projections:

Projection 0 Projection 1 Projection 2 Projection 3

What about rotations, you ask. Simply, a rotation can be seen as a sequence of two reflections.

But in general projection we lose the preservation of ratio. For example, the midpoint of a line segment is not projected to the mid point of the projected line segment:
A slice will do …..

midpoint 1midpoint 2It gets worse ! In the lower picture, as the point C is moved to the right along the line its image H moves further and further along, and, when C reaches D, H disappears altogether.
So if we look at the source plane from above, and we have a pair of lines that meet at point D, their projections must be parallel lines. (There could now follow a digression on the meaning of the word “infinity”, but it ain’t happenning).

What else ? Well, we have seen that ratios of distances are altered by projection, but when we take four points A, B, C and D on a line we can take two distance ratios (involving all four points) and take the ratio of these ratios. THIS quantity is NOT destroyed by projection. It is called the “cross-ratio” of the four points, and its value is (AB/AD)/(CB/CD), easier as (AB.CD)/(AD.CB)
(proof later)
A quick check:xratio testCopy the diagram to scale, but you can put the target line where you like.
Then measure EF, FG and GH, and calculate the cross-ratio. Should be the same, if you kept to the point matching.

Finally, and back to the quadrilateral:
arect6
Each diagonal has four points on it, two vertices and two intersection points with the other two diagonals.
If we join D to H, the point of intersection of the two diagonals BC and EF, we can see that the two sets of four points are connected by a projection from D, as H to H, B to F, G to I and C to E.
Consequently they have the same cross-ratio.

Not only that, it can be shown (one day !) that the value of this cross-ratio is -1, for all quadrilaterals, all the time.
Also, it is possible to map any quadrilateral to any other by a sequence of projections.

This is my introduction to projective geometry, a very interesting and underexposed branch of geometry. There may be a part 4 eventually !

And two more 3D gifs :

gif cube 3gif torus 2

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Quadrilaterals – a Christmas journey – part 2

It is a popular activity to join the opposite vertices of a quadrilateral (the diagonals), as in special cases they
have interesting properties. Unfortunately, for the general quadrilateral this does nothing of interest.
However, with the extended or complete quadrilateral we have six vertices to go at, and so we get another “diagonal”:

arect4

Things get more interesting when we extend the diagonals and find their points of intersection.

arect6

The three points shown circled are the points of intersection of the three pairs of diagonals.

Observe that each diagonal has four points on it, the two vertices interlaced with two points of intersection.

Now we started with “any” quadrilateral so it might be thought that nothing much can be said about measurements and
quadrilaterals in general – not so! To go any futher we need to go back to rigid motions of the plane, and their
effect on plane figures.
A rigid motion of a geometrical object just moves it to a new position, its shape and size are unchanged.
Rotations, reflections and translations (shifts would have been a simpler term)are rigid motions.
Basically what is not changed is distances between points.
The next level of transformation of plane figures adds dilations,stretches and shears. The figures change their
shapes, but one thing remains: relative distances of collinear points. Rigid no longer.
arect7arect8
Dilations can be “the same in all directions”, as in the example, and these preserve the shape of a figure but not the
size, or different in the y direction from the x direction, these are the stretches. These turns circles into ellipses !
Shears turn circles into ellipses anyway.
Notice that in both cases the mid point of the transformed line segment is the image of the midpoint of the original
line segment, and it is easy to see that ratios of distances in the same direction are preserved.
It is not quite as easy to see that any triangle can be transformed into any other triangle, with the help of these
extra transformations.
Notice that dilations have a fixed point and shears have a fixed line.

The next three pictures show the effects of
1: A stretch up and down 2: A stretch to the right and left
3. A shear horizontally

gif001gif002gif003

What use is all this, you ask. well, here is a gif showing that the medians of a triangle are concurrent, and this is preserved under stretch and shear. This means that you only have to prove it for an equilateral triangle. (which is obvious!)

gif medians2

More soon…….

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Pythagoras, triples, 3,4,5, a calculator.

How to generate Pythagorean triples (example: 3,4,5), well one way at least.

Starting with (x + y)2 = x2 + y2 + 2xy and (x – y)2 = x2 + y2 – 2xy

we can write the difference of two squares

(x + y)2  –  (x – y)2 = 4xy

and if we write  x = A2 and y = B2 the right hand side is a square as well.

Thus:

(A2  +  B2) 2 – (A2 – B2) 2 = 4A2 B2 = (2AB) 2

which can be written as

(A2  +  B2) 2 = (A2 – B2) 2 + (2AB) 2

the Pythagoras form.

Now just put in some integers for A and B

2 and 1 gives 3,4,5

Conjecture1: This process generates ALL the Pythagorean triples.

Conjecture2: Every odd number belongs to some  Pythagorean triple.

Have fun…….

My next post will be about finding the radius of the inscribed circle in a right angled triangle…..

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Geometry problem, borrowed and extended. Try it !

I borrowed this from http://fivetriangles.blogspot.com/
184. *** Overlapping sectors
overlappingsectorsspic
In the diagram is rectangle ABCD with height 10 cm. An arc with centre at point B is drawn from point A to side BC. An arc with centre at point C is drawn from point D to side BC. Given that the shaded (coloured) regions ⓐ and ⓑ have equal area, determine the length of BC.

The extension is “and where do the quarter circles cross?”
It could be a two line calculation !

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Congruence, proof, and rigid motions: The Common Core says WHAT, not HOW

With all the stuff in the high school geometry about proving congruence by rigid motions we get this sample geometry question from PARCC
PARCC geom test proofdef
(get the rest from numberwarrior here)

Numberwarrior’s concerns are about the language and the formal properties of congruence and I agree with him on this.
My concerns are about the stated claims of the CCSS to specify the “What do the need to know/understand/be able to do”, and the PARCC test which says “This is HOW you do a proof”.
In this particular example there are other ways of proving the assertion, not least those using the definition of congruence by rigid motions.

Let us do it this way:
1: vertical angles are equal, as there is a rotation of line AD to GC through the angle CBD, and then AD is on top of GC, so angle ABD ABF is also the angle of rotation, and is therefore congruent to angle CBD
2:There is a translation of line HE to line AD, as they are parallel. So the translation of H to H’ puts H’ on the line AD, and so angle H’BF is congruent to angle ABF.
3: But angles are preserved by rigid motions, so angle H’BF is congruent to HFG, and therefore angle ABF and HFG are congruent.

So, if I chose to teach about proof using this approach (my “HOW”) the students won’t even understand the question. Test items MUST be “Method Free”.

Also, the so called Reflexive, Symmetric and Transitive properties of congruence are no different from a=a, if a=b then b=a, and if a=b and b=c then a=c for numbers, and in both situations these are so STUNNINGLY obvious that it is cluttering up the minds of the learners to burden them with this sort of stuff. It is clear to me that this is a contribution to the CCSS from the sole pure mathematician on the committee.

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Lament to the Common Core geometry

Could I move this trapezoid
To that one, in the endless void?
I tried translation and rotation.
Then I had a crazy notion.
I would pass a rigid motion.
Result – a lovely hemorrhoid.

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