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Proof, EXT2 P2 EQ-Bank 4

Using mathematical induction, prove that the maximum number of diagonals of an `n`-sided plane convex polynomial is `(n(n-3))/2` for `n>=4`.  (3 marks)

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`text{Let}\ \ D_n=\ text{maximum number of diagonals of}\ n text{-sided polygon}`

`text{Prove true for}\ \ n=4:`

`D_4= 2\ \ text{(quadrilateral only has 2 possible diagonals)}`

`(4(4-3))/2 = 2`

`:.\ text{True for}\ \ n=4.`
 

`text{Assume true for}\ \ n=k:`

`text{i.e.}\ D_k=(k(k-3))/2`
 

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ D_(k+1)=((k+1)(k+1-3))/2=((k+1)(k-2))/2`
 

`text{Let}\ ktext{-sided polygon be defined by points}\ P_1,P_2,…,P_k`

`text{Let}\ (k+1)text{-sided polygon be defined by points}\ P_1,P_2,…,P_k,P_(k+1)`

`=>\ text{The extra point adds}\ (k-1)\ text{diagonals}`
 

`D_(k+1)` `=D_k+(k-1)`  
  `=(k(k-3))/2+ (k-1)`  
  `=(k^2-3k+2k-2)/2`  
  `=(k^2-k-2)/2`  
  `=((k+1)(k-2))/2`  
  `=\ text{RHS}`  

 
`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for}\ n=4,\ text{by PMI, true for integers}\ n>=4.`

Proof, EXT2 P2 EQ-Bank 3

Using mathematical induction, prove that the sum of the internal angles of an `n`-sided polynomial is  `180(n-2)°`  for  `n>=3`.  (3 marks)

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`text{Let}\ \ S_n=\ text{sum of interior angles of}\ n text{-sided polygon}`

`text{Prove true for}\ \ n=3:`

`S_3= 180°\ \ text{(sum of internal angles of a Δ)}`

`180(3-2) = 180°`

`:.\ text{True for}\ \ n=3.`
 

`text{Assume true for}\ \ n=k:`

`text{i.e.}\ S_k=(k-2)180°`
 

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ S_(k+1)=(k+1-2)180°=(k-1)180°`
 

`text{Let}\ ktext{-sided polygon be defined by points}\ P_1,P_2,…,P_k`

`text{Let}\ (k+1)text{-sided polygon be defined by points}\ P_1,P_2,…,P_k,P_(k+1)`

`S_(k+1)` `=S_k+\ text{angle sum}\ ΔP_1P_kP_(k+1)`  
  `=(k-2)180° + 180°`  
  `=(k-1)180°`  
  `=\ text{RHS}`  

 
`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for}\ n=3,\ text{by PMI, true for integers}\ n>=3.`

Proof, EXT2 P2 EQ-Bank 2

`n` lines are drawn in a 2-dimensional plane such that no three lines are concurrent and no two lines are parallel.

`S_n` is the number of regions into which these lines divide the plane with the diagram illustrating that `S_3=7`
 

  1. In a similar way, draw diagrams that illustrate `S_1, S_2` and `S_4`.  (1 mark)

  2. Using the results in part (i), make a conjecture about the general formula for `S_n`.  (2 marks)

  3. Use mathematical induction to prove the formula from part (ii).  (3 marks)

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  1. `text{See Worked Solutions}`
  2. `S_n=(n^2+n+2)/2`
  3. `text{See Worked Solutions}`
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i.       `S_1=2`                        `S_2=4`
 
             
 

   `S_4=11`


 

ii.   `text{Consider the pattern above:}`

`S_1=2,\ \ S_2=S_1+2`

`S_3=S_2+3,\ …\ , S_n=S_(n-1)+n`

`S_n` `=S_(n-1)+n`  
  `=S_(n-2)+(n-1)+n`  
  `=S_(n-3)+(n-2)+(n-1)+n`  
  `vdots`  
  `=S_1+(2+3+…+(n-1)+n)`  
  `=S_0+(1+2+…+(n-1)+n)`  
  `=1+(1+2+…+(n-1)+n)`  

 
`=>text{AP where}\ \ a=1, l=n, n=n`

`S_n` `=1+n/2(1+n)`  
  `=(2+n(n+1))/2`  
  `=(n^2+n+2)/2`  

 

iii.   `text{Prove}\ \ S_n=(n^2+n+2)/2\ \ text{for}\ \ n>=0`

`text{If}\ \ n=1`

`text{LHS}\ =S_1=2`

`text{RHS}\ = (1^2+1+2)/2=2=\ text{LHS}`

`:.\ text{True for}\ n=1.`
 

`text{Assume true for}\ \ n=k:`

`text{i.e.}\ S_k=(k^2+k+2)/2`

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ S_(k+1)=((k+1)^2+(k+1)+2)/2=(k^2+3k+4)/2`
 

`text{Consider the line}\ d_4\ text{added below that crosses 3 existing lines}`

`text{and creates 4 new regions.}`

`text{Similarly, the line}\ d_(k+1)\ text{will cross}\ k\ text{existing lines and}`

`text{create}\ (k+1)\ text{new regions.}`

`S_(k+1)` `=S_k+k+1`  
  `=(k^2+k+2)/2+k+1`  
  `=(k^2+k+2+2k+2)/2`  
  `=(k^2+3k+4)/2`  
  `=\ text{RHS}`  

 
`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for}\ n=1,\ text{by PMI, true for integers}\ n>=1`

Proof, EXT2 P2 EQ-Bank 1

Using mathematical induction, prove that the sum of the external angles of an `n`-sided plane convex polygon equals 360° for  `n>=3`.  (4 marks)

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`text{Prove true for}\ \ n=3:`

 

`text{Let}\ \ S_3=\ text{sum of exterior angles for a 3-sided polynomial (triangle)}`

`S_3=angleACF+angleCBE+angleBAD`

`angleACF+angleACB=180°\ \ (angleFCB\ text{is a straight angle})`

`angleCBE+angleCBA=180°\ \ (angleABE\ text{is a straight angle})`

`angleBAD+angleBAC=180°\ \ (angleDAC\ text{is a straight angle})`

`angleACF+angleACB+angleCBE+angleCBA+angleBAD+angleBAC=540°`

 
`text{Internal angle sum of}\ ΔABC=180°:`

`angleACF+angleCBE+angleBAD+180°` `=540°`  
`angleACF+angleCBE+angleBAD` `=360°`  

 
`:.\ text{True for}\ \ n=3`
 

`text{Assume true for}\ \ n=k`

`text{i.e.}\ \ S_k=360°`

`text{Prove true for}\ \ n=k+1:`

`text{i.e.}\ \ S_(k+1)=360°`

`text{Let}\ ktext{-sided polygon be defined by points}\ P_1,P_2,…,P_k`

`text{Let}\ (k+1)text{-sided polygon be defined by points}\ P_1,P_2,…,P_k,P_(k+1)`
 

`S_k` `=theta_1+theta_2+ … +theta_(k-1)+theta_k`  
  `=theta_1+theta_2+ … +theta_(k-1)+alpha+beta`  

 
`S_(k+1)=mu+theta_2+ … +theta_(k-1)+beta+delta`

`theta_1=mu+phi\ \ text{(vertically opposite)}`

`=>\ \ mu=theta_1-phi\ …\ (1)`

`delta=alpha+phi\ …\ (2)\ \  text{(exterior angle of triangle)}`

  
`text{Substitute (1) and (2) into}\ S_(k+1):`

`S_(k+1)` `=theta_1-phi+theta_2+ … +theta_(k-1)+beta+alpha+phi`  
  `=theta_1+theta_2+ … +theta_(k-1)+alpha+beta`  
  `=S_k`  

 

`:.\ text{True for}\ \ n=k+1`

`:.\ text{Since true for}\ n=3,\ text{true for integers}\ n>=3.`