HMMT 二月 1998 · 几何 · 第 3 题

MD 是单位圆中长为 2 的弦（即直径）。三角形 MLD 的最大面积当 L 在直径的垂直平分线与圆的交点处取得。此时 $\angle MLD = 90^{\circ}$（Thales 定理）。
$\boxed{90^{\circ}}$。

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英文解析

MD is a chord of length 2 in a circle of radius 1. The maximum area of triangle MLD occurs when the altitude from L to MD is maximized, i.e., when L is at the farthest point on the circle from chord MD.
Since the chord has length 2 and the radius is 1, the chord MD is a diameter (the only chord of length 2 in a unit circle is the diameter). The maximum distance from a point on the circle to the diameter is the radius = 1 (achieved at either endpoint of the perpendicular diameter).
For a diameter, the angle subtended at any point on the circle is $90^{\circ}$ (Thales theorem). However, we want the triangle area maximized: base = MD = 2, max height = 1, max area = 1/2 * 2 * 1 = 1.
The angle $\angle MLD$ at the optimal point: when L is at the top of the circle (perpendicular to MD), triangle MLD is isosceles with ML = LD = $\sqrt{2}$. The angle at L satisfies $\sin(\angle MLD/2) = 1/\sqrt{2}$, so $\angle MLD/2 = 45^{\circ}$, $\angle MLD = 90^{\circ}$.
Actually, for the MAXIMUM area, L should be at the top of the circle (the point on the perpendicular bisector of MD). Then $\angle MLD = 90^{\circ}$ (since MD is a diameter).
$\boxed{90^{\circ}}$.