Problem:
Find the positive integer n such that
arctan31+arctan41+arctan51+arctann1=4π
Solution:
Applying the addition formula for tangent to tan(arctanx+arctany) results in the formula arctanx+arctany=arctan1−xyx+y, which is valid for 0<xy<1. Thus arctan31+arctan41+arctan51=arctan117+arctan51= arctan2423. The left-hand side of the original equation is therefore equivalent to arctan24n−2323n+24. Because this must equal arctan1, it follows that n=47.
The problems on this page are the property of the MAA's American Mathematics Competitions