## Back to Basics: Arctangent

Recently, a student asked me about problem #8 on the 2008 AIME 1 competition, which is reproduced here for your convenience:

Find the positive integer $n$ such that

$\arctan \frac{1}{3} + \arctan \frac{1}{4} + \arctan \frac{1}{5} + \arctan \frac{1}{n} = \frac{\pi}{4}$.

She was able to solve this using the sum of arctangents identity

$\arctan x + \arctan y = \arctan \frac{x+y}{1-xy},$

but was wondering if there was another way less algebraically involved.  Also, what do you do if you can’t recall this identity?

The problem certainly suggests using the sum of arctangents formula and using that formula certainly works. If you forget the formula, you could derive it during the contest, and I’ll say more on this point further on. But, often, contest problems can be cracked by going back to basics and applying a little clear-headed thinking and a few strategic calculations.

The left hand side of the equation in the problem is a sum of four angles. Let’s recall that by definition, in a right triangle with legs of length 1 and x, the measure of the angle opposite the leg of length x is equal to the arctangent of x.  For example:

This suggests the following approach to the AIME problem (the following figures are not drawn to scale):

We then supplement that by an angle of measure arctan 1/4 by “going over 4 and up 1”:

And then we add an angle of measure arctan 1/5 by “going over 5 and up 1”:

And now, we have to figure out $n$ so that if we “go over $n$ and up 1″, we end up on the line $y=x$. This amounts to computing for which value of $n$ the point $(48n, 46n) + (-46, 48)$ has equal coordinates. That is, we seek $n$ so that:

$48n - 46 = 46n + 48$.

This is a linear equation in one variable whose solution is $n = 47$. Done!

Explaining things tends to use much more space than what is needed to actually carry out the computation, so to give an even better sense of just how short and quick this solution is, here is my actual scratch work for this problem:

It looks like the arctangent function just disappears entirely!

Often, on competitions, a problem that strongly suggests the use of some advanced identity lends itself to a simple solution if one goes back to basics. That’s because constraints on contest problems sometimes force problem designers to use specific situations whose numbers work out so conveniently that it actually saves some effort to not use the general formulas associated with the problem content. This doesn’t always happen, of course, but in any case, it’s still useful to see if you can find a solution of this elemental nature.  Doing so increases your understanding of the basics.

By the way, the solution also shows how one can recover the sum of arctangents formula (because, of course, the solution does add up angles that are specific arctangents!):

Thus $\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}.$