75th NMO Selection Tests

Question

Let ABCD be a convex quadrilateral with the property that the circles having the segments AB and CD as diameters are tangent externally at a point M, different from the intersection point of the diagonals of the quadrilateral. Let K be the second point of intersection of the circumcircle of triangle AMC with the line determined by M and the midpoint of segment AB, and let L be the second point of intersection of the circumcircle of triangle BMD with the line determined by M and the midpoint of segment CD. Prove that |MK - ML| = |AB - CD|. FIGURE_0

Accepted Answer

Let F and E be the midpoints of chords ML and MK, respectively. Then O_3F ML and O_4E MK. Since O_4O_1O_2O_3 is cyclic with O_4O_2O_3 = O_4O_1O_3 = 90^, the midpoint X of segment O_3O_4 is the center of the circumscribed circle of O_4O_1O_2O_3. Moreover, if T denotes the foot of the perpendicular from X onto O_1O_2, then TO_1 = TO_2. In the right trapezoid O_4EFO_3, since X is the midpoint of O_3O_4 and XT EF, it follows that XT EO_4 FO_3, so XT is a midline and therefore ET = TF. Consequently, using the last two equalities we get TE - TO_1 = TF - TO_2 EO_1 = FO_2. Thus, |MK - ML| = |2ME - 2MF| = 2|ME - MF| = 2|EO_1 + O_1M - MO_2 - O_2F| = |2O_1M - 2O_2M|. But AB = 2O_1M, CD = 2O_2M, so substituting into the last equality we obtain |MK - ML| = |AB - CD|, which was to be proven.