Higher Level · Geometry and Trigonometry

Voronoi diagrams and applications

Voronoi diagrams and applications

DifficultyAll difficulties
Exam modePaper 1 & 2

Question 1

MildNo calculator
[Maximum mark: 5]

The following regions contain two and three sites respectively. Construct the Voronoi diagrams by writing down on each edge the equation of the corresponding line.

(a)

Coordinate grid showing sites A and B

[2]

(b)

Coordinate grid showing sites A, B and C

[3]

Worked solution

Question 1 mark scheme

(a)

Solution for part (a)

(b)

Solution for part (b)

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Question 2

MediumNo calculator
[Maximum mark: 9]

The following region contains three sites, A(2,5)A(2,5), B(2,1)B(2,1) and C(8,1)C(8,1).

Coordinate grid showing sites A, B and C

(a) Write down the equation of the perpendicular bisector

(i) of the line segment [AB][AB],

(ii) of the line segment [BC][BC].[2]

(b) Find the equation of the perpendicular bisector of the line segment [AC][AC].[4]

(c) Hence construct the Voronoi diagram for the region above; indicate the coordinates of the vertex.[3]

Worked solution

Question 2 mark scheme

(a)

(i) y=3y=3,    (ii) x=5x=5

(b)

mAC=1−58−2=−46=−23,m⊥=32m_{AC}=\frac{1-5}{8-2}=\frac{-4}{6}=-\frac{2}{3},\qquad m_{\perp}=\frac{3}{2}

Midpoint of [AC][AC]: M(5,3)M(5,3)

Perpendicular bisector: y−3=32(x−5)⇒y=32x−92y-3=\frac{3}{2}(x-5)\Rightarrow y=\frac{3}{2}x-\frac{9}{2}

(c)

Completed Voronoi diagram with vertex at (5, 3)

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Question 3

MediumNo calculator
[Maximum mark: 8]

The Voronoi diagram for three cells, A(4,5)A(4,5), B(2,1)B(2,1) and C(8,1)C(8,1) is shown below.

Voronoi diagram for cells A, B and C

(a) Write down the equation of the edge between B and C.[1]

(b) Find the equation of the edge between A and B in the form ax+by+d=0ax+by+d=0, where a,b,d∈Za,b,d\in\mathbb{Z}.[4]

(c) Write down the coordinates of the only one vertex.[1]

(d) Given that the edge between A and C has equation y=x+cy=x+c, find the value of cc.[2]

Worked solution

Question 3 mark scheme

(a)

x=5x=5

(b)

mAB=1−52−4=−4−2=2,m⊥=−12m_{AB}=\frac{1-5}{2-4}=\frac{-4}{-2}=2,\qquad m_{\perp}=-\frac{1}{2}

Midpoint of [AB][AB]: M(3,3)M(3,3)

Perpendicular bisector: y−3=−12(x−3)⇒2y−6=−(x−3)⇒x+2y−9=0y-3=-\frac{1}{2}(x-3)\Rightarrow 2y-6=-(x-3)\Rightarrow x+2y-9=0

(c)

(5,2)(5,2)

(d)

y=x+c⇒2=5+c⇒c=−3y=x+c\Rightarrow 2=5+c\Rightarrow c=-3

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Question 4

MediumNo calculator
[Maximum mark: 8]

The Voronoi diagram for four sites, A(4,6)A(4,6), B(3,1)B(3,1), C(6,2)C(6,2) and D(4,2)D(4,2), is shown below.

Voronoi diagram with edges labelled L1 to L6

(a) Find the equation of the edge L3L_3 between cells A and C in the form y=mx+cy=mx+c.[4]

(b) Complete the table of the edges below.

Line Equation
L3L_3
L1L_1 y=−0.2x+4.2y=-0.2x+4.2
y=−x+5y=-x+5
y=−3x+15y=-3x+15
` y=4y=4
L6L_6

[4]

Worked solution

Question 4 mark scheme

(a)

mAC=2−66−4=−42=−2,m⊥=12m_{AC}=\frac{2-6}{6-4}=\frac{-4}{2}=-2,\qquad m_{\perp}=\frac{1}{2}

Midpoint of [AC][AC]: M(5,4)M(5,4)

Perpendicular bisector: y−4=12(x−5)⇒y=12x+32y-4=\frac{1}{2}(x-5)\Rightarrow y=\frac{1}{2}x+\frac{3}{2}

(b)
Line Equation
L3L_3 y=0.5x+1.5y=0.5x+1.5
L1L_1 y=−0.2x+4.2y=-0.2x+4.2
L5L_5 y=−x+5y=-x+5
L1L_1 y=−3x+15y=-3x+15
L4L_4 ` y=4y=4
L6L_6 x=5x=5

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Question 5

MildNo calculator
[Maximum mark: 8]

The diagram shows five sites are represented by the points O(0,0)O(0,0), A(2,2)A(2,2), B(2,−2)B(2,-2), C(−2,−2)C(-2,-2) and D(−2,2)D(-2,2).

Coordinate grid showing sites O, A, B, C and D

(a) On the same diagram, complete the Voronoi diagram for the five sites.[3]

The border of cell O contains four edges.

(b) Write down the equations of the four edges.[3]

(c) Find the area of the cell O.[2]

Worked solution

Question 5 mark scheme

(a)

Completed Voronoi diagram for the five sites

(b)

y=−x+2y=-x+2, y=x−2y=x-2, y=−x−2y=-x-2, y=x+2y=x+2

(c)

Area =8=8

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Question 6

SpicyCalculator
[Maximum mark: 16]

The Voronoi diagram for four sites, A(1,1)A(1,1), B(1,5)B(1,5), C(6,1)C(6,1) and D(2,3)D(2,3), is shown below.

Voronoi diagram with vertices X, Y and Z

The vertex Z is the midpoint of the line segment [AC][AC].

(a) Write down the coordinates of the vertex Z.[1]

The equation of the edge [XY][XY] is −2x+4y=13-2x+4y=13.

(b) Find the equation of the edge [ZY][ZY] in the form ax+by=dax+by=d, where a,b,d∈Za,b,d\in\mathbb{Z}.[4]

(c) Hence, find the coordinates of the vertex Y.[2]

(d) Find the exact value of the length ZY.[2]

Consider now the cell XYZ. The length XY is 1053\frac{10\sqrt{5}}{3}.

(e) Show that the edges [XZ][XZ] and [ZY][ZY] are perpendicular.[3]

(f) Let θ=XYZ^\theta=\widehat{XYZ}.

(i) Show that cos⁡θ=45\cos\theta=\frac{4}{5}.

(ii) Explain why sin⁡θ=35\sin\theta=\frac{3}{5}.[4]

(g) Find the exact value of the area of the triangle XYZ.[2]

Worked solution

Question 6 mark scheme

(a)

Z(72,1)Z\left(\frac{7}{2},1\right)

(b)

mCD=3−12−6=2−4=−12,m⊥=2m_{CD}=\frac{3-1}{2-6}=\frac{2}{-4}=-\frac{1}{2},\qquad m_{\perp}=2

Midpoint of [CD][CD]: M(4,2)M(4,2)

Perpendicular bisector: y−2=2(x−4)⇒2x−y=6y-2=2(x-4)\Rightarrow 2x-y=6

(c)

Y(376,193)Y\left(\frac{37}{6},\frac{19}{3}\right)

(d)

dZY=(72−376)2+(1−193)2=853d_{ZY}=\sqrt{\left(\frac{7}{2}-\frac{37}{6}\right)^2+\left(1-\frac{19}{3}\right)^2}=\frac{8\sqrt{5}}{3}

(e)

mZY=2m_{ZY}=2

mAD=2⇒mXZ=−12m_{AD}=2\Rightarrow m_{XZ}=-\frac{1}{2}

mZY×mXZ=−1m_{ZY}\times m_{XZ}=-1 hence perpendicular

(f)

(i)

cos⁡θ=ZYXZ=8531053=810=45\cos\theta=\frac{ZY}{XZ}=\frac{\frac{8\sqrt{5}}{3}}{\frac{10\sqrt{5}}{3}}=\frac{8}{10}=\frac{4}{5}

(ii) A right-angled triangle with sides 3,4,53,4,5 shows that sin⁡θ=35\sin\theta=\frac{3}{5}

(OR for HL, using the Pythagorean identity cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1 we obtain sin⁡θ=35\sin\theta=\frac{3}{5}**)**

(g)

Area=12×XY×YZ×sin⁡θ=12105385335=403\textit{Area}=\frac{1}{2}\times XY\times YZ\times\sin\theta=\frac{1}{2}\frac{10\sqrt{5}}{3}\frac{8\sqrt{5}}{3}\frac{3}{5}=\frac{40}{3}

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Question 7

SpicyCalculator
[Maximum mark: 14]

Part of the Voronoi diagram for four sites, A(1,1)A(1,1), B(4,0)B(4,0), C(4,4)C(4,4) and D(8,2)D(8,2) is shown below. The two vertices of the Voronoi diagram are shown by the points X and Y.

Partial Voronoi diagram with vertices X and Y

(a) Two edges are missing.

(i) Draw the two edges.

(ii) Find their equations.[6]

(b) Write down the coordinates of the two vertices X and Y.[2]

(c) Find the exact value of the distance between the sites A and D.[2]

(d) Find the area of the quadrilateral ABCD.[4]

Worked solution

Question 7 mark scheme

(a)

(i)

Completed Voronoi diagram showing the two missing edges

(ii) y=2y=2, y=−2x+13y=-2x+13

(b)

X(3,2)X(3,2), Y(5.5,2)Y(5.5,2)

(c)

dAD=(8−1)2+(2−1)2=50=52d_{AD}=\sqrt{(8-1)^2+(2-1)^2}=\sqrt{50}=5\sqrt{2}

(d)

AABCD=AABC+ADBC=12×4×3+12×4×4=14A_{ABCD}=A_{ABC}+A_{DBC}=\frac{1}{2}\times4\times3+\frac{1}{2}\times4\times4=14

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