IB Mathematics AHL 1.14 Definition of a matrix AI HL Paper 3- Exam Style Questions- New Syllabus
Question
This question explores the planning of a new town and the placement of a toxic waste dump in a metropolitan area.
A metropolitan area is represented as a square, with coordinates in kilometres where the x-coordinate indicates distance east and the y-coordinate indicates distance north from the origin at (0, 0).
- Edison is positioned at \( E(0, 40) \).
- Fermiton is positioned at \( F(40, 40) \).
- Gaussville is positioned at \( G(40, 0) \).
- Hamilton is positioned at \( H(0, 0) \).
(a) Identify conditions under which the assumption of modelling each town as a single point is reasonable. [1]
(b) Draw a Voronoi diagram illustrating the regions within the metropolitan area closest to each town. [1]
The council plans to establish a new town, Isaacopolis, at \( I(30, 20) \).
A new Voronoi diagram is to be created to include Isaacopolis. The equation of the perpendicular bisector of \([IE]\) is \[ y = \frac{3}{2}x + \frac{15}{2}. \]
(c)(i) Determine the equation of the perpendicular bisector of \([IF]\). [4]
(c)(ii) Given one vertex of the new Voronoi diagram at \( (20, 37.5) \), calculate the coordinates of the other two vertices within the metropolitan area. [4]
(c)(iii) Illustrate the new Voronoi diagram, showing the regions within the metropolitan area closest to each town. [2]
The metropolitan area is partitioned into districts based on the Voronoi regions from part (c).
(d) A car travels due east from a point directly north of Hamilton to a point directly north of Gaussville, passing through the Edison, Isaacopolis, and Fermiton districts, spending 30% of the travel time in the Isaacopolis district. Calculate the distance between Gaussville and the destination point. [4]
A toxic waste dump must be placed within the metropolitan area, as far as possible from the nearest town.
(e)(i) Determine the location of the toxic waste dump, ensuring it is not on the edge of the metropolitan area. [4]
(e)(ii) State one potential criticism of the council’s chosen location. [1]
The toxic waste dump, \( T \), is linked to the towns through a sewer network.
The connections are represented in the following matrix, \( M \), where the order of rows and columns is \( (E, F, G, H, I, T) \):
\[ M = \begin{bmatrix} 1 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \end{bmatrix} \]
A leak occurs from the toxic waste dump and travels through the sewers. The pollution takes one day to travel between directly connected locations.
The digit 1 in \( M \) represents a direct connection. The values of 1 in the leading diagonal of \( M \) indicate that once a location is polluted, it remains polluted.
(f)(i) Identify which town is the last to be polluted, and justify your answer. [3]
(f)(ii) State the number of days it takes for the pollution to reach the last town. [1]
(f)(iii) A sewer inspector must plan the shortest route to inspect each connection between different locations. Determine suitable start and end points for the inspection route. [2]
Note that connections to the same location do not correspond to sewers that need inspection.
▶️ Answer/Explanation
Markscheme
(a)
The assumption of modelling each town as a single point is reasonable if:
– Each town is small compared to the distances between towns, or
– Each town has an identifiable centre (e.g., town hall), or
– The coordinates represent the town’s centre.
Reasonable if towns are small or have defined centres R1
[1 mark]
The assumption of modelling each town as a single point is reasonable if:
– Each town is small compared to the distances between towns, or
– Each town has an identifiable centre (e.g., town hall), or
– The coordinates represent the town’s centre.
Reasonable if towns are small or have defined centres R1
[1 mark]
(b)
A Voronoi diagram for \( E(0,40) \), \( F(40,40) \), \( G(40,0) \), \( H(0,0) \) divides the square into four regions, with perpendicular bisectors of segments \( EF \), \( FG \), \( GH \), \( HE \).
Correct Voronoi diagram (no scale required)
A Voronoi diagram for \( E(0,40) \), \( F(40,40) \), \( G(40,0) \), \( H(0,0) \) divides the square into four regions, with perpendicular bisectors of segments \( EF \), \( FG \), \( GH \), \( HE \).
Correct Voronoi diagram (no scale required)
A1 [1 mark]
(c)(i)
For \( I(30,20) \), \( F(40,40) \):
1. Gradient of \( IF \): \( m_{IF} = \frac{40-20}{40-30} = 2 \).
2. Perpendicular gradient: \( m_\perp = -\frac{1}{2} \).
3. Midpoint of \( IF \): \( \left( \frac{30+40}{2}, \frac{20+40}{2} \right) = (35,30) \).
4. Equation through \( (35,30) \) with slope \( -\frac{1}{2} \):
\( y – 30 = -\frac{1}{2}(x – 35) \).
Simplify: \( y = -\frac{1}{2}x + \frac{95}{2} \).
Equation: \( y = -\frac{1}{2}x + \frac{95}{2} \) M1A1A1A1
[4 marks]
For \( I(30,20) \), \( F(40,40) \):
1. Gradient of \( IF \): \( m_{IF} = \frac{40-20}{40-30} = 2 \).
2. Perpendicular gradient: \( m_\perp = -\frac{1}{2} \).
3. Midpoint of \( IF \): \( \left( \frac{30+40}{2}, \frac{20+40}{2} \right) = (35,30) \).
4. Equation through \( (35,30) \) with slope \( -\frac{1}{2} \):
\( y – 30 = -\frac{1}{2}(x – 35) \).
Simplify: \( y = -\frac{1}{2}x + \frac{95}{2} \).
Equation: \( y = -\frac{1}{2}x + \frac{95}{2} \) M1A1A1A1
[4 marks]
(c)(ii)
The perpendicular bisector of \( EH \) (for \( E(0,40) \), \( H(0,0) \)) is: \( y = 20 \).
Intersecting bisectors give vertices: \( \left( \frac{25}{3}, 20 \right) \), \( (20, 2.5) \).
Vertices: \( \left( \frac{25}{3}, 20 \right) \), \( (20, 2.5) \) M1A1A1A1
[4 marks]
The perpendicular bisector of \( EH \) (for \( E(0,40) \), \( H(0,0) \)) is: \( y = 20 \).
Intersecting bisectors give vertices: \( \left( \frac{25}{3}, 20 \right) \), \( (20, 2.5) \).
Vertices: \( \left( \frac{25}{3}, 20 \right) \), \( (20, 2.5) \) M1A1A1A1
[4 marks]
(c)(iii)
The new Voronoi diagram includes \( I(30,20) \), with four bisectors (\( IE \), \( IF \), \( IG \), \( IH \)) intersecting original boundaries.
Correct diagram with four bisectors around \( I \)
The new Voronoi diagram includes \( I(30,20) \), with four bisectors (\( IE \), \( IF \), \( IG \), \( IH \)) intersecting original boundaries.
Correct diagram with four bisectors around \( I \)
M1A1 [2 marks]
(d)
The car travels along \( y = c \).
E–I bisector: \( y = -\frac{1}{2}x + \frac{95}{2} \), so \( x_E = 95 – 2c \).
F–G bisector: \( x = 40 \).
Isaacopolis region length: \( L_I = 40 – (95 – 2c) = 2c – 55 \).
Total length: \( L_{\text{all}} = 40 \).
Given 30% time in Isaacopolis, \( L_I = 0.3 \times 40 = 12 \).
Solve: \( 2c – 55 = 12 \Rightarrow c = 33 \).
Destination: \( (40, 33) \). Distance from \( G(40,0) \): 33 km.
Distance = 33 km M1A1A1A1
[4 marks]
The car travels along \( y = c \).
E–I bisector: \( y = -\frac{1}{2}x + \frac{95}{2} \), so \( x_E = 95 – 2c \).
F–G bisector: \( x = 40 \).
Isaacopolis region length: \( L_I = 40 – (95 – 2c) = 2c – 55 \).
Total length: \( L_{\text{all}} = 40 \).
Given 30% time in Isaacopolis, \( L_I = 0.3 \times 40 = 12 \).
Solve: \( 2c – 55 = 12 \Rightarrow c = 33 \).
Destination: \( (40, 33) \). Distance from \( G(40,0) \): 33 km.
Distance = 33 km M1A1A1A1
[4 marks]
(e)(i)
The dump must be at a Voronoi vertex to maximize distance from the nearest town.
Using (c)(ii), the interior vertex is \( \left( \frac{25}{3}, 20 \right) \).
Distance to \( E(0,40) \): \( \sqrt{\left( \frac{25}{3} \right)^2 + (20-40)^2} = \sqrt{\frac{625}{9} + 400} = \frac{65}{3} \).
Distance to \( F(40,40) \), \( G(40,0) \): \( \sqrt{\left( 40 – \frac{25}{3} \right)^2 + (40-20)^2} = \frac{\sqrt{12625}}{3} \).
Equidistant from nearest towns, maximizing minimum distance.
Location: \( \left( \frac{25}{3}, 20 \right) \) M1A1A1A1
[4 marks]
The dump must be at a Voronoi vertex to maximize distance from the nearest town.
Using (c)(ii), the interior vertex is \( \left( \frac{25}{3}, 20 \right) \).
Distance to \( E(0,40) \): \( \sqrt{\left( \frac{25}{3} \right)^2 + (20-40)^2} = \sqrt{\frac{625}{9} + 400} = \frac{65}{3} \).
Distance to \( F(40,40) \), \( G(40,0) \): \( \sqrt{\left( 40 – \frac{25}{3} \right)^2 + (40-20)^2} = \frac{\sqrt{12625}}{3} \).
Equidistant from nearest towns, maximizing minimum distance.
Location: \( \left( \frac{25}{3}, 20 \right) \) M1A1A1A1
[4 marks]
(e)(ii)
The choice ignores population or environmental risks (e.g., being upwind or upstream of a large city).
Ignores population or environmental factors A1
[1 mark]
The choice ignores population or environmental risks (e.g., being upwind or upstream of a large city).
Ignores population or environmental factors A1
[1 mark]
(f)(i)
Method 1 (Matrix Powers):
Compute powers of \( M \). Last row (for \( T \)):
\( M^2 \): \( (3, 5, 1, 6, 0, 7) \).
\( M^4 \): \( (10, 12, 4, 16, 1, 18) \).
First nonzero entry in column \( I \) appears at \( M^4 \).
Method 2 (Graph Translation):
Translate \( M \) into a step-by-step spread:
Method 1 (Matrix Powers):
Compute powers of \( M \). Last row (for \( T \)):
\( M^2 \): \( (3, 5, 1, 6, 0, 7) \).
\( M^4 \): \( (10, 12, 4, 16, 1, 18) \).
First nonzero entry in column \( I \) appears at \( M^4 \).
Method 2 (Graph Translation):
Translate \( M \) into a step-by-step spread:
| Day | Locations Polluted |
|---|---|
| 0 | \( T \) |
| 1 | \( F, H \) |
| 2 | \( E \) |
| 3 | \( G \) |
| 4 | \( I \) |
Both methods confirm Isaacopolis is the last town polluted.
Last town: \( I \) M1A1A1
[3 marks]
Last town: \( I \) M1A1A1
[3 marks]
(f)(ii)
Pollution path: \( T \to F/H (1) \to E (2) \to G (3) \to I (4) \).
4 days A1
[1 mark]
Pollution path: \( T \to F/H (1) \to E (2) \to G (3) \to I (4) \).
4 days A1
[1 mark]
(f)(iii)
Method 1 (Vertex Orders):
Vertex degrees (without self-loops):
\( E \): 2, \( F \): 1, \( G \): 2, \( H \): 2, \( I \): 1, \( T \): 2.
Method 2 (Graph Connections):
The network has edges \( E-G \), \( E-H \), \( F-H \), \( F-T \), \( H-T \), \( G-I \).
Only \( F \) and \( I \) have odd degrees (1 each), requiring the inspection route to start at one and end at the other to cover all edges.
Start: \( F \), End: \( I \) A1A1
[2 marks]
Method 1 (Vertex Orders):
Vertex degrees (without self-loops):
\( E \): 2, \( F \): 1, \( G \): 2, \( H \): 2, \( I \): 1, \( T \): 2.
Method 2 (Graph Connections):
The network has edges \( E-G \), \( E-H \), \( F-H \), \( F-T \), \( H-T \), \( G-I \).
Only \( F \) and \( I \) have odd degrees (1 each), requiring the inspection route to start at one and end at the other to cover all edges.
Start: \( F \), End: \( I \) A1A1
[2 marks]