olymmath

IR-TST 2011 Problem 6

    The circle ω\omega with center OO has given. From an arbitrary point TT outside of ω\omega draw tangents TBTB and TCTC to it. KK and HH are on TBTB and TCTC respectively.a) B′B' and C′C' are the second intersection point of OBOB and OCOC with ω\omega respectively. K′K' and H′H' are on angle bisectors of ∠BCO\angle BCO and ∠CBO\angle CBO respectively such that KK′⊥BCKK' \bot BC and HH′⊥BCHH'\bot BC . Prove that K,H′,B′K,H',B' are collinear if and only if H,K′,C′H,K',C' are collinear.b) Consider there exist two circle in TBCTBC such that they are tangent two each other at JJ and both of them are tangent to ω\omega .and one of them is tangent to TBTB at KK and other one is tangent to TCTC at HH . Prove that two quadrilateral BKJIBKJI and CHJICHJI are cyclic ( II is incenter of triangle OBCOBC ).

    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 ir-tst/2011/p6 · Tag YJS57W2YVP · Edit this page on GitHub · Keys: j next, k previous