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Question Number 74882 by ajfour last updated on 03/Dec/19

Commented by ajfour last updated on 03/Dec/19

$I f e a c h s m a l l b l u e t r i a n g l e s$ $h a v e t h e s a m e a r e a, a n d e a c h$ $s m a l l q u a d r i l a t e r a l s h a v e t h e$ $s a m e a r e a, f i n d r a t i o o f l e n g t h$ $t o b r e a d t h o f t h e o u t e r r e c t a n g l e .$
Commented by mind is power last updated on 03/Dec/19

$Op . a = 1 . CS \Rightarrow CS = a . op$ $OA . AQ = CB . CS \Rightarrow QA = CS = a . op$ $(BQ) / / (OS), QA = CS \Rightarrow OS = BQ . . . E$ $E \Rightarrow OSBQ parallelogram \Rightarrow BS = OQ$ $\Rightarrow ∠ QOP = ∠ RBS$ $sinceSame area \Rightarrow BG = OE withe same angle$ $\Rightarrow RG = EP$ $Same idea \Rightarrow CRAP parellograme$ $1 conclusion$ $O \vec{P} = R \vec{B}, P \vec{A} = C \vec{R}$ $O \vec{S} = \vec{Q B}, B \vec{G} = E \vec{O}, H \vec{S} = Q \vec{F}$ $P (p, 0) A (1, 0), S (0, s), C (0, a)$ $B (1, a)$ $A \vec{Q} = S \vec{C} \Rightarrow Q (1, a - s)$ $R (1 - p, a)$ $(OQ) equatin : y = (a - s) x$ $(SB) y = (a - s) x + s$ $(CP) : y = - \frac{a}{p} x + a$ $(RA) = - \frac{a}{p} x + \frac{a}{p}$ $E = (OQ) \cap (CP) \Rightarrow (a - s) x = - \frac{a}{p} x + a$ $\Rightarrow x = \frac{ap}{ap + a - ps}, y = \frac{ap (a - s)}{ap + a - ps}$ $F = (OQ) \cap (RA) \Rightarrow (a - s) x = \frac{- ax + a}{p}$ $\Rightarrow x = \frac{a}{ap + a - sp}, y = \frac{a (a - s)}{ap + a - sp}$ $H = (SB) \cap (CP) \Rightarrow (a - s) x + s = \frac{- a}{p} x + a$ $\Rightarrow \frac{p (a - s)}{pa + a - sp} = x, y = \frac{- a (a - s) + a (ap + a - sp)}{pa + a - sp}$ $= \frac{a (s + ap - sp)}{pa + a - sp}$ $G = (RA) \cap (SB) \Rightarrow \frac{- ax}{p} + \frac{a}{p} = (a - s) x + s$ $\Rightarrow x = \frac{a - ps}{ap + a - sp}, y = \frac{- a^{2} + aps + a^{2} p + a^{2} - aps}{p (ap - sp + a)} = \frac{a^{2}}{(ap - sp + a)}$ $\frac{1}{2} Det (Q \vec{F}, Q \vec{A}) = \frac{1}{2} Det (O \vec{P}, O \vec{E})$ $Q \vec{F} (\frac{- ap + sp}{ap + a - sp}, \frac{(a - s) (ps - ap)}{ap + a - sp})$ $Q \vec{A} = (0, s - a)$ $\Rightarrow \frac{p {(s - a)}^{2}}{ap - sp + a} = (\frac{{ap}^{2} (a - s)}{ap - sp + a})$ $\Rightarrow (s - a) = - ap$ $\Rightarrow p = \frac{a - s}{a}$ $too bee continued$ $2 other equation$ $to use$ $\det (p \vec{A}, P \vec{E}) = \det (B \vec{Q}, B \vec{G}) = \det (G \vec{H}, G \vec{F})$
Commented by ajfour last updated on 03/Dec/19

	Answered by mr W last updated on 03/Dec/19

	Commented by mr W last updated on 03/Dec/19

	$\frac{f}{e} = \frac{c}{a + b}$ $a f = c \times \frac{c}{c + d} \times e$ $\Rightarrow \frac{f}{e} = \frac{c^{2}}{a (c + d)}$ $\frac{c}{a + b} = \frac{c^{2}}{a (c + d)}$ $\Rightarrow \frac{a}{a + b} = \frac{c}{c + d}$ $\Rightarrow \frac{a}{b} = \frac{c}{d} = k$ $\frac{1}{2} [\frac{{(a + b)}^{2} f}{a} - a f] = \frac{(a + b) d}{3}$ $\Rightarrow f = \frac{2 (a + b) a d}{3 (2 a + b) b} = \frac{2 k (k + 1) d}{3 (2 k + 1)}$ $\Rightarrow e = \frac{2 {(k + 1)}^{2} b}{3 (2 k + 1)}$ $\frac{f}{a - e} = \frac{c + d}{a}$ $\Rightarrow \frac{\frac{2 k}{3 (2 k + 1)}}{k - \frac{2 {(k + 1)}^{2}}{3 (2 k + 1)}} = \frac{1}{k}$ $\Rightarrow 2 k^{2} - k - 2 = 0$ $\Rightarrow k = \frac{1 + \sqrt{17}}{4} \approx 1.28$ $\frac{k}{k + 1} = \frac{1 + \sqrt{17}}{5 + \sqrt{17}} = \frac{\sqrt{17} - 3}{2} \approx 0.56$ $i . e . t h e r a t i o o f s i d e l e n g t h e s o f t h e$ $o u t e r r e c t a n g l e c a n b e o f a n y v a l u e .$ $t h e c o n d i t i o n i s o n l y$ $\frac{a}{b} = \frac{c}{d} = \frac{1 + \sqrt{17}}{4}$ $o r$ $\frac{a}{a + b} = \frac{c}{c + d} = \frac{\sqrt{17} - 3}{2}$
	Commented by ajfour last updated on 03/Dec/19

	$s i r c o u l d i m a k e t h e q u e s t i o n$ $c l e a r ? t h e s m a l l b l u e t r i a n g l e s$ $h a v e t h e s a m e a r e a .$ $T h e f o u r p i n k q u a d r i l a t e r a l s$ $a n d t h e c e n t r a l o n e h a v e a n$ $e q u a l o t h e r a r e a .$ $W h a t i s t h e r a t i o o f l e n g t h o f$ $t h e o u t e r r e c t a n g l e t o i t s b r e a d t h ?$
	Commented by mr W last updated on 04/Dec/19

	$t h a n k y o u s i r f o r c l a r i f y i n g!$ $i h a v e m o d i f i e d m y a n s w e r .$ $t h e r a t i o o f l e n g t h e s o f t h e r e c t a n g l e$ $c a n b e o f a n y v a l u e . b u t t h e s i d e s$ $m u s t b e d i v i d e d w i t h a c e r t a i n r a t i o .$
	Commented by mr W last updated on 04/Dec/19

	$A_{r e d} = \frac{(a + b) d}{3} = \frac{(k + 1) b d}{3}$ $A_{b l u e} = \frac{a f}{2} = \frac{k^{2}}{(2 k + 1)} \times \frac{(k + 1) b d}{3}$ $\Rightarrow \frac{A_{r e d}}{A_{b l u e}} = \frac{2 k + 1}{k^{2}} = \frac{3 \sqrt{17} + 5}{8} \approx 2.17$ $i . e . t h e r a t i o o f a r e a o f a b l u e t r i a n g l e$ $t o t h a t o f a r e d z o n e i s i n d e p e n d e n t$ $f r o m t h e r a t i o o f s i d e l e n g t h e s o f t h e$ $o u t e r r e c t a n g l e . {i t}^{'} s n o t p o s s i b l e t h a t$ $a t r i a n g l e a r e a i s e q u a l t o t h a t o f a$ $r e d a r e a .$
	Commented by mr W last updated on 04/Dec/19

	Commented by mr W last updated on 04/Dec/19

	Commented by mr W last updated on 04/Dec/19

	Commented by mr W last updated on 04/Dec/19

	$ξ = \frac{\sqrt{17} - 3}{2}$ $i n e a c h c a s e w i t h d i f f e r e n t \frac{L}{B},$ $b l u e t r i a n g l e s h a v e s a m e a r e a,$ $r e d a n d y e l l o w a r e a s a r e e q u a l .$
	Commented by ajfour last updated on 04/Dec/19

	$T h a n k s s i r, t o t a l l y c o n v i n c e d .$ $c a n t h e b l u e, y e l l o w, a n d r e d$ $a r e a s a l l b e e q u a l f o r c e r t a i n$ $s i d e r a t i o o f o u t e r r e c t a n g l e, S i r ?$
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