# Category Archives: geometry

## Multiplication, the theory – by Thales’ theorem

The diagram can be simplified by using an acute triangle.

Thales’ theorem

Proof of Thales theorem :
If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.
Given : In ∆ABC , DE || BC and intersects AB in D and AC in E.
Prove that : AD / DB = AE / EC
Construction : Join BC,CD and draw EF ┴ BA and DG ┴ CA.
Statements                                                    Reasons
1) EF ┴ BA                                                      1) Construction
2) EF is the height of ∆ADE and ∆DBE     2) Definition of perpendicular
4)Area(DBE) =(DB.EF)/2                               4) Area = (Base .height)/2
6) (Area(ADE))/(Area(DEC)) = AE/EC         6) Divide (3) by Area(DEC)
7) ∆DBE ~∆DEC                                             7) Both the ∆s are on the same base and
between the same || lines.
8) Area(∆DBE)=area(∆DEC)                        8) So the two triangles have equal areas
9) AD/DB =AE/EC                                           9) From (5) and (6) and (7)

Not only this but also AD/AB = DE/BC

Some adjustments, but the Thales theorem is well done. I liked it.

## Geometry and Numbers – negative ones – “a minus times a minus is a plus”

To accommodate positive and negative numbers we need two extended number lines, with their zeros at the same place

Then multiplication of two negative numbers will always give a positive result, following the same geometrical structure.

The start, where the multiplier begins at 1

Now the 1 connects with the multiplicand -3

The multiplier is now placed at -2

6

And the parallel line from -2 connects to the 6 on the target line

This is so geometrical, and there is no “funny business”. None of the “ought to be 6”. No stuff about the distributive law.

The only geometry needs the Pythagoras theorem, and this will be the next post.

## Geometry and Numbers, not the counting sort

A number line is generally a piece of straight line with a starting point, labeled 0, and equally spaced points labeled 1 and 2 and 3 and 4 and so on till the paper runs out.

The value of a number is the distance from the zero point to the numbered point, in units of the equal spacing.

It is really much easier to draw one of these !

Two parallel number lines, same scale.

Notice that the zero points do not have to be in the same vertical line.

### Subtraction

To get the symbolic form 7 – 2 = 5 we start with 0 on the target line (now the upper line) and join it to the 2 on the subtrahend line. (arrow down) (needs a nicer word here)

Then from the 7 on the subtrahend line we produce the line from 7 parallel to the 0 to 2 line. Then “arrow up” to the target line

Magic ! The result is 5 on the target line.

I like the picture, but the subtraction words are a mess.

### Multiplication

We need two number lines, but since multiplication is  “proportional” they will now be crossing, and the common point is labeled 0.

Also, the labels are “target” and “multiplier” and each line has its own scale.

### Bonus: Nomograms, with lines.

The first is a simple calculator, with A + B = Sum

The second one calculates parallel resistances

Filed under arithmetic, education, geometry, math, nomogram, Number systems, Uncategorized

## sin(2x) less than or equal to 2sin(x), for smallish x (!)

Once upon a time, when I was deep into trigonometry, I followed the trail of sine and cosine sums, with sin(A + B) = sin(A)cos(B) + cos(A)sin(B).
Then two more:
sin(A) + sin(B) = 2sin((A + B)/2)cos((A – B)/2)
and 2sin(A)cos(B) = sin(A + B) + sin(A – B)

Being unable to remember these last formulae (there are 8 altogether) I learned the basic one and derived, over and over again, the others.
In all of this I was never able to derive the basic formula until de Moivre’s theorem appeared.
So I had another go with trig and the simple version, and this is the result:

Aim of the game : sin(2x) = 2sin(x)cos(x)
First diagram:This does not look promising !
But what about sensible labelling –
Second diagram:

The vertical from F meets AD in J
and the line from F at right angles to AB meets AB in K.
So we have two pairs of congruent triangles, BH and HD are both equal to sin(x),
and BHD is A STRAIGHT LINE
Final diagram:
Now the vertical from B is sin(2x)

so sin(2x)/(2sin(x)) is the ratio which should be cos(x)

Fix it

Filed under education, geometrical, geometry, geometry app, triangle, Uncategorized

## 1/5 is one fifth of the length of a line segment of one unit – but how?

### ….but how do you do it ?????

The mystery is solved…….

Here is the line, with 0 and 1 marked. /You chose it already !

Here is a numbered line, any size, equally spaced, at intervals of one unit.

It only has to start from zero.

Now construct the line from point 5 to the “fraction” line at point 1, and a parallel line from point 1 on the numbered line.

The point of intersection of the parallel line and the “fraction” line is then 1/5 of the distance from 0 to 1 on the “fraction” line.

1/5, 2/5, 3/5, 4/5 and 1 are equally spaced on the fraction line.

L cannot be moved in the static picture.

Filed under fractions, geometry, math, Uncategorized

## A surprise parabola and more garden

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.

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

Filed under calculus, caribbean, envelope, geometry, high school, math, Uncategorized

## “Shear”, the forgotten transformation.

Transformations of the plane are many and various.
The “nice” ones are “rigid motions”, and this term includes rotations, reflections and translations. The shape and the size of a geometrical object are not altered by a rigid motion.
There are also “shape preserving” transformations, called “dilations”, in which an object is stretched or shrunk equally in all directions.
An often overlooked transformation is the “shear”, in which there is a fixed line, and points not on that line are pushed parallel to the line in proportion to their distance from the line. Think of a stack of paper,perfectly stacked, and then pushed sideways so that the side of the pile is still flat. You will see a parallelogram at the front of the pile.
A shear will change the direction of a line, turn a rectangle into a parallelogram and turn a circle into an ellipse,
BUT
the area of any closed figure does not change at all.

Here is the static picture of a fixed point J, a fixed line, the x-axis, and a set of points on the horizontal line through A.
Also two triangles, LND and LDF, which are going to be sheared

And here is the shearing in action, for varying amounts of shear, determined by the value of k.

Notice that triangle LMN changes a lot, and its area changes, but the areas of triangles LND and LDF do not change at all.
Not shown is a rectangle and a circle, which would change into a parallelogram and an ellipse, but their areas will not change with a shear.

For more on this go back to my Christmas post:

Filed under geometry, geostruct, rigid motion, teaching, transformations

## Euclid and vertical angles

Euclid and angle between two lines

Euclid’s definition of angle:

From Euclid’s Elements, Book 1
Definition 8.
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.
Definition 9.
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
http://aleph0.clarku.edu/~djoyce/elements/bookI/cn.html

Filed under definitions, Euclid, geometry, math, teaching

## Halving a triangle, follow-up number three, bisecting an angle

While looking at the bisection of area formula (see previous posts)
a’b’ = (1/2)ab
where a’ and b’ are the distances from the vertex of points on the sides of the triangle, and a and b are the lengths of the sides I remembered another formula about triangles, the bisection of the angle formula, with a” and b” being the lengths of the two parts into which the opposite side is divided, namely
a”/b” = a/b

These are like Cuba and Puerto Rico, “Two wings of the same bird”, Jose Marti (in Spanish)
Neither involves the angle itself, and so is very general. I decided that there must be a connection, and after a futile look for some duality in the situation I suddenly saw the connection, in simple algebraic terms:
a’b’ = a’/(1/b’)
and so a triangle with sides a’ and 1/b’ will give a”/b” = a’b’
and then a”/b” = (1/2)ab as well.

This is the construction for the halving a triangle.

This is the extended construction for the bisector. The opposite side is in brown.

and this is a gif showing how as the point a’ (E) is moved the brown line (OE, opposite side) moves parallel to itself, thus preserving the value of the ratio a”/b”

And in case you got this far, some light relief. Bullfrog eats dog food, this morning. The dish is 10 inches across.

1 Comment

Filed under algebra, bisecting, geometry, halving

## Analytic (coordinate) geometry has its good points, but elegance is not one of them.

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?

In view of one of the solutions to the first problem I decided that the triangle should be chopped in half. Not as simple as chopping a rectangle in half !

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.

Small notational irritation: In the diagram below G is now C and H is now K

The visible curve is called the envelope of the lines, and it (obviously) touches the medians.

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.
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The geometric diagrams were all constructed with the web based program
http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html
√3