Two time-honoured opposite spheres of life – trade and warfare – are confronting each other. The Atlanticist civilisation of merchants is challenging the continental or Eurasianist civilisation of heroes. The former civilisation implies commercialisation of life, whereas its continental counterpart has manifested itself in militarisation of life…

(From here.)

Just another example of the odd places my brain goes sometimes, I guess.

^{1}. In the grand tradition of *Legend of the Galactic Heroes*‘s Galactic Empire and *Gundam*‘s Principality of Zeon.

In Soviet America, coffee shop patronizes customers.

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First, we convert all of these to powers of two: , so that our product is . And by the basic law of exponents , we thus have

Now, one way to find this infinite sum is to convert it into an infinite sum of infinite geometric series:

And using the infinite geometric series formula with our , we see

Which is itself a geometric series with sum

And thus our product is

There is also another way we could have found our sum, using calculus. Noting that

for , we see that

for

Now, taking the derivative of both sides with respect to *x*:

But the left hand sum becomes our desired sum when *x*=1, so making that substitution, we see ,

the same result we found above.

1. If we flip the selected coin twice, and it comes up heads both times, what is the probability that our coin is the biased one?

2. If we flip the selected coin three times, and it comes up heads all three times, what, then, is the probability that our coin is the biased one?

3. Generalize: We have *m* fair coins and one identical-looking biased coin with probability *p* of getting heads. If we select one coin at random, and obtain *k* heads in *n* flips, what is the probablility P(*m*,*p*,*n*,*k*) that we have the biased coin?

1. Let us denote selecting a biased coin by B, selecting an unbiased coin by U, and getting two heads in two flips by H^{2}. Then we are looking for P(B|H^{2}), the probability that our coin is biased given that we got two heads in two flips. According to Bayes’ Theorem,

and the law of total probability tells us that

Now, we know that and . Now, the probability of getting two heads in a row with the biased coin is , and the probability with an unbiased coin is .

Thus, we see that the total probability of getting two heads in a row is given by the law of total probability as:

Thus, Bayes’ theorem tells us:

2. Now, with H^{3} being the outcome of getting three heads in three flips, we see that the probability of H^{3} with the biased coin is , and the probability with an unbiased coin is .

Thus the total probability of getting three heads in a row is

and the probability that our coin is biased is

3. Let H[*n*,*k*] denote the outcome where we have *k* heads in *n* flips. The probability of *k* successes in *n* successive Bernoulli trials, each with probability *p*, is given by the binomial distribution as:

So, we see that the probability of H[*n*,*k*] given that our coin is biased is

and given an unbiased coin,

Now, with and , we see that the total probability of getting *k* heads in *n* flips is

Plugging these into Bayes’ theorem, we obtain:

(Or, to put it in a form with more visible parallelism between the probability *p* of the biased coin and the 1/2 probability of the fair coins,

)

Using our original *m*=3, *p*=3/5 example, we obtain

So, if we were to flip our coin 100 times, and we get 59 heads, then the probability that we have the biased coin is ≈0.625. If instead we get 55 heads, the probability drops to ≈0.247, and if we get 51 heads, then the probability that we have the biased coin is only ≈0.061.

Considering our getting all heads, as in (1) and (2), we have in the case *k*=*n* that . We can, by solving for *n*, see that to obtain , we must have *n*≥7, to get requires *n*≥13, and to get requires *n*≥19 flips all heads!

One final note: when we divided both the numerator and denomenator by , we were assuming that 0<*p*<1, so that this term is not zero. If the coin is fixed so as to always come up heads; that is to say, if p=1, then obviously if our coin ever comes up tails, we have a fair coin: P(*m*,1,*n*,*k*)=0 for any *k*<*n*. If, however, we have *n* flips all heads, then we can’t be sure we have the biased coin (rather than an unlikely “streak” on the biased coins):

For example, if we have 99 fair coins, one “fixed” coin, and get ten flips all heads, we still have a probability of 1-P(99,1,10,10)≈8.8% chance of having a fair coin.

First, we note that this is an improper integral, as the integrand is not only undefined for the limits of integration, but for the value *x*=*π*/4 as well. So, we consider the limit of the integrand as it approaches these values.

First, as *x* approaches zero (from above) the numerator approaches , while the denominator approaches -∞. Therefore, the integrand goes to zero at the lower limit of integration.

For *x*=π/4, both and equal zero, and for *x*=π/2, both approach +∞. Thus, in both cases we have indeterminate forms amenable to L’Hôpital’s rule. And since

So

and the singularity at *x*=*π*/4 is removable, and the (improper) integral is unaffected; and

And the integrand has a finite limit at the upper limit of integration as well. Thus, despite the undefined values, the integrand is sufficiently well-behaved that we need not be much concerned with the improper nature of the integral.

There does not appear to be a clear antiderivative for the integrand, and attempting a substitution like doesn’t appear to help. What we can do, though is an indirect method, which begins with the substitution . Then and . Thus,

But this last integral is just our original *I*, only with *u* as the variable of integration instead of *x*. Therefore,

To find the sum , we first note that , so that our sum . Next, we look back at this post, where it is demonstrated that

Using a similar method, we see that

And thus, for ,

which, using the quotient rule, gives us

However, we can express this more compactly using trigonometric identities. With the product-to-sum formulas, we see:

and

Substituting these into the numerator, and expanding, we see

which is much more compact.

Lastly, one could also use the half-angle identity to give equivalent form

Now, when *x* is an integer multiple of 2*π*. For these values, since *n* is an integer, , and so our expression has the indeterminate form 0/0. Thus, we can apply l’Hôpital’s rule (twice) for the limit as *x* approaches zero (and, thus due to the period, any of the other singularities):

so the singularities are removable, with a limit which matches the value expected from the sum:

for *x*=0, .

Let us construct the point *D* on the circumcircle diametrically opposed to *A*; that is to say, the point *D* such that *AD* is a diameter of the circumcenter. Then *AD*=2*R*, and *O* is the midpoint of *AD*.

Now, by Thales’ theorem, ∠*ABD* and ∠*ACD* are both right angles. Now, since *CD* and the altitude are both perpendicular to *AC*, they are parallel to each other. Similarly the altitude and segment *BD* are parallel, both being perpendicular to *AB*. Thus, the quadrilateral *BDCH* is a parallelogram.

Since the diagonals of a parallelogram bisect each other, we see that the midpoint *M _{A}* of

Now, let *P _{A}* be the midpoint of the segment

Now, let *N* be the intersection of and *HO*. By the midline-median bisection theorem proven in this post, we see that, as *HO* is the median of ∆*AHD* that crosses midline , *N* is the midpoint of both and *HO*. Thus, *NM _{A}*=

Now, consider the quadrilateral *HOM _{A}H_{A}*. Since and are both perpendicular to

By the first of the three items proven here, we see that , and so . Thus, we see that is the perpendicular bisector of , and so, by the perpendicular bisector theorem, *NH _{A}*=

Constructing diameter *BE* of the circumcircle gives us parallelogram *CEAH*, by a similar argument as above. Letting *P _{B}* be the midpoint of the segment

Therefore, we see that for any triangle, the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments connecting the vertices to the orthocenter all lie on a single circle, which, for this reason, is usually known as the nine-point circle, and its center *N* the nine-point center. We also see that the radius of the the nine-point circle is half the radius of the circumcircle, and the nine-point center is the midpoint of the segment connecting the circumcenter and orthocenter. This latter tells us that for any non-equilateral triangle, the nine-point center lies on the Euler line (for an equilateral triangle, it coincides with the circumcenter, orthocenter, and centroid).

But why do they all converge on this? Miss Nelson goes on:

Why are we girls competing to be the Queen of Pouts? Why do we scour through photos of celebrities and all our ambitious friends to find out who is the new princess of prurient poses? Even demure girls are tempted to strike sexually suggestive poses. But they must be careful, not because parents are looking but because they might not score any ”likes” and might then feel a failure, unworthy among their peers.

How confident can you appear at being lascivious? How credible is your air of lewdness? A girl who is just a try-hard will lose credibility and become an outcast. So a lot depends on how much support you can get from other girls.

She further describes the multitude of techniques young women use to increase their popularity scores. She laments the “fake flattery” young women use “to get higher on the food chain”, and that these often too-intimate pictures are posted by girls who “are seeking some sort of approval from their friends.”

However, she then attempts to blame it on “boys’ tastes”, which she calls “not always sophisticated” and based on “what they see in pornography”:

Who do we blame for this moral mess? As feminists, we correctly blame patriarchy because boys are securely at the top of the status game. Boys end up with the authority. They have their cake and eat it.

First, lamentable as the rise of pornography in our degenerate, declining culture is, it’s role here is minor at best. Science has amassed quite a bit of evidence that men’s “tastes” on women’s looks are pretty uniform and to a great extent hardwired. So, despite how Miss Nelson may wish things were, it’s unlikely that what boys like will change.

Second, one should ask why young women like Miss Nelson place so much importance and status in interest and attention from the opposite sex? THe answer is the same reason why (heterosexual) men place so much importance and status in interest and attention from the opposite sex: not “patriarchy”^{1} or “misogyny”, but *biology*. From the Darwinian perspective, finding and attracting a mate, more specifically the best quality mate one can, is literally the Most Important Thing. Thus, any accounting of social status amongst post-pubescent human beings will have sexual attractiveness as a significant component. You can rail against human nature, but biology wins out in the long run. This also strikes at the first bit; even if male tastes were different, girls would just compete over those instead, just as viciously.

Most important, though, and which Miss Nelson somewhat acknowledges when not deflecting blame, is that the most important factor is not male approval, but peer approval. The great anxiety that drives this behavior is the fear of being unpopular with other girls, of becoming an outcast. When it comes to breaking down individuality and enforcing conformity and social hierarchy, boot camp has nothing on what high school girls do to one another.

How often does a teen girl demand the latest popular album not because they like the music, but because all their friends listen to it, so she has to have it to or she’ll *just die*? See this bit from *Scientific American*, where brain scans have not only confirmed this phenomenon, but shown that the primary emotional motivation is fear of failing to fit in.

And being cast out from the pack means becoming prey to it. All the worst incidents of bullying I’ve heard or read about, in terms of psychological cruelty and viciousness, the perpetrator and victim have been teenage girls. There’s a reason the “alpha bitch” and “girl posse” tropes have such frequency and cross-cultural resonance. Or see the work of Rachel Simmons. And Paul Graham’s “Why Nerds Are Unpopular” helps further explain why high school amplifies this:

Why is the real world more hospitable to nerds? It might seem that the answer is simply that it’s populated by adults, who are too mature to pick on one another. But I don’t think this is true. Adults in prison certainly pick on one another. And so, apparently, do society wives; in some parts of Manhattan, life for women sounds like a continuation of high school, with all the same petty intrigues.

I think the important thing about the real world is not that it’s populated by adults, but that it’s very large, and the things you do have real effects. That’s what school, prison, and ladies-who-lunch all lack. The inhabitants of all those worlds are trapped in little bubbles where nothing they do can have more than a local effect. Naturally these societies degenerate into savagery. They have no function for their form to follow.

When the things you do have real effects, it’s no longer enough just to be pleasing. It starts to be important to get the right answers, and that’s where nerds show to advantage. Bill Gates will of course come to mind. Though notoriously lacking in social skills, he gets the right answers, at least as measured in revenue.

.

Given that human nature doesn’t change except on evolutionary time scales, and the prison-like nature of modern high school isn’t likely to be transformed anytime soon (save for its end with the total collapse of industrial civilization), not much can really be done. And the best thing to ameliorate the situation, parental involvement to limit the “selfie arms race”, Miss Nelson rejects in her concluding paragraph. She wants an end to girls being “compelled to act the stereotype, because those who opt out commit themselves to social leprosy”, but it is her fellow girls who provide the compulsion, and treat the non-conforming like lepers.

Miss Nelson identifies a real (if intractable) problem, but then spends the rest of her article deflecting all responsibility away from herself and her peers, blaiming “patriarchy”, and then rejecting any solution that might require trade-offs or any change in behavior on her part, instead wanting a vague, magical solution that doesn’t require any action on her part or any limits on her behavior. She wants fried ice; she is the one who wants to have her cake and eat it too. In other words, she’s a teenage girl.

1. In fact, under an *actual* patriarchy, the problem would be much less, if not non-existant, because patriarchs would not tolerate, and would curb, the female bad behavior and status games that drive this issue.

- the median of a trapezoid is parallel to the bases;
- the length of the median is half the sum of the lengths of the bases;
- the midpoints of the diagonals of a trapezoid also lie on its midline.

The key to all of these is the triangle midline theorem.

Consider the trapezoid *ABCD* with bases *AB* and *CD*. Let *M* and *N* be the midpoints of the legs *AD* and *BC*, respectively, and let *M′* be the midpoint of the diagonal *BD*.

Then, by the midline theorem with regards to ∆*ABD*, , and *MM′*=½*AB*. Applying the midline theorem to ∆*BCD*, , and *M′N*=½*CD*.

Now, since , we see .

Since and are both parallel to the same line , and pass through the same point *M′*, then we see that by Playfair’s axiom, they must be the same line; *M*, *M′* and *N* are collinear, and so the median . Further, *MN*=*MM′*+*M′N*=½*AB*+½*CD* =½(*AB*+*CD*).

Noting that we could perform the same procedure using diagonal *AC* and its midpoint *N′*, we see that the midpoints of both diagonals must lie on the median *MN*.

Let *D* and *E* be the midpoints of sides *AB* and *AC*, respectively, of ∆*ABC*. Let us extend segment *DE* past *E* to point *F* such that *DE*=*EF*, and let us draw *CF*.

Since *DE*=*EF*, *AE*=*EC*, and vertical angles ∠*AED* and ∠*CEF* are congruent, we see by the SAS condition that ∆*ADE*≅∆*CFE*. Thus, *CF*=*AD*=*BD*. Also, ∠*FCE*≅∠*DAE*; but since these are alternate interior angles for lines and cut by transversal , we see that . But then the quadilateral *BCFD* has a pair of opposite sides, *BD* and *CF*, which are of equal length and parallel, so it is therefore a parallelogram, and so . And since opposite sides of a parallelogram have equal length, *DF*=*BC*, and so *DE*=½*DF*=½*BC*.

We can also prove a similar theorem: that the line through the midpoint of one side of a triangle parallel to a second side of the triangle bisects the third side, and that the segment of that line inside the triangle is one-half as long as the parallel side. As with the midpoint theorem, it is trivial to prove with triangle similarity (this time the AA condition) and the corresponding angles postulate. So, instead, we use a similar construction as above.

Let *D* be the midpoint of side *AB* of ∆*ABC*, and let *E* be the point where the line through *D* parallel to *BC* intersects *AC*. Construct the line through *C* parallel to *AB*, and let *F* be the point where it intersects

Then *BCFD* is a parallelogram, and since opposite sides of a parallelogram have the same length, *BD*=*CF* and *BC*=*DF*. And so *CF*=*BD*=*AD*. And since ∠*ADE* and ∠*CFE* are alternate interior angles for lines and cut by transversal , they are congruent. Similarly, ∠*DAE*≅∠*CFE*, and so, by the ASA condition, ∆*ADE*≅∆*CFE*. Thus, *AE*=*EC*, and *E* is thus the midpoint of *AC*. We see also that *DE*=*EF*, and since *DF*=*BC*, thus *DE*=½*DF*=½*BC*.

Using both of these theorems together, we can prove a third: that a midline (mid-segment) of a triangle and the triangle median that intersects it bisect each other.

Let *M _{A}*,

By the midline theorem, and *M _{B}M_{C}*=½

This means, then, that , and so by our second theorem above with regards to ∆*ABM _{A}*, we see that

Similarly, our second theorem applied to triangle *AM _{A}C* establishes that

Next, construct the median from vertex *A* to the midpoint *M* of side *BC*. Then *G* lies on *AM*, with *AG*=2*GM*, as we showed here. Recalling that the circumcenter is the intersection of the perpendicular bisectors of the sides, we see that *OM*⊥*BC*.

Now, since *AG*=2*GM*, *GH*=2*OG*, and ∠*AGH*≅∠*MGO*, we see (by the SAS similarity condition) that ∆*AGH*~∆*MGO*. And since these triangles are similar, corresponding angles ∠*HAG* and ∠*OMG* are congruent. However, these are alternate interior angles for lines and cut by transversal , and therefore . And since *OM*⊥*BC*, we see , and is the triangle altitude from *A* to *BC*.

Analogous constructions show that *H* must also be on the triangle altitudes from *B* and *C*:

Thus, we see that *H* is the orthocenter of ∆*ABC*. So, we see that for any non-equilateral triangle, the circumcenter *O*, centroid *G* and orthocenter *H* are collinear, with *GH*=2*OG*; the line through these triangle centers is known as the Euler line of the triangle. (For an equilateral triangle, *O*, *G* and *H* are all the same point.)

Now, how about demonstrating that the circumcenter and orthocenter exist; that is to say, showing that the three perpendicular bisectors of the sides are concurrent, and that the three altitudes are concurrent, for any triangle.

The case for the circumcenter is simple to prove using the perpendicular bisector therorem: a point *P* is equidistant from points *A* and *B* if and only if *P* lies on the perpendicular bisector of the line segment *AB*.

So, for triangle ∆*ABC*, let point *P* be the intersection of the perpendicular bisectors of sides *AB* and *BC*. Since *P* is on the perpendicular bisector of *AB*, the perpendicular bisector therorem tells us that *AP*=*BP*; and similarly, since *P* is on the perpendicular bisector of *BC*, then *BP*=*CP*. Thus, *AP*=*CP*, and so the theorem tells us that P is on the perpendicular bisector of *AC* as well. We also have *AP*=*BP*=*CP*, and therefore that a circle with radius equal to that distance with center *P* passes through *A*, *B* and *C*.

The orthocenter case is a bit less simple. Construct through each vertex of triangle ∆*ABC* a line parallel to the opposite side, and let us label the intersections of these lines *D*, *E*, *F* as in the figure below:

Since the lines are constructed to be parallel, we see that the quadrilaterals *ABDC*, *BCEA* and *CAFB* are all three parallelograms. Since opposite sides of a parallelogram have equal length, we see from parallelogram *BCEA* that *AE*=*BC*, and from parallelogram *CAFB* that *AF*=*BC*. Thus, *AE*=*AF*, and *A* is the midpoint of *EF*. It is clear analogously that *B* is the midpoint of *DF* and *C* is the midpoint of *DE* as well.

Now, let be the altitude of ∆*ABC* from vertex *A*. By definition, , and since , we see as well. But *A* is the midpoint of *EF*, and so is the perpendicular bisector of *EF*. Similarly, the altitudes of ∆*ABC* from vertices *B* and *C* are the perpendicular bisectors of *DF* and *DE*, respectively.

But we proved above that the perpendicular bisectors of the sides of any triangle are concurrent. Thus, the altitudes of ∆*ABC*, being also the perpendicular bisectors of the sides of ∆*DEF*, must be concurrent.

Thomas Hutchinson’s anonymously published rebuttal to the Declaration of Independence (and fisking of its charges against George III).

Related, with quotes from the above, is this Radish article; also, Part 2 of Moldbug’s *A gentle introduction to Unqualified Reservations*.

•William Bernstein on the Boston Tea Party in this EconTalk podcast (circa 25 minutes in). The takeaway: it was essentially the “first anti-globalization riot”, with smugglers like Samuel Adams protesting a *reduction* in taxes that was undercutting their business.

Related: Historynet’s Debunking Boston Tea Party Myths

An exerpt:

Resistance leaders also launched a new wave of negative propaganda that played to anti-foreign sentiments: Tea from the East India Company was packed tightly in chests by the stomping of barefoot Chinese and was infested with Chinese fleas. In turn, a vast number of colonists vowed to protect American business from foreign competition, even if that business was smuggling. Beware of products from China, buy America, wage war on drugs, down with corporations—all these messages, as well as their better-known cousin, no taxation without representation—amplified the response to Parliament’s Tea Act of 1773.

•”American Loyalists“: *Buried History of the American Revolution*.

•Handle: “Suppressing Tories”

•A quote:

Sir, As the Committee of Safety is not sitting, I take the Liberty to enclose you a Copy of the Proclamation issued by Lord Dunmore; the Design and Tendency of which you will observe, is fatal to the publick Safety. An early and unremitting Attention to the Government of the SLAVES may, I hope, counteract this dangerous Attempt. Constant, and well directed Patrols, seem indispensably necessary.

Patrick Henry, in a pamphlet written to county lieutenants throughout Virginia concerning the Earl of Dunmore’s proclamation offering emancipation to slaves of Patriots who escaped and joined the Royal forces. (See also Lord Dunmore’s Ethiopian Regiment).

•*Somerset v. Stewart* — 3.5: Thirteen Colonies and United States.

•Tangentially related, particularly the comments threads: *The Economist* “Why the first world war wasn’t really” on the Seven Years’ War — and its American theater, the “French and Indian War“/”La guerre de la Conquête” — as the actual first world war. (See also the Battle of Jumonville Glen.)

Update: Nick B. Steves has created a permanent page for “Strictures upon the Declaration of Independence” over at The Reactivity Place.

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Let us call the lengths of the sides *a*, *b* and *c*, and let the distances from our interior to these sides be *x*, *y* and *z*, respectively. Lastly, let *A* be the area of the triangle.

Now, we note that the segments connecting our point to the vertices of the triangle divide it into three smaller triangles.

We see that these triangles have bases *a*, *b*, *c* and heights *x*, *y*, *z*, respectively, and thus areas of , and . But these have to add up to the area *A* of our (original) triangle:

or, multiplying by 2:

for any point in the interior of the triangle.

Thus, we want to find the point which maximizes the product *xyz* subject to the contstraint . But this is just last week’s Lagrange multiplier problem with *d*=2*A*. And thus, we see that the maximum is obtained when . However, this means that the areas of the subtriangles are equal:

But we saw two weeks ago that the point for which the triangle is divided into equal area subtriangles in this fashion is the centroid of the triangle.

Thus, our answer is that the point in the interior of a triangle for which the product of the distances from that point to the sides is maximized is the centroid of the triangle.

This is a maximization subject to constraint problem, and is thus best approached by the method of Lagrange multipliers. Thus, we find the stationary points of the function:

(where we have chosen –*λ* rather than +*λ* to make some later algebra easier).

Taking the partial derivatives of *F*, we see

Setting the first three of these equal to zero, we obtain

And setting the fourth equal to zero gives us our constraint , as expected.

Now, multiplying those three equations we obtained from setting the partial derivatives equal to zero, we obtain

and taking the square root of both sides (with *λ*≥0),

Dividing this by the first equation, we see

And analogously,

And thus,

and from our constraint , we see that

(from which we can solve for *λ*, but this is unnecessary).

Thus, the point which maximizes *f*(*x*,*y*,*z*)=*xyz* is , with the maximum value thus being .