## Four hexagons

I had to stare at this puzzle a good long time before I got it:

I had a very hard time figuring out where to start. What finally broke it for me was getting rid of unnecessary complications. For instance, in both the white and the yellow hexagons, three of the vertices don’t enter into the problem. You can replace those hexagons with triangles.

Likewise only two vertices of each of the other two hexagons are important; you can replace them with line segments.

Once you’ve done that the solution becomes obvious, in the sense of “I only had to stare at it another five minutes before saying ‘oh duhh'”. These two line segments are congruent.

As are these two.

These two angles are congruent, since both differ from 60° by the same amount.

So these triangles are congruent.

Now put back the original hexagons.

A side of one hexagon is congruent to a diameter of the other, so it’s twice a side of the other, so the areas of the two hexagons are in a 4:1 ratio. The “pink” (or I’d call it magenta) hexagon’s area is 48.

## Identity 2

This is prompted by another Catriona Shearer puzzle. Would you say it’s at all obvious that

$2 \arctan (\sqrt{3}/7) = \arccos (23/26)$ ?

Me neither. True, though. Look at this triangle:

By the law of cosines, $3 = 13 + 13 - 2\times13\cos(\theta)$ so $\theta = \arccos (23/26)$. But if you slice that triangle in half, you find its height is $7/2$ and so $\theta = 2\arctan(\sqrt{3}/7)$.

More generally, the half angle formula for the tangent is

$\tan(\theta/2) = \frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}$

implying

$\cos\theta = \frac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}$.

Then if $\theta/2 = \arctan(a/b)$,

$2\arctan\left(\frac{a}{b}\right) = \arccos\left(\frac{b^2-a^2}{b^2+a^2}\right)$

and with $a=\sqrt{3}$, $b=7$ you get $2\arctan(\sqrt{3}/7) = \arccos(46/52) = \arccos(23/26)$.

## Identity

While struggling with a Catronia Shearer problem I stumbled across a proof of the trig identity

$\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)$

which I hadn’t seen and probably wouldn’t have thought of if I’d been looking for it, though there’s nothing non-obvious about it in retrospect. Here it is. Draw a unit semicircle centered on O, with A and C the endpoints of the diameter. Draw chords AD and DC. (Angle ADC is a right angle.) Let angle CAD be $\theta$. By the Inscribed Angle Theorem, angle COD is $2\theta$. Drop a perpendicular from D to the diameter at B.

Now $\mathrm {AO} = 1$ and $\mathrm {OB} = \cos 2\theta$, so $\mathrm {AB} = 1+\cos 2\theta$. But $\mathrm {AC} = 2$ so $\mathrm {AD} = 2 \cos \theta$, so $\mathrm {AB} = 2 \cos^2 \theta$. Therefore $\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)$.

The Lego architecture photo blog Archbrick Daily recently featured this Micropolis City Swim Center by Little Brick Root:

Unusually for a Lego model (or for a real world building, for that matter) the structure is at an angle relative to the base. Of course you can do that sort of thing by covering the base with flat tiles, except for one single stud:

Then the structure can be at any angle, rotating around that stud. For that matter you could cover the entire base with flat tiles and just put the structure on top in any position or orientation, but that’s rather susceptible to jostling, isn’t it? So is the single stud connection, though less so.

(There also are Lego pieces with built in angles, such as this one:

But let’s just consider normal rectilinear bricks and plates for now.)

But maybe you want a rigid connection. If you happen to want an angle of, oh, I don’t know, let’s say 36.87°, you can do that, Make a base with two studs three rows and four columns apart:

The distance in stud units between the two studs is $\sqrt{3^2+4^2}=5$, an integer, so for instance a 1 by 6 brick can be placed on the two studs. If you want other angles you can look for other primitive Pythagorean triples to use. (Primitive meaning it’s not just another triple scaled up: (6, 8, 10), for instance, is a triple, but it’s just twice (3, 4, 5), and makes the same angle.)

The angles you get have rational values for their sine, cosine, and tangent. If you want 45°, you’re out of luck; sin(45°) is irrational. You can get pretty close, though, with for instance the triple (20, 21, 29). That gives an angle of 43.60°.

But if you want a small angle, like the Swim Center model, what do you do? Another way to think about it is, you want a triangle where the long side is nearly as long as the hypotenuse. With Pythagorean triples, the best you can get is a hypotenuse 1 unit longer than the long side.

A general formula for triples is $a=v^2-u^2, b=2uv, c=v^2+u^2$ where $u$ and $v>u$ are coprime and opposite parity (one even, one odd). Here obviously the smallest difference between $a$ and $c$ is 2, so we want $c-b = 1 = v^2+u^2-2uv = (v-u)^2$. So $v=u+1$ and $a=2u+1, b=2u^2+2u, c=2u^2+2u+1$. The first several of these are:

 u a b c angle 1 3 4 5 36.87° 2 5 12 13 22.62° 3 7 24 25 16.26° 4 9 40 41 12.68°

You can see you have to make a fairly large model if you want a rigid angle of less than 10° — more than 40 studs wide, if you use conventional plates and bricks. There are “jumper” plates, though, that give access to half integer spacing, meaning you’d need only half the width.

You can get smaller angles by stacking triples. For instance, you can mount a plate at 36.87° to the base using (3, 4, 5), then mount a plate to that at -16.26° using (7, 24, 25), then mount a plate on top of that also at -16.26° giving a net angle of 4.35°. The stacked plates can hide inside the structure so all you see is the base and the structure rotated 4.35°.

The ground floor of the Swim Center is 24 studs wide, meaning the largest distance between stud receivers is 46 half studs, accommodating no triples with angles smaller than 12.68°. The actual angle of the model is about 5°. So maybe Little Brick Root used angled Lego pieces (I don’t know of any with that small an angle, but maybe) or stacked triples, or maybe (mostly likely is my guess) only a single stud connection to the base. But not a rigid connection based on a single triple, that’s for sure.

## Euler plays Ingress

In the game Ingress you capture portals (by visiting real world locations), and then you can link pairs of portals together. You get points for doing that. If three portals are mutually linked with three links, forming a triangle, then they make a control field, and that’s a good thing if only because you get a lot of points for doing that. So if you’re trying to level up, one way is to find a lot of portals close together and circulate between them, capturing them and linking them.

One rule of linking, though, is that links can’t cross. If you want to link two portals, and a straight line segment between them intersects some other link, then tough luck: you can’t link them.

(Technically, if the Earth is regarded as a sphere, presumably links are great circle segments and control fields are spherical triangles. But for our purposes we pretend we’re working in a plane. Links are line segments, control fields are Euclidean plane triangles.)

So a question you might ask is, for some collection of portals, is there an optimal way to link them? Can you make more links or more control fields if you do it one way than if you do it another?

Answer: No. If you start creating links, and keep adding links until no more links are possible, then regardless of how you decide which links to make, you always get the same number of links, and the same number of control fields.

And how many is that?

You get the answer using the Euler characteristic. For any convex polyhedron,

$F - E + V = 2$

where $F$, $E$, and $V$ are respectively the number of faces, edges, and vertices the polyhedron has. A cube, for instance, has 6 faces, 12 edges, and 8 vertices:

$6 - 12 + 8 = 2$ .

What do polyhedra have to do with portals? Well, the same idea applies to any connected planar graph (a graph with non crossing edges), which can be thought of as the shadow of a polyhedron, provided you realize every region of the plane bounded by edges is a face, including the region extending to infinity. So this graph

also has 6 faces, 12 edges, and 8 vertices, if you count as faces the central quadrilateral, the four quadrilaterals surrounding it, and the infinite region surrounding them.

Viewed as linked Ingress portals, though, it’s not maxed out; in fact, there are no control fields because none of the polygons are triangles. So you can add more links until you have, for instance

and now there are 11 faces, 17 edges, and 8 vertices; $11-17+8=2$.

But to maximize links and control fields, all the polygons in the graph, except the external one, must be triangles. Each of the $F-1$ triangular faces has 3 edges. But $E\ne 3(F-1)$, because where two triangles adjoin, a single edge is one of the bounding edges of both triangles. It serves as two edges, in a way. On the other hand, edges of triangles along the boundary are shared with the external polyhedron. So if we regard the external polyhedron as contributing zero edges then each internal triangle (not along the boundary) contributes $3/2$ edges, and each boundary triangle contributes $2$ edges. Or

$E = \frac{3}{2}N + \frac{1}{2}N_b$

where $N=F-1$ is the number of triangles and $N_b$ is the number of boundary triangles, which is the same as the number of edges or vertices in the bounding polygon (which is the convex hull of all the vertices). Solving for $N$,

$N = \frac{2}{3}(E-\frac{N_b}{2})$

Then combining with the Euler characteristic,

$\frac{2}{3}(E-\frac{N_b}{2})+1-E+V = 2$

or

$E=3V-N_b-3$

In the above diagram, $V = 8$ and $N_b$ = 4 so $E = 24-4-3 = 17$, which is what we see. This tells us no matter how we link portals, once we’ve made all possible links, the number of links we end up with is the same. And the number of control fields, $N$, is

$N = \frac{2}{3}(3V-N_b-3-\frac{N_b}{2}) = 2V-N_b-2$

And again in the above diagram, $N = 2\times 8-4-2 = 10$ which is correct. No matter how we link a given set of portals, we end up with a fixed number of control fields. All that matters is the number of portals, and the number in the convex hull.

## Sum ideas (part 4)

So how do you sum — or, if you prefer, associate a value with — $1+1+1+...$?

One way is zeta function regularization. For this we start with the sum

$\displaystyle S(s) = \sum_{k=1}^{\infty}k^{-s}$.

For example, $S(2) = 1+1/4+1/9+1/16+...$. This series converges to the limit $\pi^2/6$. In fact $S(s)$ converges for any complex $s$ where the real part $\Re(s) > 1$.

For $s=1$, $S(1) = 1+1/2+1/3+1/4...$ and this diverges. If you approach $s=1$ along the real axis you find $S(s)$ increases without limit. Off the real axis, things are a little different. At $s=1+i$, for example, the sum fails to converge, but as you approach t $s=1+i$ from the right, $S(s)$ approaches $0.58215 + 0.92684i$. Similar behavior is found elsewhere on the $\Re(s) = 1$ line, other than $s=1$.

That suggests there might be a way to construct a function that is equal to $S(s)$ for $\Re(s)>1$ but which has well defined values elsewhere, except $s=1$. And indeed there is: analytic continuation.

Imagine I give you the following function: $f(x) = x$ for real $x \in [0,1]$. Outside that interval $f(x)$ is undefined. But you obviously could define another function $g(x) = x$ which is defined on the whole real number line and has the property that $g(x) = f(x)$ in the range where $f(x)$ is defined. Obviously $g(x)$ is continuous, and is differentiable everywhere.

On the other hand, you could instead define $g(x)$ as being quadratics grafted onto the line from $(0,0)$ to $(1,1)$:

$g(x) = x^2+x \quad x < 0$
$= x \quad 0 \le x \le 1$
$= x^2-x+1 \quad x > 1$

which has the same properties. Or you could use cubics, or quartics, or, well, anything provided it has the right value and derivative at $0$ and $1$. There’s an infinite number of ways to continue $f(x)$ to the entire real line.

In the complex plane you can do something similar. I give you a function $f(z)$ defined for $z$ within some region of the complex plane. $f(z)$ is analytic, that is, it has a complex derivative everywhere it’s defined. Then you can give me an analytic function $g(z)$ defined everywhere in the complex plane and equal to $f(z)$ everywhere $f(z)$ is defined. (I’m being sloppy and informal here; there could be poles where neither function is defined, for example.)

Here’s the thing, though: Unlike $g(x)$ on the real line, $g(z)$ is unique. There is exactly one analytic function that continues my analytic function to the entire complex plane.

So, getting back to our sum (which is analytic), we can define an analytic function $\zeta(s) = S(s)$ for $\Re(s) > 1$, whose behavior for $\Re(s) \le 1$ is given by analytic continuation. One can show

$\displaystyle \zeta(s) = \frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{e^x-1} dx$

where $\Gamma(s) = \int_0^\infty x^{s-1}e^{-x} dx$ is the usual gamma function. $\zeta(s)$ has a pole at $s=1$ but is well defined everywhere else. $\zeta(s)$ is known as the Riemann zeta function.

Now, we know $\zeta(s)$ is the value of $\sum_{k=1}^{\infty}k^{-s}$ wherever that sum converges. Zeta regularization just assigns the value of $\zeta(s)$ to that sum where it does not converge as well. For instance, when $s=0$, we have $1+1+1+1+... = \zeta(0)$, and $\zeta(0) = -1/2$.

The somewhat notorious sum of the positive integers, $1+2+3+4+...$, is $S(-1)$, to which is assigned the value $\zeta(-1) = -1/12$. If you want to start an argument on the Internet, claiming that $1+2+3+4+... = -1/12$ is a good way to do it. Of course that claim glosses over a lot.

It turns out the negative even integers are (“trivial”) zeros of the zeta function, so $\sum_{k=1}^{\infty}k^{2n} = 0$ by this summation method. Generally, for integer exponents,

$\displaystyle \sum_{k=1}^{\infty}k^{n} = \zeta(-n) = (-1)^n \frac{B_{n+1}}{n+1}$,

where $B_n$ is the nth Bernoulli number,

$\displaystyle B_n = \sum_{k=0}^n\sum_{v=0}^k (-1)^v\binom{k}{v}\frac{v^m}{k+1}$.

So

$\displaystyle \sum_{k=1}^{\infty}k^{1} = -\frac{1}{12}$
$\displaystyle \sum_{k=1}^{\infty}k^{3} = \frac{1}{120}$
$\displaystyle \sum_{k=1}^{\infty}k^{5} = -\frac{1}{252}$
$\displaystyle \sum_{k=1}^{\infty}k^{7} = \frac{1}{240}$
$\displaystyle \sum_{k=1}^{\infty}k^{9} = -\frac{1}{132}$
$\displaystyle \sum_{k=1}^{\infty}k^{11} = \frac{691}{32760}$

and on from there.

## Sum ideas (part 3)

Another summation method is Abel summation. The idea is, if we have

$\displaystyle S = \sum_{k=0}^{\infty} a_k$

then we can define

$\displaystyle f(r) = \sum_{k=0}^{\infty} a_kr^{k}$

for $r \in [0, 1)$. Then if $\lim_{r\to 1^-} f(r)$ exists and is finite, that limit is the Abel sum of $S$.

As a simple example, apply this to Grandi’s series, $1 - 1 + 1 - 1 ...$. Here $f(r) = 1 - r + r^2 - r^3 ... = 1/(1+r)$ for $|r|<1$. The limit as $r\to 1$ is $1/2$, the same result as we obtained using Cesàro summation. In fact it can be shown that Abel summation is stronger than Cesàro summation, i.e., for series that can be Cesàro summed, Abel summation give the same result, but there are additional series which can be Abel summed but not Cesàro summed. Of course Cesàro summation is consistent with ordinary summation for convergent series, and therefore so is Abel summation: that is, Abel summation is regular.

Here’s another example. Consider

$\displaystyle 1-2+3-4+... = \sum_{k=0}^{\infty} (-1)^k(k+1)$.

Not only does this series not converge, but the partial sum averages don’t converge either, so it is not Cesàro summable. But it is Abel summable. We make this sum into a function

$\displaystyle f(r) = \sum_{k=0}^{\infty} (-1)^k(k+1)r^k = r^0-2r^1+3r^2-4r^5...$

But now notice: $r^0 = -d(-r)^1/dr$, $-2r^1 = -d(-r)^2/dr$, $3r^2 = -d(-r)^3/dr$, and so on:

$\displaystyle f(r) = -\frac{d}{dr}\sum_{k=0}^{\infty} (-r)^k$

and, again, for $|r|<1$,

$\displaystyle f(r) = -\frac{d}{dr}\frac{1}{1+r} = \frac{1}{(1+r)^2}.$

Now the Abel sum is $\lim_{r\to1^-} f(r) = 1/4$.

A couple more properties (besides regularity) a summation method might have are linearity and stability. For the following let $\mathbf {A}(r)$ denote the result of applying summation method $\mathbf {A}$ to series $r$. By linearity is meant: if $\mathbf {A}(\sum a_n) = s$ and $\mathbf {A}(\sum b_n) = t$ then $\mathbf {A}(\sum ka_n+b_n) = ks+t$. By stability is meant: if $\mathbf {A}(a_0+a_1+a_2+...) = s$ then $\mathbf {A}(a_1+a_2+...) = s-a_0$, and conversely. Cesàro summation and Abel summation both are linear and stable. So is classical summation, for that matter.

You can prove that for any linear and stable summation method $\mathbf{A}$, the sum of the Grandi series $\mathbf{A}(g)$ is $1/2$, if that sum exists:

$\mathbf{A}(g) = \mathbf{A}(1-1+1-1+1...) = 1 + \mathbf{A}(-1+1-1+1...)$ (by stability)
$= 1 - \mathbf{A}(1-1+1-1...)$ (by linearity)
$= 1 - \mathbf{A}(g)$ and so
$\mathbf{A}(g) = 1/2$.

That “if that sum exists” provision is important. For instance, classical summation of the Grandi series is undefined, not $1/2$, even though classical summation is linear and stable. You can come up with similar proofs about linear and stable sums of other series, that they must always have some particular value if they have a value at all. Showing that they do indeed have a value is another matter!

Conversely, you can prove some series do not have values for any summation method that is linear and/or stable. For example, suppose $\mathbf{A}$ is stable and $\mathbf{A}(\sum_{k=0}^\infty 1)$ has value $a$. Then

$a = \mathbf{A}(1+1+1+1+1...) = 1 + \mathbf{A}(1+1+1+1...)$ (by stability)
$a = 1 + a$,

an impossibility. So $\sum_{k=0}^\infty 1$ cannot be summed by any stable summation method. There are unstable methods, however, that can sum that series.