\section*{Problem 14}

Given an isosceles triangle, find the locus of the point \(\mathrm{P}\) inside the triangle such that the distance from \(\mathrm{P}\) to the base equals the geometric mean of the distances to the sides.


\section*{Solution}

Let the triangle be \(\mathrm{ABC}\), with \(\mathrm{AB}=\mathrm{AC}\). Take the circle through \(\mathrm{B}\) and \(\mathrm{C}\) which has \(\mathrm{AB}\) and \(\mathrm{AC}\) as tangents. The required locus is the arc \(\mathrm{BC}\).

Suppose \(\mathrm{P}\) lies on the arc. Let the perpendiculars from \(\mathrm{P}\) meet \(\mathrm{BC}\) in \(\mathrm{L}, \mathrm{AB}\) in \(\mathrm{N}\) and \(\mathrm{AC}\) in \(\mathrm{M}\). Join PB and PC. The triangles PNB and PLC are similar (PNB and PLC are both 90, and NBP \(=\) LCP because NB is tangent to the circle). Hence PN/PL = PB/PC. Similarly, triangles PMC and PLB are similar and hence \(\mathrm{PM} / \mathrm{PL}=\mathrm{PC} / \mathrm{PB}\). Multiplying gives the required result \(\mathrm{PL}^{2}=\) PM.PN.

If \(\mathrm{P}\) is inside the circle and not on it, take \(\mathrm{P}^{\prime}\) as the intersection of the line \(\mathrm{AP}\) and the arc. We have \(\mathrm{PL}\mathrm{P}^{\prime} \mathrm{M}\) and \(\mathrm{PN}>\mathrm{P}^{\prime} \mathrm{N}\), hence \(\mathrm{PL}^{2}\) PM.PN.

\section*{4th ASU 1964}

