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Question Number 195470 by Shlock last updated on 03/Aug/23

	Answered by mr W last updated on 03/Aug/23

	Commented by mr W last updated on 03/Aug/23

	$s = s i d e l e n g t h o f h e x a g o n$ $c = s i d e l e n g t h o f s q u a r e$ $c^{2} = s^{2} + b^{2} + s b$ $\frac{\sin α}{s} = \frac{\sin (120 ° + α)}{b}$ $\frac{\sin α}{s} = \frac{1}{2 b} (\sqrt{3} \cos α - \sin α)$ $\Rightarrow \frac{\sqrt{3}}{\tan α} = \frac{2 b}{s} + 1 . . . (i)$ $c^{2} = {(2 s - b)}^{2} + {(2 s - a)}^{2} - (2 s - b) (2 s - a)$ $\frac{\sin β}{2 s - a} = \frac{\sin (60 ° + β)}{2 s - b}$ $\frac{\sin β}{2 s - a} = \frac{1}{2 (2 s - b)} (\sqrt{3} \cos β + \sin β)$ $\Rightarrow \frac{\sqrt{3}}{\tan β} = \frac{2 (2 s - b)}{2 s - a} - 1 . . . (i i)$ $(i) \times (i i) :$ $\frac{3}{\tan α \tan β} = (\frac{2 b}{s} + 1) (\frac{2 (2 s - b)}{2 s - a} - 1)$ $α + β = 90 ° \Rightarrow \tan α \tan β = 1$ $\Rightarrow 3 = (\frac{2 b}{s} + 1) (\frac{2 (2 s - b)}{2 s - a} - 1)$ $w i t h p = \frac{a}{s}, q = \frac{b}{s}$ $\Rightarrow 3 = (2 q + 1) (\frac{2 (2 - q)}{2 - p} - 1)$ $\Rightarrow 4 q^{2} - 2 p q - 4 p - 2 q + 4 = 0 . . . (I)$ ${(2 s - b)}^{2} + {(2 s - a)}^{2} - (2 s - b) (2 s - a) = s^{2} + b^{2} + s b$ ${(2 - p)}^{2} + {(2 - q)}^{2} - (2 - p) (2 - q) = 1 + q^{2} + q$ $\Rightarrow p^{2} - p q - 2 p - 3 q + 3 = 0 . . . (I I)$ $\Rightarrow p \approx 0.80384, q \approx 0.53591$ $\Rightarrow \frac{a}{b} = \frac{p}{q} \approx \frac{3}{2} ✓$
	Commented by Shlock last updated on 03/Aug/23
	This is a great solution, sir! Thanks!
	Answered by mr W last updated on 03/Aug/23

	Commented by mr W last updated on 03/Aug/23

	${(2 s - a)}^{2} + {(2 s - b)}^{2} - (2 s - a) (2 s - b) = s^{2} + b^{2} + s b$ $\Rightarrow a^{2} - a b - 2 s a - 3 s b + 3 s^{2} = 0$ $l e t p = \frac{a}{s}, q = \frac{b}{s}$ $\Rightarrow p^{2} - p q - 2 p - 3 q + 3 = 0 . . . (i)$ ${(2 s - a)}^{2} + s^{2} = {(2 s - b)}^{2} + b^{2}$ $\Rightarrow 2 b^{2} - a^{2} + 4 s a - 4 s b - s^{2} = 0$ $\Rightarrow 2 q^{2} - p^{2} + 4 p - 4 q - 1 = 0$ $\Rightarrow 2 q^{2} - p q + 2 p - 7 q + 2 = 0 . . . (i i)$ $l e t λ = \frac{p}{q} = \frac{a}{b}$ $λ (λ - 1) q^{2} - (2 λ + 3) q + 3 = 0$ $(2 - λ) q^{2} + (2 λ - 7) q + 2 = 0$ $\frac{3}{λ (λ - 1)} = \frac{2}{2 - λ}$ $\Rightarrow (2 λ - 3) (λ + 2) = 0$ $\Rightarrow λ = \frac{3}{2} ✓$
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