Let \(ABC\) be an equilateral triangle. Let \(P\) be a point inside, such that \(PA=6\), \(PB=12\), and \(PC=6\sqrt3\). Find the side length of the triangle.
Solution
Answer: \(6\sqrt7\)
Rotate triangle \(APC\) by \(60^\circ\) counterclockwise to triangle \(AP’B\), taking \(C\) to \(B\). In triangle \(APP’\), \(\angle P’AP=60^\circ\) and \(AP=AP’\), so \(PP’=6\).
Triangle \(PP’B\) has side lengths \(6,6\sqrt3,12\), making it a \(30\)-\(60\)-\(90\) triangle. The angle relationships give \(\angle APB=120^\circ\). By the law of cosines,
\[AB^2=6^2+12^2-2\cdot6\cdot12\cos120^\circ=252,\qquad AB=6\sqrt7.\]