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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 dd be a diameter of a circle kk, and let AA be an arbitrary point on this diameter dd in the interior of kk. Further, let PP be a point in the exterior of kk. The circle with diameter PAPA meets the circle kk at the points MM and NN.
Find all points BB on the diameter dd in the interior of kk such that \measuredangle MPA = \measuredangle BPN   \text{and}   PA \leq PB. (i. e. give an explicit description of these points without using the points MM and NN).