Solving Geometric Problems | 解决几何问题

📚 Solving Geometric Problems | 解决几何问题

Solving geometric problems at A-Level draws together coordinate geometry, vectors, trigonometry and calculus. The key is to build a clear method from the information given, rather than trying to remember every possible formula at once.

A-Level 几何问题求解综合了坐标几何、向量、三角学和微积分。关键在于根据已知条件建立清晰的方法，而不是试图一次记住所有公式。

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1. Understanding the Problem | 理解题意

Always begin by identifying the given information and the unknown that the question asks you to find. Draw a labelled diagram if the question allows, marking known lengths, angles, coordinates or vector directions.

始终先确定已知条件和题目要求求解的未知量。如果题目允许，画出带标注的示意图，标出已知长度、角度、坐标或向量方向。

Translate geometric language into algebra. For example, ‘perpendicular’ usually means two gradients multiply to −1 or a dot product is zero; ‘tangent’ means one intersection point or a repeated root; ‘midpoint’ means the average of coordinates.

把几何语言转化为代数表达。例如，’垂直’ 通常表示两个斜率乘积为 −1 或点积为零；’相切’ 表示一个交点或重根；’中点’ 表示坐标的平均值。

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2. Coordinate Geometry: Lines and Distances | 坐标几何：直线与距离

For two points A(x₁, y₁) and B(x₂, y₂), the gradient of the line AB and the distance AB are given by:

对于两点 A(x₁, y₁) 和 B(x₂, y₂)，直线 AB 的斜率和距离 AB 由下式给出：

m = (y₂ − y₁) ÷ (x₂ − x₁), d = √((x₂ − x₁)² + (y₂ − y₁)²)

The midpoint M has coordinates ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2). The equation of a straight line with gradient m through (x₁, y₁) is y − y₁ = m(x − x₁).

中点 M 的坐标为 ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)。斜率为 m 且经过 (x₁, y₁) 的直线方程为 y − y₁ = m(x − x₁)。

For parallel lines, the gradients are equal: m₁ = m₂. For perpendicular lines, the gradients satisfy m₁m₂ = −1, provided neither line is vertical.

对于平行直线，斜率相等：m₁ = m₂。对于垂直直线，斜率满足 m₁m₂ = −1，前提是两条直线都不是竖直的。

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3. Modelling Circles | 圆的建模

The standard equation of a circle with centre (a, b) and radius r is:

圆心为 (a, b)、半径为 r 的圆的标准方程为：

(x − a)² + (y − b)² = r²

An expanded form is x² + y² + 2gx + 2fy + c = 0, where the centre is (−g, −f) and the radius is √(g² + f² − c). This radius must be positive.

圆的一般方程为 x² + y² + 2gx + 2fy + c = 0，其中圆心为 (−g, −f)，半径为 √(g² + f² − c)。这个半径必须是正的。

A straight line is tangent to a circle when the perpendicular distance from the centre to the line equals the radius. For a line ax + by + c = 0 and a circle centre (h, k), the condition is:

当圆心到直线的垂直距离等于半径时，直线与圆相切。对于直线 ax + by + c = 0 和圆心 (h, k)，条件为：

|ah + bk + c| ÷ √(a² + b²) = r

This distance formula avoids solving simultaneous equations in many tangent problems.

这个距离公式可以在许多切线问题中避免解联立方程。

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4. Intersection Problems | 交点问题

To find where a line and a circle meet, substitute the line equation into the circle equation. This produces a quadratic in x or y. The discriminant Δ = b² − 4ac then tells you how many intersections exist.

要求直线与圆的交点，将直线方程代入圆的方程。这会得到一个关于 x 或 y 的二次方程。判别式 Δ = b² − 4ac 可以告诉你有多少个交点。

If Δ > 0, the line cuts the circle in two distinct places; if Δ = 0, it touches at one point and is a tangent; if Δ < 0, there is no real intersection.

如果 Δ > 0，直线与圆有两个不同交点；如果 Δ = 0，直线与圆相切于一点；如果 Δ < 0，没有实数交点。

When solving two curves simultaneously, always substitute the simpler equation into the more complex one. Check whether any solutions are extraneous because of squaring or domain restrictions.

联立求解两条曲线时，总要把较简单的方程代入较复杂的方程。检查是否有因平方或定义域限制而产生的增根。

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5. Vectors in Geometric Proof | 向量在几何证明中的应用

Vector methods are powerful for proving geometric facts without relying on coordinates. A vector equation of a line through point A with direction vector d is:

向量方法在证明几何事实时非常有效

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