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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	Determine	Question number	1	Adapted from	N/A

Question

This question asks you to investigate regular $n$ $n$ -sided polygons inscribed and circumscribed in a circle, and the perimeter of these as $n$ $n$ tends to infinity, to make an approximation for $π$ $\pi$ .

Let $P_{i} (n)$ ${P_i}\left( n \right)$ represent the perimeter of any $n$ $n$ -sided regular polygon inscribed in a circle of radius 1 unit.

Consider an equilateral triangle ABC of side length, $x$ $x$ units, circumscribed about a circle of radius 1 unit and centre O as shown in the following diagram.

Let $P_{c} (n)$ ${P_c}\left( n \right)$ represent the perimeter of any $n$ $n$ -sided regular polygon circumscribed about a circle of radius 1 unit.

Consider an equilateral triangle ABC of side length, $x$ $x$ 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 $\frac{2 π}{3}$ $\frac{{2\pi }}{3}$ 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 $3 \sqrt{3}$ $3\sqrt 3$ units.

[3]

Consider a square of side length, $x$ $x$ 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.

[3]

Find the perimeter of a regular hexagon, of side length, $x$ $x$ units, inscribed in a circle of radius 1 unit.

[2]

Show that $P_{i} (n) = 2 n sin (\frac{π}{n})$ ${P_i}\left( n \right) = 2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right)$ .

[3]

Use an appropriate Maclaurin series expansion to find $lim n \to \infty P_{i} (n)$ $\mathop {{\text{lim}}}\limits_{n \to \infty } {P_i}\left( n \right)$ and interpret this result geometrically.

[5]

Show that $P_{c} (n) = 2 n tan (\frac{π}{n})$ ${P_c}\left( n \right) = 2n\,{\text{tan}}\left( {\frac{\pi }{n}} \right)$ .

[4]

By writing $P_{c} (n)$ ${P_c}\left( n \right)$ in the form $\frac{2 tan (\frac{π}{n})}{\frac{1}{n}}$ $\frac{{2\,{\text{tan}}\left( {\frac{\pi }{n}} \right)}}{{\frac{1}{n}}}$ , find $lim n \to \infty P_{c} (n)$ $\mathop {{\text{lim}}}\limits_{n \to \infty } {P_c}\left( n \right)$ .

[5]

Use the results from part (d) and part (f) to determine an inequality for the value of $π$ $\pi$ in terms of $n$ $n$ .

[2]

The inequality found in part (h) can be used to determine lower and upper bound approximations for the value of $π$ $\pi$ .

Determine the least value for $n$ $n$ such that the lower bound and upper bound approximations are both within 0.005 of $π$ $\pi$ .

[3]

Markscheme

METHOD 1

consider right-angled triangle OCX where CX $= \frac{x}{2}$ $= \frac{x}{2}$

$sin \frac{π}{3} = \frac{\frac{x}{2}}{1}$ ${\text{sin}}\frac{\pi }{3} = \frac{{\frac{x}{2}}}{1}$ M1A1

$\Rightarrow \frac{x}{2} = \frac{\sqrt{3}}{2} \Rightarrow x = \sqrt{3}$ $\Rightarrow \frac{x}{2} = \frac{{\sqrt 3 }}{2} \Rightarrow x = \sqrt 3$ A1

$P_{i} = 3 \times x = 3 \sqrt{3}$ ${P_i} = 3 \times x = 3\sqrt 3$ AG

METHOD 2

eg use of the cosine rule $x^{2} = 1^{2} + 1^{2} - 2 (1) (1) cos \frac{2 π}{3}$ ${x^2} = {1^2} + {1^2} - 2\left( 1 \right)\left( 1 \right){\text{cos}}\frac{{2\pi }}{3}$ M1A1

$x = \sqrt{3}$ $x = \sqrt 3$ A1

$P_{i} = 3 \times x = 3 \sqrt{3}$ ${P_i} = 3 \times x = 3\sqrt 3$ AG

Note: Accept use of sine rule.

[3 marks]

$sin \frac{π}{4} = \frac{1}{x}$ ${\text{sin}}\frac{\pi }{4} = \frac{1}{x}$ where $x$ $x$ = side of square M1

$x = \sqrt{2}$ $x = \sqrt 2$ A1

$P_{i} = 4 \sqrt{2}$ ${P_i} = 4\sqrt 2$ A1

[3 marks]

6 equilateral triangles ⇒ $x$ $x$ = 1 A1

$P_{i} = 6$ ${P_i} = 6$ A1

[2 marks]

in right-angled triangle $sin (\frac{π}{n}) = \frac{\frac{x}{2}}{1}$ ${\text{sin}}\left( {\frac{\pi }{n}} \right) = \frac{{\frac{x}{2}}}{1}$ M1

$\Rightarrow x = 2 sin (\frac{π}{n})$ $\Rightarrow x = 2\,{\text{sin}}\left( {\frac{\pi }{n}} \right)$ A1

$P_{i} = n \times x$ ${P_i} = n \times x$

$P_{i} = n \times 2 sin (\frac{π}{n})$ ${P_i} = n \times 2\,{\text{sin}}\left( {\frac{\pi }{n}} \right)$ M1

$P_{i} = 2 n sin (\frac{π}{n})$ ${P_i} = 2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right)$ AG

[3 marks]

consider $lim n \to \infty 2 n sin (\frac{π}{n})$ $\mathop {{\text{lim}}}\limits_{n \to \infty } 2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right)$

use of $sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots$ ${\text{sin}}\,x = x - \frac{{{x^3}}}{{3{\text{!}}}} + \frac{{{x^5}}}{{5{\text{!}}}} - \ldots$ M1

$2 n sin (\frac{π}{n}) = 2 n (\frac{π}{n} - \frac{π^{3}}{6 n^{3}} + \frac{π^{5}}{120 n^{5}} - \dots)$ $2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right) = 2n\left( {\frac{\pi }{n} - \frac{{{\pi ^3}}}{{6{n^3}}} + \frac{{{\pi ^5}}}{{120{n^5}}} - \ldots } \right)$ (A1)

$= 2 (π - \frac{π^{3}}{6 n^{2}} + \frac{π^{5}}{120 n^{4}} - \dots)$ $= 2\left( {\pi - \frac{{{\pi ^3}}}{{6{n^2}}} + \frac{{{\pi ^5}}}{{120{n^4}}} - \ldots } \right)$ A1

$\Rightarrow lim n \to \infty 2 n sin (\frac{π}{n}) = 2 π$ $\Rightarrow \mathop {{\text{lim}}}\limits_{n \to \infty } 2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right) = 2\pi$ A1

as $n \to \infty$ $n \to \infty$ polygon becomes a circle of radius 1 and $P_{i} = 2 π$ ${P_i} = 2\pi$ R1

[5 marks]

consider an $n$ $n$ -sided polygon of side length $x$ $x$

2 $n$ $n$ right-angled triangles with angle $\frac{2 π}{2 n} = \frac{π}{n}$ $\frac{{2\pi }}{{2n}} = \frac{\pi }{n}$ at centre M1A1

opposite side $\frac{x}{2} = tan (\frac{π}{n}) \Rightarrow x = 2 tan (\frac{π}{n})$ $\frac{x}{2} = {\text{tan}}\left( {\frac{\pi }{n}} \right) \Rightarrow x = 2\,{\text{tan}}\left( {\frac{\pi }{n}} \right)$ M1A1

Perimeter $P_{c} = 2 n tan (\frac{π}{n})$ ${P_c} = 2n\,{\text{tan}}\left( {\frac{\pi }{n}} \right)$ AG

[4 marks]

consider $lim n \to \infty 2 n tan (\frac{π}{n}) = lim n \to \infty (\frac{2 tan (\frac{π}{n})}{\frac{1}{n}})$ $\mathop {{\text{lim}}}\limits_{n \to \infty } 2n\,{\text{tan}}\left( {\frac{\pi }{n}} \right) = \mathop {{\text{lim}}}\limits_{n \to \infty } \left( {\frac{{2\,{\text{tan}}\left( {\frac{\pi }{n}} \right)}}{{\frac{1}{n}}}} \right)$

$= lim n \to \infty (\frac{2 tan (\frac{π}{n})}{\frac{1}{n}}) = \frac{0}{0}$ $= \mathop {{\text{lim}}}\limits_{n \to \infty } \left( {\frac{{2\,{\text{tan}}\left( {\frac{\pi }{n}} \right)}}{{\frac{1}{n}}}} \right) = \frac{0}{0}$ R1

attempt to use L’Hopital’s rule M1

$= lim n \to \infty (\frac{- \frac{2 π}{n^{2}} se c^{2} (\frac{π}{n})}{- \frac{1}{n^{2}}})$ $= \mathop {{\text{lim}}}\limits_{n \to \infty } \left( {\frac{{ - \frac{{2\pi }}{{{n^2}}}{\text{se}}{{\text{c}}^2}\left( {\frac{\pi }{n}} \right)}}{{ - \frac{1}{{{n^2}}}}}} \right)$ A1A1

$= 2 π$ $= 2\pi$ A1

[5 marks]

$P_{i} < 2 π < P_{c}$ ${P_i} < 2\pi < {P_c}$

$2 n sin (\frac{π}{n}) < 2 π < 2 n tan (\frac{π}{n})$ $2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right) < 2\pi < 2n\,{\text{tan}}\left( {\frac{\pi }{n}} \right)$ M1

$n sin (\frac{π}{n}) < π < n tan (\frac{π}{n})$ $n\,{\text{sin}}\left( {\frac{\pi }{n}} \right) < \pi < n\,{\text{tan}}\left( {\frac{\pi }{n}} \right)$ A1

[2 marks]

attempt to find the lower bound and upper bound approximations within 0.005 of $π$ $\pi$ (M1)

$n$ $n$ = 46 A2

[3 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » AHL 5.13—Limits and L’Hopitals

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