Problem
GEO-B3-M01-P001 Inverse Point on a Ray
#1
★☆☆☆☆ Level 1 of 5
An inversion has center \(O\) and radius \(6\). Point \(A\) satisfies \(OA=4\). Find \(OA^*\), and prove that applying the inversion again returns point \(A\).
Inspired by Prasolov inversion method
Hint 1. Use \(OA\cdot OA^*=R^2\).
Hint 2. For the second inversion, apply the same formula to \(A^*\).
E. Full solution. By definition, \(OA\cdot OA^*=6^2=36\). Hence \(OA^*=9\). Point \(A^*\) lies on ray \(OA\). If we invert \(A^*\), its image \(A^{**}\) lies on the same ray and satisfies \(OA^*\cdot OA^{**}=36\). Thus \(9\cdot OA^{**}=36\), so \(OA^{**}=4\). There is only one such point on ray \(OA\), hence \(A^{**}=A\).
A basic check of the definition: the student must confidently use the product of distances and understand that inversion is its own inverse.