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Question Number 201790 by ajfour last updated on 12/Dec/23

Commented by ajfour last updated on 12/Dec/23

$C a n w e f i n d s i d e s o f ◻ P Q R S i n$ $t e r m s o f a, b, c ?$
	Answered by a.lgnaoui last updated on 14/Dec/23

	${(R - a)}^{2} + b^{2} = R^{2} \Rightarrow R = \frac{a^{2} + b^{2}}{2 a}$ $BD = 2 R = DT + TB \Rightarrow TB = 2 R - a$ $alors PQ = SR = (\frac{a^{2} + b^{2}}{a} - a) = \frac{b^{2}}{a}$ $FS = TS - b ET = TF = b$ $Δ OTF / OTS ∡ FOT = θ ∡ SOT = λ$ $OF = R {OS}^{3} = {(b + FS)}^{2} + {(R - a)}^{2}$ $Posons FS = x$ ${OS}^{2} = {(b + x)}^{2} + \frac{(b^{2} - a^{2})}{4 a^{2}}$ $OS = \frac{4 a^{2} {(b + x)}^{2} + b^{2} - a^{2}}{4 a^{2}}$ $\cos θ = \frac{(b^{2} - a^{2})}{a^{2} + b^{2}} \sin θ = \frac{2 ab}{a^{2} + b^{2}}$ $\cos λ = \frac{(b^{2} - a^{2})}{\sqrt{4 a^{2} (b + x) + b^{2} - a^{2}}}$ $\sin λ = \frac{2 a (b + x)}{\sqrt{4 a^{2} (b + x) + b^{2} - a^{2}}}$ $OF \times OS = \frac{(a^{2} + b^{2})}{4 a^{2}} \frac{\sqrt{4 a^{2} {(x + b)}^{2} + b^{2} - a^{2}}}{}$ $Δ OFS$ ${FS}^{2} = {OF}^{2} + {OS}^{2} - 2 (OF \times OS) \cos (λ - θ)$ $\cos (λ - θ) = \frac{{(b^{2} - a^{2})}^{2}}{(a^{2} + b^{2}) \sqrt{4 a^{2} (b + x) + b^{2} - a^{2}}}$ $+ \frac{4 a^{2} b (b + x)}{(a^{2} + b^{2}) \sqrt{4 a^{2} (b + x) + b^{2} - a^{2}}}$ $=$ $x^{2} = \frac{{(a^{2} + b^{2})}^{2}}{4 a^{2}} + \frac{4 a^{2} {(b + x)}^{2} + b^{2} - a^{2}}{4 a^{2}}$ $- 2 [\frac{(a^{3} + b^{2})}{2 a} \times \frac{\sqrt{4 a^{2} {(b + x)}^{2} + b^{2} - a^{2}}}{2 a}] \times$ $\frac{{(b^{2} - a^{2})}^{2} + 4 a^{2} b (b + x)}{(a^{2} + b^{2}) \sqrt{4 a^{2} + {(b + x)}^{2} + b^{2} - a^{2}}}$ $x^{2} = \frac{a^{2} + b^{2} + 4 a^{2} {(b + x)}^{2} + b^{2} - a^{2}}{4 a^{2}}$ $- \frac{{(b^{2} - a^{2})}^{2}}{2 a^{2}} - 2 b (b + x)$ $0 = \frac{a^{2} + b^{2}}{4 a^{2}} + {(b + x)}^{2} + \frac{b^{2} - a^{2}}{4 a^{2}} - \frac{{(b^{2} - a^{2})}^{2}}{2 a^{2}}$ $- 2 b (b + x)$ $x = FS \Rightarrow T S = b + x = X$ $X^{2} - 2 bX + \frac{b^{2} - {(b^{2} - a^{2})}^{2}}{2 a^{2}} = 0$ $Δ^{'} = b^{2} + \frac{{(b^{2} - a^{2})}^{2} - b^{2}}{2 a^{2}}$ $= \frac{b^{4} + a^{4} - 2 a^{2} b^{2} - b^{2}}{2 a^{2}}$ $X = b \pm \frac{1}{a} \sqrt{\frac{a^{4} + b^{4} - b^{2} (2 a^{2} + 1)}{2}}$ $alors TS = b + \frac{1}{a} \sqrt{\frac{a^{4} + b^{4} - b^{2} (2 a^{2} + 1)}{2}}$ $de meme pour PT cas (PE = FS)$ $On a alors$ $PS = PT + TS$ $C o n c l u s i o n :$ $PS = QR = 2 b + \frac{2}{a} \sqrt{\frac{a^{4} + b^{4} - b^{2} (2 a^{2} + 1)}{2}}$ $PQ = SR = \frac{b^{2}}{a}$
	Commented by a.lgnaoui last updated on 14/Dec/23

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