This Fermi question is probably easy for some people: I’m guessing there are some women for whom the answer is zero (am I right?), and some men whose necktie-wearing is dominated by “every day at work, 5 days a week, 50 days a year.” For me it’s much harder. I guess I’d say — three or four times a year, as an adult? More when I was younger and went to more weddings. I’m gonna estimate I’ve put on a tie 150 times in my life.

Followup question: how many ties do you own? I think I probably have about 8, but 5 are in the “will never wear again” category, and one of the 3 “sometimes wear” is the American flag tie I only wear on July 4.

In a 100-friend scenario, the average white person has 91 white friends; one each of black, Latino, Asian, mixed race, and other races; and three friends of unknown race. The average black person, on the other hand, has 83 black friends, eight white friends, two Latino friends, zero Asian friends, three mixed race friends, one other race friend and four friends of unknown race.

Going back to Chris Rock’s point, the average black person’s friend network is eight percent white, but the average white person’s network is only one percent black. To put it another way: Blacks have ten times as many black friends as white friends. But white Americans have an astonishing 91 times as many white friends as black friends.

100 friends and only one black person! That’s pretty white!

It’s worth taking a look at the actual study they’re writing about. They didn’t ask people to list their top 100 friends. They said to list at most seven people, using this prompt:

From time to time, most people discuss important matters with other people. Looking back over the last six months – who are the people with whom you discussed matters important to you?

The white respondents only named 3.3 people on average, of whom 1.9 were immediate family members. So a better headline wouldn’t be “75% of white people have no black friends,” but “75% of whites are married to another white person, have two white parents, and have a white best friend, if they have a best friend” As for the quoted paragraph, it should read

In a 100-friend scenario, the average white person has 57 immediate family members.

A Mylar balloon I bought weeks ago for Dr. Mrs. Q’s birthday has been holding its helium astonishingly well, but has now lost enough that its density is almost exactly that of air. It doesn’t hug the ceiling or the floor, but slowly traverses the room, carried by air currents too weak for me to feel, I guess. And here’s the thing: as I sit here working, it keeps — well — visiting me. It’s kind of creepy, when it slowly floats into my peripheral vision and then taps me once or twice. Hey. Balloon here. OK, moving on. See you later.

“I got a message for you, if I could only remember. I got a message for you, but you’re gonna have to come and get it.” Kardyhm Kelly gave me a tape of Zopilote Machine in 1995 and I played nothing but for a month. “Sinaloan Milk Snake Song” especially. Nobody but the Mountain Goats ever made do-it-yourself music like this, nobody else ever made it seem so believable that the things it occurred to you to say or sing while you were playing your guitar in your bedroom at home might actually be pop songs. The breakdown at the end of this!

“I’ve got a heavy coat, it’s filled with rocks and sand, and if I lose it I’ll be coming back one day (I got a message for you).” I spent a lot of 1993 thinking about the chord progression in the verse of this song. How does it sound so straight-ahead but also so weird? Also the “la la la”s (“Sinaloan Milk Snake Song” has these too.)

“Roll me in the greenery, point me at the scenery. Exploit me in the deanery. I got a message for you.”

The first of these I ever heard. Douglas Wolk used to send mixtapes to Elizabeth Wilmer at Math Olympiad training. This was on one of them. 1987 probably. I hadn’t even started listening to WHFS yet, I had no idea who Robyn Hitchcock was. It was on those tapes I first heard the Ramones, Marshall Crenshaw, the Mentors (OK, we were in high school, cut us some slack.)

(Update: Douglas denies ever putting the Mentors on a mixtape, and now that I really think about it, I believe Eric Wepsic was to blame for bringing the Mentors into my life.)

Why is this line so potent? Why is the message never explicitly presented? It’s enough — it’s better — that the message only be alluded to, never spoken, never delivered.

Love this post from Matt Baker in which he explains the tropical / 2-adic proof (in fact the only proof!) that you can’t dissect a square into an odd number of triangles of equal area. In fact, his argument proves more, I think — you can’t dissect a square into triangles whose areas are all rational numbers with odd denominator!

The space of quadrilaterals in R^2, up to the action of affine linear transformations, is basically just R^2, right? Because you can move three vertices to (0,0), (0,1), (1,0) and then you’re basically out of linear transformations. And the property “can be decomposed into n triangles of equal area” is invariant under those transformations. OK, so — for which choices of the “fourth vertex” do you get a quadrilateral that has a decomposition into an odd number of equal-area triangles? (I think once you’re not a parallelogram you lose the easy decomposition into 2 equal area triangles, so I suppose generically maybe there’s NO equal-area decomposition?) When do you have a decomposition into triangles whose area has odd denominator?

What if you replace the square with the torus R^2 / Z^2; for which n can you decompose the torus into equal-area triangles? What about a Riemann surface with constant negative curvature? (Now a “triangle” is understood to be a geodesic triangle.) If I have this right, there are plenty of examples of such surfaces with equal-area triangulations — for instance, Voight gives lots of examples of Shimura curves corresponding to cocompact arithmetic subgroups which are finite index in triangle groups; I think that lets you decompose the Riemann surface into a union of fundamental domains each of which are geodesic triangles of the same area.

Dream: I meet two 9-year-old boys with identical long curly hair. They’re in a band, the band is called Bugbiter. They explain to me that most of their songs are about products, and they share their songs via videos they post on Amazon.

I share this dream with you mostly because I think Bugbiter is actually a legitimately good band name.

It’s that time again — March Math Madness, where we fill out an NCAA men’s tournament bracket with the best math department winning every game. As always, this bracket was filled out by a highly trained team consisting of myself and a group of procrastinating grad students, making decisions by voice vote, and if you disapprove of one of our choices, I’m sure it’s somebody else’s fault. This is Berkeley’s first championship after falling to Harvard in 2012; meanwhile, Michigan sees its second final in three years but falls short again…

Update: In the 34th percentile at ESPN after one day of play — thanks, Yale!

Update: Down to the 5th percentile and only Duke and UVa are left out of my final 8 picks. Not gonna be the math bracket’s finest year.