Always or Never

Pentaponce

Take two ellipses, one within the other. Take a point on the outer ellipse, and draw one of the two tangents to the inner ellipse, and find its second intersection with the outer ellipse. Use this point to start the process again, and again. You will get a polygonal path in the ellipse that will most likely not close up. But in case you are lucky, something miraculous happens: If you pick any other point and repeat the game, the polygon will again close up.

This is the content of a famous theorem by Jean-Victor Poncelet.

Steiner 01

In spirit, it is similar to a theorem of Jakob Steiner that asserts that a chain of circles in an annulus bounded by two circles either always or never closes up. While Steiner’s theorem follows immediately by inverting the circles into a pair of concentric circles, such a simple proof is not available for Poncelet’s theorem. Until recently, all proofs I know of were, let’s say, advanced.

At the core of a new proof by Lorenz Halbeisen and Norbert Hungerbühler are some fundamental theorems from projective geometry.

Let’s first recall that five points, no three collinear, determine a unique conic.

Conic4points

This is because through four points, you can find two different degenerate conics consisting each of a pair of lines, and by forming linear combinations, accommodate a fifth point. Below we will need the dual theorem: Given five lines, no three concurrent, there is a unique conic tangent to them.

Pascal Hyperbola 01

Pascal’s theorem is a condition for six points to lie on a conic: They do if and only if opposite sides intersect in collinear points. Above you see this for six points on the two branches of a hyperbola.

Brianchon 01
Dual to this is Brianchon’s theorem (illustrated above): The sides of a hexagons are tangent to a conic if and only of its diagonals are concurrent.

Ponce1

As an application, Halbeisen and Hungerbühler show: If the six vertices of two triangles a1,a2,a3 and b1,b2,b3 lie on a conic, than there is a conic tangent to the six sides of the triangles. The proof is easy: Applying Pascal to the hexagon a1,b2,a3,b1,a2,b3 gives us three collinear points c12,c13,c23.

Ponce2

Then applying Brianchon to the hexagon a1,c12,b1,b3,c23,a3 shows that it is tangent to a conic. But the sides of this hexagon are the same as the sides of the two triangles, so we are done

Ponce3

From here, we obtain Poncelet’s theorem for triangles: Suppose you have two ellipses inside each other, and a triangle whose vertices lie on the outer ellipse and whose sides are tangent to the inner. Take another point on the outer ellipse, and form a second triangle by drawing the tangents to the inner ellipse. We have to show that the third side of the triangle is also tangent to the inner ellipse.

By the theorem by Halbeisen and Hungerbühler, the two triangles have an inscribed common ellipse. The given inner ellipse touches five of the same six lines by construction. But a conic is uniquely determined by five tangent lines.

The general case follows of n-gons the same idea, but requires more bookkeeping.

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Too Wide?

When I started using an SLR, I had just two lenses: A 28-85mm zoom, and a 20mm wide angle lens. That was too wide for me, back then,
and it took me a while to appreciate it.

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When I moved on to a DSLR, one of the first new lenses I bought was Nikon’s 14-24mm zoom. That was something else, and again it took me many years to make use of the wider end of it.

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This year, I decided to push myself again, and I acquired an 11mm lens.

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This lens works like the news these days: It shows a distorted reality. If you want the truth, look elsewhere.

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But, as with the news these days, the distortion is so extreme, that we are never tricked into believing it is real. It is more a provocation.

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The benefit? Maybe we can learn to resist to undergo this distortion ourselves. Or is the remaining path too narrow?

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Odd Angles

For a while, this will be my last post about conformal spiderwebs. Today, we will still look at circular quadrilaterals that are conformal images of squares, but allow the angles to be multiples of 90 degrees. Like so:

2quarter a 01

Let’s call this a square of type (1,1,3,3). Multiply the numbers by 90 and you get the angles at the vertices. I have again employed Möbius to place three corners at (1,0), (0,1), and (-1,0). The fourth vertex is again moving cautiously along the unit circle. Below is a square of type (1,3,3,3), and here the fourth vertex is on the x-axis, the second possible case we noticed for right angled circular quadrilaterals.

Mixed31b 0

Similarly, here is a square of type (1,1,1,3), also with the fourth vertex on the x-axis.

1quarter a 01

Missing are squares of type (1,3,1,3). While there are quadrilaterals of this type, all conformally correct squares I could find were only immersed (i.e. overlapping).

Then one can also have squares of type (2,2,2,2), for instance. The circle would be an example, with artificial vertices at (1,0),(0,1),(-1,0) and (0,-1), but there are also bean shaped squares like the one below.

Bean0

Finally, the square with zero angles, in its most regular form.

Zero 4 01

Out of Order

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I don’t get often to Nevada. The last time was in March 2015, flying into Las Vegas to drive north to Zion. We spent a night and a day in Moapa Valley to visit the Valley of Fire State Park. Both town and state park completely cured me of all prejudices I had about Nevada. Moapa Valley is a small relaxed community with lots of local artists and friendly people, and the Valley of Fire State Park is an amazing piece of landscape that is on my list of places to return to.

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Typical views are like the ones above, and full of interesting details. Can you spot the head below in the image above?

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Everything here is created by light and shadow, and changes within minutes.

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One can go instantly from harsh contrast to soft pastell. You would think a landscape like this can cure every ailment of the soul.

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From there, on our way north, we passed through another little town: Mesquite, Nevada, not even 40 miles northeast. Little did we know about what had taken residence there a few months earlier, brooding, breeding the incomprehensible.

In Memoriam, once more: Las Vegas, October 1, 2017.

More Spiderwebs: Conformal Squares with Circular Edges

The Schwarzian derivative of a map f(z) from the upper half plane to a right angled circular quadrilateral that sends the points -1, 0, 1 and infinity to the vertices of the quadrilateral (and thus making it conformally a square) is given by the expression

Schwarzian

From it, one can find this map by taking the quotient of two solutions of the linear ordinary differential equation u”(z) + 2 Sf(z) u(z)=0. This is one step more complicated that the hypergeometric differential equation needed for triangles.

The parameter “a” is the accessory parameter. We have seen last week that there is a two-parameter family of right angled circular quadrilaterals, and the parameter a singles out the 1-dimensional subfamily of those quadrilaterals that are conformally squares.

Square 0

In these images I have used a Möbius transformation to move three of the vertices to the points (1,0), (0,1) and (-1,0). The fourth vertex is then somewhere on the lower unit circle.

Square 1

This is somewhat remarkable: First, it shows that we can find a conformal square for any such choice of four points (the first three normalized, the fourth on the half-circle). Secondly, it appears that the second family of right circular quadrilaterals we found last week where the fourth corner would move on line through (-1,0) and (1,0) does not contain any conformal squares. Thirdly, remember again from last week that for any such choice of four vertices, there is a 1-parameter family of right circular quadrilaterals with these points as vertices, but only one of them is conformally a square.

Square 0 5

Of course one can also play with the angles. As a teaser for what’s to come next week, below is an anti-square.

3quarter b

Another Form of Recovery

After a long summer, when heat, humidity and bugs are slowly retreating, it is time to visit some old friends, like here in the canyon of McCormick’s Creek State Park.

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The plan was to capture some of the spots I remembered so that they appear like my memories. The lens of choice was the Lensbaby Velvet which wide open blurs the landscape into oblivion. There is, for instance, the wonderfully maturing tree trunk which I had first seen as a still healthy but otherwise unremarkable tree,
or the overhanging tree that (against all odds) has survived this year one of T.

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There are also the long views into the canyon that seem to support direction and focus, but instead limit choice.

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Another trunk (near the sacred spring) has been sprouting new life – not a miracle, but symbol for resistance.

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Of course there are also the frames I have written about before I am sure but can’t find anymore – – –

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Things become clear when I encounter new friends, like this unlikely pile of rocks in the middle of the stream.

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Not Being Square

I meant to post today a sequel to the circular triangles from last week, but I got carried away looking at right angled quadrilaterals bounded by circular arcs. Like the pillows, but more general. Like so:

Circulon 2

The question arises for what choices of four points we can find a right angled quadrilaterals bounded by circular arcs?

By the way, how do we call these? I thought about circulons (taken) and horny squares (oops). For now, I call them circulions (like centurions), to avoid a lawsuit about trade marks. Above you see a solution that is not a square but where the vertices are at the corners of a square. There are more like these, in fact a 1-parameter family.

Circulon 1

Below you can see the entire family at once, you just have to follow all dots with the same color.

Squaregradient

Can we do that for any choice of four points? Not so, but: Möbius allows us to move three points anywhere we like (and he will send circulions to circulions), so we can ask: where are we allowed to place a fourth point so that there is a circulion through all of them?

Möbius also tells us that this is easy if we place all four points on a circle (by sending that circle to a line, and then connecting the four points on the line alternatingly by segments of the real line and half circles, for instance). Here is an example where the first three points are at the corners of an equilateral triangle, and the fourth point is on the circle through them.

Circulon 3

Again, there is a 1-parameter family of such circulions through these points.

Trianglegradient

Pretty, isn’t it?

Now, surprisingly to me, for each choice of three points, there is a second circle on which the fourth point can reside: Take the circle that contains the given three points, and construct the circle orthogonal through it that passes through the two points between which we want to put the fourth vertex. You can put the fourth point anywhere on that new circle. Here is an example, with the first three points again at the corners of an equilateral triangle.

Circulon 4

Below is again an entire family, color coded and adorned with moiré.

Trianglegradientb

Next week you’ll see conformally correct squares. Promised.