Tag Archives: mirrors

Red, Alert

Geometry, Math Foundations, Proofs, ,

[This is the 4th and last installment in the current series on mini-golf and ellipse geometry. See the previous ones here: #1, #2, #3.]

We must settle one more question to round out our elliptical arc: Why does light, when shot from one focus of an ellipse-shaped mirrored room, reflect back to the other focus? To answer this question, we’ll need a Fact, a Formalism, and a Fairy Tale.

The Fact

Recall that, in the previous post, we saw that ellipses can be described by distances: Any ellipse has two focus points F1 and F2 so that the total length of broken path F1XF2 is the same for every point X on the ellipse; let d be this common total distance. In fact, more is true: the length of F1XF2 is smaller than d if X is inside the ellipse, and larger than d if X is outside.

[Ellipse Distances]
Left: The distance of the broken path is greater, equal, or less than d depending on whether the central vertex is outside, on, or inside the ellipse. Right: A proof of the “outside” case.

To prove this fairly intuitive fact, we’ll use the “straight line principle”: the shortest distance between two points is a straight line. Indeed, when X is outside the ellipse (see right diagram above), straight-line path YF2 is shorter than the path YXF2 that detours through X, and so \(\)d = F_1 Y F_2 < F_1 Y X F_2[/latex]. See if you can fill in the case where X is inside the ellipse.

The Formalism

Recall that when light bounces off a straight mirror, the angle of incidence equals the angle of reflection. But here we’re discussing light bouncing off an ellipse, which is decidedly not straight. So we need to formally describe how light reflects off curved mirrors.

[Reflection off a Curve]
Reflection off of a curved mirror behaves like reflection off of the tangent line.

If we zoom into where the incoming light ray strikes a curved mirror (illustrated above), the mirror closely resembles a straight line, specifically its tangent line. This suggests that the light should behave as if it is reflecting off of this line, with equal angles as marked. This is indeed the rule governing ideal reflections on curved mirrors: the angles of incidence and reflection, as measured from the tangent line, should be equal.

The Fairy Tale

The last ingredient involves Little Red Riding Hood and her thirsty grandmother. Red is delivering cake and wine from her mother (point M) to her grandmother (point G), but she must first fill a bucket of water at the nearby stream S, which is conveniently shaped like a straight line. She was warned by her Brothers to watch out for a big bad wolf, so she must minimize her total walking distance. Where on the stream should she fill her bucket to minimize this distance?

[Little Red's Geometry Problem]
Left: Paths from M to S to G transform into paths from M’ to G, the shortest of which is straight. Right: This straight path behaves like a mirror.

To answer this, imagine reflecting the first leg of Red’s journey across line S, so her path from M to S to G gets reflected to a path from M’ to G. The reverse may be done as well: any path from M’ to G turns into a path from M to G that stops somewhere along S. So we just need to find the shortest path from M’ to G. But this is easy: it’s just the straight path M’G. So Red’s shortest path from M to S to G is the one that stops at Z.

Notice that this shortest path is the one with equal angles as marked. This means Red’s best strategy is to pretend the stream is a mirror and to follow the light ray that bounces directly to grandma’s house. This neatly exemplifies Fermat’s principle, which says that light tends to follow the fastest routes.

The Proof

With these pieces in place, we can finish today’s question in a flash. Let’s say light from focus F1 hits an ellipse at point X, as illustrated below. Why does this ray bounce off the ellipse toward F2? If we draw the tangent line L at point X, by The Formalism above, this question is equivalent to: why does the light ray bounce off of line L toward F2?

[Reflection Off an Ellipse]
Why are the two marked angles equal? Because the white path solves Red’s Fairy Tale problem.

Let’s reimagine this Grimm scenario by thinking of F1 as Red’s mother’s house, F2 as grandma’s house, and L as the stream. I claim F1XF2 is the shortest path for Red to take. Why? If Y is any other point on line L, then Y is outside the ellipse, so by The Fact above, F1YF2 has distance longer than d. So F1XF2 is indeed the shortest. But by The Fairy Tale, we know that this shortest route behaves like light bouncing off of line L, i.e., the marked angles are indeed equal. So we’re done!

More Putting Predicaments

Games/Experiments, Geometry, , ,

Last time we discussed some rather challenging holes of (mathematical) mini-golf, uncovering Tokarski’s construction of some un-hole-in-one-able holes. By contrast, here is the all-time easiest hole of mini-golf, namely, a guaranteed hole-in-one:

Reflections in an Ellipse
In an ellipse, any golf shot from focus \(F_1\) will bounce directly to the other focus, \(F_2\).

By some miracle of geometry, if we take an ellipse (i.e., a stretched or squashed circle) and place the ball and cup at two special points \(F_1\) and \(F_2\) called the foci of the ellipse, then any golf shot leaving \(F_1\) will bounce off a wall and proceed directly to \(F_2\).[1] You can’t miss![2]

Curiously, this easiest golf hole can be used to construct some even more challenging ones. For starters, consider a mushroom-shaped room as illustrated below (left), where the “mushroom head” is half of an ellipse with foci \(F_1\) and \(F_2\). Then the same reflection principle mentioned above tells us that any shot entering the mushroom head via segment \(F_1F_2\) will be reflected straight back through \(F_1F_2\) and sent back down the stem. This means that no shot from P can ever reach the triangular “fang” containing Q, and in particular, no hole-in-one from P to Q is possible. In the mirrored-room setting (described last time), if a light source is placed at P, then the whole triangular fang remains unilluminated! This is stronger than Tokarski’s room, where only a single point remained unlit. Similar constructions are possible even with rooms that have no corners[3], such as the “curvy” mushroom on the right.

Mushroom-shaped mini-golf holes
Left: Any path entering the mushroom head from segment \(F_1F_2\) will be reflected back down through this segment, so the triangle containing Q is not reachable from P. Right: The same phenomenon is possible even when the room has no corners.

With this idea, we can construct a mirrored room that has dark patches no matter where a light bulb is placed. In the image below, if a light source is placed anywhere in the top half of the room, the lower triangular “fang” will be completely dark. Similarly, the upper fang is not illuminated from anywhere in the bottom half of the room. This room (or one quite like it) was originally designed by Roger Penrose in 1958.

The Penrose Room
This room will have dark regions no matter where the light source is placed.

Back in the mini-golf setting, we can do something even more devious: by chaining multiple Penrose rooms together, we can construct golf holes that require as many shots as we wish! The golf hole below cannot be completed with fewer than 7 shots. Indeed, no single shot can cross two dashed segments with different numbers, so 7 shots are required.[4]

Minigolf hole needing 7 shots
At least 7 shots are required to get from P to Q.

What about polygons?

The golf holes last time were all polygons, whereas we have allowed curved boundaries here. Can we construct a polygonal golf hole, using only flat walls, that still requires at least 7 shots? Or even 3? Unfortunately, the answer this time is “we don’t know, but we think not.” In particular, it has been conjectured that, no matter where the ball and cup are placed in a polygonal golf hole, a single shot can place the ball as close to the hole as desired, so a short second putt can finish the job.[5]

Notes

  1. In fact, even more is true: all of these paths from \(F_1\) to \(F_2\) have the same length! We will not prove these facts today. [↩]
  2. Assuming you hit hard enough, of course. [↩]
  3. For the analysts out there, the room’s boundary can be made smooth. [↩]
  4. This too can be accomplished with smooth boundary. [↩]
  5. More strongly, O’Rourke and Petrovici conjecture that with a single light source in a polygonal room, the set of unilluminated points has measure 0. Reference: Joseph O’Rourke and Octavia Petrovici. Narrowing Light Rays with Mirrors. Proceedings of the 13th Canadian Conference on Computational Geometry, 137–140, 2001. [↩]

How Unilluminating!

Games/Experiments, Geometry, Proofs, , ,

You are standing in a large, oddly-shaped room, all of whose walls are lined with mirrors. Everything is dark except for a single light bulb somewhere else in the room. The light from this bulb spreads out in all directions and bounces off the mirrors, possibly many times over. Is it possible that, despite all the mirrors, you could still be standing in the dark, unable to see light from this bulb from any direction?

Said differently, consider the room as a large hole of mathematical mini-golf, where the locations of the ball (just one point) and cup (also just a point) are predetermined. Your goal is to pick a direction to hit the ball, hoping that it will reflect off the walls and eventually find its way to the cup. Is a hole-in-one always possible?

Perfect Reflection
When a light ray hits a wall, it bounces off in such a way that the angle of incidence (top left) equals the angle of reflection (top right). In other words, the dashed trajectory is reflected through the line of the mirror.

We assume that when a ray of light (i.e., the golf ball) hits a mirror, it reflects perfectly, as illustrated above. Let’s also assume that it never loses energy, so it will continue as far as necessary to reach its destination. There’s one corner case to consider, namely, what happens if the ray hits a corner of the room? Well, it’s (usually) ambiguous as to where it should reflect next, so let’s say that the ray just dies if it hits a corner. In the mini-golf formulation, our hole-in-one must avoid the hole’s corners.

With these rules in place, it turns out that it is possible to be left in the dark, or to make a hole-in-one impossible! One of the first examples, provided by Tokarski in 1995[1], is illustrated below; let’s call this room R.

An unilluminable room
No light ray from P can exactly reach Q.

Let’s see why this room R works, i.e., why point Q is not illuminated by a light source at P.

Notice that R is built from identical square cells. As our ray of light continues through the room, we can keep track of where it is in its current cell—but not which cell it is in—by imagining that the ray is just bouncing around in a single square, as illustrated below. Call this single square S. We are essentially folding up the light ray’s trajectory into S.

Shooting a ray in Tokarski's room
An example light ray reflecting off the walls in Tokarski’s room, and the corresponding “folded” trajectory in a single cell (top middle). This ray gets very close to Q, but it will never reach it exactly. The grid is colored in a repeating pattern where each cell has 1 light yellow corner and three dark purple corners.

Color the grid points with light yellow and dark purple as illustrated above; in particular, P and Q are both yellow. Tokarski’s room has the interesting property that every dark purple point is in fact a corner of the room, so if the ray hits there, it dies. This means that the corresponding ray in square S would die when hitting one of the purple corners. So, in order for the light to reach Q in Tokarski’s room, the corresponding path in S must start at the yellow corner, end back at this same corner, and completely avoid the purple corners of the square. I claim this is impossible. Why?

Just as we folded room R into a single square, let’s consider unfolding square S to cover the whole plane. Because we’ve assumed perfect reflections, our trajectory in cell S unfolds to a straight line!

Unfolding the Ray
“Unfolding” the the light’s trajectory in the square produces a straight path in an infinite grid of squares.

This means that we would need to find a straight trajectory from the initial yellow point to some other yellow point that does not hit any purple points. But this is impossible: all the yellows are blocked by purples![2] So the light can’t get from P to Q in Tokarski’s room, as claimed.

This room R is not the only example of an unilluminable room; with similar methods, many such rooms may be constructed. For example, in an effort to find an unilluminable room with the smallest number of edges, Tokarski provided the 26-sided polygon below on the left, which was then improved by Castro[3] to a 24-sided polygon on the right.[4] Both are built from 45-45-90 triangles instead of square cells.

Tokarski's and Castro's "minimal" unilluminable rooms
Tokarski’s 26-sided unilluminable polygon (left) and a 24-sided modification by Castro (right).

Tokarski also provides the gem below (with no right angles!), and his paper gives recipes for a wide variety of other shapes.

Unilluminable room with no right angles
Another of Tokarski’s unilluminable rooms, this time with no right angles. The basic unit here is a triangle with angles 9, 72, and 99 degrees.

Here’s a challenge: so far, we’ve seen that it is possible to design a mathematical mini-golf hole that requires at least 2 shots. Can we make an even harder hole? Can we design one that needs at least, say, 10 shots? We’ll discuss this next time. See you then!

Notes

  1. George W. Tokarsky. Polygonal Rooms Not Illuminable from Every Point. American Mathathematical Monthly, 102:867–879, 1995. [↩]
  2. Here’s a proof: given any yellow-yellow segment, its midpoint lies on the grid. If this midpoint is purple, then we’re done. Otherwise, take half of the original segment and repeat this argument. [↩]
  3. D. Castro. Corrections. Quantum 7:42, 1997. [↩]
  4. It is not known whether 24 is minimal. [↩]