The length of two sides of an equilateral triangle are doubled, creating an isosceles triangle, as shown in the diagram. The two longer sides are doubled again creating a third isosceles triangle (all three triangles have the same base BC). This process is continued indefinitely. If the measure of the vertex angle of each triangle is represented by A1, A2, A3, ......... , and \sum_{n = 1}^{\infty} \left(1 - cos ⁡ A_{n}\right) = \frac{p}{q} (p, q ∈ N), then find the minimum value of (p + q).

Engineering

Mathematics

Basic Rules of Properties of Triangle

Question

The length of two sides of an equilateral triangle are doubled, creating an isosceles triangle, as shown in the diagram. The two longer sides are doubled again creating a third isosceles triangle (all three triangles have the same base BC). This process is continued indefinitely. If the measure of the vertex angle of each triangle is represented by A₁, A₂, A₃, ......... , and $\sum_{n = 1}^{\infty} (1 - \cos A_{n}) = \frac{p}{q}$ (p, q ∈ N), then find the minimum value of (p + q).

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$\sin \frac{A_{1}}{2} = \frac{1}{2}$ (A₁ = 60°)

$\sin \frac{A_{2}}{2} = \frac{a}{2 \times 2 a} = \frac{1}{4}$

$\sin \frac{A_{3}}{2} = \frac{a}{2 \times 4 a} = \frac{1}{8}$

$⋮$ $⋮$ $⋮$

$\sin \frac{A_{n}}{2} = \frac{a}{2 \times 2^{n - 1} a} = \frac{1}{2^{n}}$

$\sum_{n = 1}^{\infty} (1 - {cosA}_{n}) = 2 \sum_{n = 1}^{\infty} (\sin^{2} \frac{A_{n}}{2}) = {Lim}_{n \to \infty} 2 (\frac{1}{2^{2}} + \frac{1}{4^{2}} + \frac{1}{8^{2}} + \dots \dots \dots + + \frac{1}{2^{2 n}}) = \frac{2 \times \frac{1}{2^{2}}}{1 - \frac{1}{4}} = \frac{2}{3}$

Hence, (p + q) = 2 + 3 = 5.

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