49th Mathematical Olympiad in Ukraine

Question

Let a given triangle A_1A_2A_3 and a point X which doesn't lie on each of lines A_1A_2, A_1A_3 and A_2A_3 be given. Denote by l_i line which passes through the point A_i and is normal to the chord line XA_i, i=1,2,3. Let are the points B_1 = l_2 l_3, B_2 = l_3 l_1, B_3 = l_1 l_2, and H_1, H_2, H_3 be the orthocenters the triangles B_1A_2A_3, B_2A_3A_1, B_3A_1A_2, respectively. Prove that the triangle H_1H_2H_3 is congruent to the triangle A_1A_2A_3. FIGURE_0 Fig.10

Accepted Answer

Since A_3H_2 A_1X as the lines which are normal to the l_i, analogously A_1H_2 A_3X then the quadrangle A_1XA_3H_2 is a parallelogram. Then we have that A_1XA_2H_3 and A_3XA_2H_1 are the parallelogram too. Therefore the A_3H_2 and A_2H_3 are the equal and parallel segments. Then quadrangle A_2A_3H_2H_3 is a parallelogram, therefore A_2A_3 = H_2H_3. Analogously for the rest pairs of sides of triangles A_1A_2A_3 and H_1H_2H_3, hence they are the congruent triangles.