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(This message was added in version 6.7.0.) in /home4/scienrds/scienceandnerds/wp-includes/functions.php on line 6114Source:https:\/\/www.quantamagazine.org\/the-mysterious-math-of-billiards-tables-20240215\/#comments<\/a><\/br> In Disney\u2019s 1959 film Donald in Mathmagic Land<\/em>, Donald Duck, inspired by the narrator\u2019s descriptions of the geometry of billiards, energetically strikes the cue ball<\/a>, sending it ricocheting around the table before it finally hits the intended balls. Donald asks, \u201cHow do you like that for mathematics?\u201d<\/p>\n Because rectangular billiard tables have four walls meeting at right angles, billiard trajectories like Donald\u2019s are predictable and well understood \u2014 even if they\u2019re difficult to carry out in practice. However, research mathematicians still cannot answer basic questions about the possible trajectories of billiard balls on tables in the shape of other polygons (shapes with flat sides). Even triangles, the simplest of polygons, still hold mysteries.<\/p>\n Is it always possible to hit a ball so that it returns to its starting point traveling in the same direction, creating a so-called periodic orbit? Nobody knows. For other, more complicated shapes, it\u2019s unknown whether it\u2019s possible to hit the ball from any point on the table to any other point on the table.<\/p>\n Although these questions seem to fit snugly within the confines of geometry as it\u2019s taught in high school, attempts to solve them have required some of the world\u2019s foremost mathematicians to bring in ideas from disparate fields including dynamical systems, topology and differential geometry. As with any great mathematics problem, work on these problems has created new mathematics and has fed back into and advanced knowledge in those other fields. Yet despite all this effort, and the insight modern computers have brought to bear, these seemingly straightforward problems stubbornly resist resolution.<\/p>\n Here\u2019s what mathematicians have learned about billiards since Donald Duck\u2019s epically tangled shot.<\/p>\n They typically assume that their billiard ball is an infinitely small, dimensionless point and that it bounces off the walls with perfect symmetry, departing at the same angle as it arrives, as seen below.<\/p>\n Without friction, the ball travels indefinitely unless it reaches a corner, which stops the ball like a pocket. The reason billiards is so difficult to analyze mathematically is that two nearly identical shots landing on either side of a corner can have wildly diverging trajectories.<\/p>\n A key method for analyzing polygonal billiards is not to think of the ball as bouncing off the table\u2019s edge, but instead to imagine that every time the ball hits a wall, it keeps on traveling into a fresh copy of the table that is flipped over its edge, producing a mirror image. This process (seen below), called the unfolding of the billiard path, allows the ball to continue in a straight-line trajectory. By folding the imagined tables back on their neighbors, you can recover the actual trajectory of the ball. This mathematical trick makes it possible to prove things about the trajectory that would otherwise be challenging to see.<\/p>\n For example, it can be used to show why simple rectangular tables have infinitely many periodic trajectories through every point. A similar argument holds for any rectangle, but for concreteness, imagine a table that\u2019s twice as wide as it is long.<\/p>\n Suppose you want to find a periodic orbit that crosses the table n<\/em> times in the long direction and m<\/em> times in the short direction. Since each mirror image of the rectangle corresponds to the ball bouncing off a wall, for the ball to return to its starting point traveling in the same direction, its trajectory must cross the table an even number of times in both directions. So m <\/em>and n <\/em>must be even. Lay out a grid of identical rectangles, each viewed as a mirror image of its neighbors. Draw a line segment from a point on the original table to the identical point on a copy n<\/em> tables away in the long direction and m<\/em> tables away in the short direction. Adjust the original point slightly if the path passes through a corner. Here\u2019s an example where n<\/em>\u00a0=\u00a02 and m<\/em>\u00a0=\u00a06. When folded back up, the path produces a periodic trajectory, as shown in the green rectangle.<\/span><\/p>\n Billiards in triangles, which do not have the nice right-angled geometry of rectangles, is more complicated. As you might remember from high school geometry, there are several kinds of triangles: acute triangles, where all three internal angles are less than 90 degrees; right triangles, which have a 90-degree angle; and obtuse triangles, which have one angle that is more than 90 degrees.<\/p>\n Billiard tables shaped like acute and right triangles have periodic trajectories. But no one knows if the same is true for obtuse triangles.<\/p>\n To find a periodic trajectory in an acute triangle, draw a perpendicular line from each vertex to the opposite side, as seen to the left, below. Join the points where the right angles occur to form a triangle, as seen on the right.<\/p>\n This inscribed triangle is a periodic billiard trajectory called the Fagnano orbit, named for Giovanni Fagnano, who in 1775 showed that this triangle has the smallest perimeter of all inscribed triangles.<\/p>\n In the early 1990s, Fred Holt at the University of Washington and <\/span>Gregory Galperin<\/a><\/span> and his collaborators at Moscow State University <\/span>independently<\/a><\/span> showed<\/a><\/span> that every right triangle has periodic orbits. One simple way to show this is to reflect the triangle about one leg and then the other, as shown below. <\/span><\/p>\n Start with a trajectory that\u2019s at a right angle to the hypotenuse (the long side of the triangle). The hypotenuse and its second reflection are parallel, so a perpendicular line segment joining them corresponds to a trajectory that will bounce back and forth forever: The ball departs the hypotenuse at a right angle, bounces off both legs, returns to the hypotenuse at a right angle, and then retraces its route.<\/p>\n But obtuse triangles remain a mystery. In their 1992 paper, Galperin and his collaborators came up with a variety of methods of reflecting obtuse triangles in a way that lets you create periodic orbits, but the methods only worked for some special cases. Then, in 2008, Richard Schwartz<\/a> at Brown University showed that all obtuse triangles with angles of 100 degrees or less<\/a> contain a periodic trajectory. His approach involved breaking the problem down into multiple cases and verifying each case using traditional mathematics and computer assistance. In 2018, Jacob Garber, Boyan Marinov, Kenneth Moore<\/a> and George Tokarsky at the University of Alberta extended this threshold<\/a> to 112.3 degrees. (Tokarsky and Marinov had spent more than a decade<\/a> chasing this goal.)<\/p>\n Another approach has been used to show that if all the angles are rational \u2014 that is, they can be expressed as fractions \u2014 obtuse triangles with even bigger angles must have periodic trajectories. Instead of just copying a polygon on a flat plane, this approach maps copies of polygons onto topological surfaces, doughnuts with one or more holes in them.<\/p>\n If you reflect a rectangle over its short side, and then reflect both rectangles over their longest side, making four versions of the original rectangle, and then glue the top and bottom together and the left and right together, you will have made a doughnut, or torus, as shown below. Billiard trajectories on the table correspond to trajectories on the torus, and vice versa.<\/p>\n In a landmark 1986 article, Howard Masur<\/a> used this technique to show that all polygonal tables with rational angles have periodic orbits. His approach worked not only for obtuse triangles, but for far more complicated shapes: Irregular 100-sided tables, say, or polygons whose walls zig and zag creating nooks and crannies, have periodic orbits, so long as the angles are rational.<\/p>\n Somewhat remarkably, the existence of one periodic orbit in a polygon implies the existence of infinitely many; shifting the trajectory by just a little bit will yield a family of related periodic trajectories.<\/p>\n Shapes with nooks and crannies give rise to a related question. Rather than asking about trajectories that return to their starting point, this problem asks whether trajectories can visit every point on a given table. This is called the illumination problem because we can think about it by imagining a laser beam reflecting off mirrored walls enclosing the billiard table. We ask if, given two points on a particular table, you can always shine a laser (idealized as an infinitely thin ray of light) from one point to the other. To put it another way, if we placed a light bulb, which shines in all directions at once, at some point on the table, would it light up the whole room?<\/p>\n There have been two main lines of research into the problem: finding shapes that can\u2019t be illuminated and proving that large classes of shapes can be. Whereas finding oddball shapes that can\u2019t be illuminated can be done through a clever application of simple math, proving that a lot of shapes can be illuminated has only been possible through the use of heavy mathematical machinery.<\/p>\n In 1958, Roger Penrose<\/a>, a mathematician who went on to win the 2020 Nobel Prize in Physics<\/a>, found a curved table in which any point in one region couldn\u2019t illuminate any point in another region. For decades, nobody could come up with a polygon that had the same property. But in 1995, Tokarsky used a simple fact about triangles to create a blockish 26-sided polygon with two points that are mutually inaccessible, shown below. That is, a laser beam shot from one point, regardless of its direction, cannot hit the other point.<\/p>\n The key idea that Tokarsky used when building his special table was that if a laser beam starts at one of the acute angles in a 45\u00b0-45\u00b0-90\u00b0 triangle, it can never return to that corner.<\/p>\n His jagged table is made of 29 such triangles, arranged to make clever use of this fact. In 2019 Amit Wolecki<\/a>, then a graduate student at Tel Aviv University, applied this same technique to produce a shape<\/a> with 22 sides (shown below), which he proved was the smallest possible number of sides for a shape that had two interior points that don\u2019t illuminate each other.<\/p>\n Proving results in the other direction has been a lot harder. In 2014, Maryam Mirzakhani, a mathematician at Stanford University, became the first woman to win the Fields medal<\/a>, math\u2019s most prestigious award, for her work on the moduli spaces of Riemann surfaces \u2014 a sort of generalization of the doughnuts that Masur used to show that all polygonal tables with rational angles have periodic orbits. In 2016, Samuel Leli\u00e8vre<\/a> of Paris-Saclay University, Thierry Monteil<\/a> of the French National Center for Scientific Research and Barak Weiss<\/a> of Tel Aviv University applied a number of Mirzakhani\u2019s results to show<\/a> that any point in a rational polygon illuminates all points except finitely many. There may be isolated dark spots (as in Tokarsky\u2019s and Wolecki\u2019s examples) but no dark regions as there are in the Penrose example, which has curved walls rather than straight ones. In Wolecki\u2019s 2019 article<\/a>, he strengthened this result by proving that there are only finitely many pairs of unilluminable points.<\/p>\n
\nUnfolding the Mysteries of Polygonal Billiards<\/br>
\n2024-02-16 21:58:44<\/br><\/p>\nA Triangle Inequality<\/h2>\n
A Topological Turn<\/strong><\/h2>\n
The Illumination Problem<\/strong><\/h2>\n