Date | May Specimen paper | Marks available | 2 | Reference code | SPM.3.AHL.TZ0.1 |
Level | Additional Higher Level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Find | Question number | 1 | Adapted from | N/A |
Question
This question asks you to investigate regular n-sided polygons inscribed and circumscribed in a circle, and the perimeter of these as n tends to infinity, to make an approximation for π.
Let Pi(n) represent the perimeter of any n-sided regular polygon inscribed in a circle of radius 1 unit.
Consider an equilateral triangle ABC of side length, x units, circumscribed about a circle of radius 1 unit and centre O as shown in the following diagram.
Let Pc(n) represent the perimeter of any n-sided regular polygon circumscribed about a circle of radius 1 unit.
Consider an equilateral triangle ABC of side length, x units, inscribed in a circle of radius 1 unit and centre O as shown in the following diagram.
The equilateral triangle ABC can be divided into three smaller isosceles triangles, each subtending an angle of 2π3 at O, as shown in the following diagram.
Using right-angled trigonometry or otherwise, show that the perimeter of the equilateral triangle ABC is equal to 3√3 units.
Consider a square of side length, x units, inscribed in a circle of radius 1 unit. By dividing the inscribed square into four isosceles triangles, find the exact perimeter of the inscribed square.
Find the perimeter of a regular hexagon, of side length, x units, inscribed in a circle of radius 1 unit.
Show that Pi(n)=2nsin(πn).
Use an appropriate Maclaurin series expansion to find limn→∞Pi(n) and interpret this result geometrically.
Show that Pc(n)=2ntan(πn).
By writing Pc(n) in the form 2tan(πn)1n, find limn→∞Pc(n).
Use the results from part (d) and part (f) to determine an inequality for the value of π in terms of n.
The inequality found in part (h) can be used to determine lower and upper bound approximations for the value of π.
Determine the least value for n such that the lower bound and upper bound approximations are both within 0.005 of π.
Markscheme
METHOD 1
consider right-angled triangle OCX where CX =x2
sinπ3=x21 M1A1
⇒x2=√32⇒x=√3 A1
Pi=3×x=3√3 AG
METHOD 2
eg use of the cosine rule x2=12+12−2(1)(1)cos2π3 M1A1
x=√3 A1
Pi=3×x=3√3 AG
Note: Accept use of sine rule.
[3 marks]
sinπ4=1x where x = side of square M1
x=√2 A1
Pi=4√2 A1
[3 marks]
6 equilateral triangles ⇒x = 1 A1
Pi=6 A1
[2 marks]
in right-angled triangle sin(πn)=x21 M1
⇒x=2sin(πn) A1
Pi=n×x
Pi=n×2sin(πn) M1
Pi=2nsin(πn) AG
[3 marks]
consider limn→∞2nsin(πn)
use of sinx=x−x33!+x55!−… M1
2nsin(πn)=2n(πn−π36n3+π5120n5−…) (A1)
=2(π−π36n2+π5120n4−…) A1
⇒limn→∞2nsin(πn)=2π A1
as n→∞ polygon becomes a circle of radius 1 and Pi=2π R1
[5 marks]
consider an n-sided polygon of side length x
2n right-angled triangles with angle 2π2n=πn at centre M1A1
opposite side x2=tan(πn)⇒x=2tan(πn) M1A1
Perimeter Pc=2ntan(πn) AG
[4 marks]
consider limn→∞2ntan(πn)=limn→∞(2tan(πn)1n)
=limn→∞(2tan(πn)1n)=00 R1
attempt to use L’Hopital’s rule M1
=limn→∞(−2πn2sec2(πn)−1n2) A1A1
=2π A1
[5 marks]
Pi<2π<Pc
2nsin(πn)<2π<2ntan(πn) M1
nsin(πn)<π<ntan(πn) A1
[2 marks]
attempt to find the lower bound and upper bound approximations within 0.005 of π (M1)
n = 46 A2
[3 marks]