XIX OM - I - Problem 4

In the plane, given are parallel lines $ a $ and $ b $ and a point $ M $ lying outside the strip bounded by these lines. Determine points $ A $ and $ B $ on lines $ a $ and $ b $, respectively, such that segment $ AB $ is perpendicular to $ a $ and $ b $, and the angle $ AMB $ is the largest.
Let $ AB $ be a segment with endpoints $ A \in a $ and $ B \in b $, perpendicular to the lines $ a $ and $ b $, and $ N $ - the point symmetric to point $ M $ with respect to a line parallel to lines $ a $ and $ b $ and equidistant from them (Fig. 4).
The circle with diameter $ AB $ lies within the strip bounded by the lines $ a $ and $ b $, so point $ M $ lies outside this circle, which means $ \measuredangle AMB = \alpha $ is less than $ 90^\circ $. Therefore, the maximum of angle $ \alpha $ coincides with the maximum of $ \sin \alpha $, since for angles from $ 0^\circ $ to $ 90^\circ $, the sine of the angle increases with the angle.
When segment $ AB $ lies on line $ MN $, then $ \sin \alpha = 0 $.
When points $ A $, $ B $, and $ M $ are vertices of a triangle, and $ r $ denotes the radius of the circumcircle of this triangle, then $ \sin \alpha = \frac{AB}{2r} $. Segment $ MN $ is a chord of this circle, so $ MN \leq 2r $, hence




From this, it follows that the maximum of $ \sin \alpha $, and thus the maximum of $ \alpha $, occurs when $ \sin \alpha = \frac{AB}{MN} $, i.e., when




which means that points $ A $ and $ B $ lie on the circle with diameter $ MN $.