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Question Number 141171 by iloveisrael last updated on 16/May/21

Commented by iloveisrael last updated on 16/May/21

$i n s t e r e s t i n g q u e s t i o n$
Commented by mr W last updated on 16/May/21

$i f A C i s f i x e d, t h e n s i s a l s o f i x e d .$ $t h e r e i s n o m a x i m u m v a l u e o f s,$ $b e c a u s e i t h a s a f i x e d v a l u e .$
Commented by iloveisrael last updated on 16/May/21

$n o$
Commented by mr W last updated on 16/May/21

$i f R a n d 1 a r e f i x e d v a l u e s,$ $t h e n s i s a l s o f i x e d . o r y o u m e a n R$ $i s n o t c o n s t a n t .$
Commented by EDWIN88 last updated on 16/May/21

$s depent to r . not a constant$
Commented by mr W last updated on 16/May/21

$b u t i f r i s c o n s t a n t, t h e n s i s a l s o$ $c o n s t a n t . t h e q u e s t i o n s e e m s t o s h o w$ $t h a t r i s g i v e n, i . e . r i s c o n s t a n t .$
	Answered by EDWIN88 last updated on 16/May/21
	$let { ((A(0,0) )),((B(r,0))) :}C(0,1) eq of BC⇒ y=−(x/r)+1 eq of cifcle⇒(x−r)^2 +y^2 = r^2 y=(√(2rx−x^2 )) coordinates { ((T(s, −(r/s)+1))),((Q(s,(√(2rs−s^2 )) ))) :} . Since PQTU is a square we have s = −(s/r)+1−(√(2rs−s^2 )) rs = −s+r−r(√(2rs−s^2 )) r(√(2rs−s^2 )) = −s+r−rs r^2 (2rs−s^2 ) = s^2 +r^2 +r^2 s^2 −2rs−2r^2 s+2rs^2 2r^3 s−r^2 s^2 = s^2 +r^2 +r^2 s^2 −2rs−2r^2 s+2rs^2 2r^3 s+2rs+2r^2 s−2r^2 s^2 −2rs^2 −s^2 −r^2 =0 (−2r^2 −2r−1)s^2 +(2r^3 +2r^2 +2r)s−r^2 =0 (2r^2 +2r+1)s^2 −2r(r^2 +r+1)s +r^2 = 0 s = (r/(2r^2 +2r+1)) (ds/dr) = ((1(2r^2 +2r+1)−r(4r+2))/((2r^2 +2r+1)^2 )) = 0 ⇒ 2r^2 +2r+1−4r^2 −2r = 0 ⇒ 2r^2 = 1 →r = (1/( (√2))) max s = ((1/( (√2)))/(1+(√2)+1)) = (1/((2+(√2))(√2))) s_(max) = (1/(2(√2) +2)) × ((2(√2)−2)/(2(√2)−2)) = ((2(√2)−2)/4) s_(max) = (((√2)−1)/2) ⋇$
	$let {\begin{cases} A (0, 0) \\ B (r, 0) \end{cases} C (0, 1)$ $eq of BC \Rightarrow y = - \frac{x}{r} + 1$ $eq of cifcle \Rightarrow {(x - r)}^{2} + y^{2} = r^{2}$ $y = \sqrt{2 rx - x^{2}}$ $coordinates {\begin{cases} T (s, - \frac{r}{s} + 1) \\ Q (s, \sqrt{2 rs - s^{2}}) \end{cases} . Since$ $PQTU is a square we have$ $s = - \frac{s}{r} + 1 - \sqrt{2 rs - s^{2}}$ $rs = - s + r - r \sqrt{2 rs - s^{2}}$ $r \sqrt{2 rs - s^{2}} = - s + r - rs$ $r^{2} (2 rs - s^{2}) = s^{2} + r^{2} + r^{2} s^{2} - 2 rs - {2 r}^{2} s + {2 rs}^{2}$ ${2 r}^{3} s - r^{2} s^{2} = s^{2} + r^{2} + r^{2} s^{2} - 2 rs - {2 r}^{2} s + {2 rs}^{2}$ ${2 r}^{3} s + 2 rs + {2 r}^{2} s - {2 r}^{2} s^{2} - {2 rs}^{2} - s^{2} - r^{2} = 0$ $(- {2 r}^{2} - 2 r - 1) s^{2} + ({2 r}^{3} + {2 r}^{2} + 2 r) s - r^{2} = 0$ $({2 r}^{2} + 2 r + 1) s^{2} - 2 r (r^{2} + r + 1) s + r^{2} = 0$ $s = \frac{r}{{2 r}^{2} + 2 r + 1}$ $\frac{ds}{dr} = \frac{1 ({2 r}^{2} + 2 r + 1) - r (4 r + 2)}{{({2 r}^{2} + 2 r + 1)}^{2}} = 0$ $\Rightarrow {2 r}^{2} + 2 r + 1 - {4 r}^{2} - 2 r = 0$ $\Rightarrow {2 r}^{2} = 1 \to r = \frac{1}{\sqrt{2}}$ $\max s = \frac{\frac{1}{\sqrt{2}}}{1 + \sqrt{2} + 1} = \frac{1}{(2 + \sqrt{2}) \sqrt{2}}$ $s_{\max} = \frac{1}{2 \sqrt{2} + 2} \times \frac{2 \sqrt{2} - 2}{2 \sqrt{2} - 2} = \frac{2 \sqrt{2} - 2}{4}$ $s_{\max} = \frac{\sqrt{2} - 1}{2} ⋇$
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