Question-102769

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} \dots (1)

\sin 30^{o} = \frac{r}{r + x} = \frac{1}{2} \Rightarrow x = r \dots (2)

(1) & (2) \Rightarrow 3 r + R = \frac{s \sqrt{3}}{2} \dots (3)

r + R = R \sqrt{2} \Rightarrow r = R (\sqrt{2} - 1) \dots (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 \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}