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Question Number 150944 by ajfour last updated on 16/Aug/21

Commented by mr W last updated on 17/Aug/21

Commented by ajfour last updated on 17/Aug/21
okay find minimum side length, sir.
Commented by mr W last updated on 17/Aug/21

$t h e g r e e n o n e i s t h e s m a l l e s t$ $e q u i l a t e r a l . i c a n d e t e r m i n e i t o n l y$ $n u m e r i c a l l y .$
	Answered by ajfour last updated on 17/Aug/21
	$sin α=(a/(b−a)) A[(√((b−a)^2 −a^2 )), a]≡[c,a] let T(p,q) let s be circumradius of such an equilateral triangle. P[p−scos θ, q+ssin θ] Q[p+scos (60°−θ), q+ssin (60°−θ)] R[p+scos (60°+θ), q−ssin (60°+θ)] now (p−scos θ−c)^2 +(q+ssin θ−a)^2 = a^2 ...(i) [p+scos (60°−θ)]^2 +[q+ssin (60°−θ)]^2 = b^2 ...(ii) q=ssin (60°+θ) ⇒ (p−scos θ−c)^2 +[((s(√3))/2)cos θ+((3s)/2)sin θ−a]^2 =a^2 ........(I) [p+(s/2)cos θ+((s(√3))/2)sin θ]^2 +[s(√3)cos θ]^2 =b^2 ....(II) (II)−(I) {((3s)/2)cos θ+((s(√3))/2)sin θ+c} ×{2p−(s/2)cos θ+((s(√3))/2)sin θ−c} + {((s(√3))/2)cos θ−((3s)/2)sin θ+a} ×{((3(√3)s)/2)cos θ+((3s)/2)sin θ−a} = b^2 −a^2 from here p=f(s,θ) now using (I) or (II) h(s,θ)=0 from here we get s_(min) . Minimum side length of such an equilateral △ would then be s_(min) (√3).$
	$\sin α = \frac{a}{b - a}$ $A [\sqrt{{(b - a)}^{2} - a^{2}}, a] \equiv [c, a]$ $l e t T (p, q)$ $l e t s b e c i r c u m r a d i u s o f s u c h a n$ $e q u i l a t e r a l t r i a n g l e .$ $P [p - s \cos θ, q + s \sin θ]$ $Q [p + s \cos (60 ° - θ), q + s \sin (60 ° - θ)]$ $R [p + s \cos (60 ° + θ), q - s \sin (60 ° + θ)]$ $n o w$ ${(p - s \cos θ - c)}^{2} + {(q + s \sin θ - a)}^{2}$ $= a^{2} . . . (i)$ ${[p + s \cos (60 ° - θ)]}^{2} + {[q + s \sin (60 ° - θ)]}^{2}$ $= b^{2} . . . (i i)$ $q = s \sin (60 ° + θ)$ $\Rightarrow {(p - s \cos θ - c)}^{2}$ $+ {[\frac{s \sqrt{3}}{2} \cos θ + \frac{3 s}{2} \sin θ - a]}^{2} = a^{2}$ $. . . . . . . . (I)$ ${[p + \frac{s}{2} \cos θ + \frac{s \sqrt{3}}{2} \sin θ]}^{2}$ $+ {[s \sqrt{3} \cos θ]}^{2} = b^{2} . . . . (I I)$ $(I I) - (I)$ ${\frac{3 s}{2} \cos θ + \frac{s \sqrt{3}}{2} \sin θ + c}$ $\times {2 p - \frac{s}{2} \cos θ + \frac{s \sqrt{3}}{2} \sin θ - c}$ $+$ ${\frac{s \sqrt{3}}{2} \cos θ - \frac{3 s}{2} \sin θ + a}$ $\times {\frac{3 \sqrt{3} s}{2} \cos θ + \frac{3 s}{2} \sin θ - a}$ $= b^{2} - a^{2}$ $f r o m h e r e$ $p = f (s, θ)$ $n o w u s i n g (I) o r (I I)$ $h (s, θ) = 0$ $f r o m h e r e w e g e t s_{m i n} .$ $M i n i m u m s i d e l e n g t h o f$ $s u c h a n e q u i l a t e r a l △ w o u l d$ $t h e n b e s_{m i n} \sqrt{3} .$
	Answered by ajfour last updated on 17/Aug/21
	$center of semicircle origin. center of circle A[(√((b−a)^2 −a^2 )), a]≡(c,a) let bottom corner be B(p,0) let side of △ be s. x_C ^2 +y_C ^2 =b^2 (x_C −p)^2 +y_C ^2 =s^2 subtracting 2px_C −p^2 =b^2 −s^2 x_C =((b^2 −s^2 +p^2 )/(2p)) ((x_c +p)/2)−((s(√3))/2)sin φ=x_D (y_C /2)+((s(√3))/2)cos φ=y_D (x_D −c)^2 +(y_D −a)^2 =a^2 cos φ=((x_C −p)/s) ; sin φ=(y_C /s) x_C +p−s(√3)(((√(b^2 −x_C ^2 ))/s))=2x_D y_C +s(√3)(((x_C −p)/s))=2y_D Now {x_C +p−s(√3)((y_C /s))−2c}^2 +{y_C +s(√3)(((x_C −p)/s))−a}^2 =a^2 but x_C =((b^2 −s^2 +p^2 )/(2p)) ; y_C =(√(b^2 −x_C ^2 )) =(√(b^2 −(((b^2 −s^2 +p^2 )/(2p)))^2 )) ⇒ {((b^2 −s^2 +p^2 )/(2p))+p−(√3)(√(b^2 −(((b^2 −s^2 +p^2 )/(2p)))^2 ))−2c}^2 +{(√(b^2 −(((b^2 −s^2 +p^2 )/(2p)))^2 ))+(√3)(((b^2 −s^2 +p^2 )/(2p)))−(√3)p−a}^2 =a^2 eq. is implicit in s and p. we can graph it and can find the s_(min) .$
	$c e n t e r o f s e m i c i r c l e o r i g i n .$ $c e n t e r o f c i r c l e$ $A [\sqrt{{(b - a)}^{2} - a^{2}}, a] \equiv (c, a)$ $l e t b o t t o m c o r n e r b e B (p, 0)$ $l e t s i d e o f △ b e s .$ $x_{C}^{2} + y_{C}^{2} = b^{2}$ ${(x_{C} - p)}^{2} + y_{C}^{2} = s^{2}$ $s u b t r a c t i n g$ $2 {p x}_{C} - p^{2} = b^{2} - s^{2}$ $x_{C} = \frac{b^{2} - s^{2} + p^{2}}{2 p}$ $\frac{x_{c} + p}{2} - \frac{s \sqrt{3}}{2} \sin ϕ = x_{D}$ $\frac{y_{C}}{2} + \frac{s \sqrt{3}}{2} \cos ϕ = y_{D}$ ${(x_{D} - c)}^{2} + {(y_{D} - a)}^{2} = a^{2}$ $\cos ϕ = \frac{x_{C} - p}{s}; \sin ϕ = \frac{y_{C}}{s}$ $x_{C} + p - s \sqrt{3} (\frac{\sqrt{b^{2} - x_{C}^{2}}}{s}) = 2 x_{D}$ $y_{C} + s \sqrt{3} (\frac{x_{C} - p}{s}) = 2 y_{D}$ $N o w$ ${x_{C} + p - s \sqrt{3} (\frac{y_{C}}{s}) - 2 c}^{2}$ $+ {y_{C} + s \sqrt{3} (\frac{x_{C} - p}{s}) - a}^{2} = a^{2}$ $b u t x_{C} = \frac{b^{2} - s^{2} + p^{2}}{2 p};$ $y_{C} = \sqrt{b^{2} - x_{C}^{2}}$ $= \sqrt{b^{2} - {(\frac{b^{2} - s^{2} + p^{2}}{2 p})}^{2}}$ $\Rightarrow$ ${\frac{b^{2} - s^{2} + p^{2}}{2 p} + p - \sqrt{3} \sqrt{b^{2} - {(\frac{b^{2} - s^{2} + p^{2}}{2 p})}^{2}} - 2 c}^{2}$ $+ {\sqrt{b^{2} - {(\frac{b^{2} - s^{2} + p^{2}}{2 p})}^{2}} + \sqrt{3} (\frac{b^{2} - s^{2} + p^{2}}{2 p}) - \sqrt{3} p - a}^{2} = a^{2}$ $e q . i s i m p l i c i t i n s a n d p .$ $w e c a n g r a p h i t a n d c a n f i n d$ $t h e s_{m i n} .$
	Commented by mr W last updated on 17/Aug/21

	$i a l s o c a n o n l y s o l v e i n t h i s w a y t o f i n d$ $t h e s_{m i n} .$
	Answered by ajfour last updated on 17/Aug/21
	$O(center of semicircle) z_A =(√((b−a)^2 −a^2 ))+ia=c+ia z_B =p z_C =p+s(cos θ+isin θ) z_D =p+s[−cos (((2π)/3)−θ)+isin (((2π)/3)−θ)] ∣z_D −z_A ∣=a ∣z_C ∣=b (p+scos θ)^2 +s^2 sin^2 θ=b^2 ⇒ p^2 +2spcos θ+s^2 =b^2 ....(i) {p+scos (((2π)/3)−θ)−c}^2 +{ssin (((2π)/3)−θ)−a}^2 =a^2 ..(ii) say ((2π)/3)−θ=φ , then ⇒ p^2 +s^2 +c^2 +2spcos φ =2cp+2cscos φ+2assin φ ⇒ b^2 +2sp(cos φ−cos θ)+c^2 = 2cp+2s(ccos φ+asin φ) p=((b^2 +c^2 −2s(ccos φ+asin φ))/(2c−2s(cos φ−cos θ))) using this in p^2 +2spcos θ+s^2 =b^2 ....(i) ⇒ {((b^2 +c^2 −2s(ccos φ+asin φ))/(2c−2s(cos φ−cos θ)))+scos θ}^2 +s^2 sin^2 θ=b^2 ⇒ {((b^2 +c^2 −2s[c+atan (((2π)/3)−θ)])/(2c−2s[1−((cos θ)/(cos (((2π)/3)−θ)))]))+scos θ}^2 +s^2 sin^2 θ=b^2 for a=2, b=5 c=(√5) {((30−2s[(√5)+2tan (((2π)/3)−θ)])/(2(√5)−2s[1−((cos θ)/(cos (((2π)/3)−θ)))]))+scos θ}^2 +s^2 sin^2 θ=25 {((30−2s(√5)−2s((((√3)cos θ−sin θ)/(cos θ−(√3)sin θ))))/(2(√5)−2s+4s(((cos θ)/( (√3)sin θ+cos θ)))))+scos θ}^2 +s^2 sin^2 θ=25 graph of this f(s,θ)=25 should give s_(min) . sir can u help?$
	$O (c e n t e r o f s e m i c i r c l e)$ $z_{A} = \sqrt{{(b - a)}^{2} - a^{2}} + i a = c + i a$ $z_{B} = p$ $z_{C} = p + s (\cos θ + i \sin θ)$ $z_{D} = p + s [- \cos (\frac{2 π}{3} - θ) + i \sin (\frac{2 π}{3} - θ)]$ $∣ z_{D} - z_{A} ∣= a$ $∣ z_{C} ∣= b$ ${(p + s \cos θ)}^{2} + s^{2} \sin^{2} θ = b^{2}$ $\Rightarrow$ $p^{2} + 2 s p \cos θ + s^{2} = b^{2} . . . . (i)$ ${p + s \cos (\frac{2 π}{3} - θ) - c}^{2}$ $+ {s \sin (\frac{2 π}{3} - θ) - a}^{2} = a^{2} . . (i i)$ $s a y \frac{2 π}{3} - θ = ϕ, t h e n$ $\Rightarrow p^{2} + s^{2} + c^{2} + 2 s p \cos ϕ$ $= 2 c p + 2 c s \cos ϕ + 2 a s \sin ϕ$ $\Rightarrow b^{2} + 2 s p (\cos ϕ - \cos θ) + c^{2}$ $= 2 c p + 2 s (c \cos ϕ + a \sin ϕ)$ $p = \frac{b^{2} + c^{2} - 2 s (c \cos ϕ + a \sin ϕ)}{2 c - 2 s (\cos ϕ - \cos θ)}$ $u s i n g t h i s i n$ $p^{2} + 2 s p \cos θ + s^{2} = b^{2} . . . . (i)$ $\Rightarrow$ ${\frac{b^{2} + c^{2} - 2 s (c \cos ϕ + a \sin ϕ)}{2 c - 2 s (\cos ϕ - \cos θ)} + s \cos θ}^{2}$ $+ s^{2} \sin^{2} θ = b^{2}$ $\Rightarrow$ ${\frac{b^{2} + c^{2} - 2 s [c + a \tan (\frac{2 π}{3} - θ)]}{2 c - 2 s [1 - \frac{\cos θ}{\cos (\frac{2 π}{3} - θ)}]} + s \cos θ}^{2}$ $+ s^{2} \sin^{2} θ = b^{2}$ $f o r a = 2, b = 5$ $c = \sqrt{5}$ ${\frac{30 - 2 s [\sqrt{5} + 2 \tan (\frac{2 π}{3} - θ)]}{2 \sqrt{5} - 2 s [1 - \frac{\cos θ}{\cos (\frac{2 π}{3} - θ)}]} + s \cos θ}^{2}$ $+ s^{2} \sin^{2} θ = 25$ ${\frac{30 - 2 s \sqrt{5} - 2 s (\frac{\sqrt{3} \cos θ - \sin θ}{\cos θ - \sqrt{3} \sin θ})}{2 \sqrt{5} - 2 s + 4 s (\frac{\cos θ}{\sqrt{3} \sin θ + \cos θ})} + s \cos θ}^{2}$ $+ s^{2} \sin^{2} θ = 25$ $g r a p h o f t h i s f (s, θ) = 25$ $s h o u l d g i v e s_{m i n} .$ $s i r c a n u h e l p ?$
	Answered by mr W last updated on 17/Aug/21

	Commented by mr W last updated on 18/Aug/21

	$s = s i d e l e n g t h o f t r i a n g l e$ $O C = c = \sqrt{{(b - a)}^{2} - a^{2}} = \sqrt{b (b - 2 a)}$ $s a y A (u, 0)$ $u = c - s \cos θ + \sqrt{b^{2} - s^{2} \sin^{2} θ}$ $e q n . o f s m a l l c i r c l e :$ $x^{2} + {(y - a)}^{2} = a^{2}$ $x_{D} = u + s \cos (\frac{π}{3} + θ)$ $y_{D} = s \sin (\frac{π}{3} + θ)$ ${(u + s \cos (\frac{π}{3} + θ))}^{2} + {(s \sin (\frac{π}{3} + θ) - a)}^{2} = a^{2}$ $u^{2} + 2 s u \cos (\frac{π}{3} + θ) + s^{2} - 2 a s \sin (\frac{π}{3} + θ) = 0$ $u = - s \cos (\frac{π}{3} + θ) + \sqrt{s (2 a - s \sin (\frac{π}{3} + θ)) \sin (\frac{π}{3} + θ)}$ $c - s \cos θ + \sqrt{b^{2} - s^{2} \sin^{2} θ} = - s \cos (\frac{π}{3} + θ) + \sqrt{s (2 a - s \sin (\frac{π}{3} + θ)) \sin (\frac{π}{3} + θ)}$ $\Rightarrow s [\cos θ - \cos (\frac{π}{3} + θ)] - \sqrt{b^{2} - s^{2} \sin^{2} θ} + \sqrt{s [2 a - s \sin (\frac{π}{3} + θ)] \sin (\frac{π}{3} + θ)} = c$ $λ [\cos θ - \cos (\frac{π}{3} + θ)] - \sqrt{1 - λ^{2} \sin^{2} θ} + \sqrt{λ [μ - λ \sin (\frac{π}{3} + θ)] \sin (\frac{π}{3} + θ)} = \sqrt{1 - μ}$ $w i t h μ = \frac{2 a}{b} ⩽ 1, λ = \frac{s}{b}$ $f r o m t h i s r e l a t i o n s h i p b e t w e e n s a n d$ $θ w e c a n f i n d t h e s_{m i n} g r a p h i c a l l y .$ $e x a m p l e : a = 2, b = 5$ $s_{m i n} = 3.8157 a t θ = 1.4075 (80.64 °)$
	Commented by mr W last updated on 17/Aug/21

	Commented by mr W last updated on 17/Aug/21

	Commented by mr W last updated on 17/Aug/21

	Commented by mr W last updated on 18/Aug/21

	Commented by mr W last updated on 18/Aug/21

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