Following up on my earlier posts about overrated things and underrated things, I'll complete the sequence with four correctly rated things:

Ben Affleck. Correctly admired for his best work, correctly jeered at for his worst, and lusted after by many women, who in that respect are also correct.

Starbucks. The views of people mistakenly devoted to Starbucks perfectly balance the views of people mistakenly disdainful of it, which averages out to: correct.

Scotch whisky. A product so good—and, in addition, so correctly rated—that even when people tell you they don't like it, they often add that they wish they did.

Tuscany. In the villages you have food and wine and long views, and in the cities you have art and architecture: and all of it is wonderful. There's no contrarian take on Tuscany.

Bonus correctly rated thing: Tornadoes. They are as fascinating as they are awful…as blind as they are malevolent…as mathematical as they are biblical. Nobody ever had an original thought about a tornado—or an incorrect one.

Given three distinct obstacle points located anywhere inside the unit disk, we established previously that a greatest-area feasible triangle must satisfy one among the following three conditions:

(i) All three vertices of the triangle are on the boundary of the disk.

(ii) Exactly two vertices are on the boundary of the disk, and both sides meeting the interior vertex have one or two obstacle points on them.

(iii) Each proper side has one obstacle point on it.

Previously we also investigated scenario (ii) for the particular case of the conjectured optimum trio of obstacle points—let's call these obstacle points O_{1}, O_{2}, O_{3} from now on. By an explicit calculation, we determined that for these particular obstacles, all triangles in scenario (ii) have area less than 0.828…, the area of the spotlight triangle (and the edge triangle). Thus, the global maximizer isn't in class (ii).

Today we'll investigate scenario (iii). This is an interesting scenario, because it relates to a famous set of problems about triangles inscribed in triangles. We won't use any of that machinery, however.

We'll handle scenario (iii) in three subcases:

(iii)(a). All three vertices of the triangle are in the interior of the disk.

(iii)(b). Exactly two vertices of the triangle are in the interior of the disk.

(iii)(c). Exactly one vertex of the triangle is in the interior of the disk.

Scenario (iii)(a). All three vertices of the triangle are in the interior of the disk.

Given triangle ABC with obstacles X on BC, Y on CA, and Z on AB, we can (for example) slide vertex B slightly along AB (away from A), and simultaneously slide vertex C slightly along CA (toward A), using obstacle X as a "hinge" of sorts.

The triangle remains feasible throughout the sliding operation, because the obstacles remain on the proper sides. We also notice that if X is not the midpoint of BC, then the area of the triangle does not remain stationary. If XB is longer than CX, then the area increase due to sliding B exceeds the area decrease due to sliding C. Hence, if ABC is a greatest-area feasible triangle, then X must be the midpoint of BC.

We could have performed the sliding operation using any obstacle point as the "hinge," thus each obstacle must be the midpoint of the side it belongs to. (That is, the obstacle triangle XYZ is the medial triangle of ABC.)

So far we've made no assumption about the obstacle points other than non-collinearity, but now let's specialize to the case of the conjectured optimum trio of obstacles. If O_{1} be the midpoint of BC, O_{2} be the midpoint of CA, and O_{3} be the midpoint of AB, it follows from the fact that O_{1}O_{2}O_{3} is equilateral that ABC is also equilateral.

Thus, if ABC is a greatest-area feasible triangle with one obstacle on each side, and if all three vertices of ABC are in the interior of the disk, then up to congruence, ABC could not be other than this triangle:

However, this triangle only has area 0.53529..., which is less than the spotlight and edge triangles (0.82863...). This shows that the global maximizer isn't in class (iii)(a).

Scenario (iii)(b). Exactly two vertices of the triangle are in the interior of the disk.

By reasoning similar to the above (using O_{1} as a hinge), O_{1} must be the midpoint of BC. This allows us to parametrize triangle ABC by a single variable—say, the angle θ in this diagram.

The length r is actually a function of θ because, given the value of θ, r is to be chosen so that CO_{2} and BO_{3}, extended, meet on the boundary of the disk.

Here is an animation showing how the family of triangles evolves over the range of θ values (0 to 60 degrees).

It's clear from the animation that r(θ) grows too slowly to offset the rapid decrease in altitude that results from rotating the top side. Thus, it makes sense that the area numbers decrease throughout the motion.

At this point the natural thing to do would be to express the area of triangle ABC as a function of θ. However, I did not actually derive r(θ) or the coordinates of vertex A as functions of θ. (The graphics above were generated using a combination of explicit functions and a root-finding routine.) Instead of calculating the area of triangle ABC, I decided to bound the area by calculating instead the area of a larger region, G, consisting of triangles C*O_{1}O_{2}, B*O_{1}O_{3}, O_{1}O_{2}O_{3}, and O_{2}O_{3}Q, where vertices C*, B*, and Q lie on the boundary of the disk as shown.

For any angle θ, triangle C*O_{1}O_{2} exceeds triangle CO_{1}O_{2} in area, triangle B*O_{1}O_{3} exceeds triangle BO_{1}O_{3} in area, and triangle O_{2}O_{3}Q exceeds triangle O_{2}O_{3}A in area (the altitude from O_{2}O_{3} is as great as possible because Q is at the bottom of the disk).

The lengths O_{1}C* and O_{1}B* are the positive roots of the quadratic equation r^{2} ± 0.64197 sin(θ) r − 0.896982. These are easily expressed as functions of θ, r_{±}(θ), which makes the area of region G easy to express in closed form:

where s = 0.555925… is the known side of equilateral triangle O_{1}O_{2}O_{3}.

And here's a graph of area(G). The maximum value is 0.823….

Since area(G)≤ 0.823… and since the area of triangle ABC is less than area(G), we have that the area of triangle ABC is less than 0.823…. That value in turn is less than the area of the spotlight and edge triangles (0.82863…), which shows that the global maximizer isn't in class (iii)(b).

Scenario (iii)(c). Exactly one vertex of the triangle is in the interior of the disk.

This family of possibilities is also parametrized by a single variable—say, the y-coordinate of vertex C, as shown in this animation:

By drawing a careful diagram and repeatedly applying of the law of cosines, the law of sines, and the angle sum of a triangle, I was able to solve triangle ABC in terms of the variable y.

The graph of the area function as a function of y agrees with the behavior illustrated in the animation.

The greatest area occurs at the maximum y value, at which point the triangle ABC is isosceles.

The best area in class (iii)(c), 0.645389…, is less than the area of the spotlight and edge triangles (0.82863…), which shows that the global maximizer isn't in class (iii)(c).

And that completes the analysis of scenario (iii), at least for the particular obstacles O_{1}, O_{2}, and O_{3}.

Together with the previous analysis of scenario (ii), this leads to our conclusion for today: Conclusion

For the three obstacles O_{1}, O_{2}, and O_{3}, the greatest-area feasible triangle has all three of its vertices on the boundary of the disk.

I began this groundbreaking series with Book 4, but beginning with Book 2 would also be good. In fluent prose, Knausgaard recounts the events of his mid- to late-thirties, addressing such themes as love, parenting, masculinity, obligation, and modern values.

In Book 2 we learn how My Struggle arose from Knausgaard's life, from his painful artistic struggles, and from his growing dissatisfaction with fiction and the traditional novel form. The unstated subject of this series appears to be reality itself—or the means by which art achieves verisimilitude.

Modiano convincingly conveys character, period, and place as he examines a critical year in the formation of two lives. Louis and Odile meet at age nineteen and drift together through their early adulthood in postwar Paris. Young Once offers enough plot to maintain the reader's interest, but plot isn't the reason people read Modiano, an author whose simple prose imbues certain moments with an almost unbearable emotional depth.

Honorable mention: Chris Bachelder, The Throwback Special.

This poignantly comic, well written novel would make an excellent gift for your once-athletic husband (or from your once-athletic husband). Here is the NY Times review.

Browsing in a used bookstore while on vacation, I found Abel's Island by William Steig (1907–2003), author-illustrator of Sylvester and the Magic Pebble, Amos & Boris, Shrek!, and other beloved children's books. Abel's Island is about a wealthy, married mouse who becomes separated from his wife and his comforts by a violent storm—and by a fateful act that ultimately fuses the material with the transcendent. About 120 pages long, the book has charming illustrations, some terrific writing, and the author's usual aversion to baby-talk.

The edition I bought was the 1987 ninth printing in hardcover from Farrar/Straus/Giroux. I don't see that version available online, but there have been many editions published since then.

Dominick Tyler, Uncommon Ground: A Word-Lover's Guide to the British Landscape (2015). Tyler, a photographer, traveled Britain to produce this treasury of words that name features of the landscape. Often ancient, sometimes unpronounceable, his lexicon evokes by turns the bucolic, the desolate, the sylvan, and even the sinister. Tyler is interested in the way words allow us to attach to reality in general and to the landscape in particular. His pensive mini-essays, and his artful yet unpretentious photographs, offer pure browsing pleasure while advancing his proposition that "rebuilding our landscape vocabulary might enable more complicated conversations about nature to take place." Here is the Guardian review including a few of the photos. See my full review and buy the book on Amazon.

William M. Kramer, A Lone Traveler: Einstein in California (2004). A concise and pleasurable account of Einstein's three visits to California in the early 1930s. When the book opens, Einstein is still a German citizen and still an outspoken pacifist. By the time of the book's close, Einstein is a U.S. citizen and a proponent of military action against the Nazis. Kramer rightly judged that this transitional period in Einstein's life was worth a closer look. See my full review and buy the book on Amazon.

J. Kenji López-Alt, The Food Lab: Better Home Cooking Through Science (2015). The Food Lab has quickly earned its place on our shelf alongside The Joy of Cooking, The New Vegetarian Epicure, Isa Does It, Everyday Italian, and Marcella Says. I credit this book for some of the best dishes I've ever prepared. Don't take my word for it; try this chili con carne recipe, and then buy the book online: The Food Lab: Better Home Cooking Through Science.

Best Short Stories

Ann Beattie, "Panthers." Paris Review, Fall 2016.

David Szalay, "Lascia amor e siegui Marte." Paris Review, Winter 2015.

Best Poem

April Bernard, "Cold Morning." New York Review of Books, November 24, 2016. In this poem by a former Bennington colleague, the speaker is telling us about a morning when she saw a horse lying dead on the ground surrounded by its puzzled brethren. Later that same morning, she heard a strange news item about some buffalo that had blundered onto the interstate. Between these two recounted anecdotes, the speaker tells of her grief over the death of a friend. The emotion of the subject matter is controlled by a stately rhythm and lightened just a little by wit.

Best Essay

Ta-Nahesi Coates, "What O.J. Simpson Means to Me." The Atlantic, October 2016. An exemplary personal essay because the self is not the subject, but rather the lens on the subject. Read it online.

Patrick Deneen, "After Liberalism," the Nineteenth Annual Paul Holmer Lecture. A critique of Enlightenment liberalism from a political scientist at the University of Notre Dame. Read the transcript online.

Daniel Mendelsohn, "How Greek Drama Saved the City." New York Review of Books, June 23, 2016. One of America's foremost essayists describes the theater in ancient Greece and how it differs from theater in today's society. Reading this was like being in college again, but in a good way. Read it online.

Luc Sante, "The Invisible Man." New York Review of Books, May 10, 1984. A William S. Burroughs critical biography in miniature. Read it online.

Best Long-Form Journalism

Ariel Sabar, "The Unbelievable Tale of Jesus's Wife."

A hotly contested, supposedly ancient manuscript suggests Christ was married. But believing its origin story—a real-life Da Vinci Code, involving a Harvard professor, a onetime Florida pornographer, and an escape from East Germany—requires a big leap of faith.

Author-journalist Ariel Sabar took it upon himself to investigate the provenance of 'The Gospel of Jesus's Wife.' What he discovered, you couldn't make up. Read it online.

Christopher Goffard, "Framed: A Mystery in Six Parts."

She was the PTA mom everybody knew. Who would want to harm her?

This is crack-cocaine in newspaper form. Read it online.

Best Music, Best Theater

No awards this year. We saw one concert and one play, and both were OK.

Best Movies

(Links point to reviews.)

The Assassin – Mythic tale with sumptuous period details and sublime scenery; I wasn't always clear on the plot.

The Lobster – Not sure that dystopian-quirky-black-comedy-parables are my thing, but the film succeeds very well on its own terms.

Moana – A pretty good road movie, suffused with the joy the makers must have felt in expressing such astonishing visual creativity.

Everybody Wants Some!! – Yet another Linklater gem. "Few filmmakers have so fully embraced the bittersweet joy of living in the moment."

The Hunter (2011) – A contemplatively beautiful quest film, badly misrepresented by a thriller-style marketing poster.

Creed – A great sports flick. Director Ryan Coogler did for the Rocky franchise what J. J. Abrams ought to have done for Star Wars.

Which leads us to:

Biggest disappointment: Star Wars: The Force Awakens. There's some great movie-making in the scenes where Finn becomes Finn, and in the scenes that introduce Rae, but the film grows workmanlike as it enters recycling mode. As soon as I met Rae, I wanted the film to tell her story. Instead, the movie pastes Rae and everybody else into a story we've all heard before.

Best Meal in an Airport

Tie between Chelsea's Kitchen in PHX (the 2014 winner) and Columbia Cafe in TPA.

Columbia Cafe is the TPA outpost of Ybor City's famous Columbia Restaurant (said to be Florida's oldest). The whole staff was friendly and fast, and I had a delicious lechon asado (roasted pulled pork marinated with garlic and citrus topped with Mojo onions).

At Chelsea's Kitchen with my sister and her husband, I had a very good roast chicken before hopping on a red-eye flight. Something is working with the decor at Chelsea's, because while you're there it's easy to forget you're in an airport terminal.

Best Solo Drive

Quebec City to Halifax via the Gaspé Peninsula.

One day in 2015 I was perusing my Times Atlas of the World for ideas, and my eye was drawn to a strange offshoot of Quebec I'd never noticed before: the Gaspé Peninsula. Gazing into the map, I could almost see little fishing boats sheltering between headlands in the cold waters of the Gulf of St. Lawrence. Maybe a rustic lodge in the highlands.... I made up my mind to drive the Gaspé. Sometimes, a place is exactly what you think it will be. Here's a brief overview of the region, and if you want to research the drive, this page is a good starting point. Note: Buy a French phrasebook! I met almost nobody in the Gaspé who spoke English.

Best Single-Artist Exhibition

Portraits by Alex Katz at the Metropolitan Museum of Art.

Great portraitists work on many fronts, including color, composition, and clothing. But what matters in portraiture is the face. In great portraits, the sitter's expression is often ambiguous or enigmatic. Mr. Katz achieves this effect through abstraction—yet he doesn't abstract away the individuality of his subjects. What the portraits lack in painterliness, they make up for in boldness of design. Here is a review of the exhibition I saw, and here is a profile of the 88-year-old artist.

Best of the Year—Period.

This summer in Williamstown, my wife and I escaped from my college reunion to visit the Clark Art Institute for the opening of Splendor, Myth, and Vision: Nudes from the Prado. The exhibition included 28 pictures by Titian, Veláquez, Rubens, and other painters from the 1500s and 1600s. Twenty-four of the pictures were being shown in the United States for the first time. Some of these masterpieces were breathtaking—almost too overwhelming to look at.

The theme of the exhibition was the tension between Catholic culture and artistic representations of the body. However, the painting I lingered over the longest was not a nude, but the royal portrait of Philip IV from ca. 1653, considered one of Velázquez's greatest portraits.

The art alone would have been worth the trip, but adding to the experience was the opening event, which was held in the early evening. As the summer sun descended behind the Berkshire hills, my wife and I strolled in the new museum expansion by architect Tadao Ando. Then we enjoyed drinks and dancing with a DJ. All in all, a fabulous night.

***

I like ending the list with something to watch or listen to, so here is a scene from Creed in which the main character is going for a training run. The action takes place in a poor neighborhood of Philadelphia. This is one of those times when you're watching a B-movie and suddenly genius erupts.

In-room hotel coffee makers. Good coffee is a pleasure, but so is the penny-pinching brew that gurgles out of an in-room hotel coffee maker. Someday, hotels are going to figure out that these little machines are costing them money, and then it's good-bye in-room hotel coffee makers. Enjoy them while they last. (Pro-tip: when you check in, remember to ask for more whitening powder.)

McDonald's. Everybody loves McDonald's breakfast, but how much do they love it? Do they love it as much as it deserves? I don't think so. That part is underrated, then. And the double-quarter-pounder-with-cheese meal (aka #4) is also good. The Coke at McDonald's is always perfectly mixed, with wide-gauge straws that prevent excessive foaming. If the last time you ate McDonald's was around the time of Supersize Me, give them another try because the company has made impressive changes in food quality since that time.

Canned and frozen fruits and vegetables. Frozen snap peas are snappier than the ones in the produce aisle. Frozen blueberries are more flavorful for cooking than supermarket blueberries, which to be honest are hardly worth eating at all. I love corn on the cob, but canned corn tastes crunchy and sweet year-round, and there's no husking—you just dump it into a saucepan and turn on the heat (or substitute for fresh corn in this recipe to enjoy the taste of grilled corn in winter).

An expensive umbrella. Everybody puts up with crappy umbrellas! But nothing cheers you up on a rainy day like the impregnable foomp of a real Burberry umbrella. Sadly, I'm not allowed to have one, because on average I lose 0.18 umbrellas every time it rains. But I did have a Burberry umbrella back when I was a starving student in England, and in a rainy country, that umbrella was worth every skipped meal.

Bonus underrated thing:

Going the whole morning without talking or making eye contact.See: In-room hotel coffee makers.

Place N points in the unit disk in such a way as to minimize the greatest possible area of a triangle in the disk that doesn't contain any of the points.

In this post, I'll reduce the possibilities somewhat for the N = 3 case, making no assumptions about the locations of the obstacle points. At the end, I'll apply what we learn to the particular obstacles in the conjectured optimum configuration.

Reducing the possibilities

Given three distinct obstacle points located anywhere inside the unit disk, I think it isn't hard to show that a greatest-area feasible triangle must satisfy one among the following three conditions:

(i) All three vertices of the triangle are on the boundary of the disk.

(ii) Exactly two vertices are on the boundary of the disk, and both sides meeting the interior vertex have one or moretwo obstacle points on them.

(iii) No vertices are on the boundary of the disk, and Each proper side has one obstacle point on it.

(A triangle's "proper sides" are what remains after you delete the three vertices. This is a term I made up.)

I established that one of the conditions (i)–(iii) must hold by considering the following ten cases:

3 obstacles on vertices of the triangle, 0 obstacles on proper sides, 0 obstacles outside the triangle

2 obstacles on vertices of the triangle, 1 obstacles on proper sides, 0 obstacles outside the triangle

2 obstacles on vertices of the triangle, 0 obstacles on proper sides, 1 obstacles outside the triangle

1 obstacles on vertices of the triangle, 2 obstacles on proper sides, 0 obstacles outside the triangle

1 obstacles on vertices of the triangle, 1 obstacles on proper sides, 1 obstacles outside the triangle

1 obstacles on vertices of the triangle, 0 obstacles on proper sides, 2 obstacles outside the triangle

0 obstacles on vertices of the triangle, 3 obstacles on proper sides, 0 obstacles outside the triangle

0 obstacles on vertices of the triangle, 2 obstacles on proper sides, 1 obstacles outside the triangle

0 obstacles on vertices of the triangle, 1 obstacles on proper sides, 2 obstacles outside the triangle

0 obstacles on vertices of the triangle, 0 obstacles on proper sides, 3 obstacles outside the triangle

(That there are ten cases checks out, since we are effectively putting three indistinguishable balls into three distinguishable boxes, and there are _{3 + 3 − 1}C_{3 − 1} = _{5}C_{2} = 10 ways to do that.)

(It isn't clear that you have to go to the trouble of distinguishing between obstacles at vertices and obstacles on proper sides—everything seems to work out continuously in practice—but I didn't want to assume so.)

Some of these cases have two or more sub-cases. For example, here are the sub-cases when all three obstacle points lie on proper sides of the triangle. There are sub-cases because the greatest number of points on a single proper side could be three, two, or one. (The obstacle points are colored blue; the reason for the orange vertices is explained shortly.)

In the upper-left configuration, maximality implies that none of the three vertices are in the interior of the disk, because if any vertex were in the interior, that vertex could be perturbed in such a way as to increase area while maintaining feasibility. But in the other three configurations, there are triangle vertices (colored orange) that can't be perturbed, at least not in the same way. A simple perturbation strategy can't drive the orange vertices to the disk boundary.

The possibilities (i), (ii), (iii) listed earlier jointly summarize the state of affairs after all of the ten cases and their associated subcases have been examined, with the easy-to-perturb configurations ruled out.

In the rest of this post, I'll analyze possibility (ii): "Exactly two vertices are on the boundary of the disk, and both sides meeting the interior vertex have one or more obstacle points on them." We'll label the interior vertex C.

Given the three obstacle points, some two of them are together on one of the proper sides of the triangle. (Generically there are three choices for this pair.) The line through those two obstacle points thus passes through vertex C, and the line through those two obstacle points meets the boundary of the disk in another vertex of the triangle, label it B. The line through C and the remaining obstacle point meets the boundary of the disk in the final vertex of the triangle, label it A. For a specific configuration of obstacle points, and for a specific choice of the pair of obstacles to be with one another on a side, the situation looks like this:

There was one degree of freedom in drawing this figure: it could be described as the freedom to choose the distance between vertex C and the obstacle on CB that is closer to C. That is, once you've decided which two obstacle points will be together on a side, you get to decide how far C will be from B.

Sliding C along its degree of freedom makes side CA "swing" on the pivot of the third obstacle point, thus changing the area as shown in this video:

In scenario (ii), then, given the obstacle points, and given a choice of which two obstacle points to pair on a side of the triangle, the area of the triangle can be determined as a function of a single variable. If the function is found explicitly, it can be analyzed to find the maximum area. The maximizer so discovered can be pitted against the maximizers that result from the other two choices for the paired obstacles, and the overall winner becomes the maximizer for scenario (ii).

Observe from the video that the best triangle for this family of possibilities occurs when vertex C has moved about halfway to the disk boundary, resulting in an area of about 0.803. It seems the winning triangle from scenario (ii) could indeed have a vertex in the interior of the disk, depending on the arrangement of the obstacle points.

Applying this to the particular configuration of obstacles conjectured to be optimal

I'll end with some concrete progress on the N = 3 problem. I carried out the above analysis for the particular case of the conjectured optimum triple of obstacle points, and because the geometry in that case is tractable, I could produce the explicit function giving the triangle area as a function of the variable distance parameter. Here's a relevant page from my notebook (the notation doesn't match the above, this is just for fun).

The area function is long and complicated to write down, but it's a smooth function, and so I just graphed the thing. The graph of the area as a function of the variable distance is the blue curve here:

(There are also some black curves, because I wanted to show the smoothness of the result as we allow the configuration of obstacles to "breathe" a little bit by moving in unison towards or away from the center of the disk.)

The blue curve is monotonically increasing (more obvious to the eye when you graph the derivative, which never gets close to zero). Hence the greatest area is obtained by allowing point C to move all the way out to the boundary, resulting in the spotlight triangle from this post.

Thus, scenario (ii) has no triangles that exceed the spotlight triangle and the edge triangle in area, and we can proceed to analyze scenarios (i) and (iii). This moves us closer to showing that for the conjectured triple of obstacles, the greatest possible area is the area 0.828... attained by spotlight triangle and edge triangle. By the way, if the greatest possible area really is less than 1, it rules out any possibility of recycling the argument for N = 2 in the N = 3 case.

P.S.

For the three obstacle points shown in the video, the greatest possible scenario-(ii) area was 0.803. That's less than the greatest possible area for the conjectured optimum configuration of obstacles (0.828...). However, the obstacle points shown in the video don't unseat the conjectured optimum configuration of obstacles, because for the obstacle points shown in the video, there is a scenario-(i) triangle with area 1.037. (I found it using a numerical solver I made a while back.)

Thank you to everybody who followed along with my election simulations this year. In the past few months, I've enjoyed some interesting conversations with readers and friends about the intersection of math and politics. I was also able to use what I learned to make an informed decision about where to volunteer—Lehigh County, Pennsylvania, where I spent the last two days of October knocking on doors and committing Hillary supporters to vote on election day. (Hillary did win Lehigh County, but Trump won the state overall).

I thought Hillary Clinton was the better choice, but not enough people agreed. Democrats are the opposition party now. I don't pretend to know how it will play out, but I do want it to be every patriot's mission over the coming years to defend our bedrock liberties, some of which President-elect Trump is overtly hostile to. I'm thinking in particular of the freedom of speech, of the press, and of peaceable assembly; due process of law; the right to vote; the people's right to be secure in their persons and free from unreasonable searches and seizures; and the freedom from cruel and unusual punishments.

If you'll forgive the emphasis, I feel a need this morning to stress that these liberties are not just for White people. I feel I have to say that because some of Trump's core supporters seem not to believe it. For them, race and culture have everything to do with being an American, to the extent (for example) that recent immigrants are by that circumstance less American.

For me on the other hand, what makes us all American is the Constitution that structures our great republic. In this post from last year, written in reaction to the Obergefell same-sex marriage decision, I wrote:

I think that what we are as countrymen has less to do with cultural things like religion or language, which we inherited from Europe, and everything to do with the Constitution, which set us apart from Europe.

One reason many commentators are concerned about a Trump presidency is that in comparison with the power of place and tribe (a power that becomes concentrated in the one entrusted to speak for the tribe), the principles set down in the Constitution are—in one of the president-elect's favorite phrases—"just words." And they are just words. Maybe the Founders cared so much about education because they recognized that they were building a nation with a quill pen, and only the educated could ever pledge allegiance to a mere piece of writing.

***

Adding to this later, because I don't want to be misunderstood. There are threads of paleoconservatism that make sense to me. I don't consider myself a progressive, and I agree with key parts of this critique of progressivism. Like its author, I don't consider 'attachment to the local and the particular' to be automatically racist, xenophobic, or 'primitive.' However, I do insist that the Bill of Rights is for every American. Where that precept clashes with culture, I want culture to lose.

1. She has a better temperament. The President's job is stressful, so I think temperament matters. Donald Trump seems excitable and impulsive, which in a President makes me nervous. Hillary Clinton seems like a calmer, more deliberate person.

2. Her base is more manageable. The alt-right is nasty enough as a fringe movement. If Trump wins, the alt-right will have a seat at the table in policy discussions.

Bad things would also happen if the Marxist left were to grow in power, but the situation isn't symmetrical. The relationship between Clinton and progressive activists isn't nearly as cozy as the relationship between Trump and the alt-right. And the Republicans in Congress won't roll over for the Marxist left the way they will for the alt-right.

3. She is more fact-based and can listen better. I don't like all of Hillary's policies, and I especially don't like her interventionist instincts. But I think she is capable of weighing the arguments of critics and capable of adjusting her plans in response to facts. Donald Trump can't take any criticism, and facts don't inform his plans at any stage.

4. Criticism of her seems overblown, while criticism of him is just.Matt Yglesias's take on the email thing is pretty much how I look at it, and for every bad trait of Hillary's, Trump has the same trait except worse (1, 2). The record of Trump's bad character spans decades. He is a small, insecure person with a cruel streak and a terrible lusting after power for its own sake. Lots of people are trying to explain how it happened—but however it happened, this year the Republican party nominated a candidate who deserves to lose.

Triangle ABC has area 780 square units. Side c is 78 units long. Angles A and B satisfy

cot(A) − cot(B) = 6.

Find the lengths of sides a and b and the measures of all three angles.

I tried to devise a scary-sounding problem, but the problem isn't as bad as it sounds. Given that the area is 780 square units and side c is 78 units, the altitude of the triangle is easily found: 20 units. Now suppose first that the altitude meets side c and separates it into two parts x and y, so that x + y = 78. From the given information, we would have x − y = 120. But if two positive numbers sum to 78, then they can't have a difference of 120...so in fact, the altitude doesn't meet side c. Our triangle "overhangs" side c (out past vertex B, say).

Suppose the triangle overhangs vertex B by distance x. Then cot(A) = (78 + x)/20 and cot(B) = − x/20, so from the given information (78 + x)/20 − (−x/20) = 6. Solve for x = 21.

Knowing the altitude and the overhang, we can use the Pythagorean theorem to find one side of our triangle as 29 units. Then a second application of the Pythagorean theorem (to the largest right triangle) gives the last side as 101 units.

With all three sides of the triangle known, finding the angles is straightforward. The answers are A = sin^{−1}(20/101) ≈ 11.4°, B = π − sin^{−1}(20/29) ≈ 136.4°, and C = π − A − B ≈ 32.2°.

A devious way to solve the problem would be to play a hunch: namely, that all three side lengths are going to be whole numbers. Then because the area is also a whole number, the triangle will be what's known as a Heronian triangle, after Heron of Alexandria (AD c. 10– c. 70). A little googling then turns up this table of Heronian triangles, and CTRL-F finds the solution without much trouble.

In fact, I created this triangle by looking for Pythagorean triples that share a number in common. The triples that I chose were (20, 21, 29) and (20, 99, 101). There is no reason why the side lengths of the triangle had to be integers, but I wanted the difference of cotangents to look dramatic.

I'll end with one more method for solving the problem: first derive an area formula, ½(cot A + cot B)^{−1}c^{2}, applicable to any triangle. (This is the ASA analogue of ½ a b sin C for the SAS case.) With this we can write down the value of sum of cotangents, which in conjunction with the difference of the cotangents determines the two cotangents (angles) individually. Then with all three angles and one side known, the law of sines gives the other two sides.