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.

]]>?

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

By the law of cosines, so . But if you slice that triangle in half, you find its height is and so .

More generally, the half angle formula for the tangent is

implying

.

Then if ,

and with , you get .

]]>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 . By the Inscribed Angle Theorem, angle COD is . Drop a perpendicular from D to the diameter at B.

Now and , so . But so , so . Therefore .

]]>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 , 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 where and are coprime and opposite parity (one even, one odd). Here obviously the smallest difference between and is 2, so we want . So and . 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.

]]>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,

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

.

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; .

But to maximize links and control fields, all the polygons in the graph, except the external one, must be triangles. Each of the triangular faces has 3 edges. But , 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 edges, and each boundary triangle contributes edges. Or

where is the number of triangles and 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 ,

Then combining with the Euler characteristic,

or

In the above diagram, and = 4 so , 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, , is

And again in the above diagram, 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.

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

.

For example, . This series converges to the limit . In fact converges for any complex where the real part .

For , and this diverges. If you approach along the real axis you find increases without limit. Off the real axis, things are a little different. At , for example, the sum fails to converge, but as you approach t from the right, approaches . Similar behavior is found elsewhere on the line, other than .

That suggests there might be a way to construct a function that is equal to for but which has well defined values elsewhere, except . And indeed there is: analytic continuation.

Imagine I give you the following function: for real . Outside that interval is undefined. But you obviously could define another function which is defined on the whole real number line and has the property that in the range where is defined. Obviously is continuous, and is differentiable everywhere.

On the other hand, you could instead define as being quadratics grafted onto the line from to :

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

In the complex plane you can do something similar. I give you a function defined for within some region of the complex plane. is analytic, that is, it has a complex derivative everywhere it’s defined. Then you can give me an analytic function defined everywhere in the complex plane and equal to everywhere 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 on the real line, 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 for , whose behavior for is given by analytic continuation. One can show

where is the usual gamma function. has a pole at but is well defined everywhere else. is known as the Riemann zeta function.

Now, we know is the value of wherever that sum converges. Zeta regularization just assigns the value of to that sum where it does not converge as well. For instance, when , we have , and .

The somewhat notorious sum of the positive integers, , is , to which is assigned the value . If you want to start an argument on the Internet, claiming that 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 by this summation method. Generally, for integer exponents,

,

where is the *n*th Bernoulli number,

.

So

and on from there.

]]>then we can define

for . Then if exists and is finite, that limit is the Abel sum of .

As a simple example, apply this to Grandi’s series, . Here for . The limit as is , 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

.

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

But now notice: , , , and so on:

and, again, for ,

Now the Abel sum is .

A couple more properties (besides regularity) a summation method might have are **linearity** and **stability**. For the following let denote the result of applying summation method to series . By linearity is meant: if and then . By stability is meant: if then , 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 , the sum of the Grandi series is , *if that sum exists*:

(by stability)

(by linearity)

and so

.

That “if that sum exists” provision is important. For instance, classical summation of the Grandi series is undefined, not , 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 is stable and has value . Then

(by stability)

,

an impossibility. So cannot be summed by any stable summation method. There are unstable methods, however, that can sum that series.

]]>A divergent series has no limit, so we can’t assign that as a value. Conventionally we just say it has no value, its value is undefined. But in fact some divergent series can be associated with a definite quantity by means other than the limit of the sequence of partial sums. For some purposes, it can be useful to regard that quantity as the value, or *a* value, of the series.

There are lots of ways to associate a value with a divergent series; lots of summation methods, as they are called. It’s unfortunate terminology, in that it suggests they’re methods for finding the unfindable sum of the infinite number of numbers. But it’s the terminology we’re stuck with.

Summation methods are kind of like technical standards: the great thing about them is there’s so many to choose from. Generally a given summation method can be used with some series, but not with others. Some methods are stronger than others, in the sense that the one can be applied to any series the other can, with the same result, but it can also be applied to some series the other can’t handle.

Perhaps the simplest summation method applicable to a divergent series is Cesàro summation. In its simplest form, this is just finding the limit not of the partial sums of a series, but the average of the partial sums. For example, Grandi’s series is

.

The first partial sum is , the second is , the third is , the fourth is , and so on — they alternate between and . They don’t converge. But the average of the first one partial sum is , the average of the first two is , the average of the first three is , the average of the first four is , and so on, forming the sequence

and that sequence *does* converge, to . This value is the Cesàro summation of Grandi’s series.

Now, we know if we add together a finite number of integers, we get an integer, and it seems crazy to think you could sit down and add an infinite number of integers and get a fraction. Then again, it’s crazy to think you could sit down and add an infinite number of integers. And that’s not what we’re doing. But we know the partial sums alternate between and , so the value halfway between, , in some sense does characterize the behavior of the infinite series.

A reasonable question to ask is, what’s the Cesàro summation of a *convergent* series? It doesn’t take too much thinking to realize intuitively that if a series converges, then the average of the partial sums also should converge and to the same value. For instance, the partial sums of

converge to , and so do the averages of the partial sums. Granted, the partial sums converge much faster: after just 12 terms the partial sum is while after 10000 terms the average of the partial sums is still . But it’s getting there. A summation method that gives the conventional limit when applied to a convergent series is called **regular**. Cesàro summation is regular, and clearly that’s a nice attribute to have: it means Cesàro summation is consistent with ordinary summation, but is stronger in the sense that it also gives results for some series which have no classical value.

But does it make sense to say

?

It’s an unconventional and probably misleading use of the equal sign, but if it’s understood you’re talking about a value assigned using a summation method, specifically Cesàro summation, you maybe can get away with it. But you can also make the sense more explicit. Hardy again:

We shall make systematic use of the following notations. If we define the sum of , in some new sense, say the ‘Pickwickian’ sense, as , we shall say that is *summable* (P), call the P *sum* of , and write

(P).

We shall also say that is the P *limit* of the partial sum , and write

(P).

[I]t is broadly true to say that mathematicians before Cauchy asked not “How shall we *define* 1 – 1 + 1 – …?” but “What *is* 1 – 1 + 1 – …?”, and that this habit of mind led them into unnecessary perplexities and controversies which were often really verbal.

What is the value of an infinite series?

I mean, what does “value” mean here?

With a finite series, you can (in principle) just add all the numbers together. You take the result and call that the value of the series. The value of is . No problem. But you can’t do that with an infinite series. You’d never complete the process — and so you can’t get its result.

You learn about infinite series in school, and what you learn is that some series converge to a limit. That is, if you have the infinite series

it converges if, loosely speaking, the sequence of partial sums

approaches some value arbitrarily closely as gets larger; that value is called the limit and we can write

.

And then there’s a bit of a leap as we regard as not just the limit of the partial sums, but as the value of the infinite series:

But not all series have limits, and if the limit doesn’t exist, then we regard the value of the infinite series as undefined.

That all can be learned in an intuitive way, though it can be made much more formal, and on the surface it makes sense. But there’s a sort of swindle going on here. You’re thinking of the “value” of the series as “what you would get if you added up all the infinite number of numbers”… but you can’t add them all up, so how can you assert what you would get if you did?

Here’s a demonstration of why that way of thinking about it is deceptive: the Riemann series theorem. Consider the series

You can show this converges to the limit . But you can take the same series and rearrange the terms like this:

and that converges to . You’re adding up “the same” numbers in the second series as in the first, and addition is commutative, so you should get the same answer if you could add up all the numbers in the series (which you can’t)… but the limits are different! In fact, Riemann proved that any conditionally convergent series (one that has a limit, but the sum of the absolute values of the terms does not) can be rearranged to give you *any* limit, including or , or no limit at all. It seems the “value” of a conditionally convergent series is a property not just of the numbers being summed, but the order they’re being summed in, and that’s not at all true of the value of a finite sum. Really the limit is a number that can be unambiguously associated with an infinite series, and we can *define* it as that series’s value, but “value” here means something different than in the case of a finite series where it’s just what you get if you add all the terms together. In fact, what the value means is just what it is: the limit of the sequence of the finite sums.

That being said, what about a divergent series?

Stay tuned.

]]>Define the zeroth ramp number R(0) to be 123456789. Then define the nth ramp number R(n) to be R(n-1) with a 1 prepended and a 9 appended: R(1) = 11234567899, R(2) = 1112345678999, and so on.

Likewise, the antiramp numbers R'(n) are R'(0) = 987654321, R'(1) = 99876543211, R'(2) = 9998765432111, and so on.

Then R(17), R(19), and R'(38) are (if the Python primefac library is to be believed) primes. And they are the *only* ramp/antiramp primes for a good long time. I checked up through R(1000) and R'(1000)… and just when it seemed there were no more, R'(926) turns out to be prime:

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