Did anyone here take the Putnam exam this weekend (or administer it)?

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  • #1
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Did anyone here take the Putnam exam this weekend (or administer it)? What did you think?
 

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  • #2
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It had an unusually hard A1 and an unusually easy B1
 
  • #3
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Part B was notably harder than A.

I made way too many dumb mistakes. I could've tripled my score if I had just thought more clearly during the test.
 
  • #4
morphism
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Part B was overly difficult IMO. I was also surprised they had a real group theory question (A5)! (Which, IIRC, is an easy exercise in Dummit & Foote.)
 
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  • #5
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Personally I thought it was a reasonable test. I kinda burned out on the first half though and I screwed up part B. Spent 2 hours on B2 and didn't really work on B4 or B5 at all...
 
  • #6
Huh. Sorry for making a new thread. I made my thread earlier this afternoon and hit submit, but my internet froze. I just now resubmitted my thread, then saw this thread.
 
  • #7
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It's ok. By the way, semi-official solutions are http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml" [Broken]. I say "semi-official" because they're not official, they're done by Kiran Kedlaya and his compatriots, but everyone regards them like they would official solutions.
 
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  • #8
Integral
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I have moved this thread to GD because it is asking about the test and NOT the math. The other thread was looking at the MATH so it is still in General Math.
 
  • #9
morphism
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Personally I thought it was a reasonable test. I kinda burned out on the first half though and I screwed up part B. Spent 2 hours on B2 and didn't really work on B4 or B5 at all...
I had similar problems, also due to B2. Initially I could only get 1/2 in the upper bound, and not 1/8. But an hour later I realized that there was some vital piece of information in my construction that I wasn't using. Namely, I defined F(x) = [itex]\int_0^x f(t) \, dt[/itex], then because F(1)=F(0)=0, we get (by Rolle's theorem) a c in (0,1) such that F'(c)=0=f(c) (the last equality being the FTC), so now we can use the MVT or a local approximation to proceed from here. What I missed - for a very long time - was that c is an extreme point of F!

After seeing the two-line solution to B5 I'm kicking myself.
 
  • #10
Gokul43201
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No opinions on A2?
 
  • #11
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A2 was easy if you know a way to find the area of a quadrilateral (or even a triangle) given the coordinates. You just pick four appropriate example points, compute, and apply AM-GM to prove the lower bound. (If you don't know such a formula, I think it becomes pretty hard.)
 
  • #12
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The question doesn't even ask for a proof. A good guesser (or a lucky one, like me) will see the square as the likely solution, write down the area, and move on. Total time: 1 minute, tops!

Going out on a limb, the convex set with maximal area probably has an area of 8.
 
  • #13
morphism
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The question doesn't even ask for a proof. A good guesser (or a lucky one, like me) will see the square as the likely solution, write down the area, and move on. Total time: 1 minute, tops!
And get 1/10!
 
  • #14
Gokul43201
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And get 1/10!
:rofl: So that's how it works! If they don't ask me for a proof, I wouldn't think to give 'em one! :biggrin:
 
  • #15
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Probably 0 for that problem as the answer isn't particularly hard to guess.

What do you mean by maximal? Do you mean maximal without going to the other side of the hyperbolas? Because the original question just asked for the set to hit all four branches of the hyperbolas, and you can make arbitrarily large convex sets with that property. But if you mean the set has to be contained within the hyperbolas, then you might be right. (Then it's easy to see that the set may contain at most one point on each branch of the hyperbola.) That problem seems harder to me.
 
  • #16
Gokul43201
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Probably 0 for that problem as the answer isn't particularly hard to guess.

What do you mean by maximal? Do you mean maximal without going to the other side of the hyperbolas?
Yes, that's what I meant. So, I'd have scored a 0 on that one, damn!
 
  • #17
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Yeah, two years ago there was an integral for A5, and one of my friends said he guessed it although the answer was something like [tex]\frac{\pi \ln(2)}{8}[/tex] and it was completely non-obvious that this would be the case. That might have been worth a point. (But he didn't turn anything in!)
 
  • #18
What about for problem B3 for this year? I had a friend who was able to derive a legitimate formula for x(2007), but couldn't prove it. He wrote down his formula (which was correct), and wrote a little bit about how he derived it before time was up.

Do you think he'll get a point?
 
  • #19
morphism
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I think if the final answer is correct, but the method is sketchy, you'll usually get 1/10.
 

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