2008 AIME I 第 8 题

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 $n = 47.$