Category Archives: Proofs

Red, Alert

[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!

What Makes Ellipses… Ellipses?

Last time we used wild properties of ellipses to build some really easy—and some really devilish—golf courses. Specifically, I claimed that every ellipse has two magical points F1 and F2 (called foci) such that a ray from F1 always bounces off the ellipse and lands precisely at F2, and furthermore, this path always has the same length. Why does this happen? And how do we find these foci?

Let’s focus(!) on the last question first. Recall that an ellipse is a stretched circle. In other words, an ellipse is what forms when you slice a tall, circular tube (cylinder) along a slant:

Ellipses from Slicing Tubes
Left: An ellipse is formed by slicing a cylinder. Right: Fitting spheres above and below the cut locates the ellipse’s foci.

Take a sphere that snugly fits inside the tube, and drop it down until it touches the ellipse-slice at a single point F1. Do the same with a sphere underneath, touching the slice at F2. These points turn out to be the foci of the ellipse. Let’s see why.

We can use this tubular setup to answer one mystery from earlier: for any point X on the ellipse, the sum of distances of XF1 and XF2 is always the same! The proof lies in the following animation. Segments XF1 and XA have the same length because they’re both tangent to the upper sphere from X, and similarly, XF2=XB. So the sum XF1+XF2 is just the length of segment AB, the height between the spheres’ equators.

A Proof of Ellipse Path Lengths
The sum F1X+XF2 equals the length of AB and therefore does not change when X moves.

This has two neat consequences. First, it provides an elementary method for drawing ellipses (in real life!): all you need are two push pins and a loop of string, as illustrated below. The string ensures that the sum XF1+XF2 stays fixed while you trace the pen around, as long as you’re careful to keep the string taut throughout.

Drawing an Ellipse with String
How to draw an ellipse with push-pins and string

Second, what happens if we slice a cone instead of a cylinder? Perhaps surprisingly, we still get an ellipse! Indeed, as above, we can create a sphere on either side of the slice that snugly fits against the slice and the walls of the cone (the so-called Dandelin spheres), and exactly the same proof shows that XF1+XF2 stays constant as X moves around the edge.

Ellipses from Slicing Cones
Slicing a cone also produces an ellipse, by the same argument.

But wait, there’s still an unanswered question! We’ve seen that the path F1X+XF2 has a fixed length, but why does light bounce off an ellipse along such a path? This is what we really cared about for mini-golf! Come back next time for the answer, and in the meantime, have a great 2 weeks.

How Unilluminating!

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!


  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. []

Bucky’s Dozen

Buckyballs are remarkable structures, and not just to mathematicians. In chemistry, a buckyball—more correctly, a spherical fullerene—is a molecule of carbon atoms forming a hollow spherical “shell.” Many sizes and configurations are possible (and stable!), but due to Carbon’s chemical properties they always have a few common characteristics: the faces are rings of 5 or 6 carbon atoms, and each carbon atom bonds to 3 others. The most commonly occurring one, buckminsterfullerene, is made with 60 carbon atoms in the shape of a soccer ball:

The buckminsterfullerene molecule has sixty carbon atoms (green spheres) forming a “soccer ball” polyhedron.

These molecules were named after architect Richard Buckminster Fuller due to their resemblance to his famous geodesic domes. Fuller’s domes are also based on spheres built from hexagons and pentagons, and he discovered that these shapes naturally distribute forces and tensions evenly throughout the structure, thus enabling light, sturdy, and enormous constructions, such as the Climatron in St. Louis or the Montreal Biosphère.

Wait, where’s the math?

Mathematically, a fullerene is a convex polyhedron[1] where all faces are either hexagons or pentagons (these need not be regular), and 3 faces/edges meet at each vertex. The smallest of these mathematical fullerenes is the regular dodecahedron, which has no hexagons at all. The next smallest has two hexagons separated by twelve pentagons.

Smallest Fullerenes
Left: The smallest fullerene polyhedron is the dodecahedron, with 20 vertices and 12 pentagons. Right: The next smallest fullerene is the hexagonal truncated trapezohedron, having 24 vertices, 12 pentagons, and 2 hexagons.

Large and unwieldy fullerenes also exist. Some are highly symmetric with pentagons arranged at the corners of an icosahedron; others form long tubes of hexagons closed at the ends; still others display very little discernible structure whatsoever.

Larger Fullerene Examples
Left: A larger, icosahedrally-symmetric fullerene with 140 vertices and 72 faces. Middle: A tube-like fullerene with 66 vertices and 35 faces whose center is formed by a long spiral of hexagons. Right: a highly asymmetric fullerene with 44 vertices and 24 faces.

Despite their variety of appearances, they all have one thing in common:

All fullerenes have exactly 12 pentagons!

Why is this true?

One explanation comes from Euler’s beautiful formula that relates the number of vertices, edges, and faces in a polyhedron. This formula says, quite simply, that if there are v vertices, e edges, and f faces in any convex polyhedron (not necessarily a fullerene), then \(v-e+f=2\), always! For example, a cube has 8 vertices, 12 edges, and 6 faces, and indeed \(8-12+6=2\).

Let’s apply this to fullerenes. If a fullerene has \(p\) pentagons and \(h\) hexagons, then, since each edge is shared by two faces, there are \(e=\frac{5p+6h}{2}\) edges. Similarly, each vertex is shared by three faces, so there are \(v=\frac{5p+6h}{3}\) vertices. And since there are \(f=p+h\) faces, Euler’s formula tells us that

$$\frac{5p+6h}{3} – \frac{5p+6h}{2} + (p+h) = 2,$$

which indeed simplifies to \(p=12\).

So there are always 12 pentagons in a fullerene, but how many hexagons can be present? The two smallest examples tell us that 0 or 2 hexagons are possible, but 1 is not. Can we get 3 hexagons? (Yes. Try this!) What about 33 hexagons? (Harder, but yes. Try this too!) In fact, any number of hexagons is possible, with the one exception we’ve already mentioned. See if you can prove this![2] Furthermore, the number of possible fullerenes grows rapidly as the number of hexagons grows. For example, there are 1812 different fullerenes that, like the soccer ball, have exactly 20 hexagons or, equivalently, 60 vertices. With 50 hexagons (120 vertices), this number jumps to more than 52 billion different fullerenes![3]


  1. more correctly, the planar graph made by the vertices and edges of such a polyhedron []
  2. The “tube-like” fullerene from the last image provides a hint: the spiral of hexagons can have any desired length. []
  3. The exact number is 52628839448: OEIS Sequence A007894. []

Just Beatty It

[This is the 6th post in the current series about Wythoff’s game: see posts #1, #2, #3, #4, and #5. Caveat lector: this post is a bit more difficult than usual. Let me know what you think in the comments!]

Our only remaining task from last week was to prove the mysterious Covering Theorem: we must show that there is exactly one dot in each row and column of the grid (we already covered the diagonal case). Since the rows and columns are symmetric, let’s focus on columns.

The columns really only care about the x-coordinates of the points, so let’s draw just these x-coordinates on the number-line. We’ve drawn \(\phi,2\phi,3\phi,\ldots\) with small dots and \(\phi^2,2\phi^2,3\phi^2,\ldots\) with large dots. We need to show that there’s exactly one dot between 1 and 2, precisely one dot between 2 and 3, just one between 3 and 4, and so on down the line. For terminology’s sake, break the number line into length-1 intervals [1,2], [2,3], [3,4], etc., so we must show that each interval has one and only one dot:

Multiples of phi and phi^2 on a number line
Multiples of \(\phi\) (small blue dots) and of \(\phi^2\) (large green dots) perfectly interleave the integers on the number line. This is proved below.

Why is this true? One explanation hinges on a nice geometric observation: Take any small dot s and large dot t on our number-line above, and cut segment st into two parts in the ratio \(1:\phi\) (with s on the shorter side). Then the point where we cut is always an integer! For example, the upper-left segment in the diagram below has endpoints at \(s=2\cdot\phi\) and \(t=1\cdot\phi^2\), and its cutting point is the integer 3:

Splitting blue and green dots at integers
When a segment formed by a small and a large dot is cut into a \(1:\phi\) ratio (closer to the small dot), the cutting point is always an integer. In fact, the segment between \(j\cdot \phi\) and \(k\cdot \phi^2\) is cut at the integer j+k.

In general, if s is the jth small dot—i.e., \(s=j\cdot\phi\)—and \(t=k\cdot\phi^2\) is the kth large dot, then the cutting point between s and t is \(\frac{1}{\phi}\cdot s+\frac{1}{\phi^2}\cdot t = j+k\) (Why?![1]). But more importantly, this observation shows that no interval has two or more dots: a small dot and a large dot can’t be in the same interval because they always have an integer between them![2]

So all we have to do now is prove that no interval is empty: for each integer n, some dot lies in the interval [n,n+1]. We will prove this by contradiction. What happens if no dot hits this interval? Then the sequence \(\phi,2\phi,3\phi,\ldots\) jumps over the interval, i.e., for some j, the jth dot in the sequence is less than n but the (j+1)st is greater than n+1. Likewise, the sequence \(\phi^2,2\phi^2,3\phi^2,\ldots\) jumps over the interval: its kth dot is less than n while its (k+1)st dot is greater than n+1:

Showing no interval is empty
Illustrating the hypothetical situation where interval [n, n+1] contains no dot. This is used in a proof by contradiction to show that the interval in fact cannot be empty.

By our observation above on segment \(s=j\phi\) and \(t=k\phi^2\), we find that the integer j+k is less than n, so \(j+k\le n-1\). Similarly, \(j+k+2 > n+1\), so \(j+k+2 \ge n+2\). But together these inequalities say that \(n\le j+k\le n-1\), which is clearly absurd! This is the contradiction we were hoping for, so the interval [n,n+1] is in fact not empty. This completes our proof of the Covering Theorem and the Wythoff formula!

It was a long journey, but we’ve finally seen exactly why the Wythoff losing positions are arranged as they are. Thank you for following me through this!

A Few Words on the Column Covering Theorem

Using the floor function \(\lfloor x\rfloor\) that rounds x down to the nearest integer, we can restate the Column Covering Theorem in perhaps a more natural context. The sequence of integers $$\lfloor\phi\rfloor = 1, \lfloor 2\phi\rfloor = 3, \lfloor 3\phi\rfloor = 4, \lfloor 4\phi\rfloor = 6, \ldots$$ is called the Beatty sequence for the number \(\phi\), and similarly, $$\lfloor\phi^2\rfloor = 2, \lfloor 2\phi^2\rfloor = 5, \lfloor 3\phi^2\rfloor = 7, \lfloor 4\phi^2\rfloor = 8,\ldots$$ is the Beatty sequence for \(\phi^2\). Today we proved that these two sequence are complementary, i.e., together they contain each positive integer exactly once. We seemed to use very specific properties of the numbers \(\phi\) and \(\phi^2\), but in fact, a much more general theorem is true:

Beatty’s Theorem: If \(\alpha\) and \(\beta\) are any positive irrational numbers with \(\frac{1}{\alpha}+\frac{1}{\beta}=1\), then their Beatty sequences \(\lfloor\alpha\rfloor, \lfloor 2\alpha\rfloor, \lfloor 3\alpha\rfloor,\ldots\) and \(\lfloor\beta\rfloor, \lfloor 2\beta\rfloor, \lfloor 3\beta\rfloor,\ldots\) are complementary sequences.

Furthermore, our same argument—using \(\alpha\) and \(\beta\) instead of \(\phi\) and \(\phi^2\)—can be used to prove the more general Beatty’s Theorem!


  1. Hint: use the identity \(\frac{1}{\phi}+\frac{1}{\phi^2}=1\). []
  2. To be thorough, we should also check that no interval has two small or two large dots. Why can’t this happen? []

Putting the Why in Wythoff

Last week we drew dots at multiples of the vectors \(v=(\phi,\phi^2)\) and \(w=(\phi^2,\phi)\), and we colored grid-cells green or yellow according to whether they contain a dot or not. Our remaining task was to prove these three facts:

  • Endgame condition: The cell (0,0) is green.
  • Greens lose: From a green cell, there are no other green cells accessible with a single Wythoff’s game move.
  • Yellows win: From any yellow cell, it is possible to move to a green cell with one Wythoff’s game move.
Our guess at the solution to Wythoff's game
Our guess at the solution to Wythoff's game: multiples of vectors v and w (dotted) and their containing cells (dark green).

The “Endgame” condition is easy: the dot at \(0\cdot v = (0,0)\) makes cell (0,0) green. One condition down, two to go!

The “Greens lose” condition asks us to show that no two green cells are in the same row, column, or (slope 1) diagonal—otherwise, you could get from one to the other with a Wythoff move. In fact, we will prove something stronger:

Covering Theorem: Every row, column, and diagonal contains exactly one green cell.

Pictorially, this says that each thick line in the diagrams below will hit one and only one green cell:

Covering the columns and diagonals
All columns (left), diagonals (right), and rows (not pictured) contain exactly one green cell. (Some of these green cells are beyond the image boundaries.) Because the green cell locations are symmetric through the line y=x, the diagram for the rows is redundant and therefore omitted.

It is not difficult to describe why each diagonal is covered, i.e. has exactly one green cell. In the diagram on the right, the central diagonal line has equation \(y-x=0\); the one above it has equation \(y-x=1\); the next is \(y-x=2\); and so on. And because \(\phi^2-\phi=1\), the vector v jumps from one diagonal line to the next at each step. Vector w similarly jumps from line to line in the other direction.

By contrast, I find the fact that each column is covered quite surprising. It says that the upper and lower branches of the “V” perfectly complement each other, each filling exactly the columns that the other skips! Before we prove this fact, let’s see why it helps us.

If we assume the covering theorem for now, we can finish the three conditions needed for our proof. We already saw that (0,0) is green, and the covering theorem certainly tells us that each row, column, and diagonal has at most one green cell, so the “Greens lose” condition holds. We just have to prove the “Yellows win” property: that you can always move to a green cell from any yellow cell. The idea is simple: if your yellow cell is above the “V”[1] then move down to the unique green cell in your column (which may be on either the upper or lower “V” branch). If you are below the “V”, move left to the green cell in your row. Finally, if you are inside the “V”, move diagonally. In this way, you can always move to a green cell with one Wythoff move.

Proving the "Yellows win" condition
All yellow cells can reach a green cell with one Wythoff move. Yellow cells above the "V" can move down to a green cell; yellow cells below the "V" can move left to a green cell; yellow cells inside the "V" can move diagonally to a green cell.

With this third and last property verified, we’re finally done with our proof! …except that we still need to prove the mysterious covering theorem. We’ll discuss this next week. Come back then for the exciting QED!


  1. i.e., its lower-left corner is above the line \(y=\phi\cdot x\) []

A Golden Observation

Last week we saw that the best strategy for Wythoff’s game is to always move to one of the blue squares in the diagram below (if you can). We found the locations of the blue squares with an iterative algorithm that builds outward from (0,0), but we haven’t yet discovered the underlying pattern in how these blue cells are arranged. Let’s go pattern hunting!

As we noticed last week, the blue cells are arranged in a symmetric “V” shape, formed from two seemingly straight lines. What are the slopes of these lines? This is the question du jour.

Jumps in losing positions for Wythoff's game
Jumps between blue cells for Wythoff's game come in just two sizes: a=(1,2) and b=(2,3).

Looking more carefully, the upper branch of the “V” has only two types of “gaps”: consecutive blue cells are separated by either a “short” (1,2) step (a.k.a. a knight’s move) or a “long” (2,3) step. So, if we believe that this “V” branch is a perfect line, then that line’s slope should be somewhere between the slopes of these two steps, namely \(3/2=1.5\) and \(2/1=2\). Knowing more about how these steps are arranged will tell us more about the “V”.

Specifically, label these “short” and “long” steps as a and b respectively. What we care about is the ratio of bs to as. Why? For example, if there were an equal number of as and bs on average, then the slope of the line would be the same as the vector \(a+b=(3,5)\), i.e., the slope would be 5/3. If there were, say, two bs for each a in the long run, then the line would fall in the direction of \(a+2b=(5,8)\), with slope 8/5. Unfortunately, the ratio we seek is neither 1 nor 2; what is this magic number?

Let’s write down the sequence of jumps along our line. From the diagram above we see that it starts a b a b b a …, and using last week’s iterative algorithm, we can compute farther into this infinite sequence:

a b a b b a b a b b a b b a b a b b a b a b b a b b a b a b b a b b …

This sequence seems to be composed of “blocks” of (a b) and (a b b):

(a b)(a b b)(a b)(a b b)(a b b)(a b)(a b b)(a b)(a b b)(a b b)(a b)(a b b)(a b b)…

So for every a there are either one or two bs, meaning the ratio of bs to as is somewhere between 1 and 2. So the slope should be between 5/3 and 8/5. But what is it exactly? To answer this, we need to know how these “short” and “long” blocks are arranged.

Write A = (a b) and B = (a b b), so we can rewrite the sequence of blocks as

A B A B B A B A B B A B B …

Hold on; this sequence looks familiar. It’s exactly the same pattern as our original jump sequence! So it seems that the pattern of blocks is exactly the same as the pattern of the letters themselves! This is a “self-generating” sequence!

In particular, if the ratio of bs to as is some number r, then the ratio of (a b b) blocks to (a b) blocks is also r. But if we have one block of (a b) and r blocks of (a b b) then we have a total of \(1+2r\) bs and \(1+r\) as, so the ratio of bs to as is \(\frac{1+2r}{1+r}=r\). This simplifies to \(1+r=r^2\), and the solution is the shiny number \(r=\frac{1+\sqrt{5}}{2}\), known as the golden ratio and denoted \(\phi\) (Greek letter phi).

Now we know that our purported line should be in the direction of \(a+b\cdot\phi = (1+2\phi,2+3\phi)\), so the slope is \((2+3\phi)/(1+2\phi)=\phi\). And since our “V” shape is symmetric, the other line should have slope \(1/\phi = \frac{\sqrt{5}-1}{2}\). Done and done!

Well, yes and no. We found the slopes of the lines in the “V”, but why do they form lines at all? And what does all this have to do with the Fibonacci numbers?

Many Morley Triangles

Given my blog title, how have I gone this long without discussing triangle geometry? I will rectify this gross negligence in the next few weeks. Let’s begin with a seemingly impossible fact due to Frank Morley.

Start with a triangle ABC, with any shape you wish. Now cut its angles into three equal parts, and extend these angle trisectors until they meet in pairs as illustrated below, forming triangle PQR. Morley’s amazing theorem says that this Morley triangle, PQR, will always be equilateral!

Morley's theorem

Why would this be true? Well, Morley’s theorem tells us that this diagram has three nice 60-degree angles in the middle, but we may suspect that, in fact, all of the angles are nice! This key insight lets us piece together the following argument, where we build up the diagram backwards from its constituent pieces. Draw seven separate “puzzle piece” triangles with angles and side-lengths as shown below, where the original triangle ABC has angles \(3\alpha,3\beta,3\gamma\) respectively. (Be sure to check that puzzle pieces with these specifications actually exist! Hint: use the Law of Sines.)

Proof of Morley's theorem
Triangle ABC is built out of seven "puzzle pieces" with angles and side-lengths as specified here. In this diagram, \(\alpha^\prime=\alpha+60\) and \(\alpha^{\prime\prime}=\alpha+120\), and similarly for \(\beta\) and \(\gamma\).

Now, fit the pieces together: all matching edge-lengths are equal (by design), and the angles around vertex P add up to \(60+(\gamma+60)+(\alpha+120)+(\beta+60)=360\) and similarly for Q and R, so the puzzle fits together into a triangle similar to our original triangle ABC. But now these pieces must make up the Morley configuration, and since we started with an equilateral in the middle, we’re done with the proof![1]

But there’s more to the Morley story. Let’s push A and C to make them swap positions, dragging the trisectors and Morley triangle along for the ride (a process known as extraversion). We end up using some external angle trisectors instead of only internal ones, but the Morley triangle remains equilateral throughout. This gives us a new equilateral Morley triangle for our original ABC!

New Morley triangles through extraversion

In fact, each vertex of ABC has six angle trisectors (three pairs of two), and if you continue applying extraversions you’ll soon uncover that our original triangle ABC has 18 equilateral Morley triangles arranged in a stunning configuration! (Click for larger image.)

All 18 Morley triangles

Here’s a challenge: the diagram above has 27 equilateral triangles, but only 18 of them are Morley triangles (i.e., are made from a pair of trisectors from each vertex). Which ones are they?


  1. This is a slight modification of a proof by Conway and Doyle. []

Logic Under Construction

In last week’s discussion of proofs by contradiction and nonconstructive proofs, we showed:

Theorem: There exist irrational numbers \(x\) and \(y\) with the property that \(x^y\) is rational.

However, our proof was nonconstructive: it did not pinpoint explicit values for \(x\) and \(y\) that satisfy the condition, instead proving only that such numbers must exist. Would a more constructive proof be more satisfying? Let’s see! I claim \(x=\sqrt{2}\) and \(y=\log_2 9\) work, because \(\sqrt{2}\) we already know to be irrational, \(y=\log_2 9\) can be similarly proved to be irrational (try this!), and $$x^y = \sqrt{2}^{\log_2 9} = \sqrt{2}^{\log_{\sqrt{2}}3}=3,$$ which is rational.

Let’s further discuss why last week’s proof was less satisfying. The following rephrasing of this proof may help shed some light on the situation:

Proof: Assume the theorem were false, so that any time \(x\) and \(y\) were irrational, \(x^y\) would also be irrational. This would imply that \(\sqrt{2}^{\sqrt{2}}\) would be irrational, and by applying our assumption again, \(\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}\) would also be irrational. But this last number equals 2, which is rational. This contradiction disproves our assumption and thereby proves the theorem, QED.

So perhaps this argument seems less satisfactory simply because it is, at its core, a proof by contradiction. It does not give us evidence for the positive statement “\(x\) and \(y\) exist”, but instead only for the negative statement “\(x\) and \(y\) don’t not exist.” (Note the double negative.) This distinction is subtle, but a similar phenomenon can be found in the English language: the double negative “not bad” does not mean “good” but instead occupies a hazy middle-ground between the two extremes. And even though we don’t usually think of such a middle-ground existing between logic’s “true” and “false”, proofs by contradiction fit naturally into this haze. In fact, these ideas motivate a whole branch of mathematical logic called Constructive logic that disallows double negatives and proofs by contradiction, instead requiring concrete, constructive justifications for all statements.

But wait; last week’s proof that \(\sqrt{2}\) is irrational used contradiction, and therefore is not acceptable in constructive logic. Can we prove this statement constructively? We must show that \(\sqrt{2}\) is not equal to any rational number; what does it even mean to do this constructively? First, we turn it into a positive statement: we must show that \(\sqrt{2}\) is unequal to every rational number. And how do we constructively prove that two numbers are unequal? By showing that they are measurably far apart. So, here is a sketch of a constructive proof: \(\sqrt{2}\) is unequal to every rational number \(a/b\) because $$\left|\sqrt{2} – \frac{a}{b}\right| \ge \frac{1}{3b^2}.$$ See if you can verify this inequality![1]

PS. In case you are still wondering whether \(\sqrt{2}^{\sqrt{2}}\) is rational or irrational: It is irrational (moreover, transcendental), but the only proof that I know uses a very difficult theorem of Gelfond and Schneider.


  1. This inequality would also have to be proven in a constructive manner. See these Wikipedia articles for more information: Intuitionistic logic (another name for Constructive logic) and Square root of 2: Constructive proof. []

Methods of Irrationality

Being a mathematician requires you to think in strange ways.

For starters, you might think that mathematicians spend all day pondering purely theoretical things that only exist in Mathematical abstraction, right? Well, it can get even stranger than that: sometimes we have to think about things that don’t exist, even in the abstraction! It’s called a proof by contradiction, and here’s an example:

Theorem: The number \(\sqrt{2}\) is irrational.

Proof: To prove that \(\sqrt{2}\) is irrational, we have to show that we can never write it as a ratio of integers, \(\sqrt{2}=a/b\). So let’s assume we can find such a fraction and see where it leads us.

Let’s cancel common factors in the numerator and denominator so that \(a/b\) is in lowest terms. Squaring our equation shows that \(a^2=2b^2\), so \(a^2\) is even and so \(a\) itself is even. Since \(a/b\) is in lowest terms, \(b\) must be odd.

Since \(a\) is even, \(a^2\) is in fact divisible by 4. On the other hand, since \(b\) is odd, \(2b^2\) is not divisible by 4. But then the integer \(a^2=2b^2\) is both divisible by 4 and not divisible by 4, which is absurd! The only explanation for this contradiction is that our original assumption—that \(\sqrt{2}\) is rational—is false. So we’re done with the proof! Notice that we spent most of our effort reasoning about integers that never existed (specifically \(a\) and \(b\)).

But that’s not the only twisted thing about Mathematical thinking. Here’s a delightfully short yet aggravatingly unsatisfying proof:

Theorem: There exist irrational numbers \(x\) and \(y\) such that \(x^y\) is rational.

Proof: We already know that \(\sqrt{2}\) is irrational, so maybe we can use \(\sqrt{2}^{\sqrt{2}}\). Is this number rational? If it is, then we’re done: \(x=\sqrt{2},y=\sqrt{2}\) solves the problem. But what if \(\sqrt{2}^{\sqrt{2}}\) is irrational? In this case, I claim \(x=\sqrt{2}^{\sqrt{2}},y=\sqrt{2}\) works: indeed, $$\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}} = \sqrt{2}^{\sqrt{2} \cdot\sqrt{2}} = \sqrt{2}^2 = 2,$$ which is rational. In either case the required numbers \(x\) and \(y\) exist, so this completes the proof.

But wait; which is it? The proof shows us that one of the pairs \(x=\sqrt{2},y=\sqrt{2}\) or \(x=\sqrt{2}^{\sqrt{2}},y=\sqrt{2}\) works, but it doesn’t tell us which one! So, have we proven the theorem? Yes, technically, but only nonconstructively.

Frustrating, isn’t it?