42 Points

From the 2007 Australian Maths Competition:

There are 42 points $P_1, P_2, ..., P_{42}$ , placed in order on a straight line so that each distance from $P_i$ to $P_{i+1}$ is  $\dfrac{1}{i}$  for $1\leq i\leq 41$. What is the sum of the distances between every pair of these points?

Upon first glance, this problem is begging for a clever shortcut, and the fun will no doubt come from finding it.

We are asked to sum up all possible distances between pairs of points, but instead of considering pairs one by one (which will take too long), we can consider just the ‘chunks’ $P_iP_{i+1}$, since every line segment joining two points is just a combination of such chunks.

So in adding up all possible distances, how many times must we count the section $P_iP_{i+1}$ ? It will contribute to every line segment whose left endpoint is to the left of (or equal to) $P_i$ , and whose right endpoint is to the right of (or equal to) $P_{i+1}$ . There are $i$ choices for the left endpoint and $42-i$ choices for the right endpoint, thus the section $P_iP_{i+1}$ must be counted $i(42-i)$ times in total. Recalling that this section has length $\dfrac{1}{i}$ , it follows that the contribution made to the total distance by this particular section is

$\dfrac{1}{i} \times i(42-i) = 42 - i$ .

This is much friendlier than the problem originally appeared; now we just have to sum $42-i$ from $i = 1$ to $41$, to obtain a final answer of

$41 + 40 + \cdots + 1 = \frac{41\times 42}{2} = \boxed{861}$ .

A fantastic problem, designed in such a manner that with the right shift of focus, most of the arduous calculation melts away.