Tag Archives: envelope

Halving a triangle, follow-up number two, pursuing the hyperbola

Halving the triangle, any triangle, led to the equation XY = 2 as the condition on the points on two sides of the triangle, distant X and Y from the vertex.
The envelope of this set of lines turned out to be a hyperbola.
But XY = 2 defines a hyperbola – what is the connection ?

I took xy = 1 for the condition, on a standard xy grid, and wrote it as representing a function x —-> y, namely y = 1/x
The two points of interest are then (x,0) on the x-axis and (0,1/x) on the y-axis.
We need the equation of the line joining these two points, so first of all we have to see that our x, above, is telling us which line we are talking about, and so it is a parameter for the line.
We had better give it a different name, say p.
Now we can find the equation of the line in x,y form, using (p,0) and (0,1/p) for the two points:
(y – 0)/(1/p – 0) = (x – p)/(0 – p)
which is easier to read as yp = -x/p + 1, and easier to process as yp2 = -x + p

Now comes the fun bit !
To find the envelope of a set of straight lines we have to find the points of intersection of adjacent lines (really? adjacent?). To do this we have to find the partial derivative (derivative treating almost everything as constant) of the line equation with respect to the parameter p. A later post will reveal all about this mystifying procedure).
So do it and get  2yp = 1

And then eliminate p from the two equations, the line one and the derived one:
From the derived equation we get p = 1/(2y), so substituting in the line equation gives 1 = 2xy
This is the equation of the envelope, and written in functional form it is
y = 1/(2x), or (1/2)(1/x)
Yes ! Another rectangular hyperbola, with the same asymptotes.
(write it as xy = 1/2 if you like)

Now I thought “What will this process do with y = x2 ?”
So off I go, and to cut a long story short I found the following:
For y = x2 the envelope was y = (-1/4)x2, also a multiple of the original, with factor -1/4
parabola by axa track point
parabola by axa track point and line

Some surprise at this point, so I did it for 1/x2 and for x3
Similar results: Same function, with different factors.
Try it yourself ! ! ! ! ! ! !

This was too much ! No stopping ! Must find the general case ! (y = xk)
Skipping the now familiar details (left to the reader, in time honoured fashion) I found the following:

Original equation: y = xk

Equation of envelope: y = xk multiplied by -(k-1)k-1/kk

which I did think was quite neat.
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The next post will be the last follow-up to the triangle halving.

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Constructions made with GEOSTRUCT, an online browser application:

To get geostruct from the net click
http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html

and to download the .doc instructions file
http://www.mathcomesalive.com/geostruct/geostruct basics.doc

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Filed under algebra, calculus, construction, envelope, geometrical

Halving a triangle, follow-up number one, ellipse

The previous post is “Analytic (coordinate) geometry has its good points, but elegance is not one of them”, in which a formula for any line cutting a triangle in half was found. The envelope of the cutting line for one section (of three) of the triangle was found to be always a hyperbola, which got me thinking “How do I get an ellipse?”. Clearly not by cutting a triangle in half, which involved taking two points A’ and B’on adjacent sides of the triangle, and making the product of their distances from the point of intersection of the sides equal to half the product of the lengths of the two sides.

So we cannot take two distances along two lines from the same point, lets try two distances from separate points on two lines, and keep the product of the distances constant. Magic:
envelope ellipse cropped
The base line is AF. Line DL is set parallel to AF. B and G are the two points of interest, where AB and DG are the two distances, and the envelope of the line BG is an ellipse which touches AF and DL. The really interesting thing about this is that the lines do not have to be parallel, and that as the points A and F are placed nearer and nearer to the point of intersection the ellipse becomes more and more hyperbolic at the nearby end.

The next post will be a different follow-up to the triangle halving.

Constructions made with GEOSTRUCT, an online browser application:

To get geostruct from the net click
http://www.mathcomesalive.com/geostruct/geostructforbrowser1.html

and to download the .doc instructions file
http://www.mathcomesalive.com/geostruct/geostruct basics.doc

 

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Filed under conic sections, construction, geometrical

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.
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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