Saturday, December 31, 2016

Free Speech Among The Deplorables

1.

Carl Paladino, a wealthy Buffalo businessman, likes to send racist emails. Several of these were published (NSFW) when he ran for Governor of New York in 2010. This month's stir was bigger. Paladino apparently wrote a private reply to a newspaper's year-end quiz, intending to amuse his friends with it, but then he mistakenly sent the reply to the newspaper instead. Here are Paladino's published New Year's wishes:


In response to these comments, many in Buffalo have demanded that Paladino be removed from the city's school board, to which he was re-elected in May. The board has now voted 6–2 to ask the state Education Commissioner to remove him.

Personally, I believe Carl Paladino lacks the baseline character required of an educator. But whether the state's regulations, and the political forces arrayed on all sides, will actually permit his removal is something Commissioner Elia will have to judge. Jim Heaney, a Buffalo journalist and apparently no friend of Paladino's, has argued that executive action isn't the best approach in the present case. Among other ideas, Heaney suggests calling a special election for Paladino's board seat. If that were feasible, it sounds to me like a good idea: voters gave Paladino his authority, and ideally voters should be the ones to take it away. And I do think Paladino would lose…in May, he was nearly unseated by a high school student.

2.

Pamela Ramsey Taylor, director of a county nonprofit group in West Virginia, was removed by her board of directors in connection with a state investigation that began when Taylor posted a racist slur about Michelle Obama to her Facebook page. Mayor Beverly Whaling of Clay, West Virginia, who had commented approvingly on Taylor's post, quickly resigned her office as well.


What these people wrote is cruel and disgusting, and their apologies were insufficient. (Taylor's is here.) Yet was it fair for Whaling and Taylor to lose their jobs over this? Should the p.c. police be trawling Facebook for thought-crimes to punish people for IRL?

While I do object to p.c. thought-policing, that isn't a good description of what happened here. Mayor Whaling resigned under public pressure: those are the breaks. When you're an elected politician, you serve at the whims of the Mob, and your speech has a lot to do with it. And as I understand it, Taylor was removed in the first instance not for the content of what she posted, but rather because her post raised reasonable concerns in the Governor's office about the quality of service that Taylor's agency was providing to county residents under her leadership. And let's be real here: if you're a county-level services provider and the Governor of the state knows you by name, you done fucked up.

3.

Milo Yiannopoulus, a controversial right-wing pundit, recently signed a lucrative publishing contract with Threshold Editions, an imprint of Simon & Schuster. This has led some people on the Left to contemplate a publisher boycott. The Chicago Review of Books tweeted this:


I'll be interested to hear what the American Library Association says about this. The ALA hosts an annual campaign called "Banned Books Week", and although the word "banned" usually overstates what the ALA is criticizing, the basic idea that animates Banned Books Week is excellent:
Banned Books Week brings together the entire book community; librarians, booksellers, publishers, journalists, teachers, and readers of all types, in shared support of the freedom to seek and to express ideas, even those some consider unorthodox or unpopular.
By (literally) banning all of Simon & Schuster's books from its pages, the Chicago Review of Books directly contradicts these aims.

I hope this decision is reversed; it isn't even consistent with the journal's stated mission:
We seek to explore the connections between literature, current events, and pop culture. Our contributors are as diverse as the books we cover, from established authors, journalists, and writing professors to students and freelance critics.
What happens when even our literary journals lose faith in the power of words? The Chicago Review ought to fight Yiannopoulos's book by doing what it does best: writing a review of it.

Friday, December 23, 2016

Matthew 2:1-12

My sister has sent us a beautiful trifold Christmas card that opens up to show a 15th century painting of the adoration of the magi. As my wife and I were admiring the painting, we had to admit that neither of us really knew the details of this Bible story. So we consulted our King James Version. We were surprised by what we read. For example, I hadn't known that before visiting Jesus, the wise men checked in with Herod when they arrived; nor had I known that the news they brought raised an alarm in all of Jerusalem. I hadn't known that Herod tried to use the wise men to locate the infant, nor that the wise men sneaked out of town rather than tell Herod where Jesus lived. The story is far more ominous than I had understood. (Herod's instruction to the wise men is especially chilling to read.)

In case you prefer a more accurate translation than the KJV, below I have collected together the ESV, NET, and NRSV. These differ among themselves only in small ways.

And for your enjoyment as we head into the Christmas weekend, here is the beautiful painting from my sister's card: Adoration of the Magi, by Gentile da Fabrano (1385–1427). A larger image is here.



Now after Jesus was born in Bethlehem of Judea in the days of Herod the king, behold, wise men from the east came to Jerusalem, 2saying, "Where is he who has been born king of the Jews? For we saw his star when it rose and have come to worship him." 3When Herod the king heard this, he was troubled, and all Jerusalem with him; 4and assembling all the chief priests and scribes of the people, he inquired of them where the Christ was to be born. 5They told him, "In Bethlehem of Judea, for so it is written by the prophet

6"'And you, O Bethlehem, in the land of Judah,
    are by no means least among the rulers of Judah;
for from you shall come a ruler
    who will shepherd my people Israel.'"

7Then Herod summoned the wise men secretly and ascertained from them what time the star had appeared. 8And he sent them to Bethlehem, saying, "Go and search diligently for the child, and when you have found him, bring me word, that I too may come and worship him." 9After listening to the king, they went on their way. And behold, the star that they had seen when it rose went before them until it came to rest over the place where the child was. 10When they saw the star, they rejoiced exceedingly with great joy. 11And going into the house, they saw the child with Mary his mother, and they fell down and worshiped him. Then, opening their treasures, they offered him gifts, gold and frankincense and myrrh. 12And being warned in a dream not to return to Herod, they departed to their own country by another way.

        —ESV, available online.


1After Jesus was born in Bethlehem in Judea, in the time of King Herod, wise men from the East came to Jerusalem 2saying, "Where is the one who is born king of the Jews? For we saw his star when it rose and have come to worship him." 3When King Herod heard this he was alarmed, and all Jerusalem with him. 4After assembling all the chief priests and experts in the law, he asked them where the Christ was to be born. 5"In Bethlehem of Judea," they said, "for it is written this way by the prophet: 

   6'And you, Bethlehem, in the land of Judah,
   are in no way least among the rulers of Judah,
   for out of you will come a ruler who will shepherd my people Israel.'"

7Then Herod privately summoned the wise men and determined from them when the star had appeared. 8He sent them to Bethlehem and said, "Go and look carefully for the child. When you find him, inform me so that I can go and worship him as well." 9After listening to the king they left, and once again the star they saw when it rose led them until it stopped above the place where the child was. 10When they saw the star they shouted joyfully. 11As they came into the house and saw the child with Mary his mother, they bowed down and worshiped him. They opened their treasure boxes and gave him gifts of gold, frankincense, and myrrh. 12After being warned in a dream not to return to Herod, they went back by another route to their own country.

        —NET, available online.


1In the time of King Herod, after Jesus was born in Bethlehem of Judea, wise men from the East came to Jerusalem, 2asking, "Where is the child who has been born king of the Jews? For we observed his star at its rising, and have come to pay him homage." 3When King Herod heard this, he was frightened, and all Jerusalem with him; 4and calling together all the chief priests and scribes of the people, he inquired of them where the Messiah was to be born. 5They told him, "In Bethlehem of Judea; for so it has been written by the prophet: 6"And you, Bethlehem, in the land of Judah, are by no means least among the rulers of Judah; for from you shall come a ruler who is to shepherd my people Israel.' " 7Then Herod secretly called for the wise men and learned from them the exact time when the star had appeared. 8Then he sent them to Bethlehem, saying, "Go and search diligently for the child; and when you have found him, bring me word so that I may also go and pay him homage." 9When they had heard the king, they set out; and there, ahead of them, went the star that they had seen at its rising, until it stopped over the place where the child was. 10When they saw that the star had stopped, they were overwhelmed with joy. 11On entering the house, they saw the child with Mary his mother; and they knelt down and paid him homage. Then, opening their treasure chests, they offered him gifts of gold, frankincense, and myrrh. 12And having been warned in a dream not to return to Herod, they left for their own country by another road. 

        —NRSV, available online

Tuesday, December 20, 2016

Holiday Challenge 2016

Welcome to the 2016 Holiday Challenge! This year, I offer a menu of choices—try any or all of them. Email your answers to zimblogzimblog@gmail.com.

Challenge 1

Let's play rock/paper/scissors! I have made my throw already; it's written down in a google doc. To play, just decide what your throw will be and send it in an email. (Easy, right?)

Challenge 2

Find the word CANDY in this candy cane word search. Email a photo of your solution.

Challenge 3 (quoted verbatim from Lewis Carroll)

A and B begin, at 6 a.m. on the same day, to walk along a road in the same direction, B having a start of 14 miles, and each walking from 6 a.m. to 6 p.m. daily. A walks 10 miles, at a uniform pace, the first day, 9 the second, 8 the third, and so on: B walks 2 miles, at a uniform pace, the first day, 4 the second, 6 the third, and so on. When and where are they together?

Challenge 4

Like You'd Understand, Anyway. That's the title of a 2007 book of short stories by Jim Shepard. It's an unusual title because its shortest word is four letters long. Can you think of any other published books (fiction or nonfiction) in which the shortest word in the title is at least four letters long? The more words in the title, the better.


PRIZES!

I have made a new puzzle book this year, Word Games 5I'll send a copy to:

  • The first person who beats me in rock/paper/scissors on their first try;
  • The first person who finds CANDY in the word search;
  • The first person who solves Lewis Carroll's problem;
  • The first person who sends another published book title of at least four words with each word at least four letters long. 

Plus, everybody who attempts any of the challenges will be entered into a drawing to win a copy of the book.

Enjoy! Contest ends December 31st at 11:59:59 p.m. Eastern time.

Sunday, December 11, 2016

Points In A Circle Problem: Summary To Date

The problem we've been considering is,
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.
Note that the sense of 'containing a point' is strict: if a point lies on the boundary of a triangle, the triangle is not considered to contain the point. (Again I take the extrema to exist; this seems safe given the closed and bounded features of the problem.)

Here is a summary of where things stand:
  • For N = 1, placing the obstacle at the center of the disk results in a greatest feasible triangle area of 1. For any other location of the obstacle, there exists a feasible triangle of area greater than 1. Thus, the center of the disk is the unique solution to the N = 1 problem.
  • For N = 2, placing one obstacle at the center of the disk and the other obstacle anywhere in the disk results in a greatest feasible triangle area of 1. For any other arrangement of the obstacles, there exists a feasible triangle of area greater than or equal to 1. Thus, the center of the disk and an arbitrarily located second point together represent a solution (not unique) to the N = 2 problem. 
  • For N = 3, placing the three obstacles in the locations we have been denoting O1, O2, and O3 results in a greatest feasible triangle area of 0.8286369…. Our conjecture, not yet proven, is that no arrangement of three obstacles results in a smaller value than this for the greatest feasible triangle area.
A more economical way to summarize is to generalize the notation we used in the N = 2 solution. Given N obstacle points X1, …, XN in the closed unit disk, let

aN(X1, …, XN)

denote the greatest possible area of a triangle in the disk that doesn't contain any of the Xi. Let

aN* = minimum possible value of aN(X1, …, XN).

Then for each given value of N, our problem is to calculate aN* and find N points Xi* for which aN(X1*, …, XN*) = aN*.

Using this notation, our summary is:

  • a1* = 1 and a1(X) = 1 if and only if X = O, the center of the disk.
  • a2* = 1 and a2(X1, X2) = 1 if X1 = O or X2 = O.
  • a3* ≤ a3(O1, O2, O3) = 0.8286369….

It's been a lot of work for just a handful of equations! Still, getting to these results has been fun, and has required growing amounts of insight about the optimization. Some of this likely generalizes to arbitrary numbers of obstacles in arbitrary locations—such as the idea that if you know two vertices in a greatest-area feasible triangle, then the third vertex has to be one of a finite number of possibilities.

I will keep thinking about the N = 3 problem, although it's hard not to cheat and start exploring N = 4. Would any of my readers like to make a conjecture for the best obstacles in the N = 4 case? I'll stay out of the way while you work on it. Email any thoughts to zimblogzimblog@gmail.com.

***

A note about N = 3. The three obstacles O1, O2, and O3 are each a distance 0.32096… from the center of the disk. I have been using decimals and dot-dot-dot to express this distance, but it can also be expressed exactly as the real root of the cubic polynomial 5x3 − 4x2 + 7x − 2, namely


And the resulting greatest area (0.8286369…) can also be expressed exactly as the positive root of the polynomial 160000x6 − 265824x4 +  412857x2 − 209952, namely


To calculate these values, I first generalized the coordinates of O1, O2, and O3 to be functions of a variable radius r. Then I calculated the vertices of the spotlight triangle and the edge triangle in terms of r. Because the spotlight triangle and the edge triangle share a common base, the two triangles have equal areas when their altitudes are equal. The equal-altitude condition turns out to be equivalent to 5r3 − 4r2 + 7r − 2 = 0.

(Note, even Mathematica won't be able to solve the equal-altitude equation unless you are careful to orient your diagram so that the common base is horizontal. Doing so leads to much simpler expressions in r.)

Book Review: A Great Reckoning

A Great Reckoning

(the 12th Inspector Gamache novel)

By Louise Penny

Minotaur Books, 2016

Hardcover, 386 pages











In 2014, I called Louise Penny's How the Light Gets In "a triumph in the genre and an absorbing novel altogether." A Great Reckoning is a regression to the mean for the Inspector Gamache series, but that's still a pretty high mean.

It is easy to draw parallels between Louise Penny's Inspector Gamache series and the Adam Dalgleish series of books by P.D. James. James's hero and Penny's are similarly principled, brave, charismatic, and soulful. Both command teams of younger investigators who receive lessons in life and work from their mentors. The cases described in the books take both detectives from their city offices to more intimate settings—Gamache drawn somehow repeatedly to the cozy village of Three Pines, Dalgleish posted to a variety of English locations, typically with a touch of the gothic about them.

Phlegmatic, Francophone, and fulfilled in monogamy, Armand Gamache is also an homage to George Simenon's Inspector Jules Maigret. The Maigret stories broke with the "puzzle story" tradition of early mystery novels by putting emotional motivations and psychological insight at the center of sleuthing. Penny's books are also intensely psychological, though her practice of spelling everything out could not be more different from Simenon's show-don't-tell minimalism. So much explicitness on Penny's part leaves her readers with little need, or room, to discover meaning on their own.

The emotional themes can be treacly, verging on self-help—the kind of thing one hears in a second-rate commencement speech. But I don't mind it much. Sometimes we need help, and sometimes it can only come from ourselves. Read the Acknowledgments section at the end of this book before you judge.

A good sentence in a novel might be complex (P.D. James) or simple (George Simenon); but either way, sentences are the best way to write literature. The writing in A Great Reckoning includes many staccato sentence-fragments. This is a difficult technique that sometimes succeeds in enhancing drama but sometimes fails by sounding mawkish and theatrical.
But on the village green itself stood the the three tall pines from which the village took its name. Vibrant, straight and strong. Evergreen. Immortal. Pointing to the sky. Daring it to do its worst. Which it planned to do. 
I can't read this and not hear a campily ominous "Soon" as fragment number seven.
He found them in the weight room, where Leduc worked out, searching the lockers. For clues.
How about, "searching the lockers for clues"? Or just: "searching the lockers"? From the context—remember, this is a crime novel—it's pretty clear that these these people are searching for clues. That makes the extra emphasis not just melodramatic, but, again, campy. The temptation to add a silent "duh-duh-duuuuuh" to the end is, to me, irresistible.
A cup of tea sat on his desk beside a couple of chocolate chip cookies. Uneaten.
Well, sure—had the cookies been eaten, they would be in Gamache's stomach, not on his desk. Untouched is the word Penny wanted here. 
Amelia sat forward, leaning toward the stage. Even after the Commander had disappeared. She stared at the empty space once occupied by him.
A good sentence here would have been more effective and more considerate of the reader.

There are some other problems with the writing, such as when two people walk into a bistro brushing wet snow from their coats, and we are confusingly told that outside the bistro it is sleeting, not snowing. In the otherwise effective opening scene, a careful description of Gamache's study omits to tell us that his dog Henri is the room with him, so that Henri bursts awkwardly into the reader's mind mid-stride as he follows Gamache out of the study. One never has such experiences when reading George Simenon or P.D. James.

Of course the writing is also good in places. I laughed when it was said of a profusely sweating man, "If people were mostly water, then this young man was more human than most." And here is a good passage from an exchange between Gamache and his former boss, the corrupt Michel Brébeuf:
Michel Brébeuf looked at Gamache with undisguised tenderness.
But what, Armand asked himself, did that tenderness itself disguise? What was lurking, swishing its tail, in those depths?
(The "swishing its tail" is nice.) The character of Brébeuf exerts strong forces on Gamache, not only because of his dark power, but also because the two were childhood friends. Gamache's vulnerability to Brébeuf's dangerous pull is elegantly conveyed here:
Gamache listened, but didn't nod. Didn't agree or disagree. He was bending much of his will to disengaging from Brébeuf, while still listening closely.
It shouldn't be passed over that Penny, like James a female author, succeeds very well at drawing her main character, with all his complex masculine traits. Relations between men are also well observed, as in this brief exchange between Gamache and Paul Gélinas, a Deputy Commissioner of the Royal Canadian Mounted Police, recently widowed.
(Gélinas speaking) "Brébeuf wouldn't kill the only person not just willing but happy to keep him company. What did he call Leduc?"
(Gamache answering) "His life raft. And now? Are you still lonely?"
     "I was talking about Brébeuf."
     "Oui."
He paused to let Gélinas know he was listening, if he wanted to talk. The RCMP officer said nothing more, but his lips compressed, and Gamache turned away to give the man at least the semblance of privacy.

Of the half-dozen or so Inspector Gamache novels that I've read, few match the artistic standard set by most Simenon and James novels. Some ingredients are there, with lifelike characters and cases that put interesting themes on the table. But the books also have limitations that prevent them from rising above the status of genre fiction. Penny might be perfectly comfortable as an excellent genre author, but I wonder if a more forceful literary editor could help her transition from writing good books to writing great ones.


Read this review on Amazon.

In case you missed it: My list of favorite genre fiction.

Sunday, December 4, 2016

Points In A Circle Problem: The Best Triangles for the Conjectured Best Obstacles

Previously we established that for the obstacles O1, O2, and O3 shown here, the greatest-area feasible triangle must have all three of its vertices on the boundary of the disk.




(This needn't be the case for larger numbers of obstacles, as suggested by this N = 400 example.)



In today's post, we'll (finally!) solve the problem of finding the greatest-area triangle that doesn't strictly contain O1, O2, and O3. Later on, I'll write another post summarizing where things stand for the Points In A Circle Problem as a whole.

The starting point for today generalizes an observation from the solution to the N = 1 problem, as follows:

If two vertices are known in a greatest-area feasible triangle, then the remaining vertex must be one of a finite number of possibilities.

To see this, let obstacle points X1, …, XN be given in any arrangement. (The figure below shows an example for N = 3.) Given two vertices A and B in a greatest-area feasible triangle ABC (recall that A and B are on the disk boundary by hypothesis), define A1 as the projection of A through X1 to the disk boundary on the other side, and define B1 as the projection of B through X1 to the disk boundary on the other side; and similarly for A2, …, AN and B2, …, BN.


If vertex C (also on the disk boundary, by hypothesis) were located inside the proper arc A1B1, then triangle ABC would contain obstacle X1. Conversely, if vertex C were not inside the proper arc A1B1, then triangle ABC would not contain obstacle X1. The same is true for any other obstacle, and hence triangle ABC strictly contains obstacle Xi if and only if C belongs to the proper arc AiBiThe set of feasible locations for vertex C is therefore the complement of the union of all the proper arcs AiBi. This is, itself, a set of (closed) arcs on the boundary. 

Now, given a segment AB in the disk, we maximize the area of a triangle on base AB by putting the third vertex as far as possible perpendicularly from AB, subject to whatever constraints operate.

For example, if the only constraint is that the triangle must belong to the disk, then we maximize the area of triangle ABC by making vertex C one of the two points Q and Q' on the disk boundary where the tangent to the boundary is parallel to AB. These points are locally furthest from base AB (measuring distances perpendicularly).



Note that diameter QQ' bisects chord ab, so that aQ is congruent to bQ and aQ' is congruent to bQ'. Triangles abQ and abQ' are both isosceles.

The feasible region for vertex C is a union of closed arcs. Suppose first that C is in the interior of one of these arcs. Then, we can slide C back and forth by a small amount within that arc.


By the altitude property, either the area of the triangle will increase and decrease to first order as C slides back and forth, or else vertex C occupies one of the two points Q, Q' on the boundary where the sliding motion runs parallel to AB and thus doesn't change the area to first order. But if triangle ABC is a greatest-area feasible triangle, then the area can't increase to first order by sliding vertex C. Therefore, if triangle ABC is a greatest-area feasible triangle, then C must be one of the two points Q and Q', or else an endpoint of one of the feasible arcs. This is a finite set of possibilities for vertex C, determined by A, B, and the obstacles X1, …, XN.

Next we generalize another observation from the N = 1 solution:

If a greatest-area feasible triangle doesn't have any obstacle points on its boundary, then the triangle is an inscribed equilateral triangle.

This observation is corollary of the previous one, since the only possibilities for the third vertex are the two special places where sliding the third vertex doesn't gain you any area. The upshot is that the triangle must be isosceles on each of its bases, and therefore equilateral.

With the two italicized conclusions in hand, we can specialize to the case of obstacles O1, O2, and O3. To begin with, an inscribed equilateral triangle is clearly infeasible for these obstacles, because it would contain all three obstacle points. Thus, a greatest-area feasible triangle for O1, O2, and O3. must have some side touching some obstacle point. Let's say without loss of generality that side AB touches O1.

This condition is enough to allow us to parametrize the space of possibilities by a single variable, say the angle θ in this diagram, which ranges from 0 to 90 degrees.


Given a θ value, A and B are determined, as are the finitely many possible candidates for the third vertex C. These candidates are denoted A2, A3, B2, B3, Q, and Q'. Here's an animation showing how the players move as the angle θ changes. Points leading to infeasible triangles are crossed out. 


For each angle θ, there are in principle six candidate triangles to consider. It isn't hard to determine their areas as explicit functions of θ. I began by writing the coordinates of vertex B as functions of θ:

Bx = cos θ [[1 − a02cos2 θ]] − a0 sin θ cos θ

By = cos θ [[1 − a02cos2 θ]] + a0 cos2 θ

where [[ … ]] stands for a radical (i.e., square root).

Then I defined a function on points as follows. For a point P on the boundary of the disk and a point X in the interior, let F(P, X) denote the point on the boundary obtained by projecting from P through X all the way to the boundary. This is a useful function because given vertex B, we can use the function F to generate all but two of the other points needed for the problem.

A = F(B, O1)
A2 = F(A, O2)
A3 = F(A, O3)
B2 = F(B, O2)
B3 = F(B, O3).

To generate the remaining two points, Q and Q', define the projection onto the disk boundary for points M ≠ 0 by G(M) = M/||M||. Then for θ < 90°, Q = G(½(A + B)) and Q' = G(−½(A + B)). For θ = 90°, set Q = (−1, 0) and Q' = (1, 0).

To obtain an explicit formula for F(P, X), solve the equation ||P + t(X − P)||2 = 1 for t (the equation is linear after you discard the root t = 0; don't forget that ||P|| = 1), and plug the result into ||P + t(X − P)|| to obtain

F(P, X) = P + 2(1 − X•P)||X − P||−2(X − P)

where X•P denotes the inner product. With this explicit expression for F, the coordinates of all the points can be calculated explicitly as functions of θ, and hence all the candidate triangle areas can also be calculated.

If we simply graph the areas of triangles ABA2, ABA3, ABB2, ABB3, ABQ, and ABQ' as functions of θ on a single set of axes, we get a complicated picture:


Some of these areas are quite large; however, the largest areas belong to infeasible triangles. In order to sort all this out, let's chunk the θ range into parts, each with their own state of affairs.

1. When you watched the animation, you might have noticed the three points A, O2, and O3 becoming collinear at around 26°—actually, 26.0016…°—a value we'll denote by θ*. At this angle, points A2 and A3 coincide, and the feasible triangles are ABQ, ABB3, and ABA3



Triangle ABA3 is none other than the spotlight triangle, with area 0.828637…. The other two feasible triangles, ABB3 and ABQ, have smaller areas (0.452… and 0.681…, respectively). 

So for θ = θ*, the best triangle is the spotlight triangle.


2. For 0 ≤ θ < θ*, the order of the points around the disk boundary is stable (B, Q, A, B3, B2, Q', A3, A2), and the feasible triangles are ABQ, ABB3, and ABA2. The figure shows the example of 17°.



Here is a graph of how the three feasible triangle areas vary over the range 0 ≤ θ < θ*.


Over this range, none of the feasible triangles reaches or exceeds the benchmark 0.828637… (shown as a dashed line).


3. For θ* < θ < 60°, the ordering is stable (B, Q, A, B3, B2, Q', A2, A3), and the feasible triangles are ABQ, ABB3, and ABA3. The figure shows the example of 48°.



Here is a graph of how the three feasible triangle areas vary over the range θ* < θ < 60°.



Over this range, none of the feasible triangles reaches or exceeds the benchmark 0.828637…. 


4. At θ = 60°, A coincides with B3 and B coincides with A3, making B, A, O1, and O3 collinear, and giving nontrivial feasible triangles ABQ, ABB2, and ABA2.



The feasible triangles for θ = 60° are none other than the spotlight triangle and the edge triangle, which (thanks to the particular choice of a0) have equal areas, namely the benchmark area 0.828637….


5. For 60° < θ < 90°, the ordering is stable (B, A3, Q, B3, A, B2, Q', A2), and the feasible triangles are ABA2, ABA3, ABB2, and ABB3.



Here is a graph of how the four feasible triangle areas vary over the range 60° < θ < 90°.


Over this range, none of the feasible triangles reaches or exceeds the benchmark 0.828637…. 


6. At θ = 90°, the feasible triangles are ABA2, ABA3, ABB2, and ABB3



Triangles ABA2 and ABA3 have equal areas 0.596…, and triangles ABB2 and ABB3 have equal areas 0.453…. None of them bests the benchmark 0.828637….


Altogether then, among feasible triangles over the entire θ range, the spotlight triangles and edge triangles are best. 

Conclusion

The triangle of greatest area that doesn't strictly contain any of the obstacles O1, O2, or O3 is either a spotlight triangle or an edge triangle, with area 0.828637….

This settles the question for these particular obstacles, and it provides an upper bound for the problem of placing three 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.

Saturday, December 3, 2016

Two Rights and a Wrong

My kids liked these puzzles. In each group of three sentences, two sentences are right and one sentence is wrong. Which sentence is wrong?

1.

A flying fox is a fox.
A flying fish is a fish.
A flying squirrel is a squirrel.

2.

A spider monkey is a monkey.
A rhinoceros beetle is a beetle.
A tarantula hawk is a hawk.

3.

A glowworm is not a worm.
A whale shark is not a shark.
A butterfly is not a fly.

4.

Bluegrass is a kind of grass.
Acorn squash is a kind of squash.
Venus fly-traps are from Venus.

5.

An anglerfish is a fish.
A silverfish is a fish.
A puffer fish is a fish.