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Question Number 165389 by ajfour last updated on 31/Jan/22

Commented by ajfour last updated on 31/Jan/22

$I f A i s c e n t e r o f u p p e r c i r c l e, f i n d$ $i t s r a d i u s, i f r_{B} = 1 .$
	Answered by mahdipoor last updated on 31/Jan/22

	$r_{a} = a r_{b} = b$ $d : 0 = a + x - y$ $B = (- b, - \sqrt{2 a b + a^{2}})$ $BH = b = \frac{∣ a - b + \sqrt{2 a b + a^{2}} ∣}{\sqrt{2}} \Rightarrow$ $\pm \sqrt{2} b = a - b + \sqrt{2 a b + a^{2}} \Rightarrow (1 \pm \sqrt{2}) b - a = \sqrt{2 a b + a^{2}}$ $(5 \pm 2 \sqrt{2}) b^{2} + a^{2} - (2 \pm 2 \sqrt{2}) a b = 2 a b + a^{2} \Rightarrow$ $b [(5 \pm 2 \sqrt{2}) b - (4 \pm 2 \sqrt{2}) a] = 0 \Rightarrow$ $a = \frac{5 \pm 2 \sqrt{2}}{4 \pm 2 \sqrt{2}} b$
	Answered by mr W last updated on 01/Feb/22

	$A B = a + b$ $A D = \frac{b}{\tan 22.5 °} - a$ ${(a + b)}^{2} = b^{2} + {(\frac{b}{\tan 22.5 °} - a)}^{2}$ $2 a (1 + \frac{1}{\tan 22.5 °}) = \frac{b}{\tan^{2} 22.5 °}$ $\frac{a}{b} = \frac{1}{2 (1 + \tan 22.5 °) \tan 22.5 °}$ $\frac{a}{b} = \frac{1}{2 (1 - \tan 22.5 °)} = \frac{1}{2 (2 - \sqrt{2})}$ $\frac{a}{b} = \frac{2 + \sqrt{2}}{4} \approx 0.853$
	Commented by mr W last updated on 01/Feb/22

	Commented by ajfour last updated on 01/Feb/22

	$S i r h o w ∠ A C B i s 22.5 ° ?$ $i d o n t f o l l o w . . h e l p . .$
	Commented by mr W last updated on 01/Feb/22

	$A C = A F = r a d i u s$ $∠ F A C = 90 ° a s g i v e n$ $\Rightarrow ∠ A F C = ∠ A C F = 45 °$ $∠ D C B = ∠ E C B = \frac{45 °}{2} = 22.5 °$ $C D = \frac{B D}{\tan 22.5 °}$
	Commented by ajfour last updated on 01/Feb/22

	$E x c e l l e n t w a y, s i r! h a t s o f f!$
	Commented by Tawa11 last updated on 01/Feb/22

	$Weldone sir$
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