IB Math AI HL Voronoi Diagrams — Complete Cheatsheet
Every Voronoi skill and trap for IB Mathematics Applications & Interpretation HL — distance and midpoint, perpendicular-bisector edges, the nearest-site rule, Voronoi vertices and the toxic-waste-dump problem. Hand-built by an IBO-certified Singapore tutor, with a print-ready PDF to download.
Voronoi diagrams are the most visual topic in IB Mathematics Applications & Interpretation HL, and a perfect fit for AI's modelling-first philosophy. From a scatter of "sites" — shops, hospitals, mobile masts, rainfall gauges — you carve the plane into cells so that every location is grouped with its nearest site. The machinery is coordinate geometry you already own (distance, midpoint, gradient, the equation of a line), so a calculator is always allowed and the marks live in choosing the right boundary and reading the diagram correctly, not in heavy algebra.
This cheatsheet condenses the whole of SL 3.5 — the distance and midpoint engines, perpendicular-bisector edges, the nearest-site (nearest-neighbour) rule, Voronoi vertices, adding a new site, and the flagship "toxic-waste dump" / largest-empty-circle problem — onto one page, and flags the traps that quietly cost method marks. The print-ready PDF is at the bottom, free to download.
§1 — Anatomy of a Voronoi diagram SL 3.5
A Voronoi diagram partitions the plane into cells, one per site, so that every point in a cell is closer to its own site than to any other. Four pieces of vocabulary carry the whole topic:
Site (seed):one of the given points $A,B,C,\dots$ that the diagram is built around
Cell:the region of all points closer to that site than to any other
Edge:a boundary between two cells — the perpendicular bisector of the two sites; every point on it is equidistant from them
Vertex:a point where three edges (three cells) meet — equidistant from three sites
A Voronoi diagram for three sites. Each edge is the perpendicular bisector of two neighbouring sites, so every point on it is equidistant from them — and the three edges meet at the Voronoi vertex, equidistant from all three sites.
NoteThe defining rule is nearest site wins: a location belongs to whichever site is closest. Edges are the two-way ties (equidistant from two sites) and vertices are the three-way ties (equidistant from three).
§2 — The two engines: distance & midpoint SL 3.1
Every Voronoi construction reduces to two coordinate-geometry formulas — the distance between two points and the midpoint of the segment joining them.
Distance:$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Midpoint:$M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$ — every boundary passes through $M$
TrickFor distance, square the gaps, add, then square-root once at the very end. For the midpoint, average the coordinates — add and halve, never subtract.
TrapTwo classic slips: stopping before the square root, and adding $\Delta x+\Delta y$ instead of squaring each gap. The distance is $\sqrt{(\Delta x)^2+(\Delta y)^2}$, never $\Delta x+\Delta y$.
§3 — The boundary is a perpendicular bisector: its gradient SL 3.5
A Voronoi edge between sites $A$ and $B$ is the perpendicular bisector of $[AB]$, so its gradient is the negative reciprocal of the gradient of $AB$.
Gradient of $AB$:$m_{AB}=\dfrac{y_B-y_A}{x_B-x_A}$
Boundary gradient:$m_\perp=-\dfrac{1}{m_{AB}}$ — flip the fraction and change the sign
TrapThe perpendicular gradient is $-\dfrac{1}{m}$, not $-m$ and not $\dfrac{1}{m}$ — you must do both steps. If $m_{AB}=\tfrac{1}{3}$ then $m_\perp=-3$.
NoteRead the two special cases straight off: if $[AB]$ is horizontal ($m_{AB}=0$) the boundary is vertical, $x=x_M$; if $[AB]$ is vertical the boundary is horizontal, $y=y_M$. The negative-reciprocal rule is only awkward in these two cases.
§4 — Equation of a Voronoi edge SL 3.5
With the midpoint $M$ and the perpendicular gradient $m_\perp$, the edge is a straight line $y=m_\perp x+c$. Substitute $M$ to pin down the intercept.
Order of operations: gradient of $AB$ $\to$ negative reciprocal $\to$ midpoint $\to$ substitute for $c$. This "bisector builder" is the single most-marked sub-skill in AI Voronoi questions.
TrickPoint-gradient form $y-y_M=m_\perp(x-x_M)$ skips solving for $c$ separately — plug in the midpoint and the perpendicular gradient, then expand.
TrapThe line must pass through the midpoint of $[AB]$, not through $A$ or $B$. Using an endpoint gives a parallel line in the wrong place and loses every follow-on mark.
§5 — The nearest-site (nearest-neighbour) rule SL 3.5
Every point in the plane belongs to the cell of its nearest site. To decide which site "owns" a point $P$, compute the distance from $P$ to each site and take the smallest — the engine behind delivery zones, catchment areas and nearest-hospital maps (nearest-neighbour interpolation).
Owner of $P$:the site $S_i$ that minimises $\sqrt{(x_P-x_i)^2+(y_P-y_i)^2}$
TrickTo find only the winner, compare squared distances — skip the square roots, since the smallest squared distance is the smallest distance. Take the root only if the question asks for the actual distance.
TrapDo not eyeball which site "looks" closest on the diagram — a point near an edge can deceive you. Store the site coordinates on the GDC and compute every distance.
§6 — Voronoi vertices: the equidistant point SL 3.5
A Voronoi vertex is equidistant from three sites — geometrically the circumcentre of the triangle they form, where two perpendicular bisectors cross. Find it by solving two edge equations simultaneously.
Method:build the bisectors of $[AB]$ and $[AC]$, then solve them together for $(x,y)$
Check:the vertex must satisfy $|VA|=|VB|=|VC|$ — equal distances to all three sites
TrickTwo bisectors are enough — the third passes through the same point automatically. On the GDC, solve the two linear equations as a system.
TrapThe vertex is where the perpendicular bisectors meet — not the medians, and not the triangle's centroid. If your three site-distances come out unequal, the point is wrong.
§7 — Adding a new site SL 3.5
Add a site $D$ and it carves a fresh cell out of its neighbours: a point switches to $D$ exactly when $D$ becomes its closest site. The new cell is bounded by the perpendicular bisectors of $D$ with each nearby site.
New boundaries:draw the perpendicular bisector of $D$ with each neighbouring site $S_i$
Re-assign:a point joins $D$'s cell when $D$ is closer to it than every original site
TrapAfter adding $D$, recompute the nearest site from scratch — do not trust the old winner. Only the cells around $D$ change; distant cells are untouched.
NoteAdding a site can only shrink existing cells, never grow them — the new cell is stolen from whatever used to be nearest there.
§8 — The largest empty circle & the "toxic-waste dump" SL 3.5
The flagship AI application: to place a new facility as far as possible from every existing site — a toxic-waste dump far from towns, or a new shop far from competitors — sit it at a Voronoi vertex. That vertex centres the largest empty circle: the biggest circle you can draw with no site inside.
Optimal spot:a Voronoi vertex (equidistant from its three surrounding sites)
Safety radius:$r=$ distance from the vertex to any one of those three sites
The largest empty circle is centred at the Voronoi vertex and just touches the three nearest sites. Its radius r — the equal distance from the vertex to each site — is the safety distance for the new facility.
NoteFor the tidy right-angle layout $A(0,0),\ B(2p,0),\ C(0,2q)$ the bisectors are simply $x=p$ and $y=q$, so the vertex is $(p,q)$ and the safety radius is $\sqrt{p^2+q^2}$ — worth spotting to save time.
TrapThe optimum must also lie inside the region of interest (e.g. within the map). A point on the boundary of the region can sometimes sit further from every site than any interior vertex — always check the candidates the question allows.
§9 — Synthesis: safety radius from side lengths SL 3.2
Sometimes you are given the three town-to-town distances instead of coordinates — there is nothing to bisect. The equidistant point is still the circumcentre, and its equal distance is the circumradius $R$ (the largest-empty-circle radius). Read one angle from the cosine rule, then the extended sine rule delivers $R$ directly.
Cosine rule:$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$ — recover an angle from the three sides
Extended sine rule:$\dfrac{a}{\sin A}=2R\ \Rightarrow\ R=\dfrac{a}{2\sin A}$
TrickEquidistant point $=$ Voronoi vertex $=$ circumcentre; the equal distance is the circumradius $R=\dfrac{a}{2\sin A}$, with $a$ the side opposite the angle $A$ you found.
TrapIt is $2R$, not $R$, that equals $\dfrac{a}{\sin A}$ — dropping the factor of 2 halves your radius. And $a$ must be the side opposite the angle used.
§10 — Exam attack plan All sections
Question cue
What to do
Watch for
"Distance between two sites"
$\sqrt{(\Delta x)^2+(\Delta y)^2}$
Root once at the end; don't add the gaps
"Equation of the boundary / edge"
Midpoint $+$ perpendicular gradient $-1/m$, then $y=mx+c$
Through the midpoint, not an endpoint
"Which site is a point nearest?"
Compare distances (or squared distances) to each site
Smallest wins; don't eyeball the diagram
"Point equidistant from three sites"
Solve two perpendicular bisectors simultaneously
It is the circumcentre; check equal distances
"Best place for a new facility / dump"
A Voronoi vertex — centre of the largest empty circle
Also test the region's boundary candidates
"Safety distance / radius"
Distance from the vertex to any one site
All three site-distances must be equal
"Add a new site"
Bisect $D$ with each neighbour; re-assign points
Recompute the nearest site from scratch
"Given only the side lengths"
Cosine rule for an angle, then $R=\dfrac{a}{2\sin A}$
$2R=a/\sin A$; $a$ opposite $A$
Worked Example — AI HL-Style Toxic-Waste Dump
Question (AI HL Paper 2 style — 6 marks)
Three towns lie on a coordinate map, with distances in km: $A(2,2)$, $B(8,4)$ and $C(2,10)$. A waste-processing site is to be built equidistant from all three towns and as far from them as possible. (a) Find the equation of the perpendicular bisector of $[AB]$. (b) The perpendicular bisector of $[AC]$ is $y=6$. Find the coordinates of the point $D$ equidistant from all three towns. (c) Find the distance of $D$ from each town, giving the safety radius to 3 significant figures.
Solution
(a) Midpoint of $[AB]=\left(\tfrac{2+8}{2},\tfrac{2+4}{2}\right)=(5,3)$; gradient $m_{AB}=\dfrac{4-2}{8-2}=\dfrac{1}{3}$, so $m_\perp=-3$. (M1) Edge: $y-3=-3(x-5)\Rightarrow y=-3x+18$. (A1)
(b) $D$ lies on both bisectors, so solve $y=-3x+18$ with $y=6$: $\;6=-3x+18\Rightarrow 3x=12\Rightarrow x=4$. Hence $D=(4,6)$. (M1)(A1)
(c) $|DA|=\sqrt{(4-2)^2+(6-2)^2}=\sqrt{4+16}=\sqrt{20}=4.47$ km (3 s.f.). (M1) Check: $|DB|=\sqrt{(4-8)^2+(6-4)^2}=\sqrt{20}$ and $|DC|=\sqrt{(4-2)^2+(6-10)^2}=\sqrt{20}$ — all equal, confirming $D$ is the Voronoi vertex. Safety radius $=\sqrt{20}\approx 4.47$ km. (A1)
Examiner's note: the equal distances $|DA|=|DB|=|DC|$ are the built-in check that $D$ is correct — always verify them before you commit. Keep the units (km) and round only at the very end, to 3 s.f. unless told otherwise. In a full "toxic-waste dump" question you would also confirm $D$ lies inside the region of interest, since a point on the map's boundary can occasionally sit even further from the towns.
Common Student Questions
What is the difference between a Voronoi edge and a Voronoi vertex?
An edge is a boundary between two neighbouring cells — the perpendicular bisector of those two sites, so every point on it is equidistant from two sites. A vertex is where edges meet, equidistant from three sites at once; it is the circumcentre of the triangle those three sites form. In short: edges are two-way ties, vertices are three-way ties.
Why is a Voronoi boundary always the perpendicular bisector?
Because the set of all points equidistant from two fixed points $A$ and $B$ is exactly the perpendicular bisector of $[AB]$. That equidistant line is precisely where a location stops being nearer to $A$ and starts being nearer to $B$ — which is the definition of the boundary between their cells. So "boundary between two sites" and "perpendicular bisector of the two sites" mean the same thing.
How do I find the point equidistant from three towns?
Build the perpendicular bisector of two of the pairs — say $[AB]$ and $[AC]$ — and solve the two equations simultaneously (the GDC's simultaneous-equation solver is quickest). The intersection is the Voronoi vertex, equivalently the circumcentre of triangle $ABC$. Always finish by checking $|VA|=|VB|=|VC|$; if the three distances are not equal, you have made a slip somewhere.
When can I compare squared distances instead of the actual distances?
Whenever you only need to know which site is nearest, not how far. Squaring is order-preserving for non-negative numbers, so the smallest squared distance belongs to the same site as the smallest distance — and you save yourself several square roots. Only take the root at the end if the question actually asks for a distance, radius or length.
What exactly is the "toxic-waste dump" problem?
It asks you to place a new point as far as possible from every existing site — a dump far from towns, a new store far from rivals. The best interior spot is a Voronoi vertex, because it centres the largest empty circle (equidistant from its three surrounding sites, with none nearer). Compare the candidate vertices and take the one with the biggest radius — but also check the boundary of the allowed region, which can occasionally be even further from every site.
What's NOT in this cheatsheet
This page gives you the formulas and traps for free. The full Photon Academy IB Math AI HL library (enrolled students, or the lifetime resource library) adds:
Notes PDF — every Voronoi method in full: distance, midpoint, perpendicular-bisector edges, vertices and the toxic-waste-dump problem, with the exact GDC steps.
Tutorial booklet — 25+ AI HL-style questions from distance-formula fluency to full Voronoi constructions and optimal-placement applications.
Tutorial Solutions — mark-scheme-style solutions with M1/A1 annotations for every perpendicular-bisector and Voronoi-vertex step.
Predicted-paper questions — the exact Voronoi and toxic-waste-dump question types most likely in the next AI HL session.
