Problem:
Find the positive integer n such that
arctan31β+arctan41β+arctan51β+arctann1β=4Οβ
Solution:
Applying the addition formula for tangent to tan(arctanx+arctany) results in the formula arctanx+arctany=arctan1βxyx+yβ, which is valid for 0<xy<1. Thus arctan31β+arctan41β+arctan51β=arctan117β+arctan51β= arctan2423β. The left-hand side of the original equation is therefore equivalent to arctan24nβ2323n+24β. Because this must equal arctan1, it follows that n=47β.
The problems on this page are the property of the MAA's American Mathematics Competitions