XVI OM - I - Problem 7

On the side $ AB $ of triangle $ ABC $, a point $ K $ is chosen, and points $ M $ and $ N $ are determined on the lines $ AC $ and $ BC $ such that $ KM \perp AC $, $ KN \perp BC $. For what position of point $ K $ is the area of triangle $ MKN $ the largest?
Let $ BC = a $, $ AC = b $, $ \measuredangle ACB = \gamma $, $ KM = p $, $ KN = q $ (Fig. 2).
Notice that $ \measuredangle MKN = 180^\circ - \gamma $.
Indeed, if from a point $ K_1 $ chosen on the bisector of angle $ ACB $ we drop perpendiculars $ K_1M_1 $ and $ K_1N_1 $ to the sides of this angle, then the rays $ KM $ and $ K_1M_1 $ are parallel and similarly directed since points $ K $ and $ K_1 $ lie on the same side of the line $ AC $.
The same applies to the rays $ KN $ and $ K_1N_1 $. Hence, angles $ MKN $ and $ M_1K_1N_1 $ are equal, as one can be obtained from the other by a parallel shift. Angles $ M_1K_1N_1 $ and $ ACB $ are opposite angles of a convex quadrilateral, whose two other angles are right angles. Therefore, $ \measuredangle M_1K_1N_1 = 180^\circ - \gamma $ and similarly $ \measuredangle MKN = 180^\circ - \gamma $, thus




From this equality, we conclude that the area of $ MKN $ is maximized when the product $ pq $ reaches its maximum value; simultaneously, the product $ abpq $ also reaches its maximum. However, the sum of the numbers $ bp $ and $ aq $ is constant, since




Therefore, the maximum of the product $ bp \cdot aq $ occurs when $ bp = aq $, i.e., when $ \mathrm{area}\ AKC = \mathrm{area}\ BKC $, which happens when point $ K $ is the midpoint of side $ AB $.
The reasoning above can be slightly modified by calculating $ p $ and $ q $ from triangles $ AKM $ and $ BKN $:






The maximum of $ pq $ occurs when the product $ AK \cdot KB $ is the largest. Since the sum $ AK + KB = AB $ is constant, this happens when $ AK = KB $.