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Question Number 166779 by ajfour last updated on 27/Feb/22

	Answered by mr W last updated on 27/Feb/22

	Commented by mr W last updated on 28/Feb/22

	$A^{'} (h, a)$ $B^{'} (b, k)$ ${(h - b)}^{2} + {(k - a)}^{2} = {(a + b)}^{2}$ $k^{2} - 2 a k + h^{2} - 2 b (h + a) = 0$ $\Rightarrow k = a + \sqrt{(a + 2 b - h) (h + a)}$ $x_{A} = \frac{s}{2} (\sqrt{3} \cos θ + \sin θ)$ $y_{A} = \frac{s}{2} (\sqrt{3} \sin θ - \cos θ)$ $x_{B} = \frac{s}{2} (\sqrt{3} \cos θ - \sin θ)$ $y_{B} = \frac{s}{2} (\sqrt{3} \sin θ + \cos θ)$ ${(\frac{s}{2} (\sqrt{3} \cos θ + \sin θ) - h)}^{2} + {(\frac{s}{2} (\sqrt{3} \sin θ - \cos θ) - a)}^{2} = a^{2}$ $s^{2} - [(\sqrt{3} h - a) \cos θ + (\sqrt{3} a + h) \sin θ] s + h^{2} = 0 . . . (i)$ ${(\frac{s}{2} (\sqrt{3} \cos θ - \sin θ) - b)}^{2} + {(\frac{s}{2} (\sqrt{3} \sin θ + \cos θ) - k)}^{2} = b^{2}$ $s^{2} - [(\sqrt{3} b + k) \cos θ + (\sqrt{3} k - b) \sin θ] s + k^{2} = 0 . . . (i i)$ $(\sqrt{3} h - a) \cos θ + (\sqrt{3} a + h) \sin θ = \frac{s^{2} + h^{2}}{s}$ $\cos (θ - \tan^{- 1} \frac{\sqrt{3} a + h}{\sqrt{3} h - a}) = \frac{s^{2} + h^{2}}{2 s \sqrt{h^{2} + a^{2}}}$ $\Rightarrow θ = \tan^{- 1} \frac{\sqrt{3} a + h}{\sqrt{3} h - a} + \cos^{- 1} \frac{s^{2} + h^{2}}{2 s \sqrt{h^{2} + a^{2}}}$ $s^{2} - 2 s \sqrt{k^{2} + b^{2}} \cos (θ - \tan^{- 1} \frac{\sqrt{3} k - b}{\sqrt{3} b + k}) + k^{2} = 0$ $s^{2} - 2 s \sqrt{k^{2} + b^{2}} \cos (\cos^{- 1} \frac{s^{2} - h^{2}}{2 s \sqrt{h^{2} + a^{2}}} + \tan^{- 1} \frac{\sqrt{3} a + h}{\sqrt{3} h - a} - \tan^{- 1} \frac{\sqrt{3} k - b}{\sqrt{3} b + k}) + k^{2} = 0$ $. . .$
	Commented by ajfour last updated on 28/Feb/22
	$(h−scos α)^2 +(a−ssin α)^2 =a^2 (b−ssin β)^2 +(k−scos β)^2 =b^2 ⇒ h=scos α+(√(a^2 −(a−ssin α)^2 )) k=scos β+(√(b^2 −(b−ssin β)^2 )) α+β=(π/6) ; α−β=2θ ⇒ α=(π/(12))+θ , β=(π/(12))−θ (h−b)^2 +(k−a)^2 =(a+b)^2 ⇒ {scos ((π/(12))+θ) −b+(√(a^2 −[a−ssin ((π/(12))+θ)]^2 ))}^2 +{scos ((π/(12))−θ)−a+(√(b^2 −[b−ssin ((π/(12))−θ)]^2 ))}^2 =(a+b)^2$
	${(h - s \cos α)}^{2} + {(a - s \sin α)}^{2} = a^{2}$ ${(b - s \sin β)}^{2} + {(k - s \cos β)}^{2} = b^{2}$ $\Rightarrow$ $h = s \cos α + \sqrt{a^{2} - {(a - s \sin α)}^{2}}$ $k = s \cos β + \sqrt{b^{2} - {(b - s \sin β)}^{2}}$ $α + β = \frac{π}{6}; α - β = 2 θ$ $\Rightarrow α = \frac{π}{12} + θ, β = \frac{π}{12} - θ$ ${(h - b)}^{2} + {(k - a)}^{2} = {(a + b)}^{2}$ $\Rightarrow$ ${s \cos (\frac{π}{12} + θ) - b + \sqrt{a^{2} - {[a - s \sin (\frac{π}{12} + θ)]}^{2}}}^{2}$ $+ {s \cos (\frac{π}{12} - θ) - a + \sqrt{b^{2} - {[b - s \sin (\frac{π}{12} - θ)]}^{2}}}^{2}$ $= {(a + b)}^{2}$
	Commented by mr W last updated on 28/Feb/22

	$v e r y n i c e w a y!$
	Commented by Tawa11 last updated on 28/Feb/22

	$Great sir$
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