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Question Number 214297 by ajfour last updated on 04/Dec/24

Commented by ajfour last updated on 04/Dec/24

$t h e t a n g e n t s a r e a t r i g h t a n g l e s t o$ $o n e a n o t h e r a n d c o r n e r m a y n o t b e$ $a t s e m i c i r c l e c e n t r e .$
Commented by mr W last updated on 05/Dec/24
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Commented by ajfour last updated on 05/Dec/24
https://youtu.be/ys6xN29xNUM?si=9rk6nXD0GcmHTp9Z
	Answered by mr W last updated on 04/Dec/24

	Commented by mr W last updated on 06/Dec/24
	$a=1 R=3 b=r (b/(tan (β/2)))=x+(√((R−b)^2 −b^2 ))=x+(√(R^2 −2Rb)) ⇒tan (β/2)=(b/(x+(√(R^2 −2Rb)))) similarly ⇒tan (α/2)=(a/(−x+(√(R^2 −2Ra)))) (β/2)=(π/4)−(α/2) tan (β/2)=((1−tan (a/2))/(1+tan (a/2))) (b/(x+(√(R^2 −2Rb))))=((1−(a/(−x+(√(R^2 −2Ra)))))/(1+(a/(−x+(√(R^2 −2Ra)))))) (b/(x+(√(R^2 −2Rb))))=((−x+(√(R^2 −2Ra))−a)/(−x+(√(R^2 −2Ra))+a)) let α=(a/R), β=(b/R), ξ=(x/R) (β/(ξ+(√(1−2β))))=((−ξ+(√(1−2α))−α)/(−ξ+(√(1−2α))+α)) let λ=(√(1−2β)), p=(√(1−2α))+α, q=(√(1−2α))−α ((1−λ^2 )/(2(ξ+λ)))=((q−ξ)/(p−ξ)) (p−ξ)λ^2 +2(q−ξ)λ+(2q+1)ξ−[2ξ^2 −(2q+1)ξ+p]=0 λ=((ξ−q+(√((ξ−q)^2 +(p−ξ)[2ξ^2 −(2q+1)ξ+p])))/(p−ξ)) b={1−{((ξ−q+(√((ξ−q)^2 +(p−ξ)[2ξ^2 −(2q+1)ξ+p])))/(p−ξ))}^2 }(R/2) b_(max) ≈0.66471561 at ξ=(x/R)≈−0.14031961 =================$
	$a = 1$ $R = 3$ $b = r$ $\frac{b}{\tan \frac{β}{2}} = x + \sqrt{{(R - b)}^{2} - b^{2}} = x + \sqrt{R^{2} - 2 R b}$ $\Rightarrow \tan \frac{β}{2} = \frac{b}{x + \sqrt{R^{2} - 2 R b}}$ $s i m i l a r l y$ $\Rightarrow \tan \frac{α}{2} = \frac{a}{- x + \sqrt{R^{2} - 2 R a}}$ $\frac{β}{2} = \frac{π}{4} - \frac{α}{2}$ $\tan \frac{β}{2} = \frac{1 - \tan \frac{a}{2}}{1 + \tan \frac{a}{2}}$ $\frac{b}{x + \sqrt{R^{2} - 2 R b}} = \frac{1 - \frac{a}{- x + \sqrt{R^{2} - 2 R a}}}{1 + \frac{a}{- x + \sqrt{R^{2} - 2 R a}}}$ $\frac{b}{x + \sqrt{R^{2} - 2 R b}} = \frac{- x + \sqrt{R^{2} - 2 R a} - a}{- x + \sqrt{R^{2} - 2 R a} + a}$ $l e t α = \frac{a}{R}, β = \frac{b}{R}, ξ = \frac{x}{R}$ $\frac{β}{ξ + \sqrt{1 - 2 β}} = \frac{- ξ + \sqrt{1 - 2 α} - α}{- ξ + \sqrt{1 - 2 α} + α}$ $l e t λ = \sqrt{1 - 2 β}, p = \sqrt{1 - 2 α} + α, q = \sqrt{1 - 2 α} - α$ $\frac{1 - λ^{2}}{2 (ξ + λ)} = \frac{q - ξ}{p - ξ}$ $(p - ξ) λ^{2} + 2 (q - ξ) λ + (2 q + 1) ξ - [2 ξ^{2} - (2 q + 1) ξ + p] = 0$ $λ = \frac{ξ - q + \sqrt{{(ξ - q)}^{2} + (p - ξ) [2 ξ^{2} - (2 q + 1) ξ + p]}}{p - ξ}$ $b = {1 - {\frac{ξ - q + \sqrt{{(ξ - q)}^{2} + (p - ξ) [2 ξ^{2} - (2 q + 1) ξ + p]}}{p - ξ}}^{2}} \frac{R}{2}$ $b_{m a x} \approx 0.66471561 a t ξ = \frac{x}{R} \approx - 0.14031961$ $=================$
	Commented by mr W last updated on 06/Dec/24

	Commented by ajfour last updated on 06/Dec/24
	$(a/(R−a))=sin θ (R−a)cos θ=acot (α/2)+x R+x=bcot (β/2) (α/2)+(β/2)=(π/4) ((A+B)/(1−AB))=1 [ for ((tan (α/2)+tan (β/2))/(1−tan (α/2)tan (β/2)))=1] A+B=1−AB (a/((R−a)cos θ−x))+(b/(R+x)) +((ab)/((R+x){(R−a)cos θ−x}))=1 or a(R+x)+b(R−a)cos θ−bx+ab +x^2 +{R−(R−a)cos θ}x−R(R−a)cos θ=0 or x^2 +{a−b+R−(R−a)cos θ}x+a(b+R) +(b−R)(R−a)cos θ=0 diff. 2x+{a−b+R−(R−a)cos θ}=0 or (R−a)cos θ−x=R+x+a−b x=((b−a−R+(R−a)cos θ)/2) b{(1/(R+x))+(a/((R+x){(R−a)cos θ−x))} =1−(a/({(R−a)cos θ−x}))$
	$\frac{a}{R - a} = \sin θ$ $(R - a) \cos θ = a \cot \frac{α}{2} + x$ $R + x = b \cot \frac{β}{2}$ $\frac{α}{2} + \frac{β}{2} = \frac{π}{4}$ $\frac{A + B}{1 - A B} = 1 [f o r \frac{\tan \frac{α}{2} + \tan \frac{β}{2}}{1 - \tan \frac{α}{2} \tan \frac{β}{2}} = 1]$ $A + B = 1 - A B$ $\frac{a}{(R - a) \cos θ - x} + \frac{b}{R + x}$ $+ \frac{a b}{(R + x) {(R - a) \cos θ - x}} = 1$ $o r$ $a (R + x) + b (R - a) \cos θ - b x + a b$ $+ x^{2} + {R - (R - a) \cos θ} x - R (R - a) \cos θ = 0$ $o r$ $x^{2} + {a - b + R - (R - a) \cos θ} x + a (b + R)$ $+ (b - R) (R - a) \cos θ = 0$ $d i f f .$ $2 x + {a - b + R - (R - a) \cos θ} = 0$ $o r (R - a) \cos θ - x = R + x + a - b$ $x = \frac{b - a - R + (R - a) \cos θ}{2}$ $b {\frac{1}{R + x} + \frac{a}{(R + x) {(R - a) \cos θ - x}}$ $= 1 - \frac{a}{{(R - a) \cos θ - x}}$
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