Mon 9 Nov 2009
GREEN CHICKEN
Posted by Ronit under Academics, Quasi-Academic Topics at 10:51 am
Excuse the all caps headline. This news comes to us via EphBlog math correspondent “12”
Williams recovered the Green Chicken from Middlebury in a hard fought battle Saturday, November 7, 2009, winning 199-180 (maximum possible score was a 240). The top four scores for each side made up the team; the Williams team was led by Nick Arnosti (tied for top score overall), Carlos Dominguez, Ralph Morrison, and Wei Sun. There were 40 participants, 19 from Williams and 21 from Middlebury (making this one of the largest competitions on Middlebury soil). Pictures and a copy of the test. Celebration dinner this Wednesday, November 11, 5:45-7 pm, Mission (Dennett Room); all welcome. Congratulations to all participants and fans, including Coach Steven Miller and Professors Beeson and Stoiciu.
For extra credit, solve it yourself:
(thanks to nuts for the image)
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12 Responses to “GREEN CHICKEN”
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David says:
1) Shout out to one of my former interns (Hai!) for participating in this contest, although, to judge by the pictures, he was there in 2008 (when Williams lost) and not in 2009 . . .
2) All good Ephs who want to avoid any “uncomfortable learning” today should not discuss the gender/racial breakdown of the participants . . .
Diana says:
Steve Miller is also a professor.
JeffZ says:
Good stuff. The Green Chicken certainly belongs on any list of the quirkiest / coolest Eph traditions (Mountain Day, The Walk, Trivia, etc. …).
nuts says:
Ronit here’s an image of the PDF if you want to post it.
David says:
It would be the geekiest EphBlog thread ever if we started to work this out together. So, let’s go!
3) This can’t be as simple as X = -1, can it? It has been a long time since I took calculus, but don’t you just take the derivative and set it equal to zero? That gives you X as either 1 or -1. Both work in the restricted set of allowed answers, so then you just plug in and see which yields the biggest answer.
Surely the problem is much harder then that . . .
Ronit says:
I was a miserable failure at math, but I did enable latex on ephblog a while ago when we did this thread… so feel free to post your work here.
syntax:
[latex]e^{\i \pi} + 1 = 0[/latex]
will display as…
dm '10 says:
David: setting derivative to zero gives you the points where the function “turns around” (i.e. local maxima and minima), but to find a maximum you also need to check the edges of the domain in case there’s some point at which the function keeps increasing without turning around. In this case the allowed answers are [-2,-1] and [1, 2], so you’d need to check 2 and -2 in addition to 1 and -1. It turns out that the maximum value occurs at f(2)=22.
Anyway, you’re right that that particular problem isn’t terribly difficult – it’s all fairly straightforward calculation and just remembering to check all the possibilities. Some of the others have more interesting tricks.
David says:
dm: Thanks! I did forget to check the endpoints. I realize that this should be obvious, but how did you determine that the endpoints are 2 and -2? (My 8th grader daughter is just starting to do math like this, so I need to get a clue!)
David says:
4) Let me try b first.
You want to have every possible pair of numbers produce a unique average. The number of different pairs is just
It is easy to have a set of numbers that produces this many pairs, just ensure that no two pairs sum to the same total. There are plenty of sets that have this feature. Easiest (for me) ensure that the numbers are of radically different sizes). So, how about:
for
David says:
Whoops. This is wrong because I don’t account for the fact that averages don’t care about order, so, the combination A and B is the same as the combination B and A. So, I need something like:
dm '10 says:
David: If you take the constraint inequality x^4+4 <= 5x^2 and rearrange it, you get x^4-5x^2+4<=0 which can be written (x^2-4)(x^2-1)<=0. So the roots are -2,-1,1, and 2 and by testing a few values you can see that the inequality holds from -2 to -1 and from 1 to 2.
Dick Swart says:
In the hubbub of Mr Moore, a phrase has been used by David to capture and provide a rationalization for the continuing back-and-forth with some vehemence, denial, reversal of fields, and the burning of straw men (a penny for the old Guy) to arrive at conclusions.
The phrase is “iterate to agreement”. I believe I catch the sense in which David applies the phrase: If we argue long enough, and you keep proving me wrong, it isn’t that I’m a bad person, it just means we are coming closer and closer to a correct answer that may not be what I was positive it was, but means that I helped you arrive at your answer.
Now certainly mathiness is next to Godliness, and Kepler’s Equation has been proven through iteration and agile software seem to include iteration in the defining process.
Can you who laugh in the face of the Green Chicken questions, develop a proof for the semantic process posited by David?
Beat Amherst!