Question-87939

Question Number 87939 by ajfour last updated on 07/Apr/20

Commented by naka3546 last updated on 07/Apr/20

{w h a t}^{'} s t h e q u e s t i o n, s i r ?

Commented by ajfour last updated on 07/Apr/20

I f e a c h c o l o u r e d r e g i o n h a s u n i t

a r e a, n a m e t h e c o l o u r e d r e g i o n s

t h a t h a v e t h e l a r g e s t p e r i m e t e r,

a n d t h e s m a l l e s t .

(w o n t e d i t a n y m o r e!)

Commented by ajfour last updated on 07/Apr/20

o v e r a l l f i g u r e i s a s q u a r e .

Commented by Tony Lin last updated on 07/Apr/20

Commented by Tony Lin last updated on 07/Apr/20

\frac{1}{2} \times \sqrt{3} (\sqrt{3} - \frac{2 \sqrt{3}}{3}) = \frac{1}{2}

= h a l f o f t h e g r e e n a r e a

t h e h y p o t e n u s e o f b r o w n a r e a i s

\sqrt{{(\sqrt{3})}^{2} + {(\frac{2 \sqrt{3}}{3})}^{2}} = \frac{\sqrt{39}}{3}

t h e b l u e o n e a n d h a l f o f t h e g r e e n o n e

s h a r e t h e s a m e h e i g h t

\Rightarrow r a t i o o f t h e b o t t o m s i d e i s 2

\Rightarrow b o t t o m s i d e o f t h e b l u e o n e i s \frac{2 \sqrt{39}}{9}

t h e g r e e n o n e i s \frac{\sqrt{39}}{9}

b l u e a r e a = 1 = \frac{1}{2} \times \sqrt{3} \times \frac{2 \sqrt{39}}{9} s i n θ

\Rightarrow s i n θ = \frac{3}{\sqrt{13}} \to c o s θ = \frac{2}{\sqrt{13}}

l^{2} = {(\sqrt{3})}^{2} + {(\frac{2 \sqrt{39}}{9})}^{2} - 2 \times \sqrt{3} \times \frac{2 \sqrt{39}}{9} \times \frac{2}{\sqrt{13}}

= \frac{61}{27}

\Rightarrow l = \frac{\sqrt{183}}{9}

b r o w n p e r i m e t e r

= \sqrt{3} + \frac{2 \sqrt{3}}{3} + \frac{\sqrt{39}}{3} = \frac{5 \sqrt{3} + \sqrt{39}}{3} \approx 4.9684

g r e e n p e r i m e t e r

= \sqrt{3} + (\sqrt{3} - \frac{2 \sqrt{3}}{3}) + \frac{\sqrt{39}}{9} + \frac{\sqrt{183}}{9}

= \frac{12 \sqrt{3} + \sqrt{39} + \sqrt{183}}{9} \approx 4.5064

b l u e p e r i m e t e r

= \sqrt{3} + \frac{2 \sqrt{39}}{9} + \frac{\sqrt{183}}{9}

= \frac{9 \sqrt{3} + 2 \sqrt{39} + \sqrt{183}}{9} \approx 4.6229

Commented by ajfour last updated on 07/Apr/20

T h a n k s S i r, p e r f e c t!

Commented by ajfour last updated on 07/Apr/20

Commented by ajfour last updated on 07/Apr/20

s q u a r e s i d e s = \sqrt{3}

B E = \frac{2}{\sqrt{3}} = G M = p

A E = \sqrt{3 + \frac{4}{3}} = \frac{\sqrt{39}}{3}

A M = \frac{p}{s} (A E) = \frac{2}{3} (\frac{\sqrt{39}}{3}) = \frac{2 \sqrt{39}}{9}

\frac{q}{p} = \frac{B E}{s} \Rightarrow q = \frac{2}{3} \times \frac{2}{\sqrt{3}} = \frac{4 \sqrt{3}}{9}

M D = \sqrt{p^{2} + {(s - q)}^{2}}

= \sqrt{\frac{4}{3} + {(\sqrt{3} - \frac{4 \sqrt{3}}{9})}^{2}}

= \frac{\sqrt{183}}{9}

C E = \sqrt{3} - \frac{2}{\sqrt{3}} = \frac{\sqrt{3}}{3}

M E = A E - A M = \frac{\sqrt{39}}{9}

p e r i m e t e r (b r o w n) = \sqrt{3} + \frac{2}{\sqrt{3}} + \frac{\sqrt{39}}{3}

\approx 4.9684

(b l u e) = \sqrt{3} + \frac{2 \sqrt{39}}{9} + \frac{\sqrt{183}}{9}

\approx 4.6229

(g r e e n) = \sqrt{3} + \frac{\sqrt{3}}{3} + \frac{\sqrt{39}}{9} + \frac{\sqrt{183}}{9}

\approx 4.5064

s o p e r i m e t e r s o f

(b r o w n) > (b l u e) > (g r e e n)

Commented by $@ty@m123 last updated on 07/Apr/20

a r (A B C D) = 3 s q . u n i t

∴ e a c h s i d e = \sqrt{3} u n i t \dots (i)

a r (A E B) = 1 s q . u n i t

\frac{1}{2} \times A B \times E B = 1

\frac{1}{2} \times \sqrt{3} \times E B = 1

E B = \frac{2}{\sqrt{3}} \dots . (i i)

\frac{1}{2} \times A D \times G M = 1

G M = \frac{2}{\sqrt{3}} \dots (i i i)

(P l . i g n o r e t h i s c o m m e n t)