MathDB

2014.22

Part of Ukrainian TYM Qualifying - geometry

Problems(1)

orthocenter of ABC is incenter of XYZ, symmetric wrt lines, altitudes, midpoints

Source: 2014 XVII All-Ukrainian Tournament of Young Mathematicians named after M. Y. Yadrenko, Qualifying p22

5/7/2021
In △ABC\vartriangle ABC on the sides BC,CA,ABBC, CA, AB mark feet of altitudes H1,H2,H3H_1, H_2, H_3 and the midpoint of sides M1,M3,M3M_1, M_3, M_3. Let HH be orthocenter △ABC\vartriangle ABC. Suppose that X2,X3X_2, X_3 are points symmetric to H1H_1 wrt BH2BH_2 and CH3CH_3. Lines M3X2M_3X_2 and M2X3M_2X_3 intersect at point XX. Similarly, Y3,Y1Y_3,Y_1 are points symmetric to H2H_2 wrt C3HC_3H and AH1AH_1.Lines M1Y3M_1Y_3 and M3Y1M_3Y_1 intersect at point Y.Y. Finally, Z1,Z2Z_1,Z_2 are points symmetric to H3H_3 wrt AH1AH_1 and BH2BH_2. Lines M1Z2M_1Z_2 and M2Z1M_2Z_1 intersect at the point ZZ Prove that HH is the incenter △XYZ\vartriangle XYZ .
geometryincenterorthcenterSymmetricUkrainian TYM