Question-27771

Question Number 27771 by ajfour last updated on 14/Jan/18

Commented by ajfour last updated on 14/Jan/18

I f a n e q u i l a t e r a l t r i a n g l e o f

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

a b o u t i t s v e r t e x A b y a n a n g l e θ

f i n d t h e a r e a c o m m o n i n i t s

n e w a n d p r e v i o u s p o s i t i o n s .

(t h e p i n k a r e a) .

Answered by mrW2 last updated on 14/Jan/18

Commented by mrW2 last updated on 14/Jan/18

2 α + θ = \frac{π}{3}

\Rightarrow α = \frac{π}{6} - \frac{θ}{2}

\frac{B P}{\sin θ} = \frac{A B}{\sin (π - \frac{π}{3} - θ)} = \frac{a}{\sin (\frac{π}{3} + θ)}

\Rightarrow B P = \frac{\sin θ}{\sin (\frac{π}{3} + θ)} \times a

s i m i l a r l y

\frac{C M}{\sin α} = \frac{a}{\sin (\frac{π}{3} + α)}

\Rightarrow C M = \frac{\sin α}{\sin (\frac{π}{3} + α)} \times a = \frac{\sin (\frac{π}{6} - \frac{θ}{2})}{\sin (\frac{π}{2} - \frac{θ}{2})} \times a = \frac{\sin (\frac{π}{6} - \frac{θ}{2})}{\cos \frac{θ}{2}} \times a

P M = a - \frac{\sin θ}{\sin (\frac{π}{3} + θ)} \times a - \frac{\sin (\frac{π}{6} - \frac{θ}{2})}{\cos \frac{θ}{2}} \times a

= a [1 - \frac{\sin θ}{\frac{\sqrt{3}}{2} \cos θ + \frac{1}{2} \sin θ} - \frac{\frac{1}{2} \cos \frac{θ}{2} - \frac{\sqrt{3}}{2} \sin \frac{θ}{2}}{\cos \frac{θ}{2}}]

= a [\frac{1}{2} - \frac{2 \tan θ}{\sqrt{3} + \tan θ} + \frac{\sqrt{3}}{2} \tan \frac{θ}{2}]

A_{Δ A P M} = \frac{1}{2} \times P M \times \frac{\sqrt{3} a}{2} = \frac{\sqrt{3} a^{2}}{4} [\frac{1}{2} - \frac{2 \tan θ}{\sqrt{3} + \tan θ} + \frac{\sqrt{3}}{2} \tan \frac{θ}{2}]

\Rightarrow A_{A P M Q} = 2 A_{Δ A P M} = \frac{\sqrt{3} a^{2}}{4} [1 - \frac{4 \tan θ}{\sqrt{3} + \tan θ} + \sqrt{3} \tan \frac{θ}{2}]

Commented by ajfour last updated on 15/Jan/18

t h a n k y o u s i r . p l e a s e v i e w m y

w a y i f y o u {h a v e n}^{'} t .

Commented by mrW2 last updated on 16/Jan/18

Y o u r w a y i s v e r y n i c e t o o . I {d i d n}^{'} t c o m e

t o i t .

Answered by ajfour last updated on 14/Jan/18

Commented by ajfour last updated on 14/Jan/18

e q . o f B P :

y = - \sqrt{3} (x - a)

e q . o f A P : y = x \tan θ

e q . o f A M : y = x \tan (θ + α)

= x \tan (\frac{π}{6} + \frac{θ}{2})

y_{P} = - \sqrt{3} (x_{P} - a) = x_{P} \tan θ

\Rightarrow y_{P} = \frac{a \sqrt{3} \tan θ}{\sqrt{3} + \tan θ}

s i m i l a r l y y_{M} = \frac{a \sqrt{3} \tan (\frac{π}{6} + \frac{θ}{2})}{\sqrt{3} + \tan (\frac{π}{6} + \frac{θ}{2})}

= \frac{a \sqrt{3}}{\sqrt{3} (\frac{1 - \frac{\tan (θ / 2)}{\sqrt{3}}}{\frac{1}{\sqrt{3}} + \tan \frac{θ}{2}}) + 1}

= \frac{a \sqrt{3}}{\sqrt{3} [\frac{\sqrt{3} - \tan (θ / 2)}{1 + \sqrt{3} \tan (θ / 2)}] + 1}

= \frac{a \sqrt{3} [1 + \sqrt{3} \tan (θ / 2)]}{4}

r e q d . a r e a = 2 \times \frac{a}{2} (y_{M} - y_{P})

= \frac{a^{2} \sqrt{3}}{4} [1 + \sqrt{3} \tan \frac{θ}{2} - \frac{4 \tan θ}{\sqrt{3} + \tan θ}] .

Facebook Tweet Pin