Note: All polygons are regular with side length \( x \), centered at \( O \), in a circle of radius \( \text{1} \) unit. Use \( \pi \approx \text{3.1415926535} \), and numerical answers are exact or rounded to \( \text{3} \) significant figures unless specified.

(a)

Consider triangle \( OCX \), right-angled at \( X \), where \( CX = \frac{x}{\text{2}} \), radius \( OC = \text{1} \), and angle at \( O = \frac{\pi}{\text{3}} \) (see third diagram).

Apply sine: \[ \sin \left( \frac{\pi}{\text{3}} \right) = \frac{\frac{x}{\text{2}}}{\text{1}} \] (M1).

\[ \frac{x}{\text{2}} = \sin \left( \frac{\pi}{\text{3}} \right) = \frac{\sqrt{\text{3}}}{\text{2}} \implies x = \text{2} \cdot \frac{\sqrt{\text{3}}}{\text{2}} = \sqrt{\text{3}} \] (A1).

Perimeter: \[ P_i(\text{3}) = \text{3} \cdot x = \text{3} \sqrt{\text{3}} \] (A1).

Alternative: In triangle \( OAB \), use cosine rule: \[ x^2 = \text{1}^2 + \text{1}^2 – \text{2} \cdot \text{1} \cdot \text{1} \cdot \cos \left( \frac{\text{2}\pi}{\text{3}} \right) = \text{1} + \text{1} – \text{2} \cdot \left( -\frac{\text{1}}{\text{2}} \right) = \text{3} \implies x = \sqrt{\text{3}} \] (M1)(A1).

[3 marks]

(b)

Divide the square into four isosceles triangles, each subtending \( \frac{\text{2}\pi}{\text{4}} = \frac{\pi}{\text{2}} \) at \( O \). In right-angled triangle \( OCX \), \( CX = \frac{x}{\text{2}} \), radius \( OC = \text{1} \), angle at \( O = \frac{\pi}{\text{4}} \):

\[ \sin \left( \frac{\pi}{\text{4}} \right) = \frac{\frac{x}{\text{2}}}{\text{1}} \implies \frac{x}{\text{2}} = \frac{\sqrt{\text{2}}}{\text{2}} \] (M1).

\[ x = \text{2} \cdot \frac{\sqrt{\text{2}}}{\text{2}} = \sqrt{\text{2}} \] (A1).

Perimeter: \[ P_i(\text{4}) = \text{4} \cdot \sqrt{\text{2}} \] (A1).

[3 marks]

(c)

A regular hexagon inscribed in the circle forms \( \text{6} \) equilateral triangles. Each triangle has vertices at \( O \) and two adjacent hexagon vertices, with radius \( OC = \text{1} \) equal to side length \( x = \text{1} \) (A1).

Perimeter: \[ P_i(\text{6}) = \text{6} \cdot \text{1} = \text{6} \] (A1).

[2 marks]

(d)

For an \( n \)-gon, divide into \( n \) isosceles triangles, each subtending \( \frac{\text{2}\pi}{n} \) at \( O \). In right-angled triangle \( OCX \), angle at \( O = \frac{\pi}{n} \), \( CX = \frac{x}{\text{2}} \), radius \( OC = \text{1} \):

\[ \sin \left( \frac{\pi}{n} \right) = \frac{\frac{x}{\text{2}}}{\text{1}} \implies \frac{x}{\text{2}} = \sin \left( \frac{\pi}{n} \right) \implies x = \text{2} \sin \left( \frac{\pi}{n} \right) \] (M1)(A1).

Perimeter: \[ P_i(n) = n \cdot x = \text{2} n \sin \left( \frac{\pi}{n} \right) \] (A1).

[3 marks]

(e)

Find: \[ \lim_{n \to \infty} P_i(n) = \lim_{n \to \infty} \text{2} n \sin \left( \frac{\pi}{n} \right) \].

Use Maclaurin series for sine: \[ \sin x = x – \frac{x^3}{\text{6}} + \frac{x^5}{\text{120}} – \cdots \] for small \( x \) (M1).

Let \( x = \frac{\pi}{n} \): \[ \sin \left( \frac{\pi}{n} \right) = \frac{\pi}{n} – \frac{\left( \frac{\pi}{n} \right)^3}{\text{6}} + \frac{\left( \frac{\pi}{n} \right)^5}{\text{120}} – \cdots \] (A1).

\[ \text{2} n \sin \left( \frac{\pi}{n} \right) = \text{2} n \left( \frac{\pi}{n} – \frac{\pi^3}{\text{6} n^3} + \frac{\pi^5}{\text{120} n^5} – \cdots \right) = \text{2} \pi – \frac{\pi^3}{\text{3} n^2} + \frac{\pi^5}{\text{60} n^4} – \cdots \] (A1).

As \( n \to \infty \), higher-order terms vanish: \[ \lim_{n \to \infty} \left( \text{2} \pi – \frac{\pi^3}{\text{3} n^2} + \cdots \right) = \text{2} \pi \] (A1).

Geometrically, as \( n \to \infty \), the inscribed \( n \)-gon’s vertices approach the circle’s circumference, yielding the circle’s perimeter \( \text{2} \pi \cdot \text{1} = \text{2} \pi \) (R1).

[5 marks]

(f)

For a circumscribed \( n \)-gon (see first diagram), divide into \( \text{2} n \) right-angled triangles, each with angle \( \frac{\pi}{n} \) at \( O \), opposite side \( \frac{x}{\text{2}} \), adjacent side (radius) \( \text{1} \):

\[ \tan \left( \frac{\pi}{n} \right) = \frac{\frac{x}{\text{2}}}{\text{1}} \implies \frac{x}{\text{2}} = \tan \left( \frac{\pi}{n} \right) \implies x = \text{2} \tan \left( \frac{\pi}{n} \right) \] (M1)(A1).

Perimeter: \[ P_c(n) = n \cdot x = n \cdot \text{2} \tan \left( \frac{\pi}{n} \right) = \text{2} n \tan \left( \frac{\pi}{n} \right) \] (A1)(A1).

[4 marks]

(g)

Rewrite: \[ P_c(n) = \text{2} n \tan \left( \frac{\pi}{n} \right) = \frac{\text{2} \tan \left( \frac{\pi}{n} \right)}{\frac{\text{1}}{n}} \].

Evaluate: \[ \lim_{n \to \infty} \frac{\text{2} \tan \left( \frac{\pi}{n} \right)}{\frac{\text{1}}{n}} \]. As \( n \to \infty \), \( \frac{\pi}{n} \to \text{0} \), yielding \( \frac{\text{0}}{\text{0}} \). Apply L’Hôpital’s rule (M1):

Numerator derivative: \[ \text{2} \sec^2 \left( \frac{\pi}{n} \right) \cdot \left( -\frac{\pi}{n^2} \right) \].

Denominator derivative: \[ -\frac{\text{1}}{n^2} \].

\[ \lim_{n \to \infty} \frac{\text{2} \sec^2 \left( \frac{\pi}{n} \right) \cdot \left( -\frac{\pi}{n^2} \right)}{-\frac{\text{1}}{n^2}} = \text{2} \pi \cdot \lim_{n \to \infty} \sec^2 \left( \frac{\pi}{n} \right) = \text{2} \pi \cdot \text{1} = \text{2} \pi \] (A1).

[2 marks]

(h)

The inscribed \( n \)-gon’s perimeter is less than the circle’s (\( \text{2} \pi \)), and the circumscribed \( n \)-gon’s is greater:

\[ P_i(n) < \text{2} \pi < P_c(n) \implies \text{2} n \sin \left( \frac{\pi}{n} \right) < \text{2} \pi < \text{2} n \tan \left( \frac{\pi}{n} \right) \] (M1).

Divide by \( \text{2} \): \[ n \sin \left( \frac{\pi}{n} \right) < \pi < n \tan \left( \frac{\pi}{n} \right) \] (A1).

[2 marks]

(i)

Find the least \( n \) such that the bounds from (h), \( n \sin(\pi/n) < \pi < n \tan(\pi/n) \), are within \( \text{0.005} \) of \( \pi \):

• Conditions: \( |\pi – n \sin(\pi/n)| < \text{0.005} \), \( |n \tan(\pi/n) – \pi| < \text{0.005} \).
• Approximate for large \( n \): Use Maclaurin series, \( \sin(\pi/n) \approx \pi/n – \pi^3/(\text{6} n^3) \).
  Then, \( n \sin(\pi/n) \approx \pi – \pi^3/(\text{6} n^2) \), so \( \pi – n \sin(\pi/n) \approx \pi^3/(\text{6} n^2) \).
  Require: \( \pi^3/(\text{6} n^2) < \text{0.005} \).
  With \( \pi^3 \approx \text{31.00627668} \), compute: \( \text{31.00627668}/(\text{6} \cdot \text{0.005}) \approx \text{1033.542556} \).
  Thus, \( n^2 > \text{1033.542556} \), so \( n > \sqrt{\text{1033.542556}} \approx \text{32.148} \) (M1).
• Upper bound: Use \( \tan(\pi/n) \approx \pi/n + \pi^3/(\text{3} n^3) \).
  Then, \( n \tan(\pi/n) \approx \pi + \pi^3/(\text{3} n^2) \), so \( n \tan(\pi/n) – \pi \approx \pi^3/(\text{3} n^2) \).
  Require: \( \pi^3/(\text{3} n^2) < \text{0.005} \).
  Compute: \( \text{31.00627668}/(\text{3} \cdot \text{0.005}) \approx \text{2067.085112} \).
  Thus, \( n^2 > \text{2067.085112} \), so \( n > \sqrt{\text{2067.085112}} \approx \text{45.466} \).
• Least \( n =46\)