Number theory problem from 11th Iberoamerican olympiads Given a number $n \in \mathbb{N}$, such that $n>1$, let us consider all the fractions of the form $1 \over{ab}$, where $a$ and $b$ are coprime natural numbers such that $0<a<b \leq n$ and $a+b>n$. Prove that the sum of all such fractions is equal to $1 \over{2}$.

Induction on $n$. Let $S_n$ be your sum. The base case is easy:
$$S_2=\frac{1}{1\cdot 2}$$
In general, 
$$S_{n+1} = S_n-
   \left(
      \sum
      _{\stackrel{a=1}{(a,n+1-a)=1}}
      ^{\left\lfloor \frac n2 \right\rfloor} 
      \frac{1}{a(n+1-a)}\right) 
      + \left(
         \sum_{\stackrel{a=1}{(a,n+1)=1}}^{n}\frac{1}{a(n+1)}
        \right)$$
That is, we remove the pairs $a<b$ with $a+b=n+1$, since they no longer belong in the sum, and we add in the pairs $a,b$ with $b=n+1$.
Now $$\frac{1}{a(n+1-a)} = \frac{1}{n+1}\left(\frac{1}{a} + \frac{1}{n+1-a}\right)\tag{1}$$
From that, it is not hard to show that:
$$\sum_{\stackrel{a=1}{(a,n+1-a)=1}}^{\left\lfloor\frac{n}2\right\rfloor} \frac{1}{a(n+1-a)} = \sum_{\stackrel{a=1}{(a,n+1)=1}}^{n}\frac{1}{a(n+1)}$$
Essentially, $(a,n+1-a)=1$ if and only if $(a,n+1)=1$, and then use $(1)$ on the left hand side.
I'd love to see a more geometric proof of this, or one using probability.
