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Question Number 111187 by ajfour last updated on 02/Sep/20

Commented by ajfour last updated on 03/Sep/20

$F i n d s i d e s A B a n d B C o f r e c t a n g l e .$
	Answered by ajfour last updated on 03/Sep/20
	$let center of circle be origin. P(−1,0) ; B(p,−1) ; D(−1,q) ⇒ BQ=p let ∠ABQ=2θ tan θ=(1/p) Q(p−pcos 2θ , psin 2θ−1) Q{p(1−((1−(1/p^2 ))/(1+(1/p^2 )))), (2/(1+(1/p^2 )))−1} ⇒ Q(((2p)/(p^2 +1)), ((p^2 −1)/(p^2 +1))) PQ=BQ ⇒ (((2p)/(p^2 +1))+1)^2 +(((p^2 −1)/(p^2 +1)))^2 =p^2 ⇒ (p+1)^4 +(p−1)^2 (p+1)^2 =p^2 (p^2 +1)^2 ⇒ 2(p+1)^2 =p^2 (p^2 +1) ⇒ p^4 −p^2 −4p−2=0 ⇒ p ≈ 1.90897 AB = p+1 ≈ 2.90897 ....$
	$l e t c e n t e r o f c i r c l e b e o r i g i n .$ $P (- 1, 0); B (p, - 1); D (- 1, q)$ $\Rightarrow B Q = p$ $l e t ∠ A B Q = 2 θ$ $\tan θ = \frac{1}{p}$ $Q (p - p \cos 2 θ, p \sin 2 θ - 1)$ $Q {p (1 - \frac{1 - \frac{1}{p^{2}}}{1 + \frac{1}{p^{2}}}), \frac{2}{1 + \frac{1}{p^{2}}} - 1}$ $\Rightarrow Q (\frac{2 p}{p^{2} + 1}, \frac{p^{2} - 1}{p^{2} + 1})$ $P Q = B Q$ $\Rightarrow {(\frac{2 p}{p^{2} + 1} + 1)}^{2} + {(\frac{p^{2} - 1}{p^{2} + 1})}^{2} = p^{2}$ $\Rightarrow {(p + 1)}^{4} + {(p - 1)}^{2} {(p + 1)}^{2} = p^{2} {(p^{2} + 1)}^{2}$ $\Rightarrow 2 {(p + 1)}^{2} = p^{2} (p^{2} + 1)$ $\Rightarrow p^{4} - p^{2} - 4 p - 2 = 0$ $\Rightarrow p \approx 1.90897$ $A B = p + 1 \approx 2.90897$ $. . . .$
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