Geometry and trigonometry

📚The mathematics

3.1 Three dimensions

Distance in space extends Pythagoras by one term: \( d = \sqrt{(x_2-x_1)^{2} + (y_2-y_1)^{2} + (z_2-z_1)^{2}} \), and the midpoint is the average of each coordinate. The volumes and surface areas of the pyramid, cone, sphere, hemisphere and their combinations are all in the formula booklet.

At standard level, all three-dimensional trigonometry reduces to right-angled triangles. The skill is finding the triangle inside the solid, then redrawing it flat, in two dimensions, with its lengths marked. Doing trigonometry directly on the perspective sketch is how people talk themselves into using a face diagonal where they needed a space diagonal.

3.2 & 3.3 Solving triangles, and applications

Choose by counting what you have. The sine rule needs a matched pair — a side and the angle opposite it. With no such pair, you need the cosine rule: the first form for two sides and the angle between them, the rearranged form for three sides. The area formula needs two sides and the included angle.

Angles of elevation and depression are measured from the horizontal, and the angle of depression from A to B equals the angle of elevation from B to A. Bearings are three figures measured clockwise from north — east is 090°, south-west is 225°. Draw a north line at every point you take a bearing from; most bearing errors are drawing errors.

3.4 Arcs and sectors

At standard level these are done in degrees — radians are AHL only in AI. Both formulae are just fractions of the whole circle, so if you forget them, work out what fraction of 360° the angle is and take that fraction of the circumference or the area.

3.5 Perpendicular bisectors

The perpendicular bisector of the segment joining two points is the set of points equidistant from both. To find it: take the midpoint, find the gradient of the segment, take the negative reciprocal, and use point–gradient form. This is a three-step recipe and it is worth drilling, because everything in the next section depends on it.

3.6 Voronoi diagrams

Given a set of sites, a Voronoi diagram divides the plane into cells, one per site, where every point in a cell is closer to that site than to any other. The edges between cells are perpendicular bisectors of the segments joining neighbouring sites, and the vertices are where edges meet — each vertex is equidistant from three sites.

What you are asked to do:

• Identify the nearest site to a given point — read off which cell the point lies in, or compute distances and compare.
• Find the equation of a boundary — a perpendicular bisector calculation, with the two sites always given as coordinates.
• Add a site to an existing diagram — the new cell takes territory from every neighbouring cell, bounded by new perpendicular bisectors.
• Calculate the area of a region — usually a polygon you can split into triangles.
• Nearest neighbour interpolation — estimate a value at a point by assigning it the value of the site whose cell it falls in. Useful for rainfall or air-quality estimates, and worth criticising: it produces sharp jumps at the boundaries that the real quantity does not have.
• The “toxic waste dump” problem — find the point in a region furthest from all sites. In examinations the answer is always at a Voronoi vertex, so compute the distance from each vertex to its nearest site and take the largest. Boundary points of the region can matter in real applications, but not in the exam.

Contexts include siting a new shop, hospital or mobile mast, mapping the spread of disease from known cases, and defining catchment areas.

AHL 3.7 & 3.8 Radians and the unit circle

\( \pi \) radians is 180°. In radians the arc and sector formulae simplify to \( l = r\theta \) and \( A = \tfrac{1}{2}r^{2}\theta \) — and both are wrong in degrees, so check your calculator’s mode. On the unit circle the point at angle \( \theta \) is \( (\cos\theta, \sin\theta) \), giving \( \cos^{2}\theta + \sin^{2}\theta = 1 \) and \( \tan\theta = \dfrac{\sin\theta}{\cos\theta} \).

The ambiguous case of the sine rule: because \( \sin\theta = \sin(180^\circ - \theta) \), finding an angle with the sine rule may admit two triangles. If the question gives two sides and a non-included angle, check whether the obtuse alternative also produces a valid triangle — that is, whether the three angles still sum to less than 180°.

AHL 3.9 Matrix transformations

A transformation of the plane can be written \( \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} e \\ f \end{pmatrix} \) — a matrix for the reflection, stretch, enlargement or rotation, plus a vector for the translation. Composing transformations means multiplying the matrices, and because matrix multiplication is not commutative, the order is part of the answer. The matrix applied first goes on the right.

The key geometric fact:

A determinant of 1 preserves area (rotations and reflections do), and a negative determinant indicates that orientation has been flipped. Iterating a set of these transformations generates fractals — the Sierpinski triangle and the Koch snowflake being the standard examples, and both connect back to the infinite geometric series in Topic 1.

AHL 3.10–3.13 Vectors

\( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} \), with magnitude \( |\mathbf{v}| = \sqrt{v_1^{2}+v_2^{2}+v_3^{2}} \) and unit vector \( \dfrac{\mathbf{v}}{|\mathbf{v}|} \). The displacement from A to B is \( \overrightarrow{AB} = \mathbf{b} - \mathbf{a} \) — destination minus start.

A line is \( \mathbf{r} = \mathbf{a} + \lambda\mathbf{b} \). In kinematics this becomes

with \( \mathbf{r}_0 \) the initial position, \( \mathbf{v} \) the constant velocity and \( |\mathbf{v}| \) the speed. To find when two objects are closest, write the displacement between them as a function of \(t\), form the square of its magnitude — a quadratic in \(t\) — and minimise it. Minimising the square avoids differentiating a square root and gives the same answer, since the two are minimised at the same time.

The scalar product \( \mathbf{v}\cdot\mathbf{w} = v_1w_1 + v_2w_2 + v_3w_3 = |\mathbf{v}||\mathbf{w}|\cos\theta \) gives angles, and is zero exactly when the vectors are perpendicular. The vector product gives a vector perpendicular to both, with \( |\mathbf{v}\times\mathbf{w}| = |\mathbf{v}||\mathbf{w}|\sin\theta \) equal to the area of the parallelogram they span — so half of it is the area of the triangle.

AHL 3.14 & 3.15 Graph theory

A graph is vertices joined by edges. Learn the vocabulary precisely, because the questions are written in it: adjacent vertices are joined by an edge; the degree of a vertex is how many edges meet it; a simple graph has no loops or repeated edges; a complete graph \( \kappa_n \) joins every pair; a weighted graph carries a number on each edge (cost, distance, time); a directed graph has one-way edges, with in-degree and out-degree; a tree is connected with no cycles.

The adjacency matrix \( A \) has a 1 (or the number of edges) in position \( (i,j) \) when vertex \(i\) joins vertex \(j\). The useful theorem:

so to count walks of length at most \(k\), sum \( A + A^{2} + \cdots + A^{k} \). A transition matrix is the weighted, probabilistic version, and it is the same object that drives Markov chains in Topic 4 and the Google PageRank algorithm.

AHL 3.16 The graph algorithms

• Eulerian trails and circuits traverse every edge exactly once. A connected graph has an Eulerian circuit if every vertex has even degree, and an Eulerian trail if exactly two vertices have odd degree — and the trail must start at one odd vertex and finish at the other. This is a stated condition you check, not something you search for by trial.
• Hamiltonian paths and cycles visit every vertex exactly once. Note the contrast with Eulerian: edges versus vertices. There is no simple test for existence.
• Minimum spanning tree — connect all vertices at least total weight. Kruskal’s algorithm repeatedly adds the cheapest remaining edge that does not create a cycle. Prim’s algorithm grows outwards from a starting vertex, each time adding the cheapest edge from the tree so far to a vertex not yet in it, and has a tidy matrix form.
• The Chinese postman problem — the shortest closed route using every edge at least once. If all degrees are even the graph is Eulerian and the answer is the total weight. If there are odd vertices (up to four in examinations), pair them up, find the shortest path between each pair, choose the pairing with the smallest total, and add that to the total weight. You should be able to explain why the method works: the repeated edges are exactly what makes every degree even.
• The travelling salesman problem — the cheapest Hamiltonian cycle. There is no efficient exact algorithm, so you bound it. The nearest neighbour algorithm gives an upper bound: from the start, always go to the closest unvisited vertex, then return. The deleted vertex algorithm gives a lower bound: delete a vertex, find the minimum spanning tree of what remains, then add back the two shortest edges from the deleted vertex. Practical problems must first be converted to the classical problem with a table of least distances.