2008 AIME I Problem 8

Question

Find the positive integer

n

such that

\begin{aligned} &\arctan\frac{1}{3} + \arctan\frac{1}{4} \\ &\quad {}+ \arctan\frac{1}{5} + \arctan\frac{1}{n} = \frac{\pi}{4}. \end{aligned}

Answer

Combine two terms at a time using

\arctan x + \arctan y

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

(valid here since all products

xy

are small)

The first three arctangents sum to

\arctan\frac{23}{24};

the last term must top it up to

\arctan 1

For positive

x, y

with

xy \lt 1,

the tangent addition formula gives

\arctan x + \arctan y

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

Applying it twice:

\begin{aligned} &\arctan\frac{1}{3} + \arctan\frac{1}{4} \\ &= \arctan\frac{\frac{1}{3} + \frac{1}{4}}{1 - \frac{1}{12}} \\ &= \arctan\frac{7}{11}, \\ &\arctan\frac{7}{11} + \arctan\frac{1}{5} \\ &= \arctan\frac{\frac{7}{11} + \frac{1}{5}}{1 - \frac{7}{55}} \\ &= \arctan\frac{23}{24}. \end{aligned}

The equation becomes

\arctan\frac{23}{24}

+ \arctan\frac{1}{n} = \arctan 1,

so

\frac{\frac{23}{24} + \frac{1}{n}}{1 - \frac{23}{24n}} = 1.

Clearing denominators,

23n + 24 = 24n - 23,

giving

n = 47.

Problem 8 in other years