Russian problems Problem 40 Given an isosceles triangle, find the locus of the point P inside the triangle such that the distance from P to the base equals the geometric mean of the distances to the sides.
Let the triangle be ABC, with AB=AC. Take the circle through B and C which has AB and AC as tangents. The required locus is the arc BC. Suppose P lies on the arc. Let the perpendiculars from P meet BC in L, AB in N and AC in 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 PM/PL = PC/PB. Multiplying gives the required result PL 2 = PM.PN. If P is inside the circle and not on it, take P' as the intersection of the line AP and the arc. We have PLP'M and PN>P'N, hence PL 2 PM.PN. Russian 31-40 (C) John Scholes jscholes@kalva.demon.co.uk 19 Sep 1998