Macedonian Mathematical Olympiad

Question

A segment AB and its midpoint K are given. An arbitrary point C, different from K, is chosen on the perpendicular to AB through K. Let N be the intersection of AC and the line passing through B and the midpoint of the segment CK. Let U be the intersection of AB with the line that passes through C and the midpoint L of the segment BN. Prove that the ratio of the areas of the triangles CNL and BUL doesn't depend on the choice of point C. A segment *AB* and its midpoint *K* are given. An arbitrary point *C*, different from *K*, is chosen on the perpendicular to *AB* through *K*. Let *N* be the intersection of *AC* and the line passing through *B* and the midpoint of the segment *CK*. Let *U* be the intersection of *AB* with the line that passes through *C* and the midpoint *L* of the segment…

Accepted Answer

Let M be the midpoint of the segment CK. From Menelaus' theorem for the triangle AKC and the line BN we have CNNA ABBK KMMC = 1. From this we get NA = 2NC, from which it follows that AC = 3NC. Hence P_BNC = 13P_ABC. From Menelaus' theorem for the triangle ABN and the line CU we have AUUB BLLN NCCA = 1. FIGURE_0 Therefore we get AU = 3UB. Therefore U is the midpoint of the segment BK. It follows that P_BUC = 14P_ABC. Let x = P_CNL and y = P_BLU. Since L is the midpoint of BN, we have P_BLC = x. Now x + y = P_BLC + P_BLU = P_BUC = 14P_ABC, on the other hand we have 2x = P_CNL + P_BLC = P_BNC = 13P_ABC. If we divide these two equalities we get 12 + y2x = 34, hence yx = 12, from where we get the required statement.