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Question Number 102769 by ajfour last updated on 11/Jul/20

Commented by bramlex last updated on 11/Jul/20

Commented by bramlex last updated on 11/Jul/20

$(1) BG = \sqrt{\frac{s^{2}}{4} - R^{2}}$ $(2) GD =$ $\sqrt{{(r + R)}^{2} - R^{2}} = \sqrt{r^{2} + 2 rR}$ $(3) BD = \sqrt{{(r + R)}^{2} + \frac{s^{2}}{4}}$ $BD = BG + GD$ $\sqrt{{(r + R)}^{2} + \frac{s^{2}}{4}} =$ $\sqrt{\frac{s^{2}}{4} - R^{2}} + \sqrt{r^{2} + 2 rR}$
Commented by ajfour last updated on 11/Jul/20

$I f △ A B C i s e q u i l a t e r a l w i t b$ $s i d e s, f i n d r a d i i o f b o t h c i r c l e s$ $i n t e r m s o f s .$
Commented by bramlex last updated on 11/Jul/20

Commented by bramlex last updated on 11/Jul/20

$AD = x + 2 r + R = \frac{s \sqrt{3}}{2} . . . (1)$ $\sin 30^{o} = \frac{r}{r + x} = \frac{1}{2} \Rightarrow x = r . . . (2)$ $(1) & (2) \Rightarrow 3 r + R = \frac{s \sqrt{3}}{2} . . . (3)$ $r + R = R \sqrt{2} \Rightarrow r = R (\sqrt{2} - 1) . . . (4)$ $(3) & (4) \Rightarrow 3 R (\sqrt{2} - 1) + R = \frac{s \sqrt{3}}{2}$ $R (3 \sqrt{2} - 2) = \frac{s \sqrt{3}}{2} \Rightarrow R = \frac{s \sqrt{3}}{2 (3 \sqrt{2} - 2)}$ $then r = \frac{s (\sqrt{6} - \sqrt{3})}{2 (3 \sqrt{2} - 2)} . \overset{∙}{⌣} ∣ \overset{∙}{⌣}$
Commented by ajfour last updated on 11/Jul/20

$h o w c o m e R + r = R \sqrt{2} ?$
Commented by bramlex last updated on 11/Jul/20

Commented by ajfour last updated on 11/Jul/20

$t h e q u a d r i l a t e r a l {i s n}^{'} t a$ $p a r a l l e l o g r a m, w r o n g . .$
Commented by bramlex last updated on 11/Jul/20

$no .$
	Answered by mr W last updated on 11/Jul/20

	Commented by mr W last updated on 11/Jul/20

	$C F = \sqrt{\frac{s^{2}}{4} - R^{2}}$ $\frac{O D}{O C} = \frac{O F}{C F}$ $\Rightarrow O D = \frac{s R}{\sqrt{s^{2} - 4 R^{2}}}$ $r = O D - R = (\frac{s}{\sqrt{s^{2} - 4 R^{2}}} - 1) R$ $O A = R + 3 r = \frac{\sqrt{3} s}{2}$ $R + 3 (\frac{s R}{\sqrt{s^{2} - 4 R^{2}}} - R) = \frac{\sqrt{3} s}{2}$ $w i t h λ = \frac{R}{s}$ $\Rightarrow \frac{3}{\sqrt{1 - 4 λ^{2}}} = 2 + \frac{\sqrt{3}}{2 λ}$ $\Rightarrow 64 λ^{4} + 32 \sqrt{3} λ^{3} + 32 λ^{2} - 8 \sqrt{3} λ - 3 = 0$ $\Rightarrow λ = \frac{R}{s} \approx 0.3642$ $\Rightarrow \frac{r}{s} = (\frac{1}{\sqrt{1 - 4 λ^{2}}} - 1) λ \approx 0.1674$
	Commented by mr W last updated on 11/Jul/20

	Commented by ajfour last updated on 11/Jul/20

	$V e r y N i c e s o l u t i o n S i r; T h a n k s!$
	Answered by 1549442205 last updated on 11/Jul/20

	Commented by 1549442205 last updated on 11/Jul/20
	$Since DKE^(�) =30°,KD=((DE)/(sinDKE^(�) ))=(r/(sin30°)) ⇒KD=2r.We have KC=KAcos30°=((s(√3))/2) ,so KD=KC−CD.Hence, 2r=((s(√3))/2)−(R+r)⇔r=((s(√3)−2R)/6)(1) On ther other hands, AD^2 =(R+r)^2 +(s^2 /4) KE=r(√3) and AE=s−KE=s−r(√3) , AE^2 +DE^2 =AD^2 =DC^2 +AC^2 ,so (s−r(√3))^2 +r^2 =(R+r)^2 +(s^2 /4) ⇔R^2 +2Rr−3r^2 +2sr(√3)−((3s^2 )/4)=0(due to (1)) ⇔R^2 +((s(√3)−2R)/3)(R+s(√3))−((3s^2 −4Rs(√3)+4R^2 )/3)−((3s^2 )/4)=0 4R^2 −12Rs(√3) +9s^2 =0 .Δ′=108s^2 −36s^2 =72s^2 R=((6s(√3)−6s(√2))/4)=((3s((√3)−(√2)))/2).Replace into (1) we get r=((s(√3)−3s((√3)−(√2)))/6)=(((3(√2)−2(√3))s)/6) Thus, { (((R/s)=((3((√3)−(√2)))/2)≈0.47675586)),(((r/s)=((3(√2)−2(√3))/6)≈0.12975512)) :}$
	$Since \hat{DKE} = 30 °, KD = \frac{DE}{\sin \hat{DKE}} = \frac{r}{\sin 30 °}$ $\Rightarrow KD = 2 r . We have KC = KAcos 30 ° = \frac{s \sqrt{3}}{2}$ $, so KD = KC - CD . Hence,$ $2 r = \frac{s \sqrt{3}}{2} - (R + r) \Leftrightarrow r = \frac{s \sqrt{3} - 2 R}{6} (1)$ $On ther other hands,$ ${AD}^{2} = {(R + r)}^{2} + \frac{s^{2}}{4}$ $KE = r \sqrt{3} and AE = s - KE = s - r \sqrt{3},$ ${AE}^{2} + {DE}^{2} = {AD}^{2} = {DC}^{2} + {AC}^{2}, so$ ${(s - r \sqrt{3})}^{2} + r^{2} = {(R + r)}^{2} + \frac{s^{2}}{4}$ $\Leftrightarrow R^{2} + 2 Rr - {3 r}^{2} + 2 sr \sqrt{3} - \frac{{3 s}^{2}}{4} = 0 (due to (1))$ $\Leftrightarrow R^{2} + \frac{s \sqrt{3} - 2 R}{3} (R + s \sqrt{3}) - \frac{{3 s}^{2} - 4 Rs \sqrt{3} + {4 R}^{2}}{3} - \frac{{3 s}^{2}}{4} = 0$ ${4 R}^{2} - 12 Rs \sqrt{3} + {9 s}^{2} = 0 . Δ^{'} = {108 s}^{2} - {36 s}^{2} = {72 s}^{2}$ $R = \frac{6 s \sqrt{3} - 6 s \sqrt{2}}{4} = \frac{3 s (\sqrt{3} - \sqrt{2})}{2} . Replace$ $into (1) we get$ $r = \frac{s \sqrt{3} - 3 s (\sqrt{3} - \sqrt{2})}{6} = \frac{(3 \sqrt{2} - 2 \sqrt{3}) s}{6}$ $Thus, {\begin{cases} \frac{R}{s} = \frac{3 (\sqrt{3} - \sqrt{2})}{2} \approx 0.47675586 \\ \frac{r}{s} = \frac{3 \sqrt{2} - 2 \sqrt{3}}{6} \approx 0.12975512 \end{cases}$
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