MathDB
about the incenter

Source: Romanian ROM TST 2004, problem 11, from Kvant Magazine

May 3, 2004
geometryincentervectortrigonometrycircumcirclecomplex numbersromania

Problem Statement

Let II be the incenter of the non-isosceles triangle ABCABC and let A,B,CA',B',C' be the tangency points of the incircle with the sides BC,CA,ABBC,CA,AB respectively. The lines AAAA' and BBBB' intersect in PP, the lines ACAC and ACA'C' in MM and the lines BCB'C' and BCBC intersect in NN. Prove that the lines IPIP and MNMN are perpendicular. Alternative formulation. The incircle of a non-isosceles triangle ABCABC has center II and touches the sides BCBC, CACA and ABAB in AA^{\prime}, BB^{\prime} and CC^{\prime}, respectively. The lines AAAA^{\prime} and BBBB^{\prime} intersect in PP, the lines ACAC and ACA^{\prime}C^{\prime} intersect in MM, and the lines BCBC and BCB^{\prime}C^{\prime} intersect in NN. Prove that the lines IPIP and MNMN are perpendicular.