SELECTION EXAMINATION

Question

Let AB be a quadrilateral inscribed into the circle. With centers A, B, , we draw circles C_A, C_B, C_, C_ respectively, not having common points. The circle C_A intersects the sides of the quadrilateral at the points A_1, A_2, the circle C_B at the points B_1, B_2, the circle C_ at the points _1, _2 and the circle C_ at the points _1, _2. Prove that the quadrilateral created by the lines A_1A_2, B_1B_2, _1_2 and _1_2 is cyclic. (E. Psychas) FIGURE_0

Accepted Answer

Since the triangles AA_1A_2, BB_1B_2, _1_2 and _1_2 are isosceles, using small letters for their equal angles we have the equalities: FIGURE_0 Figure 4 A + x + x = 180^ x = 90^ - A2 (1) B + y + y = 180^ y = 90^ - B2 (2) + z + z = 180^ z = 90^ - 2 (3) + + = 180^ = 90^ - 2 (4). Since the quadrilateral AB is cyclic, we have: A + = 180^ x + z = 90^ (5) B + = 180^ y + = 90^ (6). Let now the lines A_1A_2, B_1B_2, _1_2 form the quadrilateral K MN. From the triangle KA_1B_2, we have: K + x + y = 180^, while from the triangle M_1_2, we have: M + z + = 180^. Summing up the last two relations and using (5) and (6) we obtain: K + M = 180^.