How do you improve at competition-type maths?

Question

I'm having trouble doing basic competition-level problems and I'm not too sure why. It's mostly the kind of problems that stem from probability, counting, and recurrence relations word problems. For example, let's take the question below as an example.

Lines $L_1$, $L_2$, $L_3$, $L_4$, $\ldots$, $L_{50}$ are distinct. All lines $L_{4n}$, where $n$ are positive integers, are parallel to each other. All lines $L_{4n-3}$, where $n$ are positive integers, pass through a given point $A$. What is the maximum number of points of intersection of pairs of lines from the complete set $\{L_{1},\space L_2,\space\ldots,\space L_{50}\}$?

I'm having lots of trouble with these kinds of problems because I'm not sure where to start and how to tackle it. Do I first draw a diagram? Should I consider simpler cases first? I'm not sure what I'm supposed to do with these kinds of counting/probability problems.

Answer 1

I've posted my thoughts on the example problem in the comments.

But in regards to the broader question of how to improve on these problems, there isn't any cookbook approach. Here are some ideas:

• As littleO mentioned in the comments, check out the artofproblemsolving.com web site.
• I also like Martin Gardner's Mathematical Games column, which I grew up with; it's an entertaining source of recreational mathematics, much of which contest problems are drawn from.
• Problems are often simple questions with complex clothing; it's the reduction to that simple question that's the meat of the original problem. Sometimes, these reductions can be seen by sketching the situation, and seeing that there are more constraints than it seems at first glance.
• I'm a fan of guessing for many problems. It's often easier to confirm or deny a guess than it is to work the other way from the problem to a solution.
• Find a community of people with whom to work. This doesn't have to be online, although online communities are often easier to find and hook up with. But if you're at a school, seek out resources in the mathematics department. Socializing a problem is a great way to figure out approaches.

Not at all an exhaustive list. Just some thoughts to get the ball rolling.

Answer 2

For this problem, if you have $n$ lines in general position you have $n \choose 2$ intersection points. The conditions they give you reduce this number. Parallel lines don't intersect and there are $12$ lines that are parallel. This is $12 \choose 2$ points that are missing. The $13$ lines that go through a common point give you one point instead of $13 \choose 2$, so the maximum number of points is ${50 \choose 2}-{12\choose 2}-{13 \choose 2}+1$