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Question Number 53071 by behi83417@gmail.com last updated on 16/Jan/19

Commented by behi83417@gmail.com last updated on 16/Jan/19

$a s s h o w n i n f i g :$ $p r o v e t h a t :$ $\to \frac{area of o u t e r hexagon}{area of i n n e r triangle} ⩾ 13$
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19

$a r e a A B C t r i a n g l e = S$ $S = \frac{1}{2} a c s i n B = \frac{1}{2} a b s i n C = \frac{1}{2} b c s i n A$ $o t h e r s t h r e e t r i a n g l e . . . b u t t e r f l y w i n g s l i k e . .$ $a r e a l s o S .$ $a r e a o f h e x a g o n = 4 S + P + Q + R$ $n o w S + Q = \frac{1}{2} (b + c) (a + b) s i n B$ $S + P = \frac{1}{2} (a + c) (a + b) s i n A$ $S + R = \frac{1}{2} (b + c) (a + c) s i n C$ $n o w w e k n o w$ $\frac{s i n A}{a} = \frac{s i n B}{b} = \frac{s i n C}{c} = k (s a y)$ $s o △ A B C a r e a = \frac{1}{2} a c s i n B = \frac{a b c k}{2}$ $a r e a h e x a g o n = 4 S + P + Q + R$ $= (S + P) + (S + Q) + (S + R) + S$ $= \frac{(a + c) (a + b) k a + (b + c) (a + b) k b + (b + c) (a + c) k c + a b c k}{2}$ $= \frac{k}{2} [(a^{2} + a b + a c + b c) a + (a b + b^{2} + a c + b c) b + (a b + b c + a c + c^{2}) c + a b c]$ $= \frac{k}{2} [a^{3} + a^{2} b + a^{2} c + a b c + {a b}^{2} + b^{3} + a b c + b^{2} c + a b c + {b c}^{2} + {a c}^{2} + c^{3} + a b c]$ $= \frac{k}{2} [4 a b c + a^{3} + b^{3} + c^{3} + a^{2} (b + c) + b^{2} (a + c) + c^{2} (a + b)]$ $= 2 a b c k + \frac{k}{2} [a^{2} (a + b + c) + b^{2} (a + b + c) + c^{2} (a + b + c)]$ $= 2 a b c k + \frac{k}{2} [(a + b + c) (a^{2} + b^{2} + c^{2})]$ $n o w \frac{a + b + c}{3} ⩾ {(a b c)}^{\frac{1}{3}}$ $\frac{a^{2} + b^{2} + c^{2}}{3} ⩾ {(a^{2} b^{2} c^{2})}^{\frac{1}{3}}$ $s o (a + b + c) (a^{2} + b^{2} + c^{2}) ⩾ 9 a b c$ $s o a r e a o f h e x a g o n$ $= 2 a b c k + \frac{k}{2} (a + b + c) (a^{2} + b^{2} + c^{2}) ⩾ 2 a b c k + \frac{9 a b c k}{2}$ $s o a r e a o f h e x a g o n ⩾ \frac{13 a b c k}{2}$ $a r e a o f h e x a g o n ⩾ 13 \times a r e a o f △ A B C$ $\frac{a r e a o f h e x a g o n}{△ A B C} ⩾ 13 p r o v e d$
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19

$s i r p l s c h e c k i h a v e p r o v e d . . .$
Commented by behi83417@gmail.com last updated on 17/Jan/19

$t h a n k y o u v e r y m u c h s i r t a n m a y .$ $i t i s a n i c e a n d s m a r t w o r k .$
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19

$t h a n k y o u s i r . . .$
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