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.
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.
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.
Putnam problems are "shorter" than IMO problems. On the Putnam you only have half an hour to solve each problem, compared to one-and-a-half-hours 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 real-valued 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 x-1 is in S; and (c) If x is in S and x is not in {0,1}, then 1/(x(x-1)) 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^3-ab^2. For what real numbers c can one travel from 0 to 1 in a finite number of jumps with total cost exactly c?
A1 have a lot to do with infinitesimals way above what you learn in high school (Specifically A1 wants you to use line integrals) while B2 wants you to do some standard techniques learned in normal analysis classes and A4 require that you know a standard proof ie that roots to prime numbers can't be rational. I don't really understand how the median can be 0 to 2 points on that test though, it isn't all that hard. No, they are not. University maths helps a lot with IMO questions but usually you would never get anything like most IMO questions on exams. Maybe as homework.
A1 and A4 have elementary combinatorial solutions. B2 doesn't require anything beyond Riemann sums and the intermediate value theorem, which should be covered in AP calculus. But now you got me curious - how you would you solve A1 with line integrals?
I meant partial derivative or gradient, sorry, but with that it is really easy. You just state that if you have a rectangle and move two of the points outwards or inwards the sum must remain unchanged so the gradient must be opposite, do this again after changing an other side and you get that the whole line need to have zero directional derivative. Now since this line was arbitrary it means that all derivatives are zero so the function is constant and must therefore be zero. 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?
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.
Well, nothing in the proof really hinges on that, just that moving the function by a distance of D means that the two points in question needs to be altered in opposite directions. But yeah I am hurting a bit from the physics syndrome, sorry. In my world if nothing else is stated then functions are nice :p 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. I meant B2, the rectangle problem is fairly straight forward no matter how you try to solve it, as evident by my strange approach. I would solve B2 by just proving that the value gets less for every point you add and the value converges to the integral over 0,1 of the function x square. Also you need to show that there are no values in between you can't find since if you place the new point arbitrarily close to an old you can always move the value by a small distance.
But can you move only two points? Since you are initially restricted to squares, it seems that you would need to move three points at a time. 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.
But I wouldn't try this approach in the first place if I saw that there were squares instead of rectangles, this is mostly trying to patch it, but I get what you are saying. Still many putnam problems are just about calculus and there is a lot of them so I don't get it how half of the students get 0, especially since the bad ones wouldn't try to do it in the first place. 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.