Ruby 18

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I used to think of Taiwan as the country of wonderfully floral, greenish Oolong teas. Only last summer I learned that they also make a few black teas, the most famous one called Ruby 18. This is one of the most intense black teas I have ever tasted: It shares strong chocolate notes with many Chinese black teas, but adds to it a malt and mint flavors. An intense experience. A second brewing is possible and more mellow. My latest delivery came from a company called Taiwan Sourcing.

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The vaccuum sealed pouches were accompanied by a handwritten note and a sample in an even smaller pouch. The leaves of the Ruby are long and thinly rolled,

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and expand quite a bit while being brewed. The history of this tea is remarkable. During the Japanese occupation, the Taiwan Tea Experimental Station researched local wild tea plants and their suitability for cultivating black tea. The crowning achievement became Ruby 18.

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I didn’t know that the Japanese were interested in black tea. Apparently, they even used to export it, but the production has shrunk to very small quantities, and I never had any of it.


There are other black teas grown in Taiwan.Very different in shape and flavor (but also excellent) is the Imperial Grade Lalashan Organic Black Tea, a high mountain tea:

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All this reminded of Twan Tan Eng’s wonderful book The Garden of Evening Mists, in which a Japanese gardener tends to a garden on a Malaysian tea plantation.







A few weeks ago, I explained that two Euclidean polygons are scissor congruent if and only of they have the same area. A scissor congruence is a dissection of the two polygons into smaller polygons (“pieces”) so that the pieces of the first polygon can be translated and rotated into the pieces of the second polygon. Then I asked whether we really need to rotate the pieces, or whether translating them is enough. For instance, can one dissect a square into pieces that can be translated into a second square that is rotated by say 45º?


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That this is possible is a consequence of a famous dissection of a square into two squares, shown above, which uses translations only, but tilts the square by an angle which we can make 22.5º. Then we need to repeat this process, tilting the other way, using a mirrored dissection.  This increases the number of pieces needed, and the question arises with how few pieces one can do this. A few years back I ran a contest about this in our department. The best solution with six pieces was found by Seth, one of our (then) undergraduates:

SolpuzzleSince then, I learned that there are solutions with only five pieces. Check them out!

Another cool example is the trans-dissection of a single pentagon into four smaller ones, by Harry Lindgren:

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So, do trans-dissections always exist? Not at all. Let’s try to trans-dissect an equilateral triangle into a copy of it that is rotated by 180º. Suppose we found such a dissection, like the one above. Look at its horizontal edges. There are two types, the ones at the bottom of the pieces, (pointing right), and the ones at the top of the pieces, pointing left. When we add these together (taking the direction into account) for the dissected pieces, the edges in the interior cancel, while the boundary edges add up to the length of the bottom edge of the initial triangle.



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Thus the oriented length of all horizontal edges of the two tiles that we want to dissect into each other need to be equal. This eliminates the possibility to rotate a triangle at all. The same argument works of course for all directions.

Therefore, in order for two polygons to be trans-scissor-congruent, they not only need to have the same area, but also have the same oriented edge lengths for all edge directions. Remarkably, these conditions are also sufficient, as was proven in 1951 by Paul Glur and Hugo Hadwiger. The argument is a little tricker than the one for general dissections, but not too bad. Maybe I’ll come back to it later.

The Red-Veined Snow Trillium (?)

Three years ago I mentioned a Snow Trillium with red veins. This year, it (or a close relative) is back, prettier than ever:

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There is no such thing as a red-veined snow trillium. But given that there are different trillium species that I have a hard time to tell apart, this one is quite distinct from the much more common trillium nivale (ordinarium). Which is also pretty:

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And this year, the red-veined (or hot-blooded?) variant was not a single appearance. Here is another specimen:

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My attempt to find more was marred by heavy sudden snowfall. After most of the 3 inches had melted away, what was left looked a bit ruffled. 

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Some protected themselves by huddling together, like the apparent snow hexillium below.

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That one is definitely a fraud. There are, however, very rare quadrilliums.

The Octagon (CLP-1)

Let’s start with an equation: y²=x⁸-1. Solving for y is easy, because for each x we appear to have just two choices for the sign, good and evil. If we do this in the complex plane, the set of solutions therefore looks like two copies of the x-plane. There is a little problem at the eighth roots of unity, because there, good and evil coalesce. 


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A good way to imagine this is to think about the (extended) complex plane as two disks, and of each disk as a regular octagon, with vertices at these eighth roots of unity. Then it takes four such octagons to build the solution space of the equation y²=x⁸-1, and we need to have four octagons at each vertex coming together, alternating between good and evil. Luckily, this can be done in the hyperbolic plane, using a tiling by regular right-angled hexagons.To get an idea how these are glued together, it helps to think about the equation in the form x⁸=y²+1. This represents the same solution space as 16 copies of a single triangle, with vertices at the octagon centers as shown above. Thus the entire solution space can also be obtained by gluing together the edges of the 16-gon above, where the identifications are indicated by the (extended) edges of the central octagon.

Wouldn’t it be nice if we could visualize this in ℝ³? This is indeed possible if we are willing to conformally bend our octagon a little so that every other edge becomes a straight segment, and the other edges lie in planes that meet the octagon orthogonally along that edge.

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This allows to extend the octagon by rotating and reflecting about its edges like above, which shows four such hexagons, i.e. the entire solution space. If you do this right, you get one of the many views of the CLP surface of Hermann Amandus Schwarz. CLP stands for crossed layers of parallels. This is once again a triply periodic minimal surface. Here is another translational fundamental piece that corresponds to the 16-gon:

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Let’s begin to rotate through the associate family. For angle π/16, we see how the touching vertices are being separated.

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At π/4, we get a nice symmetric piece, but translational copies will intersect so that the surface will not remain embedded.

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At π/2 we meet the conjugate surface of the CLP surface. The amusing point here is that it is congruent to the CLP surface, a feature it shares with the Enneper surface and one surface in the family of Riemann’s minimal surfaces.Clp 2


The Lidinoid (H 2)

All minimal surfaces can be locally bent in a 1-parameter family of associate minimal surfaces. In the right context this is just a rotation. The best known example is the deformation of the catenoid into the helicoid. 

H pieceRemarkably, the triply periodic Meeks surfaces, rotated in this sense by 90 degrees, are again in the Meeks family. Very curiously, there are two more known cases where an associate surface of a triply periodic minimal surface is again triply periodic and embedded. One of them is Alan Schoen’s Gyroid, the other is Sven Lidin’s Lidinoid. While the Gyroid occurs in the associate family of the P/D surface, the Lidinoid arises in the associate family of one particular H-surface. To see how this happens, let’s start with a top view of that H surface. When we start rotating in the associate family, the vertices of the three triangles that meet at the center of the image move apart:


But the other vertices then move towards each other, so that, after about 64.2°, they come together:

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Not only the vertices fit, but again the surface can be extended by translations in space. Here is a view of a much larger piece.

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If you are good at cross-eyed viewing, here is a stereo pair of a side view:



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Indiana is mostly flat, so there are no spectacular views from mountain peaks. Every little dropoff becomes an attraction, in particular when flowing water is involved.

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One hopes that heavy rainfall would make things better. Up above you can see a portion of the Upper Cataract Fall, which continues over a few cascades and has a total height of 45 feet. 

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Not spectacular, but quite pleasant. The site claims this is the largest waterfall by volume in Indiana. I guess this depends on when you measure. 

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Equally pretty are the flirting trees on the shore. The words “upper” suggests that there is also a Lower Cataract Fall. It is supposed to drop 30 feet. This is it:

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The second problem with Indiana is that the ground drains poorly. So, after heavy rainfall one gets heavy flooding. In this case, the lake behind the Lower Cataract Fall has risen to the height of the top of that fall, making it disappear. Bad luck?

Schwarz Hexagonal Surface (H-1)

One of the early minimal surfaces I have neglected so far is the H-surface of Hermann Amandus Schwarz.H double

Think about it as the triangular catenoids. Two copies make a translational fundamental domain, i.e. the 10 boundary edges can be identified in pairs by Euclidean translations, thus making the surface triply periodic. As a quotient surface it has genus 3, which implies that the Gauss map has 8 branched points. They occur at the triangle vertices and midpoints of triangle edges. Thus the branched values lie at the north and south pole of the sphere, and at the vertices of two horizontal equilateral triangles in parallel planes. In particular, they are not antipodal, making these surfaces the earliest examples of triply periodic minimal surfaces that lie not in the 5-dimensional Meeks family.

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Above is a larger portion of the H-surface with the triangle planes close to each other. In the limit we get parallel planes joined by tiny catenoidal necks. When we pull the planes apart, we get Scherk surfaces:


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The spiderweb for this surface looks also pretty:

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Among the crude polyhedral approximations, there is one that tiles the surface with regular hexagons. The valencies are 4 and 6, so the tiling is not platonic.

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Next week we will look at one of its more surprising features.