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Is it too late for me to do math problem solving? 
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#1
Aug1410, 03:48 PM

P: 11

Hello, I am an incoming senior. I've normally been pretty good in classroom mathematics. However, only until recently have I been exposed to "competitive problem solving". I literally have not heard of the AMCs until my junior year, when my friend from another school told me about it. I became really excited about trying it out, but looking at the curriculum, I found out that there's so much to learn and that these things aren't covered in high school math: advanced geometry, combinatorics, probability, number theory, graph theory, etc.
Now I'm self studying linear algebra and I'm doing fine. I was just wondering if it would be worth my time to study all these "problem solving" topics to try to be a good problem solver, or would it be just a waste of time? Originally I was planning on moving on to DiffEqs after finishing Linear Algebra. Then I found out theres also this thing called Putnam in college so maybe I could do that. I'm going to major in math by the way. Thanks. 


#2
Aug1410, 08:15 PM

P: 211

These extra topics are not hard to learn. Most of them are covered in a single college course called Discrete Math.
You might find it enjoyable to take the Putnam recreationally, but I wouldn't waste too much time preparing to take it competitively. One former high school math olympiad participant put it this way: In high school, math competitions were his creative outlet and an opportunity to go beyond the standard curriculum. In college, taking the Putnam competitively would hold him back from pursuing more interesting mathematics at the graduate level. 


#3
Aug1510, 10:55 AM

P: 11

Is university math the same flavor as IMO questions? I'm looking at these and boy are they hard!!! I've looked at questions from different years and I can only do 1 out of like the 10 or so questions I went through.



#4
Aug1510, 11:15 AM

P: 211

Is it too late for me to do math problem solving?
Putnam problems are "shorter" than IMO problems. On the Putnam you only have half an hour to solve each problem, compared to oneandahalfhours on the IMO. That does not mean that the Putnam is easy. The Putnam is scored out of 120 points and in most years the median score is between 0 and 2 points.
If you are curious, here are three questions from last year's exam that you should be able to solve with your current background. [A1] Let f be a realvalued function on the plane such that for every square ABCD in the plane, f(A)+f(B)+f(C)+f(D)=0. Does it follow that f(P)=0 for all points P in the plane? [A4] Let S be a set of rational numbers such that (a) 0 is in S; (b) If x is in S then x+1 is in S and x1 is in S; and (c) If x is in S and x is not in {0,1}, then 1/(x(x1)) is in S. Must S contain all rational numbers? [B2] A game involves jumping to the right on the real number line. If a and b are real numbers and b > a, the cost of jumping from a to b is b^3ab^2. For what real numbers c can one travel from 0 to 1 in a finite number of jumps with total cost exactly c? 


#5
Aug1510, 01:14 PM

P: 614




#6
Aug1510, 01:29 PM

P: 211

But now you got me curious  how you would you solve A1 with line integrals? 


#7
Aug1510, 01:56 PM

P: 614

At first I didn't see that here were squares but in the same way you can prove that if all squares fulfill this equation then all rectangles must as well. You can do this by showing that since moving the square do not change the value the sum of the derivatives in one direction must be zero, then put two squares next to each other and write out the movement things in opposite directions and you just proved that you can make the rectangle arbitrarily wider or thinner. Maybe you can do it through combinatorics but I think that it is a lot easier to do it through calculus since that is what I am the most used to. Also how do you solve problem 3 without transforming it into an integral? 


#8
Aug1510, 02:02 PM

P: 211

That's an interesting approach. I am not sure you can take partial derivatives though because you don't even know that f is continuous, let alone differentiable.
The combinatorial argument is pretty straight forward. Suppose you want to show that f(pt) = 0. Draw a square centered at pt, and then divide that into four smaller squares (each with a vertex at pt). Now you've got a picture with 6 squares. Use the condition for f on each of them and after a bit of algebra you can conclude that f(pt) = 0. 


#9
Aug1510, 02:08 PM

P: 614

Edit: Oh yeah, the rectangle thingy don't work then. You could probably show it in some other way, but yeah. Going over the river for water in this case. 


#10
Aug1510, 02:14 PM

P: 211

Now you can see why the median score is so low :) Make an extra assumption and you are down from 10 points to 1. There isn't much partial credit. 


#11
Aug1510, 02:30 PM

P: 614

Are these guys freshmen or what? I don't study in the US btw so I don't really know much about the putnam exam. 


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