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Question Number 145316 by imjagoll last updated on 04/Jul/21

Commented by imjagoll last updated on 04/Jul/21

$interesting problem$ $\neq geometry$
Commented by imjagoll last updated on 04/Jul/21

$can anyone solve this ?$
	Answered by ajfour last updated on 04/Jul/21

	Commented by ajfour last updated on 04/Jul/21
	$let B be the origin and x-axis leftwards. A(0,d) eq. of first circles x^2 +y^2 =r^2 (x−d)^2 +y^2 =1 subtracting the two for x coordinate of point of intersection. d(2p−d)=r^2 −1 p=((r^2 +d^2 −1)/(2d)) q=±(√(r^2 −p^2 )) m=((dy/dx))_r =−(p/q) s=((dy/dx))_2 =((d−p)/q) ((m−s)/(1+ms))=(√3) (=tan 60°) m−s=(√3)+(√3)ms −(p/q)−((d−p)/q)=(√3)−(√3)((p/q))(((d−p)/q)) dq=(√3)(dp−p^2 −q^2 ) d(√(r^2 −p^2 ))=(√3)(dp−p^2 −r^2 +p^2 ) ⇒ d(√(r^2 −p^2 ))=(√3)(dp−r^2 ) d^2 {r^2 −(((r^2 +d^2 −1)/(2d)))^2 } =3(((r^2 +d^2 −1)/2)−r^2 )^2 ⇒ r^2 d^2 −(((r^2 +d^2 −1)^2 )/4) =(3/4)(r^2 +d^2 −1)^2 +3r^4 −3r^2 (r^2 +d^2 −1) ⇒ (r^2 +d^2 −1)^2 +3r^4 =3r^2 (r^2 +d^2 −1)+r^2 d^2 let d^2 −1=kr^2 ⇒ (1+k)^2 +3=3(1+k)+(1/r^2 )+k ⇒ k^2 −2k+1=(1/r^2 ) or r=(1/(k−1)) d^2 =1+(1+(1/r))r^2 d^2 =1+r^2 +r d^2 =(r+(1/2))^2 +(3/4) d is minimum for r=−(1/2) d_(min) =((√3)/2)$
	$l e t B b e t h e o r i g i n a n d x - a x i s$ $l e f t w a r d s . A (0, d)$ $e q . o f f i r s t c i r c l e s$ $x^{2} + y^{2} = r^{2}$ ${(x - d)}^{2} + y^{2} = 1$ $s u b t r a c t i n g t h e t w o f o r$ $x c o o r d i n a t e o f p o i n t o f$ $i n t e r s e c t i o n .$ $d (2 p - d) = r^{2} - 1$ $p = \frac{r^{2} + d^{2} - 1}{2 d}$ $q = \pm \sqrt{r^{2} - p^{2}}$ $m = {(\frac{d y}{d x})}_{r} = - \frac{p}{q}$ $s = {(\frac{d y}{d x})}_{2} = \frac{d - p}{q}$ $\frac{m - s}{1 + m s} = \sqrt{3} (= \tan 60 °)$ $m - s = \sqrt{3} + \sqrt{3} m s$ $- \frac{p}{q} - \frac{d - p}{q} = \sqrt{3} - \sqrt{3} (\frac{p}{q}) (\frac{d - p}{q})$ $d q = \sqrt{3} (d p - p^{2} - q^{2})$ $d \sqrt{r^{2} - p^{2}} = \sqrt{3} (d p - p^{2} - r^{2} + p^{2})$ $\Rightarrow d \sqrt{r^{2} - p^{2}} = \sqrt{3} (d p - r^{2})$ $d^{2} {r^{2} - {(\frac{r^{2} + d^{2} - 1}{2 d})}^{2}}$ $= 3 {(\frac{r^{2} + d^{2} - 1}{2} - r^{2})}^{2}$ $\Rightarrow r^{2} d^{2} - \frac{{(r^{2} + d^{2} - 1)}^{2}}{4}$ $= \frac{3}{4} {(r^{2} + d^{2} - 1)}^{2} + 3 r^{4}$ $- 3 r^{2} (r^{2} + d^{2} - 1)$ $\Rightarrow$ ${(r^{2} + d^{2} - 1)}^{2} + 3 r^{4}$ $= 3 r^{2} (r^{2} + d^{2} - 1) + r^{2} d^{2}$ $l e t d^{2} - 1 = {k r}^{2} \Rightarrow$ ${(1 + k)}^{2} + 3 = 3 (1 + k) + \frac{1}{r^{2}} + k$ $\Rightarrow k^{2} - 2 k + 1 = \frac{1}{r^{2}} o r$ $r = \frac{1}{k - 1}$ $d^{2} = 1 + (1 + \frac{1}{r}) r^{2}$ $d^{2} = 1 + r^{2} + r$ $d^{2} = {(r + \frac{1}{2})}^{2} + \frac{3}{4}$ $d i s m i n i m u m f o r r = - \frac{1}{2}$ $d_{m i n} = \frac{\sqrt{3}}{2}$
	Answered by mr W last updated on 04/Jul/21

	$(1)$ $d^{2} = 1^{2} + r^{2} - 2 r \cos 60 °$ $\Rightarrow d = \sqrt{r^{2} - r + 1}$ $(2)$ $d = \sqrt{{(r - \frac{1}{2})}^{2} + \frac{3}{4}}$ $d_{m i n} = \frac{\sqrt{3}}{2} a t r = \frac{1}{2}$
	Commented by mr W last updated on 04/Jul/21

	Commented by mr W last updated on 04/Jul/21

	$(3)$ $a r e a o f i n t e r s e c t i o n :$ $\frac{π}{2} \times {(\frac{1}{2})}^{2} + \frac{1^{2}}{2} (\frac{π}{3} - \sin \frac{π}{3})$ $= \frac{7 π}{24} - \frac{\sqrt{3}}{4}$
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