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Question Number 25871 by ajfour last updated on 16/Dec/17

Commented by ajfour last updated on 16/Dec/17

$△ A B C i s e q u i l a t e r a l a n d G i t s$ $c e n t r o i d . P a r a b o l a s A G B, B G C,$ $a n d C G A e a c h h a v e t h e i r v e r t i c e s$ $a t G . F i n d r a t i o o f a r e a i n b r o w n$ $t o t h e a r e a o f t h e △ A B C .$
Commented by moxhix last updated on 16/Dec/17

$p u t l = A B = B C = C D$ $d i s t a n c e o f G a n d B C = \frac{1}{3} \times \frac{\sqrt{3}}{2} l = \frac{l}{2 \sqrt{3}}$ $f i n d B G C b y i n t e g r a t i o n$ $y = - a x (x - l) = - a {(x - \frac{l}{2})}^{2} + \frac{{a l}^{2}}{4}$ $\frac{l}{2 \sqrt{3}} = \frac{{a l}^{2}}{4}$ $∴ a = \frac{2}{l \sqrt{3}}$ $B G C = \int_{0}^{l} - a x (x - l) d x = \frac{{a l}^{3}}{6} = \frac{l^{2}}{3 \sqrt{3}}$ $△ A B C = \frac{1}{2} l \frac{\sqrt{3}}{2} l = \frac{\sqrt{3} l^{2}}{4}$ $\frac{b r o w n}{△ A B C} = \frac{3 B G C - △ A B C}{△ A B C} = (3 \frac{l^{2}}{3 \sqrt{3}} - \frac{\sqrt{3} l^{2}}{4}) / (\frac{\sqrt{3} l^{2}}{4})$ $= (\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{4}) / (\frac{\sqrt{3}}{4})$ $= \frac{1}{3}$
Commented by ajfour last updated on 16/Dec/17

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