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Question Number 206064 by mr W last updated on 06/Apr/24

Commented by mr W last updated on 06/Apr/24

$a n u n s o l v e d o l d q u e s t i o n (Q 172465)$
	Answered by A5T last updated on 06/Apr/24

	$W L O G, l e t N b e t h e o r i g i n a n d B N E c o i n c i d e$ $w i t h t h e r e a l a x i s (o r x a x i s) .$ $B (- b, 0), E (b, 0), N (0, 0), C (x, y), D (u, v)$ $\Rightarrow B C = \sqrt{{(x + b)}^{2} + y^{2}} = A B = \sqrt{x^{2} + y^{2} + b^{2} + 2 x b}$ $[A B C] = \frac{\sqrt{3}}{4} (x^{2} + y^{2} + b^{2} + 2 x b)$ $A (\frac{x - b - y \sqrt{3}}{2}, \frac{x \sqrt{3} + b \sqrt{3} + y}{2});$ $F (\frac{u + b - v \sqrt{3}}{2}, \frac{u \sqrt{3} - b \sqrt{3} + v}{2})$ $D E = \sqrt{{(u - b)}^{2} + v^{2}} = \sqrt{u^{2} + v^{2} + b^{2} - 2 u b}$ $\Rightarrow [F E D] = \frac{\sqrt{3}}{4} (u^{2} + v^{2} + b^{2} - 2 u b)$ $P (\frac{2 x + b - v \sqrt{3}}{4}, \frac{2 y + u \sqrt{3} - b \sqrt{3} + v}{4})$ $M (\frac{2 u + x - b - y \sqrt{3}}{4}, \frac{2 v + x \sqrt{3} + b \sqrt{3} + y}{4})$ $\Rightarrow [P N M] = \frac{1}{2} ∣ \frac{(2 x + b - v \sqrt{3}) (2 v + x \sqrt{3} + b \sqrt{3} + y)}{16}$ $- \frac{(2 y + u \sqrt{3} - b \sqrt{3} + v) (2 u + x - b - y \sqrt{3})}{16} ∣$ $= \frac{\sqrt{3} ∣ x^{2} + y^{2} - u^{2} - v^{2} + 2 b u + 2 b x ∣}{16}$ $∣ [A B C] - [D E F] ∣= \frac{\sqrt{3} ∣ x^{2} + y^{2} + 2 x b + 2 u b - u^{2} - v^{2} ∣}{4}$ $\Rightarrow \frac{[P N M [}{∣ [A B C - [D E F] ∣} = \frac{\frac{\sqrt{3}}{16}}{\frac{\sqrt{3}}{4}} = \frac{1}{4}$
	Commented by mr W last updated on 07/Apr/24
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	Answered by mr W last updated on 07/Apr/24

	Commented by mr W last updated on 07/Apr/24

	$s a y B N = N E = c$ $x_{A} = - c + a \cos α$ $y_{A} = a \sin α$ $x_{D} = c - b \cos β$ $y_{D} = b \sin β$ $x_{M} = \frac{x_{A} + x_{D}}{2} = \frac{a \cos α - b \cos β}{2}$ $y_{M} = \frac{y_{A} + y_{D}}{2} = \frac{a \sin α + b \sin β}{2}$ $s i m i l a r l y$ $x_{P} = \frac{a \cos (\frac{π}{3} - α) - b \cos (\frac{π}{3} - β)}{2}$ $y_{P} = - \frac{a \sin (\frac{π}{3} - α) + b \sin (\frac{π}{3} - β)}{2}$ $[M N P] =∣ \frac{(x_{M} + x_{P}) (y_{M} - y_{P}) - x_{M} y_{M} + x_{P} y_{P}}{2} ∣$ $[M N P] =∣ \frac{x_{P} y_{M} - x_{M} y_{P}}{2} ∣$ $[M N P] =∣ \frac{(a \cos (\frac{π}{3} - α) - b \cos (\frac{π}{3} - β)) (a \sin α + b \sin β) + (a \cos α - b \cos β) (a \sin (\frac{π}{3} - α) + b \sin (\frac{π}{3} - β))}{8} ∣$ $[M N P] =∣ \frac{a^{2} \sin α \cos (\frac{π}{3} - α) - a b \sin α \cos (\frac{π}{3} - β) + a b \sin β \cos (\frac{π}{3} - α) - b^{2} \sin β \cos (\frac{π}{3} - β) + a^{2} \cos α \sin (\frac{π}{3} - α) + a b \cos α \sin (\frac{π}{3} - β) - a b \cos β \sin (\frac{π}{3} - α) - b^{2} \cos β \sin (\frac{π}{3} - β)}{8} ∣$ $[M N P] =∣ \frac{a^{2} \sin \frac{π}{3} + a b \sin (\frac{π}{3} - α - β) + a b \sin (α + β - \frac{π}{3}) - b^{2} \sin \frac{π}{3}}{8} ∣$ $[M N P] =∣ \frac{(a^{2} - b^{2}) \sin \frac{π}{3}}{8} ∣$ $[M N P] =∣ \frac{\sqrt{3} (a^{2} - b^{2})}{16} ∣$ $s i n c e X = \frac{\sqrt{3} a^{2}}{4}, Y = \frac{\sqrt{3} b^{2}}{4}$ $\Rightarrow [M N P] = \frac{∣ X - Y ∣}{4} ✓$
	Answered by aleks041103 last updated on 08/Apr/24

	$H e r e i s a v e c t o r i a l w a y t o a p p o a c h t h i s .$ $F i r s t l y, w e s h o u l d n o t e t h a t t h e v e c t o r s M, N$ $a n d P a r e l i n e a r c o m b i n a t i o n s o f t h e v e c t o r s$ $o f t h e i n i t i a l 6 v e r t i c e s - A, B, C, D, E, F . M o r e p r e c i s e l y M, N, P$ $a r e t h e a v e r a g e o f s o m e 2 v e r t i c e s .$ $F o r e x a m p l e$ $N = \frac{1}{2} (B + E)$ $S e c o n d l y, b y t h e f i r s t o b s e r v a t i o n, i f w e$ $t r a n s l a t e △ D F N b y a v e c t o r v$ $t h e n t h e v e r t i c e s o f △ P M N t r a n s l a t e b y \frac{1}{2} v a n d$ $t h e r e f o r e o n l y t h e {t r i a n g l e}^{'} s p o s i t i o n w i l l c h a n g e .$ $I n p a r t i c u l a r, i t s a r e a r e m a i n s t h e s a m e .$ $T h i r d l y, t r a n s l a t e t h e p i n k △, s . t F \equiv C \equiv P \equiv O,$ $w h e r e O i s t h e b e g i n n i n g o f t h e c o o r d i n a t e$ $s y s t e m .$ $T h e n$ $M = \frac{1}{2} (A + D)$ $N = \frac{1}{2} (B + E)$ $I f \times i s t h e v e c t o r c r o s s p r o d u c t :$ $2 {s e}_{z} = M \times N = \frac{1}{4} (A + D) \times (B + E) =$ $= \frac{1}{4} (A \times B + A \times E + D \times B + D \times E)$ $A \times B = 2 {x e}_{z}$ $E \times D = 2 {y e}_{z}$ $\Rightarrow {s e}_{z} = \frac{1}{4} (x - y) e_{z} +$ $+ \frac{e_{z}}{8} (∣ D ∣∣ B ∣ s i n (∠ D O B) - ∣ A ∣∣ E ∣ s i n (∠ A O E))$ $S i n c e t h e t r i a n g l e s a r e e q u i l a t e r a l, ∣ A ∣=∣ B ∣= a,$ $∣ D ∣=∣ E ∣= b .$ $A l s o$ $∠ D O B = ∠ A O B + ∠ D O A =$ $= 60 + ∠ D O A =$ $= ∠ D O E + ∠ D O A = ∠ A O E$ $T h e r e f o r e, f i n a l l y :$ $\Rightarrow s = \frac{1}{4} ∣ x - y ∣$
	Commented by mr W last updated on 08/Apr/24
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