olymmath

PAMO 2002 Problem 2

  • Day 1

△AOB\triangle{AOB} is a right triangle with ∠AOB=90o\angle{AOB}=90^{o} . CC and DD are moving on AOAO and BOBO respectively such that AC=BDAC=BD . Show that there is a fixed point PP through which the perpendicular bisector of CDCD always passes.

Dataset text AI-MO AoPS. Quoted with source.

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Similar problems

These problems have the most words in common with this statement.

Locator pamo/2002/p2 · Tag WV1W5YGW6X · Edit this page on GitHub · Keys: j next, k previous