Category Archives: geometry

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 !
sine pic 1But what about sensible labelling –
Second diagram:
sine pic 2

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:
sine pic 3Now the vertical from B is sin(2x)

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

Fix it

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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?

This comes from the Common Core

Develop understanding of fractions as numbers.
1. ……
2. Understand a fraction as a number on the number line; represent
fractions on a number line diagram.
a. Represent a fraction 1/b on a number line diagram by defining the
interval from 0 to 1 as the whole and partitioning it into b equal
parts. Recognize that each part has size 1/b and that the endpoint
of the part based at 0 locates the number 1/b on the number line.
b. Represent a fraction a/b on a number line diagram by marking off
a lengths 1/b from 0. …….

….but how do you do it ?????

The mystery is solved…….

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

download20

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

It only has to start from zero.

download22

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.

download23

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.

download24

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.

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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. ..

parabola surprise constructio

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

parabola surpriseConstructions 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.

DSC00978
DSC00979
DSC00980

This one below is a yellow eleconia, also spreading madly.

DSC00981

And this is a cute little tree with probably poisonous berries.

DSC00982

And what the hell have they done to the post editor. It only works in BOLD

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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
shear transformation in xy plane
And here is the shearing in action, for varying amounts of shear, determined by the value of k.
gif for shear
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:
https://howardat58.wordpress.com/2015/01/02/quadrilaterals-a-christmas-journey-part-2/

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

Euclid and vertical angles

Euclid and angle between two lines

euclid pair of 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

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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.
half triangle and bisector0

This is the extended construction for the bisector. The opposite side is in brown.
half triangle and bisector1
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”

gif halving bisecting

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

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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 !
envelope half a cake

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.
half equilateral 1

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
triangle area bisector detail
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:
envelope hyperbola equilateral 2
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:
envelope hyperbola

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
http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html
√3

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Filed under envelope, geometry, math, triangle

Vertex of a parabola – language in math again

Here are some definitions of the vertex of a parabola.

One is complete garbage, one is correct  though put rather chattily.

The rest are not definitions, though very popular (this is just a selection). But they are true statements

Mathwarehouse: The vertex of a parabola is the highest or lowest point, also known as the maximum or minimum of a
parabola.
Mathopenref: A parabola is the shape defined by a quadratic equation. The vertex is the peak in the curve as shown on
the right. The peak will be pointing either downwards or upwards depending on the sign of the x2 term.
Virtualnerd: Each quadratic equation has either a maximum or minimum, but did you that this point has a special name?
In a quadratic equation, this point is called the vertex!
Mathwords: Vertex of a Parabola: The point at which a parabola makes its sharpest turn.
Purplemath: The point on this axis which is exactly midway between the focus and the directrix is the “vertex”; the vertex is the point where the parabola changes direction.
Wikibooks: One important point on the parabola itself is called the vertex, which is the point which has the smallest distance between both the focus and the directrix. Parabolas are symmetric, and their lines of symmetry pass through the vertex.
Hotmath: The vertex of a parabola is the point where the parabola crosses its axis of symmetry

Scoring is 10 points for finding the garbage definition and 5 points for the correctish definition !!!! Go for it!

When I studied parabolas, back in 1958 or so (!) the parabola had an apex. So I checked the meaning of vertex, and found that the word was frequently misused.

Here is a good account: https://en.wikipedia.org/wiki/Vertex_(curve)

Basically a vertex of a curve is a point where the curvature is a maximum or a minimum (in non math terms, most or least curved).

Here are two fourth degree polynomials, one has three vertices and the other has five. The maximum curvature points are indicated. The minimum curvature points are at the origin for the first curve, and at the points of inflexion for the second curve (curvature = zero)

01Gquartic201Gquartic0

A hyperbola has two vertices, one on each branch; they are the closest of any two points lying on opposite branches of the hyperbola, and they lie on the principal axis. On a parabola, the sole vertex lies on the axis of symmetry. On an ellipse, two of the four vertices lie on the major axis and two lie on the minor axis.

For a circle, which has constant curvature, every point is a vertex.

The center of curvature at a (nice) point on a curve is the center of the closest matching circle at that point. This circle will usually lie “outside” the curve on one side of the point, and “inside” the curve on the other side. Look carefully at the picture. It is called the osculating or kissing circle (from the Latin).

The center of curvature can be estimated by taking two point close to the point of interest, finding the tangents at these points, and then the lines at right angles to them and through the points. the center of curvature is roughly at the point of intersection of these two lines

01center of curvature

The diagram below shows this estimate, for the blue parabola, at the vertex.

02center of curvature

Finally (this has gone on further than expected!) I found this delightful gif.

01Lissajous-Curve+OsculatingCircle+3vectors_animated

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Filed under conic sections, conics, construction, geometry, language in math, teaching

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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Filed under conic sections, construction, engineering, geometry, teaching

Real problems with conic sections (ellipse, parabola)

So there is an oval hole in a metal casting. It’s supposed to be an elliptical hole. Is it ????? How can we find out ?????
A good start would be to find the line which would be the major axis if it was elliptical. This turns out to be an engineering problem, not a mathematical one (I cannot see a way!). If the oval curve has an axis of symmetry then the method below will find it:

Firstly, get a computer picture of the oval.
Take two circles, of different radii, and push them along until each one touches the oval in two places.

ellipse12

The line joining the two centers will be the axis of symmetry if there is one (this can be shown mathematically).

ellipse34

ellipse5

The equation of an ellipse uses the lengths of the major and minor axes. Do it !

The closeness to elliptic can be assessed in various ways. Think of one.

next…..finding the focus of a parabolic shape

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Filed under conic sections, conics, education, engineering, geometry, math, teaching