Saturday, December 02, 2006

Chicken with Breadcrumbs and Parsley

A year ago, I took the Putnam (a six-hour math contest) and during the break ate dinner at Cynthia and Angela's, who enforced a strict no-Putnam-discussion rule. Joel and I almost got kicked out before the meal started for our Putnam-discussion-discussion (we were trying to figure out the exact parameters of what we were allowed to say), but in the end we were allowed to eat dinner, which was chicken. Now, it is the Putnam break again, and I am eating chicken again. Can I make it through this blog post without discussing the Putnam? (No. Joel: what do you think so far? I liked #2 and I got it completely (or so I think). #1 seemed straightforward but I didn't actually compute the integral. I was on the way to getting #4 but then I ran out of time.)

Today's chicken is leftover from yesterday's dinner, which I'll describe now. This serves two and should take about an hour to make. You could make it quicker by using boneless chicken breasts and not making the breadcrumbs yourself (or replacing them with flour or cornmeal).

  • 1 bone-in chicken breast
  • 1 clove of garlic
  • parsley--more than you think, but I don't know how much--maybe 1/4 cup packed or so?
  • breadcrumbs, made from 4-5 slices of bread
  • olive oil (lots)
  • salt, pepper

Make breadcrumbs by putting bread in a 325 degree oven until it's very dry and then breaking it up. I don't have a food-processor so I put it into a ziplock bag and crushed it. It's tedious, but not that bad. Cut up the garlic and parsley very fine and mix them up with the breadcrumbs. (If you're using a food processor, you could just throw these in with the bread to be chopped.)

Take your chicken and make two fillets from it. I followed Mark Bittman's advice, which is to cut as close to the bone as possible, starting on the outside and ending on the inside. (By outside, I mean the side away from what must be the chicken's sternum.) It's a very good idea to make stock from this, because then you have an excuse for why you're leaving so much meat on the bones.

While you're doing all of that, heat a pan over medium-high heat until it's very hot and then add a good amount of olive oil (probably the more, the better). When the oil is very hot, dredge each piece of chicken in some other olive oil, salt them, dredge them in the breadcrumb mixture, and put them in the pan. (Do them one after another, not at the same time, so that the pan stays hot.) Add pepper to each side as you cook. When the outside is nicely browned on both sides (two minutes a side or so), turn the heat down to about medium and keep on cooking. Cook the chicken till it's done all the way through, probably another 2-4 minutes a side (but check often). If your fillets are very thick, consider transferring everything to the oven at 400 degrees after the chicken is browned. (I did this out of desperation, and it worked pretty well.)

Mark Bittman says that to do a good job of browning, you have to get the pan very hot before adding fat, and then get the fat very hot before adding meat. Is there really any reason to wait for the pan to get hot before adding oil? I usually just add oil to the pan when I first turn the stove on and let it heat up.

On the second half of the Putnam, I did the first two problems and didn't even work on anything past that. Joel?


Anonymous said...

Heh, I'd forgotten about that -- good times.

On the A side, that inequality actually describes a solid torus -- I thought it was my weakest solution on the first half, because I figured out it was a torus and couldn't think of how to explain it other than to intone, "Look, it's a torus!"
For #2, I used a density argument (after writing working out who would win starting with every number of stones from 1 to 27 and deciding there was no pattern) (yay prime number theorem!) -- is that what you did? (Actually, on further thought, this might have been my weakest solution because I didn't define the word "density.")
Number 4 was nice -- you could do it inductively, looking at how many maxima you get when you insert (n + 1) into a permutation of {1, 2, ..., n}.
Number 3 I did with a cute trick -- recursions like that define sequences going backwards as well as forwards, and recursions like that always yield cyclic sequences in any mod. Apparently there's some way to figure out what the sequence is explicitly in terms of binomial coefficients.
For numbers 5 and 6, I drew some pretty pictures, but that was it.

I thought "dredging" was only for coating with solids (the breadcrumbs), not liquids (the olive oil) -- sounds simple and delicious (except for having to be your own food processor).

On the B-side, I was very exasperated. I think number 5 was actually one of the easiest questions there. The two combinatorial geometry problems (3 and 4) threw me for a loop (I wrote some garbage down for the one with line partitions), which annoys me to no end. I also thought number 1 was obnoxious (it took me an hour and a half to actually consider that the "curve" might not really be curvy and I should try to draw it). I can't even remember what problem number 2 was -- have I confused A2 and B2?, and I didn't so much as look at problem number 6 (it had a limit and roots, so I gave up on it immediately). All told, I thought it was marginally easier than last year.


Toby said...

I think you're right about the word dredging. As for the Putnam, I think I screwed up the first problem, because I did not find it to be a torus. For #2, I assumed that there were only a finite number of winning numbers for Bob, and then generated a winning number not on the list (using the fact that for a number to be winning for Alice, you have to be able to subtract a prime-1 and get a winning number for Bob). I didn't use the prime number theorem, but I guess I could have--I needed a long stretch of composite numbers, which I made by considering n!+2, n!+3, etc. I worked on 3 for a while but didn't get anywhere, and I was writing up a very messy solution for #4 involving a recurrence when time ran out--hopefully I can get a few points from that, but it seems unlikely.

For the B-side, I spent the first hour and a half doing #1--my solution was fine, but it was really just a big mess of algebra (substituting a=x+y-1 and b=xy was the main step). Is there some easy way to do it (i.e., to prove tht there's only one point off the line)? #2 was proving that for any n numbers, there exists a subset of them summing to within 1/(n+1) of an integer. I spent all the rest of my time working on that, but I at least got the reward of a nice solution: Let S_0=0 and S_k be [x_1+...+x_k] (i.e., the fractional part). If any S_i and S_j are within 1/(n+1) of each other, then we have a sum within 1/(n+1) of an integer, since S_j-S_i is a sum as well (namely x_{i+1}+...+x_j). Since there's n+1 of these numbers, there must be two of them <= 1/(n+1) apart. Too bad I never got to 5, whatever it was.

I'm not sure how to compare this to last year's because I had a lot more trouble with time this year--I never even looked at about half the problems.

Anonymous said...

Ah, yes, B2 did have that nice solution! (I forgot it because I didn't agonize about it, unlike some others on that half.)

As for B1: The equivalent of the substitution a = x + y - 1 is to notice that x^3 + 3xy + y^3 - 1 factors as (x + y - 1)(x^2 +y^2 - xy + x + y + 1), so either x + y - 1 = 0 (the line) or x^2 + y^2 - xy + x + y + 1 = 0. This latter expression seems rather mysterious, but it happens to be equal to exactly one half of (x - y)^2 + (x + 1)^2 + (y + 1)^2, a sum of three squares and thus a non-negative number. It can be zero only if x = y = -1, so there's the lone point out there.

Long stretch of composite numbers sounds like a good approach (and it makes me feel better about that question, since I thought requiring the use of the Prime Number Theorem was a little absurd).

Working towards the recurrence in A4 was definitely the right approach, so you might pick up some points there.

B5 was the following: for a continuous function f (actually, I think the only necessary condition is integrable, but whatever), define I(f) to be the integral of x^2f(x) between 0 and 1, and J(f) to be the integral of xf(x)^2 between 0 and 1, and compute the maximum possible value of I - J. It can be solved more-or-less directly from the fact that y^2 + z^2 >= 2yz for all real numbers y, z, with appropriate substitutions.

Oh, I might as well talk this up, too: I almost got put on the Harvard team. Apparently I was designated first alternate, and one of the team members showed up a little bit late. I'm not sure whether I'm relieved or disappointed.


Toby said...

Sounds like you did pretty well. Maybe you'll do better than an honorable mention this year? I'm pretty sure I'll get in the thirties, which is a little bit worse than last year but not really too different.