olymmath

MK-MO 2013 Problem 5

    An arbitrary triangle ABC is given. There are 2 lines, p and q, that are not parallel to each other and they are not perpendicular to the sides of the triangle. The perpendicular lines through points A, B and C to line p we denote with pa,pb,pcp_a, p_b, p_c and the perpendicular lines to line q we denote with qa,qb,qcq_a, q_b, q_c . Let the intersection points of the lines pa,qa,pb,qb,pcp_a, q_a, p_b, q_b, p_c and qcq_c with qb,pb,qc,pc,qaq_b, p_b, q_c, p_c, q_a and pap_a are K,L,P,Q,NK, L, P, Q, N and MM . Prove that KL,MNKL, MN and PQPQ intersect in one point.

    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 mk-mo/2013/p5 · Tag M1Q24JWE6R · Edit this page on GitHub · Keys: j next, k previous