For all fans of case-distinction [< MPA = < BPN]
Source: German TST 2004, exam VI, problem 2
May 30, 2004
geometry proposedgeometryLocus problemsGeometric InequalitiesGermanyTSTTeam Selection Test
Problem Statement
Let be a diameter of a circle , and let be an arbitrary point on this diameter in the interior of . Further, let be a point in the exterior of . The circle with diameter meets the circle at the points and .Find all points on the diameter in the interior of such that
\measuredangle MPA = \measuredangle BPN \text{and} PA \leq PB.
(i. e. give an explicit description of these points without using the points and ).